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学霸学习网 这下你爽了

学霸学习网 这下你爽了

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- 重庆市2014年中考物理试题AB卷及答案合
- 2014年美国高中数学竞赛(AMC12)A卷试题
- 2014全国Ⅱ卷生物试题及答案解析
- 2014全国 I 卷生物试题及答案解析
- 2015美国数学竞赛AMC12试题及答案
- 2010-2014 AMC12 数学竞赛试题及详解
- 美国数学竞赛AMC12词汇
- 2015AMC10A试题及详解(繁体中文版)
- AMC12历年试卷_2000-2008

2010 AMC 12-A Problem 1

What is ?

Solution

.

Problem 2

A ferry boat shuttles tourists to an island every hour starting at 10 AM until its last trip, which starts at 3 PM. One day the boat captain notes that on the 10 AM trip there were 100 tourists on the ferry boat, and that on each successive trip, the number of tourists was 1 fewer than on the previous trip. How many tourists did the ferry take to the island that day?

Solution

It is easy to see that the ferry boat takes trips total. The total number of people taken to the island is

Problem 3

Rectangle shares , pictured below, shares of its area with rectangle of its area with square . What is ? . Square

Solution

If we shift equal parts: to coincide with , and add new horizontal lines to divide into five

This helps us to see that

and

, where

. Hence

.

Problem 4

If , then which of the following must be positive?

Solution

is negative, so we can just place a negative value into each expression and find the one that is positive. Suppose we use .

Obviously only

is positive.

Problem 5

Halfway through a 100-shot archery tournament, Chelsea leads by 50 points. For each shot a bullseye scores 10 points, with other possible scores being 8, 4, 2, and 0 points. Chelsea always scores at least 4 points on each shot. If Chelsea's next be guaranteed victory. What is the minimum value for ? shots are bullseyes she will

Solution

Let be the number of points Chelsea currently has. In order to guarantee victory, we must consider the possibility that the opponent scores the maximum amount of points by getting only bullseyes.

The lowest integer value that satisfies the inequality is

.

Problem 6

A , such as 83438, is a number that remains the same when its digits are and ? are three-digit and four-digit palindromes, respectively. reversed. The numbers

What is the sum of the digits of

Solution

is at most , so is at most and . The minimum value of is , which means that . is . However, must be . the only palindrome between It follows that is

, so the sum of the digits is

Problem 7

Logan is constructing a scaled model of his town. The city's water tower stands 40 meters high, and the top portion is a sphere that holds 100,000 liters of water. Logan's miniature water tower holds 0.1 liters. How tall, in meters, should Logan make his tower?

Solution

The water tower holds Therefore, Logan should make his tower tower. This is Also, the fact that meters high, or choice . times more water than Logan's miniature. times shorter than the actual

doesn't matter since only the ratios are important.

Problem 8

Triangle has . Let . Let ? and be on and , respectively, such that and , and suppose that be the intersection of segments

is equilateral. What is

Solution

Let

.

Since

, triangle

is a

triangle, so

Problem 9

A solid cube has side length inches. A -inch by -inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?

Solution

Solution 1

Imagine making the cuts one at a time. The first cut removes a box cut removes two boxes, each of dimensions the second cut, on the last two faces. Hence the total volume of all cuts is Therefore the volume of the rest of the cube is . . The second . , and the third cut does the same as

Solution 2

We can use Principle of Inclusion-Exclusion to find the final volume of the cube. There are 3 "cuts" through the cube that go from one end to the other. Each of these "cuts" has central cube twice. Hence the total volume of the cuts is Therefore the volume of the rest of the cube is . . cubic inches. However, we can not just sum their volumes, as the cube is included in each of these three cuts. To get the correct result, we

can take the sum of the volumes of the three cuts, and subtract the volume of the central

Solution 3

We can visualize the final figure and see a cubic frame. We can find the volume of the figure by adding up the volumes of the edges and corners. Each edge can be seen as a box. . box, and each corner can be seen as a

Solution 4

First, you can find the volume, which is 27. Now, imagine there are three prisms of dimensions 2 x 2 x 3. Now subtract the prism volumes from 27. We have -9. From here we add two times 2^3, because we over-removed. This is 16 - 9 = 7 (A).

Problem 10

The first four terms of an arithmetic sequence are term of this sequence? , , , and . What is the

Solution

and

are consecutive terms, so the common difference is .

The common difference is . The first term is and the

term is

Problem 11

The solution of the equation can be expressed in the form . What is ?

Solution

This problem is quickly solved with knowledge of the laws of exponents and logarithms.

Since we are looking for the base of the logarithm, our answer is

.

Problem 12

In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements. Brian: "Mike and I are different species."

Chris: "LeRoy is a frog." LeRoy: "Chris is a frog." Mike: "Of the four of us, at least two are toads." How many of these amphibians are frogs?

Solution

Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog. As Mike is a frog, his statement is false, hence there is at most one toad. As there is at most one toad, at least one of Chris and LeRoy is a frog. But then the other one tells the truth, and therefore is a toad. Hence we must have one toad and frogs.

Problem 13

For how many integer values of do the graphs of and not intersect?

Solution 1

The image below shows the two curves for clearly a circle with radius . The blue curve is . , which is , and the red curve is a part of the curve

In the special case

the blue curve is just the point

, and as

, this point is

on the red curve as well, hence they intersect. The case is symmetric to : the blue curve remains the same and the red curve is .

flipped according to the

axis. Hence we just need to focus on

Clearly, on the red curve there will always be points arbitrarily far from the origin: for example, as approaches 0, approaches . Hence the red curve intersects the blue one if . and only if it contains a point whose distance from the origin is at most

At this point we can guess that on the red curve the point where the origin, and skip the rest of this solution.

is always closest to

For an exact solution, fix the origin is least

and consider any point

on the red curve. Its distance from

. To minimize this distance, it is enough to minimize . By the Arithmetic Mean-Geometric Mean Inequality we get that this value is at , i.e., .

, and that equality holds whenever

Now recall that the red curve intersects the blue one if and only if its closest point is at most from the origin. We just computed that the distance between the origin and the closest point on the red curve is . . Therefore, we want to find all positive integers such that

Clearly the only such integer is . This is a total of

, hence the two curves are only disjoint for values.

and

Solution 2

From the graph shown above, we see that there is a specific point closest to the center of the circle. Using some logic, we realize that as long as said furthest point is not inside or on the graph of the circle. This should be enough to conclude that the hyperbola does not intersect the circle. Therefore, for each value of k, we only need to check said value to determine intersection. Let said point, closest to the circle have coordinates Then, all coordinates that satisfy After multiplying through by . We see this is a quadratic in derived from the equation. intersect the circle. Squaring, we find and rearranging, we find and consider taking the determinant, We plot this . We then are points which

which tells us that solutions are real when, after factoring: inequality on the number line to find it is satisfied for all values except: eliminate 0 because it is extraneous as both coincide. Therefore, there are a total of and values.

Problem 14

Nondegenerate has integer side lengths, is an angle bisector, , and . What is the smallest possible value of the perimeter?

Solution

By the Angle Bisector Theorem, we know that . If we use the lowest possible

integer values for AB and BC (the measures of AD and DC, respectively), then , contradicting the Triangle Inequality. If we use the next lowest values ( answer is and ), the Triangle Inequality is satisfied. Therefore, our , or choice .

Problem 15

A coin is altered so that the probability that it lands on heads is less than

and when the . What is

coin is flipped four times, the probaiblity of an equal number of heads and tails is the probability that the coin lands on heads?

Solution

Let be the probability of flipping heads. It follows that the probability of flipping tails is . The probability of flipping heads and tails is equal to the number of ways to flip it times the product of the probability of flipping each coin.

As for the desired probability the positive root, hence

both

and

are nonnegative, we only need to consider

Applying the quadratic formula we get that the roots of this equation are

. As the

probability of heads is less than

, we get that the answer is

.

Problem 16

Bernardo randomly picks 3 distinct numbers from the set and arranges them in descending order to form a 3-digit number. Silvia randomly picks 3

distinct numbers from the set than Silvia's number?

and also arranges them in descending

order to form a 3-digit number. What is the probability that Bernardo's number is larger

Solution

We can solve this by breaking the problem down into cases and adding up the probabilities.

Case : Bernardo picks . If Bernardo picks a then it is guaranteed that his number will be larger than Silvia's. The probability that he will pick a is .

Case : Bernardo does not pick . Since the chance of Bernardo picking is probability of not picking is .

, the

If Bernardo does not pick 9, then he can pick any number from to . Since Bernardo is picking from the same set of numbers as Silvia, the probability that Bernardo's number is larger is equal to the probability that Silvia's number is larger. Ignoring the for now, the probability that they will pick the same number is the number of ways to pick Bernardo's 3 numbers divided by the number of ways to pick any 3 numbers. We get this probability to be

Probability of Bernardo's number being greater is Factoring the fact that Bernardo could've picked a but didn't:

Adding up the two cases we get

Problem 17

Equiangular hexagon sum of all possible values of ?

has side lengths . The area of is

and of the area of the hexagon. What is the

Solution

It is clear that is an equilateral triangle. From the Law of Cosines, we get that . Therefore, the area of . If we extend and triangle , and so that and meet at , and meet at , , and and is

meet at

, we find that hexagon

is formed by taking equilateral is therefore

of side length

and removing three equilateral triangles,

, of side length . The area of

.

Based on the initial conditions,

Simplifying this gives us possible value of is .

. By Vieta's Formulas we know that the sum of the

Problem 18

A 16-step path is to go from coordinate or the of the square , to with each step increasing either the -coordinate by 1. How many such paths stay outside or on the boundary at each step?

Solution

Each path must go through either the second or the fourth quadrant. Each path that goes through the second quadrant must pass through exactly one of the points and . , ,

There are paths of the first kind,

paths of the second kind, and

paths of the third type. Each path that goes through the fourth quadrant must pass through exactly one of the points paths of the first kind, the third type. Hence the total number of paths is . , , and . Again, there are paths of

paths of the second kind, and

Problem 19

Each of 2010 boxes in a line contains a single red marble, and for the position also contains , the box in white marbles. Isabella begins at the first box and be the probability that Isabella stops after drawing exactly for which ?

successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let

marbles. What is the smallest value of

Solution

The probability of drawing a white marble from box a red marble from box To stop after drawing is . is , and the probability of drawing

marbles, we must draw a white marble from boxes Thus,

and draw a red marble from box

So, we must have

or

Since

increases as

increases, we can simply test values of is , since

; after some trial and but

error, we get that the minimum value of

Problem 20

Arithmetic sequences for some and have integer terms with ? and . What is the largest possible value of

Solution

Solution 1

Since and have integer terms with , we can write the terms of each sequence as

where

and

(

) are the common differences of each, respectively.

Since

it is easy to see that .

Hence, we have to find the largest

such that

and

are both integers.

The prime factorization of product of

is

. We list out all the possible pairs that have a

and soon find that the largest value is .

value is for the pair

, and so the largest

Solution 2

As above, let Now we get . Therefore . For we easily see that for it is way too large. For values for we are looking for must be divisible by a suitable solution Hence such that we get . Note that , which has the right side is less than and for any other divides and for some , hence . And as the .

second term is greater than the first one, we only have to consider the options

. We can start looking for the solution by trying the possible . . (There is no need to check anymore.)

, and we easily discover that for

is the largest possible

Problem 21

The graph of three values of lies above the line except at , where the graph and the line intersect. What is the largest of these values?

Solution

The values in which intersect at . are the same as the zeros of

Since there are zeros and the function is never negative, all zeros must be double roots because the function's degree is . Suppose we let , , and be the roots of this function, and let , , and . be the

cubic polynomial with roots

In order to find terms of .

we must first expand out the

[Quick note: Since we don't know the expansion.]

, , and , we really don't even need the last 3 terms of

All that's left is to find the largest root of

.

Problem 22

What is the minimum value of ?

Solution

Solution 1

If we graph each term separately, we will notice that all of the zeros occur at any integer from to , inclusive. , where is

The minimum value occurs where the sum of the slopes is at a minimum, since it is easy to see that the value will be increasing on either side. That means the minimum must happen at some .

The sum of the slope at

is

Now we want to minimize the slope is where We can now verify that both . and

. The zeros occur at

and

, which means

yield

.

Solution 2

Rewrite the given expression as follows: Imagine the real line. For each the coordinate imagine that there are boys standing at

. We now need to place a girl on the real line in such a way that the sum of

her distances from all the boys is minimal, and we need to compute this sum. Note that there are from 1 (the only boy placed at ) to (the last boy placed at boys in total. Let's label them .

Clearly, the minimum sum is achieved if the girl's coordinate is the median of the boys' coordinates. To prove this, place the girl at the median coordinate. If you now move her in any direction by any amount distances does not decrease. Hence the optimal solution is to place the girl at the median coordinate. Or, more precisely, as is even, we can place her anywhere on the segment formed by boy and boy , there will be boys such that she moves away from this boy. For each of the remaining boys, she moves at most closer, hence the total sum of

: by extending the previous argument, anywhere on this segment the sum of distances is the same. By trial and error, or by solving the quadratic equation boy number is the last boy placed at we get that . Hence

and the next boy is the one placed at .

the given expression is minimized for any

Common part of both solutions

To find the minimum, we want to balance the expression so that it is neither top nor bottom heavy. Now that we know that the sum of the first 84 we can plug either Note that the terms or to find the minimum. to are negative, and the terms to are . 's is equivalent to the sum of 's 85 to 119,

positive. Hence we get:

and

Hence the total sum of distances is .

Solution 3

Since the minimum exists, we want all the such that

s to cancel out. Thus, we want to find some

Then, which becomes

. The answer(expression's value) is then .

,

Problem 23

The number obtained from the last two nonzero digits of is equal to . What is ?

Solution

We will use the fact that for any integer ,

First, we find that the number of factors of Let be because If we divide by . The ,

in

is equal to , or . If . Only one of

.

we want is therefore the last two digits of , we know that , has to be a multiple of 4. by taking out all the factors of in , or

instead we find

, what we are looking for, could

these numbers will be a multiple of four, and whichever one that is will be the answer,

, we can write

as

where is

where every multiple of 5 is replaced by the number with all its factors of 5 removed. Specifically, every number in the form replaced by The number , and every number in the form can be grouped as follows: is replaced by .

Where the first line is composed of the numbers in

that aren't multiples of five, the

second line is the multiples of five and not 25 after they have been divided by five, and the third line is multiples of 25 after they have been divided by 25. Using the identity at the beginning of the solution, we can reduce to

Using the fact that

(or simply the fact that . Therefore

if you

have your powers of 2 memorized), we can deduce that . Finally, combining with the fact that yields .

Solution 2

Let be after we truncate its zeros. Notice that has exactly (floored) modulo 4 and 25, to is divisible by 4 (consider the ), and so we only need to

factors of 5; thus, number of 2s dividing consider modulo 25. we have

We shall consider minus the number of 5s dividing

determine its residue modulo 100. It is easy to prove that

Now, notice that for integers Thus, for integral a:

Using this process, we can essentially remove all the numbers which had not formerly been a multiple of 5 in from consideration.

Now, we consider the remnants of the 5, 10, 15, 20, ..., 90 not yet eliminated. The 10, 20, 30, ..., 90 becomes 1, 2, 3, 4, 1, 6, 7, 8, 9, whose product is 1 mod 25. Also, the 5, 5, 15, 25, ..., 85 becomes 1, 1, 3, 1, 7, 9, 11, 13, 3, 17 and compute that . We deduce that from . multiplying out the 1, 1, 3, 1, 7, ..., 17 is equivalent to 2 modulo 25, and so we need to . But this is simply by Fermat's Little Theorem and so the answer is . Because 12 is also a multiple of 4, we can utilize the Chinese Remainder Theorem to show

Problem 24

Let domain of with the interval is a union of . The intersection of the disjoint open intervals. What is ?

Solution

The question asks for the number of disjoint open intervals, which means we need to find the number of disjoint intervals such that the function is defined within them. We note that since all of the factors are inside a logarithm, the function is undefined

where the inside of the logarithm is equal to or less than . First, let us find the number of zeros of the inside of the logarithm.

After counting up the number of zeros for each factor and eliminating the excess cases we get zeros and intervals.

In order to find which intervals are negative, we must first realize that at every zero of each factor, the sign changes. We also have to be careful, as some zeros are doubled, or even tripled, quadrupled, etc. The first interval is obviously positive. This means the next interval is negative.

Continuing the pattern and accounting for doubled roots (which do not flip sign), we realize

that there are negative intervals from to . Since the function is symmetric, we know that there are also negative intervals from to .

And so, the total number of disjoint open intervals is

Problem 25

Two quadrilaterals are considered the same if one can be obtained from the other by a rotation and a translation. How many different convex cyclic quadrilaterals are there with integer sides and perimeter equal to 32?

Solution 1

It should first be noted that given any quadrilateral of fixed side lengths, the angles can be manipulated so that the quadrilateral becomes cyclic. Denote , , , and as the integer side lengths of the quadrilateral. Without loss of . , the Triangle Inequality implies that .

generality, let Since

We will now split into cases.

Case :

( side lengths are equal) , and no matter how the sides

Clearly there is only way to select the side lengths are rearranged only unique quadrilateral can be formed. Case : or

( side lengths are equal) to

If side lengths are equal, then each of those side lengths can only be integers from quadrilateral that can be formed from one set of side lengths, resulting in a total of quadrilaterals. Case : ( pairs of side lengths are equal)

except for (because that is counted in the first case). Obviously there is still only unique

and can be any integer from to

, and likewise and

can be any integer from to .

.

However, a single set of side lengths can form different cyclic quadrilaterals (a rectangle and a kite), so the total number of quadrilaterals for this case is Case : or or

( side lengths are equal) , which we have

If the equal side lengths are each , then the other sides must each be already counted in an earlier case. If the equal side lengths are each we find a total of

, there is possible sets of side

set of side lengths. Likewise, for side lengths of there are sets. Continuing this pattern, lengths. (Be VERY careful when adding up the total for this case!) For each set of side lengths, there are possible quadrilaterals that can be formed, so the total number of quadrilaterals for this case is Case : starting from possible side lengths is .

(no side lengths are equal) Using the same counting principles and eventually reaching . There are , we find that the total number of ways to arrange the side lengths, but there is only

unique quadrilateral for rotations, so the number of quadrilaterals for each set of side lengths is . The total number of quadrilaterals is .

And so, the total number of quadrilaterals that can be made is .

Solution 2

As with solution we would like to note that given any quadrilateral we can change its angles to make a cyclic one. Let be the sides of the quadrilateral.

There are

ways to partition

. However, some of these will not be quadrilaterals since .

they would have one side bigger than the sum of the other three. This occurs when For , . There are ways to partition . Since

could be any of the ,

four sides, we have counted for other values of

degenerate quadrilaterals. Similarly, there are . Thus, there are

non-degenerate partitions of by the hockey stick theorem. However, for or , each , and , each quadrilateral is counted times, for each rotation. Also, for quadrilateral is counted twice. Since there is quadrilateral for which for which or , there are . Thus there are quads for which total quadrilaterals.

2010 AMC 12-B Problem 1

Makarla attended two meetings during her -hour work day. The first meeting took minutes and the second meeting took twice as long. What percent of her work day was spent attending meetings?

Solution

The total number of minutes in here -hour work day is time spend in meetings in minutes is The total amount of The answer is then

Problem 2

A big is formed as shown. What is its area?

Solution

We find the area of the big rectangle to be rectangle to be , and the area of the smaller . . The answer is then

Problem 3

A ticket to a school play cost $ . How many values for dollars, where is a whole number. A group of 9th graders buys tickets costing a total of $ , and a group of 10th graders buys tickets costing a total of

are possible?

Solution

We find the greatest common factor of which is the answer . and to be . The number of factors of is

Problem 4

A month with days has the same number of Mondays and Wednesdays.How many of the seven days of the week could be the first day of this month?

Solution

so the week cannot start with Saturday, Sunday, Tuesday or Wednesday as that would result in an unequal number of Mondays and Wednesdays. Therefore, Monday, Thursday, and Friday are valid so the answer is .

Problem 5

Lucky Larry's teacher asked him to substitute numbers for , , , , and in the expression and evaluate the result. Larry ignored the parenthese but added and subtracted correctly and obtained the correct result by coincidence. The numbers Larry substituted for , , , and were , , , and , respectively. What number did Larry substitude for ?

Solution

We simply plug in the numbers

Problem 6

At the beginning of the school year, year, answered "Yes" and of all students in Mr. Wells' math class answered answered "No." At the end of the school of the students gave a "Yes" to the question "Do you love math", and

answerws "No." Altogether, ?

different answer at the beginning and end of the school year. What is the difference between the maximum and the minimum possible values of

Solution

Clearly, the minimum possible value would be would be . The difference is . The maximum possible value .

Problem 7

Shelby drives her scooter at a speed of for a total of miles in miles per hour if it is not raining, and miles per hour if it is raining. Today she drove in the sun in the morning and in the rain in the evening, minutes. How many minutes did she drive in the rain?

Solution

Let be the time it is not raining, and be the time it is raining, in hours. and We have the system:

Solving gives We want

and

in minutes,

Problem 8

Every high school in the city of Euclid sent a team of students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed

th

and

th

, respectively. How many schools are in the city?

Solution

There are schools. This means that there are is an odd number. people. Because no one's score was the , because there wouldn't be a same as another person's score, that means that there could only have been median score. This implies that cannot be less than th place if there were. . cannot be greater than either, because that would tie Andrea

and Beth or Andrea's place would be worse than Beth's. Thus, the only possible answer is

Problem 9

Let be the smallest positive integer such that is divisible by ? , is a perfect cube, and is a perfect square. What is the number of digits of

Solution

We know that . Let and , which means that whose sixth power is a multiple of and . Cubing and squaring the equalities respectively gives . Now we know is must be a perfect -th power because are and , . must be a perfect -th power. The smallest number , because the only prime factors of , with digits

. Therefore our is equal to number

Problem 10

The average of the numbers and is . What is ?

Solution

We first sum the first sum of the series is find the sum to equal numbers: . Since the average is . Setting equal . Thus, the answer is . . Then, we know that the , and there are terms, we also

Problem 11

A palindrome between divisible by ? and is chosen at random. What is the probability that it is

Solution

View the palindrome as some number with form (decimal representation): . But because the number is a palindrome, . Recombining this yields means that as long as palindromes out of 90 ( . 1001 is divisible by 7, which , then this , the palindrome will be divisible by 7. This yields 9 ) possibilities for palindromes. However, if

gives another case in which the palindrome is divisible by 7. This adds another 9 palindromes to the list, bringing our total to

Problem 12

For what value of does

Solution

Problem 13

In , and . What is ?

Solution

We note that and . Therefore, there is no other way to satisfy this and , since any other and is a equation other than making both

way would cause one of these values to become greater than 1, which contradicts our previous statement. From this we can easily conclude that and solving this system gives us triangle with and . . It is clear that

Problem 14

Let , , , , and be postive integers with , , and and let be the ? largest of the sum . What is the smallest possible value of

Solution

We want to try make , , , and as close as possible so that , the maximum of these, if smallest. Notice that . In order to express as a sum of numbers, we or

must split up some of these numbers. There are two ways to do this (while keeping the sum of two numbers as close as possible):

. We see that in both cases, the value of answer is .

is

, so the

Problem 15

For how many ordered triples two distinct elements in the set of nonnegative integers less than , where ? are there exactly

Solution

We have either For . , , this only occurs at . has five solutions between zero and nineteen, has nineteen integer solutions between zero and nineteen. So for , we have For ordered triples. For while and , this occurs at and has fifteen solutions. , while ordered triples. In total we have ordered triples . and and both have one solution ordered triples. has nineteen solutions, , or . . has only one solution, namely, , and

, again this only occurs at . has five solutions, and

has one solution, so again we have

both have one solution, namely,

has twenty solutions. So we have

Problem 16

Positive integers the set , , and are randomly and independently selected with replacement from . What is the probability that is divisible by ?

Solution

The value of

is arbitrary other than it is divisible by , so the set

can

be grouped into threes. Obviously, if event that is divisible by (which has probability ) then the sum is divisible by . In the , then the sum is divisible by if

is not divisible by (which has probability , which is the same as .

This only occurs when one of the factors or equivalent to , and threes. In total the probability is . All four events

is equivalent to ,

and the other is ,

have a probability of

because the set is grouped in

Problem 17

The entries in a array include all the digits from through , arranged so that the entries in every row and column are in increasing order. How many such arrays are there?

Solution

The value of is arbitrary other than it is divisible by , so the set can be grouped into threes. Obviously, if event that is divisible by (which has probability ) then the sum is divisible by . In the , then the sum is divisible by if

is not divisible by (which has probability , which is the same as .

This only occurs when one of the factors or equivalent to . All four events

is equivalent to ,

and the other is ,

, and threes. In total the probability is

have a probability of

because the set is grouped in

Problem 18

A frog makes jumps, each exactly meter long. The directions of the jumps are chosen independenly at random. What is the probability that the frog's final position is no more than meter from its starting position?

Solution 1

We will let the moves be complex numbers , , and , each of magnitude one. The frog has magnitude less than or equal to . Hence, the probability is . starts on the origin. It is relatively easy to show that exactly one element in the set

Solution 2

The first frog hop doesn't matter because no matter where the frog hops is lands on the border of the circle you want it to end in. The remaining places that the frog can jump to form a disk of radius 2 centered at the spot on which the frog first landed, and every point in the disk of radius 2 is equally likely to be reached in two jumps. The entirety of the circle you want the frog to land in is enclosed in this larger disk, so find the ratio of the two areas, which is .

Problem 19

A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the

fourth quarter, the Raiders had won by one point. Neither team scored more than What was the total number of points scored by the two teams in the first half?

points.

Solution

Let be the quarterly scores for the Raiders. We know that The Raiders and be quarterly scores for the Raiders. The sum of the Raiders scores is and the sum of the Wildcats scores is search for the values of integers, we can conclude that and . Now we can narrow our , and . Because points are always measured in positive are positive integers. We can also conclude that is a Wildcats both scored the same number of points in the first quarter so let

positive integer by writing down the equation:

Now we can start trying out some values of . We start at which is a contradiction. Next we try , which gives

which gives that

We need the smallest multiple of see that this is , and therefore

(to satisfy the <100 condition) that is and .

. We

So the Raiders first two scores were and

and the Wildcats first two scores were and

.

Problem 20

A geometric sequence . For what value of does has , ? , and for some real number

Solution

By defintion of a geometric sequence, we have we can rewrite this as . . Since ,

The common ratio of the sequence is

, so we can write

Since

, we have , which is

making our answer

.

Problem 21

Let , and let be a polynomial with integer coefficients such that , and . What is the smallest possible value of ?

Solution

There must be some polynomial such that

Then, plugging in values of

we get

Thus, the least value of . Solving, we receive , so our answer is .

must be the

Problem 22

Let than be a cyclic quadrilateral. The side lengths of such that are distinct integers less ? . What is the largest possible value of

Solution

Let , , we have cyclic quadrilateral, we must have that . Therefore, we have We now look at the equation equal . Suppose that . We let . . Suppose that and . . Then, we must have either or , , and . We see that by the Law of Cosines on . Also, . Now, we know that . Also, because , so . Now, adding, is a

Now,

, so it is

or

.

Problem 23

Monic quadratic polynomials and and , and have the property that has zeros at and ? has zeros at and . What

is the sum of the minimum values of

Solution

. Notice that roots of or are the roots of . Doing something similar for sum of the roots will be has roots , so that the , . We now have . . For each individual equation, the gives us . Since Thus, we have

(symmetry or Vieta's). Thus, we have is monic, the roots of , or

are "farther" from the axis of symmetry than the roots of

. Adding these gives us . Plugging this into is , and the minimum value of . is , we get . Thus, our answer is

, or , or

. The minimum value of

answer

Problem 24

The set of real numbers for which

is the union of intervals of the form intervals?

. What is the sum of the lengths of these

Solution

Because the right side of the inequality is a horizontal line, the left side can be translated horizontally by any value and the intervals will remain the same. For simplicity of calculation, we will find the intervals where . We shall say that has three vertical asymptotes at . As the

sum of decreasing hyperbolas, the function is decreasing at all intervals. Values immediately to the left of each asymptote approach negative infinity, and values immediately to the right of each asymptote approach positive infinity. In addition, the function has a horizontal asymptote at some point from . The function intersects at some point from to , and at some point to the right of to , and at . The intervals

where the function is greater than are between the points where the function equals and the vertical asymptotes. If , , and are values of x where , then the sum of the lengths of the intervals is .

And now our job is simply to find the sum of the roots of formulas, we find this to be .

. Using Vieta's

Solution 2

As in the first solution, note that the expression can be translated into without affecting the interval lengths.

This simplifies into , where length is

and so , , and are the roots of , which is the sum of the roots, or .

. Each interval is so the total

Problem 25

For every integer For example divides , let be the largest power of the largest prime that divides . What is the largest integer such that .

?

Solution

Because 67 is the largest prime factor of 2010, it means that in the prime factorization of

, there'll be

where

is the desired value we are looking for. would fit this form. However, this itself must be counted numbers (as 71 but less

Thus, to find this answer, we need to look for the number of times 67 would be incorporated into the giant product. Any number of the form number tops at because 71 is a higher prime than 67.

twice because it's counted twice as a squared number. Any non-prime number that's less than 79 (and greater than 71) can be counted, and this totals 5. We have isn't counted - 1 through 70), an additional ( than (72, 74, 75, 76, 77, and 78). Thus, ), and values just greater than

2011 AMC 12-A Problem 1

A cell phone plan costs talked for dollars each month, plus cents per text message sent, plus hours. In January Michelle sent text messages and cents for each minute used over

hours. How much did she have to pay?

Solution

The base price of Michelle's cell phone plan is costs cents per text, then she must have spent talked for hours, but additional amount for. hours dollars. If she sent text messages and it cents for texting, or dollars. She minutes. Since the price for phone calls is dollars dollars .

will give us the amount of time that she has to pay an

cents per minute, the additional amount Michelle has to pay for phone calls is cents, or dollars. Adding

Problem 2

There are coins placed flat on a table according to the figure. What is the order of the coins from

top to bottom?

Solution

By careful inspection, the answer is .

Problem 3

A small bottle of shampoo can hold

milliliters of shampoo, whereas a large bottle can hold

milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?

Solution

To find how many small bottles we need, we can simply divide Since the answer must be an integer greater than bottles, or by . This simplifies to

, we have to round up to

Problem 4

At an elementary school, the students in third grade, fourth grade, and fifth grade run an average of , , and minutes per day, respectively. There are twice as many third graders as fourth graders, and twice as many fourth graders as fifth graders. What is the average number of minutes run per day by these students?

Solution

Let us say that there are fourth graders and fifth graders. According to the given information, there must be third graders. The average time run by each student is equal to the

total amount of time run divided by the number of students. This gives us

Problem 5

Last summer herons, and of the birds living on Town Lake were geese, were swans, were were ducks. What percent of the birds that were not swans were geese?

Solution

To simplify the problem, let us say that there were a total of that are not swans is . The number of geese is

birds. The number of birds

. Therefore the percentage is just

Problem 6

The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was did they make? points. How many free throws

Solution

For the points made from two-point shots and from three-point shots to be equal, the numbers of made shots are in a ratio. Therefore, assume they made and two- and three- point shots, respectively, and thus free throws. The total number of points is

Set that equal to

, we get

, and therefore the number of free throws they made

Problem 7

A majority of the students in Ms. Demeanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than . The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was . What was the cost of a pencil in cents?

Solution

The only factor of is student bought

between

and

(

because it has to be a majority of her students) students, pencils cost cents, and each

. Therefore, we can conclude that there are pencils. Thus the answer is

Problem 8

In the eight term sequence , , , , , , , ? , the value of is and the sum of any three consecutive terms is . What is

Solution 1

Let finally . Then from . So , we find that . Continuing this pattern, we find . From , , , and , we then get that

Solution 2

Given that the sum of 3 consecutive terms is 30, we have and

It follows that Subtracting, we have that .

because

.

Problem 9

At a twins and triplets convention, there were sets of twins and sets of triplets, all from different families. Each twin shook hands with all the twins except his/her siblings and with half the triplets. Each triplet shook hands with all the triplets except his/her siblings and with half the twins. How many handshakes took place?

Solution

There are with the total twins and total triplets. Each of the twins shakes hands with the twins not in their family and of the triplets, a total of people. Each of the triplets shakes hands people. Dividing by

triplets not in their family and of the twins, for a total of

two to accommodate the fact that each handshake was counted twice, we get a total of

Problem 10

A pair of standard -sided dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference?

Solution

For the circumference to be greater than the area, we must have Now since is determined by a sum of two dice, the only possibilities for , or .

are thus and .

In order for two dice to sum to , they most both show a value of . The probability of this happening is . In order for two dice to sum to , one must show a and the

other must show a . Since this can happen in two ways, the probability of this event occurring is . The sum of these two probabilities now gives the final answer:

Problem 11

Circles and each have radius 1. Circles and share one point of tangency. Circle but has a point of tangency with the midpoint of outside circle and circle What is the area inside circle

Solution 1

The requested area is the area of Let be the midpoint of and ,

minus the area shared between circles be the other intersection of circles is of the regions between arc ) a quarter of the circle minus and

, .

and

.

Then area shared between

and

and line :

,

which is (considering the arc on circle

(We can assume this because

is 90 degrees, since )

is a square, due the

application of the tangent chord theorem at point So the area of the small region is

The requested area is area of circle

minus 4 of this area:

.

Solution 2

We can move the area above the part of the circle above the segment

down, and

similarly for the other side. Then, we have a square, whose diagonal is , so the area is then just .

Problem 12

A power boat and a raft both left dock on a river and headed downstream. The raft drifted downriver, then immediately turned and to at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock How many hours did it take the power boat to go from

Solution

Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of . In this case, when the powerboat travels from from thus to . and sum to hours, the trip from to to , the raft remains at . Thus the trip hours. The answer is takes the same time as the trip from to the raft. Since these times are equal

must take half this time, or

Problem 13

Triangle incenter of

has side-lengths parallel to intersects at

and and at

The line through the What is the perimeter of

Solution

Let be the incenter. Because and is the angle bisector, we have

It then follows due to alternate interior angles and base angles of isosceles triangles that . Similarly, . The perimeter of then becomes

Problem 14

Suppose and are single-digit positive integers chosen independently and at random. What lies above the parabola ? is the probability that the point

Solution

If lies above the parabola, then must be greater than . Solving this for gives us when that . We thus get the inequality constantly increases , we can deduce are thus , , and .

. Now note that

is positive. Then since this expression is greater than when

must be less than in order for the inequality to hold, since otherwise would be

greater than and not a single-digit integer. The only possibilities for For For , we get , we get

for our inequality, and thus can equal any integer from to . for our inequality, and thus can equal any integer from to .

For

, we get

for our inequality, and thus can equal any integer from

to .

Finally, if we total up all the possibilities we see there are out of is thus

points that satisfy the condition,

total points. The probability of picking a point that lies above the parabola

Problem 15

The circular base of a hemisphere of radius rests on the base of a square pyramid of height . The hemisphere is tangent to the other four faces of the pyramid. What is the edgelength of the base of the pyramid?

Solution

Let square be the pyramid with and the midpoint of side at . . Since the hemisphere is tangent to the . Hence is similar to . as the square base. Let and be the center of respectively. Lastly, let the hemisphere be

tangent to the triangular face Notice that triangular face

has a right angle at at , is also

The length of the square base is thus

Problem 16

Each vertex of convex polygon colorings are possible?

is to be assigned a color. There are colors to

choose from, and the ends of each diagonal must have different colors. How many different

Solution

We can do some casework when working our way around the pentagon from each stage, there will be a makeshift diagram. 1.) For , we can choose any of the 6 colors. A : 6 2.) For , we can either have the same color as and . , or any of the other 5 colors. We do this will be to . At

because each vertex of the pentagon is affected by the 2 opposite vertices, and affected by both A : 6 B:1 3.) For B:5 , we cannot have the same color as if is the same as

. Also, we can have the same color as can't be the same as

(

will be affected), or any of the other 4 colors. Because the same as A : 6 B:1 C:5 4.) C:4 B:5 C:1 and . If they are the same, then

, it can't be

, so it can be any of the 5 other colors.

is affected by

can be any of the other 5 colors.

If they are different, then A : 6 B:1 C:5 D:5 5.) C:4 D:4 B:5 C:1 D:4 and

can be any of the (6-2)=4 colors.

is affected by

. If they are the same, then

can be any of the other 5 colors.

If they are different, then A : 6 B:1 C:5 C:4 B:5 C:1

can be any of the (6-2)=4 colors.

D:5 E:4

D:4 E:4

D:4 E:5

6.) Now, we can multiply these three paths and add them:

7.) Our answer is

!

Problem 17

Circles with radii , , and are mutually externally tangent. What is the area of the triangle determined by the points of tangency?

Solution

The centers of these circles form a 3-4-5 triangle, which has an area equal to 6. The areas of the three triangles determined by the center and the two points of tangency of each circle are, by Law of Sines,

which add up to

. The area we're looking for is the large 3-4-5 triangle minus the three .

smaller triangles, or

Problem 18

Suppose that . What is the maximum possible value of ?

Solution

Plugging in some values, we see that the graph of the equation square bounded by Notice that point to point the feasible region farthest from point substituting into the function, yields and . means the square of the distance from a minus 9. To maximize that value, we need to choose the point in , which is . . Either one, when is a

Problem 19

At a competition with players, the number of players given elite status is equal to players are given elite status. What is the sum of the ? . Suppose that two smallest possible values of

Solution

We start with . Since integer is a positive integer, . From this fact, we get must be in the form of . for some positive . After rearranging, we get

If we now check integer values of N that satisfy this condition, starting from quickly see that the first values that work for and for are and , we get , respectively. Adding up these two values for

, we

, giving values of

Problem 20

Let , is ? , where , , , and are integers. Suppose that , . What for some integer

Solution

From , we know that . . Subtracting , and thus . . Again subtracting , or . It follows from . Since from must be an From the first inequality, we get this gives us integer, it follows that

Similarly, from the second inequality, we get from this gives us this that . ,

We now have a system of three equations: Solving gives us

, and

.

and from this we find that

Since

, we find that

Problem 21

Let largest value of ? , and for integers for which the domain of , let is nonempty, the domain of . If is is the . What is

Solution

The domain of is defined when .

Applying the domain of

and the fact that square roots must be positive, we get becomes .

. Simplifying, the domain of

Repeat this process for For

to get a domain of

.

, since square roots are positive, we can exclude the negative values of the as the domain of . We now arrive at a , however, since we are looking for the largest to get a domain of . Solve for

previous domain to arrive at domain with a single number that defines value for to get We add to get for which the domain of domain that is empty. We continue with

is nonempty, we must continue until we arrive at a

. Since square roots cannot be negative, this is the last nonempty domain. .

Problem 22

Let be a square region and an integer. A point in the interior or into is called n-ray partitional if there are How many points are rays emanating from that divide triangles of equal area.

-ray partitional but not

-ray partitional?

Solution

There must be four rays emanating from region. Depending on the location of that intersect the four corners of the square -ray , the number of rays distributed among these four

triangular sectors will vary. We start by finding the corner-most point that is partitional (let this point be the bottom-left-most point).

We first draw the four rays that intersect the vertices. At this point, the triangular sectors with bases as the sides of the square that the point is closest to both do not have rays dividing their areas. Therefore, their heights are equivalent since their areas are equal. The remaining rays are divided among the other two triangular sectors, each sector with triangles of equal areas. rays, thus dividing these two sectors into

Let the distance from this corner point to the closest side be . From this, we get the equation Therefore, point moving is

and the side of the square be . Solve for to get .

of the side length away from the two sides it is closest to. By

to the right, we also move one ray from the right sector to the left sector, -ray partitional point. We can continue moving -ray partitional. apart from one another. Since this grid from the same side, we have a -ray partitional, -ray partitional point to the sides closest grid, each point apart from each right and up

which determines another

to derive the set of points that are

In the end, we get a square grid of points each ranges from a distance of grid, a total of from one side to

-ray partitional points. To find the overlap from the

we must find the distance from the corner-most to it. Since the -ray partitional points form a

other, we can deduce that the and , which are , , and and

-ray partitional points form a

grid, each point

apart from each other. To find the overlap points, we must find the common divisors of . Therefore, the overlapping points will form grids with points away from

away from each other respectively. Since the grid with points grid, which has . points. Subtract from to get

each other includes the other points, we can disregard the other grids. The total overlapping set of points is a

Problem 23

Let and and , where for all for which ? and are complex numbers. Suppose that is defined. What is the difference between

the largest and smallest possible values of

Solution

By algebraic manipulations, we obtain In order for implies implies implies Since , or or , or . or . . , we must have , , and

where .

, in order to satisfy all 3 conditions we must have either . In the first case .

For the latter case note that . On the other hand, Thus . Hence the maximum value for or is be achieved for instance when the answer is . so,

and hence, . while the minimum is (which can respectively). Therefore

Problem 24

Consider all quadrilaterals quadrilateral? such that , , , and . What is the radius of the largest possible circle that fits inside or on the boundary of such a

Solution 1

Note as above that ABCD must be cyclic to obtain the circle with maximal radius. Let , and Let , and Hence Therefore, Let . Then . Using , be the points on and , , , and . Since the quadrilateral is cyclic, . Let the circle have center are right angles. , and , and , . , and we have , and . and radius . Note that , , and , , respectively where the circle is tangent.

, and

. By equating the value of . Solving we obtain so that .

from each,

Solution 2

To maximize the radius of the circle, we also need to maximize its area. To maximize the area of the circle, the quadrilateral must be tangential (have an incircle). In a tangential quadrilateral, the sum of opposite sides is equal to the semiperimeter of the quadrilateral. , so this particular quadrilateral has an incircle. By definition, given side lengths, a cyclic quadrilateral has the maximum area of any quadrilateral with those side lengths. Therefore, to maximize the area of the quadrilateral and thus the incircle, we assume that this quadrilateral is cyclic. For cyclic quadrilaterals, Brahmagupta's formula gives the area as where is the semiperimeter and equal to and are the side , the lengths. Breaking it up into triangles, we see the area of a tangential quadrilateral is also . Equate these two equations. Substituting , the semiperimeter, and .

area and solving for ,we get

Problem 25

Triangle has , , , and ? . Let , , and be the orthocenter, incenter, and circumcenter of area of pentagon , respectively. Assume that the

is the maximum possible. What is

Solution

Let , , , , it follows that is cyclic. Observe that is fixed. Since . is fixed, whence the (both are is is also fixed. Similarly, as for convenience. , and (verifiable and

It's well-known that by angle chasing). Then, as consequently pentagon circumcircle of cyclic pentagon radii), it follows that and also maximal, it suffices to maximize

Verify that

,

by angle chasing; it follows that since by Triangle Angle Sum. Similarly, (isosceles base angles are equal), whence Since ,

by Inscribed Angles. There are two ways to proceed.

Letting

and

be the circumcenter and circumradius, respectively, of cyclic pentagon , whence and, using the fact

, the most straightforward is to write

that

is fixed, maximize

with Jensen's Inequality.

A more elegant way is shown below. Lemma: is maximized only if . is maximized when the midpoint of minor arc ; similarly . Then since the altitude from is greater than that from , , so our claim follows. With our lemma( ) and from above, along with the fact that inscribed to to . Taking . Let be the

Proof by contradiction: Suppose midpoint of minor arc be and

to be the new orthocenter, incenter, respectively, this contradicts the maximality of

angles that intersect the same length chords are equal,

2011 AMC 12-B Problem 1

What is

Solution

Add up the numbers in each fraction to get subtraction yields , which equals . Doing the

Problem 2

Josanna's test scores to date are to accomplish this goal? , , , , and . Her goal is to raise her test average at least points with her next test. What is the minimum test score she would need

Solution

Take the average of her current test scores, which is This means that she wants her test average after the sixth test to be that Josanna receives on her sixth test. Thus, our equation is Let be the score

Problem 3

LeRoy and Bernardo went on a week-long trip together and agreed to share the costs equally. Over the week, each of them paid for various joint expenses such as gasoline and car rental. At the end of the trip it turned out that LeRoy had paid had paid dollars, where they share the costs equally? dollars and Bernardo . How many dollars must LeRoy give to Bernardo so that

Solution

The total amount of money that was spent during the trip was pay So each person should dollars

if they were to share the costs equally. Because LeRoy has already paid

of his part, he still has to pay

Problem 4

In multiplying two positive integers and , Ron reversed the digits of the two-digit number and ? . His erroneous product was 161. What is the correct value of the product of

Solution

Taking the prime factorization of ways to represent neither reveals that it is equal to and are Therefore, the only and and Therefore, Because Because is a twoand as a product of two positive integers is

nor is a two-digit number, we know that

digit number, we know that a, with its two digits reversed, gives Multiplying our two correct values of and yields

Problem 5

Let than be the second smallest positive integer that is divisible by every positive integer less . What is the sum of the digits of ?

Solution

must be divisible by every positive integer less than , or and . Each number . The . that is divisible by each of these is is a multiple of their least common multiple. , so each number divisible by these is a multiple of smallest multiple of is clearly , so the second smallest multiple of is is

Therefore, the sum of the digits of

Problem 6

Two tangents to a circle are drawn from a point circle into arcs with lengths in the ratio . The points of contact and divide the ? . What is the degree measure of

Solution

In order to solve this problem, use of the tangent-tangent intersection theorem (Angle of intersection between two tangents dividing a circle into arc length A and arc length B = 1/2 (Arc A° - Arc B°). In order to utilize this theorem, the degree measures of the arcs must be found. First, set A (Arc length A) equal to 3d, and B (Arc length B) equal to 2d. Setting 3d+2d = 360° will find d = 72°, and so therefore Arc length A in degrees will equal 216° and arc length B will equal 144°. Finally, simply plug the two arc lengths into the tangent-tangent intersection theorem, and the answer: 1/2 (216°-144°) = 1/2 (72°)

Problem 7

Let ratio and ? be two-digit positive integers with mean . What is the maximum value of the

Solution

If is and have a mean of . , then and . To maximize , we need to , so this

maximize and minimize which gives

. Since they are both two-digit positive integers, the maximum of cannot be decreased because doing so would increase , which is

gives the maximum value of

Problem 8

Keiko walks once around a track at exactly the same constant speed every day. The sides of the track are straight, and the ends are semicircles. The track has width meters, and it takes her seconds longer to walk around the outside edge of the track than around the inside edge. What is Keiko's speed in meters per second?

Solution

To find Keiko's speed, all we need to find is the difference between the distance around the inside edge of the track and the distance around the outside edge of the track, and divide it by the difference in time it takes her for each distance. We are given the difference in time, so all we need to find is the difference between the distances. The track is divided into lengths and curves. The lengths of the track will exhibit no difference in distance between the inside and outside edges, so we only need to concern ourselves with the curves.

The curves of the track are semicircles, but since there are two of them, we can consider both of the at the same time by treating them as a single circle. We need to find the difference in the circumferences of the inside and outside edges of the circle. The formula for the circumference of a circle is circle. Let's define the circumference of the inside circle as circle as . ) is ) is , then given the thickness of the track is 6 meters, . and the circumference of the outside where is the radius of the

If the radius of the inside circle ( the radius of the outside circle (

Using this, the difference in the circumferences is:

is the difference between the inside and outside lengths of the track. Divided by the time differential, we get:

Problem 9

Two real numbers are selected independently and at random from the interval What is the probability that the product of those numbers is greater than zero? .

Solution

For the product to be greater than zero, we must have either both numbers negative or both positive.

Both numbers are negative with a Both numbers are positive with a

chance. chance.

Therefore, the total probability is

and we are done.

Problem 10

Rectangle has and . Point is chosen on side ? so that . What is the degree measure of

Solution

Since Therefore , hence . Therefore .

Problem 11

A frog located at ends at , with both and integers, makes successive jumps of length and and always lands on points with integer coordinates. Suppose that the frog starts at . What is the smallest possible number of jumps the frog makes?

Solution

Since the frog always jumps in length and lands on a lattice point, the sum of its coordinates must change either by (by jumping parallel to the x- or y-axis), or by or (based off the 3-4-5 right triangle). Because either , , or frog to go from in two moves. However, a path is possible in 3 moves: from Thus, the answer is . to to to . to is always the change of the sum of the coordinates, the sum of the in an even number of moves. Therefore, the frog cannot reach

coordinates will always change from odd to even or vice versa. Thus, it is impossible for the

Problem 12

A dart board is a regular octagon divided into regions as shown below. Suppose that a dart thrown at the board is equally likely to land anywhere on the board. What is the probability that the dart lands within the center square?

Solution

Let's assume that the side length of the octagon is . The triangles are all . The area of the center square is just . The triangles, with a side length ratio of

area of each of the identical triangles is triangles is also the side lengths is rectangles is

, so the total area of all of the

. Now, we must find the area of all of the 4 identical rectangles. One of and the other side length is , so the area of all of the

. The ratio of the area of the square to the area of the octagon is . Cancelling from the fraction, the ratio becomes . Multiplying the

numerator and the denominator each by

will cancel out the radical, so the fraction

is now

Problem 13

Brian writes down four integers differences of these numbers are ? whose sum is . The pairwise positive and . What is the sum of the possible values of

Solution

Assume that difference, and thus it is . This means results in the greatest pairwise . must be in the set . , and must be .

The only way for 3 numbers in the set to add up to 9 is if they are either Case 1 or .

then must be the remaining two numbers which are and . The ordering of

Case 2

The sum of the two w's is

Problem 14

A segment through the focus of a parabola with vertex and . What is is perpendicular to ? and intersects the parabola in points

Solution

Name the directrix of the parabola . Define and a line . Now we remember the geometric definition of a parabola: given any line (called the directrix) and any point (called the focus), the parabola corresponding to the given and . Therefore . Now note that . Similarly, , so . We now use the directrix and focus is the locus of the points that are equidistant from . Let this distance be . Therefore on triangle Law of Cosines: ; to be the distance between a point

. We now use the Pythagorean Theorem

This shows that the answer is

.

Problem 15

How many positive two-digit integers are factors of ?

Solution

From repeated application of difference of squares:

Applying sum of cubes:

A quick check shows

is prime. Thus, the only factors to be concerned about are will make any factor too large. itself will also work. The next smallest .

, since multiplying by Multiply factor, Multiply

by or will give a two digit factor;

, gives a three digit number. Thus, there are factors which are multiples of by or will also give a two digit factor, as well as

itself. Higher numbers will

not work, giving an additional factors. Multiply by or for a two digit factor. There are no mare factors to check, as all are already counted. Thus, there are an additional factors.

factors which include Multiply by or

for a two digit factor. All higher factors have been counted already, so

there are more factors. Thus, the total number of factors is

Problem 16

Rhombus area of ? has side length and . Region consists of all points inside of the rhombus that are closer to vertex than any of the other three vertices. What is the

Solution

Suppose that of . Then and is a point in the rhombus if and only if and let be the perpendicular bisector as . The line . Let us define . The is on the same side of is equal to

divides the plane into two half-planes; let similarly . Then

be the half-plane containing

region turns out to be an irregular pentagon. We can make it easier to find the area of this region by dividing it into four triangles:

Since contains , contains and , and contains with

and . Then and

are equilateral,

so

and

has area

.

Problem 17

Let integers . What is the sum of the digits of , and ? for

Solution

Proof by induction that For Assume is true for n:

:

Therefore, if it is true for n, then it is true for n+1; since it is also true for n = 1, it is true for all positive integers n. , which is the 2011-digit number 8888...8889 The sum of the digits is 8 times 2010 plus 9, or

Problem 18

A pyramid has a square base with side of length 1 and has lateral faces that are equilateral triangles. A cube is placed within the pyramid so that one face is on the base of the pyramid and its opposite face has all its edges on the lateral faces of the pyramid. What is the volume of this cube?

Solution

We can use the Pythagorean Theorem to split one of the triangular faces into two 30-60-90 triangles with side lengths and .

Next, take a cross-section of the pyramid, forming a triangle with the top of the triangle and the midpoints of two opposite sides of the square base.

This triangle is isosceles with a base of 1 and two sides of length

.

The height of this triangle will equal the height of the pyramid. To find this height, split the triangle into two right triangles, with sides and .

The cube, touching all four triangular faces, will form a similar pyramid which sits on top of the cube. If the cube has side length , the pyramid has side length .

Thus, the height of the cube plus the height of the smaller pyramid equals the height of the larger pyramid.

.

side length of cube.

Problem 19

A lattice point in an The graph of that -coordinate system is any point where both and are integers. such passes through no lattice point with . What is the maximum possible value of ? for all

Solution

It is very easy to see that the lattice. in the graph does not impact whether it passes through

We need to make sure that graph close to ,

cannot be in the form of

for

, otherwise the very

passes through lattice point at ,

. We only need to worry about

will be the only case we need to worry about and we want the , the smallest is , so answer is

minimum of those, clearly for

Problem 20

Triangle midpoints of circumcircles of has , and and , and respectively. Let . What is . The points ? , and are the be the intersection of the

Solution 1

Answer: (C) Let us also consider the circumcircle of .

Note that if we draw the perpendicular bisector of each side, we will have the circumcenter of which is and and , Also, since all intercept at , so is . . is cyclic, similarly, , are also cyclic. With this, we know that the circumcircles of

The question now becomes calculate the sum of distance from each vertices to the circumcenter. We can do it will coordinate geometry, note that circumcenter. Let , , , because of being

Then

is on the line

and also the line with slope

and passes through

.

So

and

Solution 2

Consider an additional circumcircle on each triangle has side values: , , . After drawing the diagram, it is noticed that . Thus they are congruent, and their respective , , and are the circumdiameters,

circumcircles are. By inspection, we see that

and so they are congruent. Therefore, the solution can be found by calculating one of these circumdiameters and multiplying it by a factor of . We can find the circumradius quite easily with the formula circumradius. Since : , s.t. and R is the

After a few algebraic manipulations:

.

Problem 21

The arithmetic mean of two distinct positive integers geometric mean of is ? and and is a two-digit integer. The is obtained by reversing the digits of the arithmetic mean. What

Solution

Answer: (D) for some , .

Note that in order for x-y to be integer, Since is at most , If , , if or , is impossible. -> , , , or or and

has to be

for some perfect square

.

. In AMC, we are done. Otherwise, we need

to show that

,

,

respectively. And since , .

, but there is no integer solution for

Problem 22

Let and be a triangle with sides , and . For , if to the sides , and ? and , and , , are the points of tangency of the incircle of is a triangle with side lengths

respectively, then

, if it exists. What is

the perimeter of the last triangle in the sequence

Solution

Answer: (D) Let Then Then Hence: , , , , and and ,

Note that

and

for , then

, I claim that it is true for all

, assume for

induction that it is true for some

Furthermore, the average for the sides is decreased by a factor of 2 each time. So is a triangle with side length , ,

and the perimeter of such

is

Now we need to find when that

fails the triangle inequality. So we need to find the last

such

For

, perimeter is

Problem 23

A bug travels in the coordinate plane, moving only along the lines that are parallel to the axis or to -axis. Let and of length at most . Consider all possible paths of the bug from

. How many points with integer coordinates lie on at least one of

these paths?

Solution

Answer: (C) If a point satisfy the property that is the shortest path from to to is the shortest path from , then it , and

is in the desire range because

If here. else let

, then

satisfy the property. there are

lattice points

(and for

it is symmetrical,

,

So for for etc. For

, there are , there are

lattice points,

lattice points,

, there are lattice points. lattice points.

Hence, there are a total of

Problem 24

Let . What is the minimum perimeter among all ? the -sided polygons in the complex plane whose vertices are precisely the zeros of

Solution

Answer: (B) First of all, we need to find all such that

So or

or

Now we have a solution at one and multiply by .

if we look at them in polar coordinate, further more, the 8-

gon is symmetric (it is an equilateral octagon) . So we only need to find the side length of

So answer

distance from to

Side length Hence, answer is .

Problem 25

For every odd integer and , let integers with odd, denote by the integer closest to . For every be the probability that

for an integer

randomly chosen from the interval over the odd integers in the interval

. What is the minimum ?

possible value of

Solution

Answer:

First of all, you have to realize that

if

then So, we can consider what happen in , it is always a multiple of and it will repeat. Also since range of for . is to

. So we can just consider

LET

be the fractional part function

This is an AMC exam, so use the given choices wisely. With the given choices, and the previous explanation, we only need to consider , , , .

For

,

. 3 of the

that should consider lands in here.

For

,

, then we need

else for

,

, then we need

For

, .( , no worry for the rounding to be

So, for the condition to be true, ) , so this is always true.

For

,

, so we want

, or

For k = 67,

For k = 69, etc. We can clearly see that for this case, . So for AMC purpose, answer is (D). Now, let's say we are not given any answer, we need to consider I claim that . has the minimum , which is . Also,

If

got round down, then

all satisfy the condition along with

because if and for

and , it is the same as .

, so must

, which makes .

If

got round up, then

all satisfy the condition along with

because if

and

Case 1)

-> Case 2)

->

and for -> in this set

, since

is odd, -> , and is prime so or , which is not

, which makes . Now the only case without rounding, . It must be true.

2012 AMC 12-A Problem 1

A bug crawls along a number line, starting at . It crawls to , then turns around and crawls to . How many units does the bug crawl altogether?

Solution

Crawling from of

to

takes it a distance of units. Crawling from to get

to takes it a distance

units. Add and

Problem 2

Cagney can frost a cupcake every seconds and Lacey can frost a cupcake every seconds. Working together, how many cupcakes can they frost in minutes?

Solution

Solution 1: Cagney can frost one in they can frost one in frost Solution 2: cupcakes. seconds, and Lacey can frost one in seconds. In seconds. Working together,

seconds ( minutes), they can

In

seconds ( minutes), Cagney will frost

cupcakes, and Lacey will frost

cupcakes. Therefore, working together they will frost cupcakes

Problem 3

A box centimeters high, centimeters wide, and centimeters long can hold first box can hold grams of clay. What is ? grams of clay. A second box with twice the height, three times the width, and the same length as the

Solution

The first box has volume , and the second has volume . The second has a volume that is times greater, so it holds grams.

Problem 4

In a bag of marbles, of the marbles are blue and the rest are red. If the number of red marbles is doubled and the number of blue marbles stays the same, what fraction of the marbles will be red?

Solution

Assume that there are 5 total marbles in the bag. The actual number does not matter, since all we care about is the ratios, and the only operation performed on the marbles in the bag is doubling. There are 3 blue marbles in the bag and 2 red marbles. If you double the amount of red marbles, there will still be 3 blue marbles but now there will be 4 red marbles. Thus, the answer is .

Problem 5

A fruit salad consists of blueberries, raspberries, grapes, and cherries. The fruit salad has a total of pieces of fruit. There are twice as many raspberries as blueberries, three times as many grapes as cherries, and four times as many cherries as raspberries. How many cherries are there in the fruit salad?

Solution

So let the number of blueberries be be the number of raspberries be the number of grapes and finally the number of cherries be pieces of fruit,

Observe that since there are

Since there are twice as many raspberries as blueberries, The fact that there are three times as many grapes as cherries implies, Because there are four times as many cherries as raspberries, we deduce the following:

Note that we are looking for

So, we try to rewrite all of the other variables in terms of

The third equation gives us the value of in terms of already. We divide the fourth equation by to get that provides us with the equation this equation by to get: Finally, substituting this value of into the first equation and substituting yields: Multiply

Problem 6

The sums of three whole numbers taken in pairs are number? , , and . What is the middle

Solution

Let the three numbers be equal to , , and . We can now write three equations:

Adding these equations together, we get that and

Substituting the original equations into this one, we find

Therefore, our numbers are 12, 7, and 5. The middle number is

Problem 7

Mary divides a circle into sectors. The central angles of these sectors, measured in degrees, are all integers and they form an arithmetic sequence. What is the degree measure of the smallest possible sector angle?

Solution

If we let be the smallest sector angle and be the difference between consecutive sector . Use the formula for the sum angles, then we have the angles

of an arithmetic sequence and set it equal to 360, the number of degrees in a circle.

All sector angles are integers so must be a multiple of 2. Plug in even integers for starting from 2 to minimize We find this value to be 4 and the minimum value of to be

Problem 8

An iterative average of the numbers , , , , and is computed in the following way. Arrange the five numbers in some order. Find the mean of the first two numbers, then find the mean of that with the third number, then the mean of that with the fourth number, and finally the mean of that with the fifth number. What is the difference between the largest and smallest possible values that can be obtained using this procedure?

Solution

The minimum and maximum can be achieved with the orders and .

The difference between the two is

Problem 9

A year is a leap year if and only if the year number is divisible by divisible by but not by was Dickens born? (such as ). The , Charles Dickens was celebrated on February (such as ) or is anniversary of the birth of novelist , a Tuesday. On what day of the week

Solution

Each year we go back is one day back, because back is two days back, since so 200 years would have count as a leap year. =

. Each leap year we go

. A leap year is GENERALLY every four years, leap years, but the problem points out that 1900 does not

This would mean a total of 151 regular years and 49 leap years, so days back. Since , four days back from Tuesday would be

=

Problem 10

A triangle has area , one side of length , and the median to that side of length . Let ? be the acute angle formed by that side and the median. What is

Solution

Solution 1

is the side of length

, and

is the median of length . The altitude of . To find . This is equal to

to

is

because the 0.5(altitude)(base)=Area of the triangle. is opposite over hypotenuse with the right triangle

, just use .

Solution 2

It is a well known fact that a median divides the area of a triangle into two smaller triangles of equal area. Therefore, the area of in terms of , . Solving for in the above figure. Expressing the area gives . .

Solution 3

The area of a triangle with sides and angle between them is Therefore,

as two angles along the same line must be supplementary. This simplifies to

Problem 11

Alex, Mel, and Chelsea play a game that has rounds. In each round there is a single winner, and the outcomes of the rounds are independent. For each round the probability that Alex wins is , and Mel is twice as likely to win as Chelsea. What is the probability that Alex wins

three rounds, Mel wins two rounds, and Chelsea wins one round?

Solution

If is the probability Mel wins and is the probability Chelsea wins, and and .

From this we get

. For Alex to win three, Mel to win two, and Chelsea to . Multiply this by the number of permutations

win one, in that order, is (orders they can win) which is

Problem 12

A square region point on the side is externally tangent to the circle with equation . Vertices and are on the circle with equation at the .

What is the side length of this square?

Solution

The circles have radii of and . Draw the triangle shown in the figure above and write expressions in terms of (length of the side of the square) for the sides of the triangle. Because is the radius of the larger circle, which is equal to , we can write the Pythagorean Theorem.

Use the quadratic formula.

Problem 13

Paula the painter and her two helpers each paint at constant, but different, rates. They always start at , and all three always take the same amount of time to eat lunch. of a house, quitting at . On Tuesday, of the house and quit at On Monday the three of them painted

when Paula wasn't there, the two helpers painted only

. On Wednesday Paula worked by herself and finished the house by working until . How long, in minutes, was each day's lunch break?

Solution

Let Paula work at a rate of takes to eat lunch be , the two helpers work at a combined rate of and , and the time it , where are in house/hours and L is in hours. Then the labor

on Monday, Tuesday, and Wednesday can be represented by the three following equations:

With three equations and three variables, we need to find the value of and third equations together gives us equation from this new one gives us the second equation: , so we get

. Adding the second . Subtracting the first . Plugging into

We can then subtract this from the third equation:

Plugging Converting

into our third equation gives: minutes, which is

from hours to minutes gives us

Problem 14

The closed curve in the figure is made up of congruent circular arcs each of length hexagon of side . What is the area enclosed by the curve? , where each of the centers of the corresponding circles is among the vertices of a regular

Solution

Draw the hexagon between the centers of the circles, and compute its area . Then add the areas of the three sectors outside the hexagon ( ) and subtract the areas of the three sectors inside the hexagon ( ) to get the area enclosed in the curved figure , which is .

Problem 15

A square is partitioned into unit squares. Each unit square is painted either white or clockwise about its center, and every white square in a position formerly black with each color being equally likely, chosen independently and at random. The square is the rotated occupied by a black square is painted black. The colors of all other squares are left unchanged. What is the probability that the grid is now entirely black?

Solution

First, there is only one way for the middle square to be black because it is not affected by the rotation. Then we can consider the corners and edges separately. Let's first just consider the number of ways we can color the corners. There is case with all black squares. There are four cases with one white square and all work. There are six cases with two white squares, but only the with the white squares diagonal from each other work. There

are no cases with three white squares or four white squares. Then the total number of ways to color the corners is are also . In essence, the edges work the same way, so there ways to color them. The number of ways to fit the conditions over the number of

ways to color the squares is

Problem 16

Circle circle has its center ? lying on circle and . The two circles meet at , , and and . Point in the exterior of lies on circle . What is the radius of

Solution 5

Notice that intersection of and

as they subtend arcs of the same length. Let . We now have and

be the point of

. Furthermore, notice

that

is isosceles, thus the altitude from

to

bisects

at point

above. By

the Pythagorean Theorem,

Thus,

Problem 17

Let be a subset of with the property that no pair of distinct elements in ? has a sum divisible by . What is the largest possible size of

Solution

Of the integers from to an element that is Considering , there are six each of . No element can be . We can create . No element can be with iff there is elements. iff there is an element that is several rules to follow for the elements in subset . Ignoring those that are

, we can get a subset

, there can be one element that is so because it will only be divisible . The final answer is .

by if paired with another element that is

Problem 18

Triangle has , , and . What is ? . Let denote the intersection of the internal angle bisectors of

Solution

Inscribe circle at circle Hence . Let , of radius inside triangle , and , and so that it meets . Note that , . at , and at , and ) of . . Note that angle bisectors of triangle are concurrent at the center (also

. Subtracting the last 2 equations we have

and adding this to the first equation we have

By Heron's formula for the area of a triangle we have that the area of triangle . On the other hand the area is given by Then Since the radius of circle theorem so that is perpendicular to . at

is .

, we have by the pythagorean .

so that

Problem 19

Adam, Benin, Chiang, Deshawn, Esther, and Fiona have internet accounts. Some, but not all, of them are internet friends with each other, and none of them has an internet friend outside this group. Each of them has the same number of internet friends. In how many different ways can this happen?

Solution

Note that if inclusive. Also note that the cases of and are the same, since a map showing a solution for by simply making every is the number of friends each person has, then can be any integer from to ,

can correspond one-to-one with a map of a solution for when compared to configurations of and For

pair of friends non-friends and vice versa. The same can be said of configurations with . Thus, we have two cases to examine, , and we count each of these combinations twice. , if everyone has exactly one friend, that means there must be pairs of friends,

with no other interconnections. The first person has choices for a friend. There are people left. The next person has choices for a friend. There are two people left, and these remaining two must be friends. Thus, there are For configurations with .

, there are two possibilities. The group of can be split into two groups of , with ways to pick two

each group creating a friendship triangle. The first person has triangular configurations.

friends from the other five, while the other three are forced together. Thus, there are

However, the group can also form a friendship hexagon, with each person sitting on a vertex, and each side representing the two friends that person has. The first person may be

seated anywhere on the hexagon Without loss of generality. This person has choices for the two friends on the adjoining vertices. Each of the three remaining people can be seated "across" from one of the original three people, forming a different configuration. Thus, there are . As stated before, total of has configurations, and has configurations. This gives a . hexagonal configurations, and in total configurations for

configurations, which is option

Problem 20

Consider the polynomial

The coefficient of

is equal to

. What is ?

Solution

Every term in the expansion of the product is formed by taking one term from each factor and multiplying them all together. Therefore, we pick a power of factor. Every number, including , has a unique representation by the sum of powers of two, . from the factor, , , and . from or a power of from each

and that representation can be found by converting a number to its binary form. , meaning Thus, the the Thus, from the , and . term is . So the answer is . term was made by multiplying

factor, and so on. The only numbers not used are factors,

, , and were chosen as opposed to

Thus, the coefficient of the

Problem 21

Let , , and be positive integers with such that What is ?

Solution

Add the two equations. . Now, this can be rearranged:

and factored:

, , and are all integers, so the three terms on the left side of the equation must all be perfect squares. Recognize that , since determine by inspection whether We want to solve for terms of . is the biggest difference. It is impossible to or , or whether or .

, so take the two cases and solve them each for an expression in or . Plug these . .

. Our two cases are

values into one of the original equations to see if we can get an integer for

, after some algebra, simplifies to 2021 is not divisible by 7, so The other case gives . Thus, and the answer is . is not an integer. , which simplifies to

Problem 22

Distinct planes

intersect the interior of a cube

. Let

be the union of the

faces of

and let

. The intersection of

and

consists of the union of all . ?

segments joining the midpoints of every pair of edges belonging to the same face of What is the difference between the maximum and minimum possible values of

Solution

We need two different kinds of planes that only intersect

at the mentioned segments (we 's.

call them traces in this solution). These will be all the possible

First, there are two kinds of segments joining the midpoints of every pair of edges belonging to the same face of the same face of the same face of Suppose . of a face of . Then it must also contain some trace . So, there are three and . containing both of : long traces are those connecting the midpoint of opposite sides of , and short traces are those connecting the midpoint of adjacent sides of

contains a short trace , where

an adjacent face of possibilities for Case 1:

share a common endpoint with

, each of which determines a plane . In this case,

makes an acute angle with

is an equilateral triangle made

by three short traces. There are of them, corresponding to the vertices. Case 2: is a long trace. is a rectangle. Each pair of parallel faces of such rectangles. contributes

of these rectangles so there are

Case 3: angle with

is the short trace other than the one described in case 1, i.e. . It is possible to prove that

makes an obtuse

is a regular hexagon (See note #1 for a

proof) and there are of them. Case 4: contains no short traces. This can only make be a square enclosed by

long traces. There are such squares. In total, there are possible planes in . So the maximum of is .

On the other hand, the most economic way to generate these long and short traces is to take all the planes in case 3 and case 4. Overall, they intersect at each trace exactly once (there is a quick way to prove this. See note #2 below.) and also covered all the traces. So the minimum of ... . is . The answer to this problem is then

Problem 23

Let point that be the square one of whose diagonals has endpoints and . Let and and .A such centered at . What is chosen uniformly at random over all pairs of real numbers be a translated copy of

is the probability that the square region determined by integer coefficients in its interior?

contains exactly two points with

Solution

The unit square's diagonal has a length of parallel to the axis, the two points must be adjacent. Now consider the unit square consider only two vertices, that the point and adjacent vertices. For to contain the point , must be inside square . with vertices and ,

. Because

square is not

and

. Let us first

. We want to find the area of the region within such that it covers both

will create the translation of

. By symmetry, there will be three equal regions that cover the other pairs of

. Similarly, for

to contain

the point

, must be inside a translated square

with center at

, which we will call

. Therefore, the area we seek is Area To calculate the area, we notice that Area midpoint of within Area Noting that the side length of , so and Therefore, the area is Area , and , and along the line and

Area . Let . Let

by symmetry. Let be the and

be the intersection of

be the intersection of . Because and

outside

. Therefore, the area we seek is , they are collinear.

all have coordinate . If follows that

is (as shown above), we also see that . . that we need to count, the total . . Therefore,

Because there are three other regions in the unit square area of within such that

contains two adjacent lattice points is and

By periodicity, this probability is the same for all the answer is .

Problem 24

Let be the sequence of real numbers defined by , and in general,

Rearranging the numbers in the sequence sequence . What is the sum of all integers

in decreasing order produces a new , , such that

Solution

First, we must understand two important functions: exponential function), and ). constant. for for (decreasing (increasing exponential function for positive

is used to establish inequalities when we change the exponent and keep the base is used to establish inequalities when we change the base and keep the

exponent constant. We will now examine the first few terms. Comparing and , . Therefore, Comparing and , . Comparing and , . Therefore, Comparing and , . Comparing and , . Therefore, . . .

Continuing in this manner, it is easy to see a pattern(see Note 1). Therefore, the only . when is when . Solving gives

Problem 25

Let solutions. What is and where denotes the fractional part of . The number has at least is a real number is the real such that smallest positive integer such that the equation ? Note: the fractional part of is an integer.

Solution

Our goal is to determine how many times intersects the line the behavior of . We begin by analyzing . It increases linearly with a slope of one, then when it reaches the next . The slope of the teeth is

integer, it repeats itself. We can deduce that the function is like a sawtooth wave, with a period of one. We then analyze the function multiplied by 2 to get 2, and the function is moved one unit downward. The function can then be described as starting at -1, moving upward with a slope of 2 to get to 1, and then repeating itself, still with a period of 1. The absolute value of the function is then taken. This results in all the negative segments becoming flipped in the Y direction. The positive slope starting at -1 of the function ranging from to , where u is any arbitrary integer, is now a negative slope starting at positive 1. The function now looks like the letter V repeated within every square in the first row. It is now that we address the goal of this, which is to determine how many times the function intersects the line If the height of the function is higher than that it is multiplied by , and then fed into . Since there are two line for every integer. segments per box, the function has two chances to intersect the line

for every integer on an interval, then every . Since is a periodic function, we can chances for every chances for every integer. The , since , since the height is

chance within that interval intersects the line. Returning to analyzing the function, we note model it as multiplying the function's frequency by . This gives us

integer, which is then multiplied by 2 once more to get n. The function intersects the line of the function is lower than then therefore equal to then to

amplitude of this function is initially 1, and then it is multiplied by n, to give an amplitude of for every chance in the interval of when . The number of times the function intersects . It is easy to see that this is equal to . the function is n units high. The function ceases to intersect

. The problem , and

then simplifies to the algebraic expression , which rounds up to 32.

, which simplifies to,

2012 AMC 12-B Problem 1

Each third-grade classroom at Pearl Creek Elementary has 18 students and 2 pet rabbits. How many more students than rabbits are there in all 4 of the third-grade classrooms?

Solution 1

Multiplying and by we get students and rabbits. We then subtract:

Solution 2

In each class, there are more students than rabbits. So for all classrooms, the

difference between students and rabbits is

Problem 2

A circle of radius 5 is inscribed in a rectangle as shown. The ratio of the length of the rectangle to its width is 2:1. What is the area of the rectangle?

Solution

Note that the diameter of the circle is equal to the shorter side of the rectangle. Since the radius is , the diameter is the longer side has length . Since the sides of the rectangle are in a . Therefore the area is or ratio,

Problem 3

For a science project, Sammy observed a chipmunk and squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide?

Solution

If is the number of holes that the chipmunk dug, then , so , and . The number of acorns hidden by the chipmunk is equal to

Problem 4

Suppose that the euro is worth 1.30 dollars. If Diana has 500 dollars and Etienne has 400 euros, by what percent is the value of Etienne's money greater that the value of Diana's money?

Solution

The ratio can be simplified using conversion factors: which means the money is greater by percent.

Problem 5

Two integers have a sum of 26. when two more integers are added to the first two, the sum is 41. Finally, when two more integers are added to the sum of the previous 4 integers, the sum is 57. What is the minimum number of even integers among the 6 integers?

Solution

So, x+y=26, x could equal 15, and y could equal 11, so no even integers required here. 4126=15. a+b=15, a could equal 9 and b could equal 6, so one even integer is required here. 57-41=16. m+n=16, m could equal 9 and n could equal 7, so no even integers required here, meaning only 1 even integer is required; A.

Problem 6

In order to estimate the value of Xiaoli rounded where and are real numbers with , up by a small amount, rounded down by the same amount, and then

subtracted her rounded values. Which of the following statements is necessarily correct?

Solution

The original expression where now becomes , is a positive constant, hence the answer is (A).

Problem 7

Small lights are hung on a string 6 inches apart in the order red, red, green, green, green, red, red, green, green, green, and so on continuing this pattern of 2 red lights followed by 3 green lights. How many feet separate the 3rd red light and the 21st red light? Note: 1 foot is equal to 12 inches.

Solution

We know the repeating section is made of 2 red lights and 3 green lights. The 3rd red light would appear in the 2nd section of this pattern, and the 21st red light would appear in the

11th section. There would then be a total of 44 lights in between the 3rd and 21st red light, translating to 45 6-inch gaps. Since it wants the answer in feet, so the answer is

Problem 8

A dessert chef prepares the dessert for every day of a week starting with Sunday. The dessert each day is either cake, pie, ice cream, or pudding. The same dessert may not be served two days in a row. There must be cake on Friday because of a birthday. How many different dessert menus for the week are possible?

Solution

We can count the number of possible foods for each day and then multiply to enumerate the number of combinations. On Friday, we have one possibility: cake. On Saturday, we have three possibilities: pie, ice cream, or pudding. This is the end of the week. On Thursday, we have three possibilities: pie, ice cream, or pudding. We can't have cake because we have to have cake the following day, which is the Friday with the birthday party. On Wednesday, we have three possibilities: cake, plus the two things that were not eaten on Thursday. Similarly, on Tuesday, we have three possibilities: the three things that were not eaten on Wednesday. Likewise on Monday: three possibilities, the three things that were not eaten on Tuesday. On Sunday, it is tempting to think there are four possibilities, but remember that cake must be served on Friday. This serves to limit the number of foods we can eat on Sunday, with the result being that there are three possibilities: The three things that were not eaten on Monday. So the number of menus is The answer is

Problem 9

It takes Clea 60 seconds to walk down an escalator when it is not moving, and 24 seconds when it is moving. How many seconds would it take Clea to ride the escalator down when she is not walking?

Solution

She walks at a rate of and to each other, units per second to travel a distance , which means that . As , we find by , where is the speed of the escalator. Setting the two equations equal . Now we divide

because you add the speed of the escalator but remove the walking, leaving the final answer that it takes to ride the escalator alone as

Problem 10

What is the area of the polygon whose vertices are the points of intersection of the curves and ?

Solution

The first curve is a circle with radius 5 centered at the origin, and the second curve is an ellipse with center (4,0) and end points of (-5,0) and (13,0). Finding points of intersection, we get (-5,0) (4,3) and (4,-3), forming a triangle with height of 9 and base of 6. So 9x6x0.5 =27 ; B.

Problem 11

In the equation below, and are consecutive positive integers, and What is ? , , and represent number bases:

Solution

Change the equation to base 10:

Either expressed as .

or

, so either

or . Since A must be positive, and

. The second case has no integer roots, and the first can be re-

Problem 12

How many sequences of zeros and ones of length 20 have all the zeros consecutive, or all the ones consecutive, or both?

Solution 1

There are selections; however, we count these twice, therefore

. The wording of the question implies D not E. MAA decided to accept both D and E, however.

Solution 2

Consider the 20 term sequence of 0's and 1's. Keeping all other terms 1, a sequence of consecutive 0's can be placed in locations. That is, there are 20 strings with 1 zero, 19 strings with 2 consecutive zeros, 18 strings with 3 consecutive zeros, ..., 1 string with 20 consecutive zeros. Hence there are consecutive zeros. The same argument shows there are This yields strings with strings with consecutive 1's.

strings in all. However, we have counted twice those strings in which all , , , , , ..., , ...,

the 1's and all the 0's are consecutive. These are the cases (of which there are 19) as well as the cases

(of which there are 19 as well). This yields answer is .

so that the

Problem 13

Two parabolas have equations and , where , , , and are integers, each chosen independently by rolling a fair six-sided die. What is the probability that the parabolas will have a least one point in common?

Solution 1

Set the two equations equal to each other: squared and get x's on one side: out . Now factor : cannot equal , then there is always a solution, but if that to work, is if , a in . Now remove the x . If a , then the only way for , a in chance, leaving a

, always having at least one point in common. And if

chance, however, this can occur ways, so a in chance of , we get the simplified fraction of ; answer

this happening. So adding one sixth to .

Solution 2

Proceed as above to obtain . The probability that the parabolas have at and . The probability that is least 1 point in common is 1 minus the probability that they do not intersect. The equation has no solution if and only if while the probability that that the parabolas intersect. is . Thus we have

for the probability

Problem 14

Bernardo and Silvia play the following game. An integer between 0 and 999 inclusive is selected and given to Bernardo. Whenever Bernardo receives a number, he doubles it and

passes the result to Silvia. Whenever Silvia receives a number, she addes 50 to it and passes the result to Bernardo. The winner is the last person who produces a number less than 1000. Let N be the smallest initial number that results in a win for Bernardo. What is the sum of the digits of N?

Solution 1

The last number that Bernado says has to be between 950 and 999. Note that contains 4 doubling actions. Thus, we have . Thus, . Then, . Working backwards from 956, . So the starting number is 16, and our answer is , which is A. . If , we have

Solution 2

Work backwards. The last number Bernardo produces must be in the range means that before this, Silvia must produce a number in the range Bernardo must produce a number in the range number in the range . That . Before this,

. Before this, Silvia must produce a . Before this,

. Before this, Bernardo must produce a number in the range . Before this, Silvia must produce a is 16 with the sum of digits

. Before this, Silvia must produce a number in the range Bernardo must produce a number in the range number in the range to obtain a number in the range being .

. Bernardo could not have added 50 to any number before this , hence the minimum

Problem 15

Jesse cuts a circular paper disk of radius 12 along two radii to form two sectors, the smaller having a central angle of 120 degrees. He makes two circular cones, using each sector to form the lateral surface of a cone. What is the ratio of the volume of the smaller cone to that of the larger?

Solution

If the original radius is the smaller arc is , then the circumference is ; since arcs are defined by the central angles, the smaller arc, a degree angle, is half the size of the larger sector. so . Those two arc lengths become the two and the smaller

, and the larger is

circumferences of the new cones; so the radius of the smaller cone is and the larger cone is . Using the Pythagorean theorem, the height of the larger cone is cone is get the volume of each cone: and , and now for volume just square the radii and multiply by of the height to [both multiplied by three as ratio come

out the same. now divide the volumes by each other to get the final ratio of

Problem 16

Amy, Beth, and Jo listen to four different songs and discuss which ones they like. No song is liked by all three. Furthermore, for each of the three pairs of the girls, there is at least one song liked by those girls but disliked by the third. In how many different ways is this possible?

Solution 1

Let the ordered triple denote that . To show this, observe these are all valid conditions. Second, note that none of bigger than 3. Suppose otherwise, that can be songs are liked by Amy and Beth, songs by Beth and Jo, and songs by Jo and Amy. We claim that the only possible triples are

. Without loss of generality, say that Amy and

Beth like songs 1, 2, and 3. Then because there is at least one song liked by each pair of girls, we require either or to be at least 1. In fact, we require either or to equal 1, otherwise there will be a song liked by all three. Suppose since no song is liked by all three girls, a contradiction. Case 1:How many ways are there for to equal ? There are 4 choices for . Then we must have

which song is liked by Amy and Beth, 3 choices for which song is liked by Beth and Jo, and

2 choices for which song is liked by Jo and Amy. The fourth song can be liked by only one of the girls, or none of the girls, for a total of 4 choices. So ways. in

Case 2:To find the number of ways for

, observe there are

choices of songs for the first pair of girls. There remain 2 choices of songs for the next pair (who only like one song). The last song is given to the last pair of girls. But observe that we let any three pairs of the girls like two songs, so we multiply by 3. In this case there are ways for the girls to like the songs. That gives a total of . ways for the girls to like the song, so the answer is

Solution 2

We begin by noticing that there are four ways to assign a song liked by both Amy and Beth, three ways to assign a song liked by both Amy and Jo (because Jo may not like the song liked by both Amy and Beth), and two ways to assign a song liked by both Beth and Jo (because both Beth and Jo may not like the song liked by the previous pairs. Additionally, there are obtain an answer of two girls but not the third. We proceed again by ignoring the cases in which the fourth song is liked by two girls but not the third. There are of these cases. However, in doing so, we have committed yet another egregious error. The cases in which the fourth song is liked by two girls but not the third have not been accounted for! In performing our past two calculations, we have, however, established that the answer has a lower bound of and an upper bound of . As . is the only answer within these bounds, we conclude that the answer must be ways to assign song preferences for the fourth song. Multiplying, we . However, in doing so, we have committed an

egregious error. We have in fact over counted the cases in which the fourth song is liked by

Solution 3: A Different Way of Looking at Solution 1

Let , and denote a song that is liked by Amy and Beth (but not Jo), Beth and Jo and (but not Amy), and Amy and Jo (but not Beth), respectively. Similarly, let Since we know that there is at least , and

denote a song that is liked by only Amy, only Beth, only Jo, and none of them, respectively. , they must be songs out of the

that Amy, Beth, and Jo listened to. The fourth song can be of any type and the set Case 1: Fourth song = Note that in Case 1, all four of the choices for the fourth song are different from the first three songs. Number of ways to rearrange = Case 2: Fourth song = Note that in Case , all three of the choices for the fourth song repeat somewhere in the first three songs. Number of ways to rearrange = . rearrangements for each choice choices = . rearrangements for each choice choices = . (there is no because no song is liked by all three, as stated in the problem.) , and a song from . Therefore, we must find the number of ways to rearrange

,

Problem 17

Square square lies in the first quadrant. Points and ? and lie on lines , respectively. What is the sum of the coordinates of the center of the

Solution 1

Note that the center of the square lies along a line that has an and also along another line with intercept intercept of ,

. Since these 2 lines are parallel be

to the sides of the square, they are perpindicular (since the sides of a square are). Let the slope of the first line. Then

is the slope of the second line. We may use the pointand .

slope form for the equation of a line to write

We easily calculate the intersection of these lines using substitution or elimination to obtain

as the center or the square. Let denote the (acute) angle formed by and the that where upon we obtain Hence the answer is . axis. Note that . Using . Then . Let denote the side length of the square. Then and the axis is so (for acute ) we have . Substituting into . . On the other hand the acute angle formed by

so that the sum of the coordinates is

Solution 2 (Fast)

Suppose

where

. and and equals

Recall that the distance between two parallel lines is to them, we get .

, we have distance between and equals to

, and the distance between

. Equating

Then, the center of the square is just the intersection between the following two "mid" lines:

The solution is

, so we get the answer

.

.

Problem 18

Let either are there? or be a list of the first 10 positive integers such that for each or both appear somewhere before in the list. How many such lists

Solution

Let greater than neither

. Assume that must be nor

. If

, the first number appear after larger than must be . Similarly, one can conclude that if

that is ,

, otherwise if it is any number appearing before that is larger than

, there will be , and so forth. must be ,

the first number appear after On the other hand, if and then

, the first number appear after

that is less than

, and so forth. is given, we set up spots after . to , we get , and .

To count the number of possibilities when assign of them to the numbers less than The number of ways in doing so is choose Therefore, when summing up the cases from

and the rest to the numbers greater than

Problem 19

A unit cube has vertices adjacent to , and for vertices and and . Vertices , , , , , and , are , and are opposite to each other. A regular

octahedron has one vertex in each of the segments . What is the octahedron's side length?

Solution

Observe the diagram above. Each dot represents one of the six vertices of the regular octahedron. Three dots have been placed exactly x units from the (0,0,0) corner of the unit cube. The other three dots have been placed exactly x units from the (1,1,1) corner of the unit cube. A red square has been drawn connecting four of the dots to provide perspective regarding the shape of the octahedron. Observe that the three dots that are near (0,0,0) are each (x)( ) from each other. The same is true for the three dots that are near (1,1,1). ). There is a unique x for which the rectangle drawn in red becomes a square. This will occur when the distance from (x,0,0) to (1,1-x, 1) is (x)(

Using the distance formula we find the distance between the two points to be: = sides, we have the equation: = . Equating this to (x)( ) and squaring both

=

. ), we have a final result of . Thus, Answer choice

Since the length of each side is (x)( A is correct

Problem 20

A trapezoid has side lengths 3, 5, 7, and 11. The sums of all the possible areas of the trapezoid can be written in the form of rational numbers and and What is the greatest integer less than or equal to , where , , and ? are are positive integers not divisible by the square of any prime.

Solution

Name the trapezoid Draw a line through , where parallel to is parallel to , crossing the side , at , and . Then , , so there .

. One needs to guarantee that are only three possible trapezoids:

In the first case,

, so . Therefore the area of this trapezoid is .

In the second case,

, so . Therefore the area of this trapezoid is .

In the third case, So .

, therefore the area of this trapezoid is , which is rounded down to

.

Problem 21

Square and on the square? is inscribed in equiangular hexagon . Suppose that , and with on , on , . What is the side-length of

Solution (Long)

Extend and so that they meet at , hence . Let , , and . . Then . Also, since is congruent to , so is parallel and equal to . We now get and therefore , we get is parallel to

Drop a perpendicular line from perpendicular line from to since

to the line of

that meets line at , then

at

, and a is congruent

to the line of

that meets

is complementary to

. Then we have the following equations:

The sum of these two yields that

So, we can now use the law of cosines in

:

Therefore

Problem 22

A bug travels from to along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?

Solution

There is way to get to any of the red arrows. From the first red arrow, there are ways to get to each of the first and the second blue arrows; from the second red arrow, there are ways to get to each of the first and the second blue arrows. So there are in total ways to get to each of the blue arrows. From each of the first and second blue arrows, there are respectively ways to get to each of the first and the second green arrows; from each of the third and the fourth blue arrows, there are respectively ways to get to each of the first and the second green arrows. Therefore there are in total arrows. ways to get to each of the green

Finally, from each of the first and second green arrows, there is respectively ways to get to the first orange arrow; from each of the third and the fourth green arrows, there are to get to the first orange arrow. Therefore there are get to each of the orange arrows, hence ways to get to the point . ways ways to

Problem 23

Consider all polynomials of a complex variable, and are integers, , and the polynomial has a zero What is the sum of all values , where with

over all the polynomials with these properties?

Solution

First, assume that . Then . Also, has to be true since are . The sum of , therefore the only possible choices for , so or . does not work because , we have . Now over these cases is . In these cases, . Assume that , so gives

. Second, assume that , so for some real , therefore is a factor of , and we may assume that , . By

conjugate roots theorem we have that

for some real

. Expanding this polynomial and comparing the coefficients, we have the

following equations:

From the first and the third we may deduce that (we will consider by the end). Let

and that

, if . From the

second equation, we know that

is non-negative.

Consider the following cases: Case 1: . Then , , so ). , we have . Also, we get . has a root and that . However, . has a root . . , therefore , , . However, this has already

been found (i.e. the form of Case 2: so Last case: . Therefore the desired sum is , . Then since . This is true only when , . We have

again. In this case,

.

Problem 24

Define the function For every the sequence , let on the positive integers by setting , then . For how many unbounded? , there is a in the range is and if is the prime factorization of

Note: A sequence of positive numbers is unbounded if for every integer member of the sequence greater than .

Solution

First of all, notice that for any odd prime than but when number , the largest prime that divides is no larger

, therefore eventually the factorization of

does not contain any prime , when it stays the same or any that is divisible by also works:

larger than . Also, note that such that is divisible by multiples of .

it grows indefinitely. Therefore any number within .

makes the sequence

unbounded. There are

Now let's look at the other cases. Any first power of prime in a prime factorization will not contribute the unboundedness because to contribute. So we test primes that are less than . At least a square of prime is :

works, therefore any number are of them. could also work if satisfies , but

that are divisible by

works: there

.

does not work. works. There are no other multiples of could also work if , but within .

already. , they don't work because none of nor a multiple of 's ... . .

For number that are only divisible by these primes are such that In conclusion, there are could be a multiple of

number of

Problem 25

Let set of all right triangles whose vertices are in vertices , , and . For every right triangle , let in counter-clockwise order and right angle at . What is . Let be the with

Solution

Consider reflections. For any right triangle problem, any reflection one vertex at the line have vertices Case 1: . Then First we consider the reflection about the line do not reflect to a traingle do not reflect to a triangle reflection about the line with the right labeling described in the . that have . Only those triangles labeled that way will give us . Within those triangles, consider a that have one vertex on

. Then only those triangles . There are three cases: is impossible.

. So we only need to look at right triangles that

Case 2: They are:

. Then we look for , is

such that and .

and that .

.

The product of their values of Case 3: . Then

is impossible.

Therefore

is the answer.

2013 AMC12-A Problem 1

Square ? has side length . Point is on , and the area of is . What is

Solution

We are given that the area of The area of a triangle: is , and that .

Using

as the height of

,

and solving for b, , which is

Problem 2

A softball team played ten games, scoring

, and

runs. They lost by one

run in exactly five games. In each of the other games, they scored twice as many runs as their opponent. How many total runs did their opponents score?

Solution

To score twice as many runs as their opponent, the softball team must have scored an even number. Therefore we can deduce that when they scored an odd number of runs, they lost by one, and when they scored an even number of runs, they won by twice as much. Therefore, the total runs by the opponent is , which is

Problem 3

A flower bouquet contains pink roses, red roses, pink carnations, and red carnations. One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink. What percent of the flowers are carnations?

Solution

To score twice as many runs as their opponent, the softball team must have scored an even number. Therefore we can deduce that when they scored an odd number of runs, they lost by one, and when they scored an even number of runs, they won by twice as much. Therefore, the total runs by the opponent is , which is

Problem 4

What is the value of

Solution

We can factor a

out of the numerator and denominator to obtain

The

cancels, so we get

, which is

Problem 5

Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid $ , Dorothy paid $ , and Sammy paid $ . In order to share the dollars. What is ? costs equally, Tom gave Sammy dollars, and Dorothy gave Sammy

Solution

Add up the amounts that Tom, Dorothy, and Sammy paid to get $ get $ , the amount that each should have paid. , owes Sammy $ , and Dorothy, having paid $ , owes Sammy , and divide by 3 to

Tom, having paid $ $ .

Thus,

, which is

Problem 6

In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on attempted of her three-point shots and of her two-point shots. Shenille shots. How many points did she score?

Solution

Let the number of 3-point shots attempted be of 2-point shots attempted must be . . Since she attempted 30 shots, the number

Since she was successful on then her score must be

, or

, of her 3-pointers, and

, or

, of her 2-pointers,

, which is

Problem 7

The sequence and . What is has the property that every term beginning with the third Suppose that ? is the sum of the previous two. That is,

Solution

,

Therefore, the answer is

Problem 8

Given that and are distinct nonzero real numbers such that , what is ?

Solution

Since

, we may assume that

and/or, equivalently, .

.

Cross multiply in either equation, giving us

Problem 9

In , , respectively, such that the perimeter of parallelogram and and ? . Points are parallel to and are on sides , , and and , respectively. What is

Solution

Note that because and It follows that and are parallel to the sides of , the internal triangles are similar to . Thus, , and are therefore also isosceles triangles. . .

Since opposite sides of parallelograms are equal, the perimeter is

Problem 10

Let be the set of positive integers with for which has the repeating decimal representation ? and different digits. What is the sum of the elements of

Solution

Note that Dividing by 3 gives . , and dividing by 9 gives .

The answer must be at least the conditions, so the answer is

, but cannot be .

since no

other than

satisfies

Problem 11

Triangle on trapezoids is equilateral with and and . Points are parallel to and are on and points ? and and are such that both . Furthermore, triangle

all have the same perimeter. What is

Solution

Let , and . We want to find , , and , which is nothing but . Based on the fact that the following: have the same perimeters, we can say

Simplifying, we can find that

Since

,

.

After substitution, we find that

, and

=

.

Again substituting, we find

=

.

Therefore,

=

, which is

Problem 12

The angles in a particular triangle are in arithmetic progression, and the side lengths are . The sum of the possible values of x equals positive integers. What is ? where , and are

Solution

Because the angles are in an arithmetic progression, and the angles add up to second largest angle in the triangle must be , the . Also, the side opposite of that angle must

be the second longest because of the angle-side relationship. Any of the three sides, , , or , could be the second longest side of the triangle. The law of cosines can be applied to solve for When the second longest side is , we get that . By using the quadratic formula, . When the second longest side is . When the second longest side is , we get that . Using the quadratic formula, real, therefore the second longest side cannot equal . Adding the two other possibilities gets , which is answer choice . , with , and . . However, , therefore is not , we get that , therefore in all three cases. , therefore , therefore

Problem 13

Let points and . Quadrilateral at point is cut ,

into equal area pieces by a line passing through where these fractions are in lowest terms. What is

. This line intersects ?

Solution

If you have graph paper, use Pick's Theorem to quickly and efficiently find the area of the quadrilateral. If not, just find the area by other methods. Pick's Theorem states that = - , where is the number of lattice points in the interior of the polygon, and is

the number of lattice points on the boundary of the polygon. In this case,

= so =

- =

The bottom half of the quadrilateral makes a triangle with base and half the total area, so we can deduce that the height of the triangle must be in order for its area to be .

This height is the y coordinate of our desired intersection point.

Note that segment CD lies on the line that the x coordinate of our intersection point is .

. Substituting in

for y, we can find

Therefore the point of intersection is ( , which is .

,

), and our desired result is

Problem 14

The sequence , , , , ?

is an arithmetic progression. What is

Solution

Since the sequence is arithmetic, + = , where is the common difference.

Therefore, = = , and

=

(

)= , we just add it to the first term to find = = = = , which is :

Now that we found + =

Problem 15

Rabbits Peter and Pauline have three offspring—Flopsie, Mopsie, and Cotton-tail. These five rabbits are to be distributed to four different pet stores so that no store gets both a parent and a child. It is not required that every store gets a rabbit. In how many different ways can this be done?

Solution

There are two possibilities regarding the parents. 1) Both are in the same store. In this case, we can treat them both as a single bunny, and they can go in any of the 4 stores. The 3 baby bunnies can go in any of the remaining 3 stores. There are combinations.

2) The two are in different stores. In this case, one can go in any of the 4 stores, and the other can go in any of the 3 remaining stores. The 3 baby bunnies can each go in any of the remaining 2 stores. There are Adding up, we get combinations. combinations.

Problem 16

, and , is are three piles of rocks. The mean weight of the rocks in is is pounds, the mean and is weight of the rocks in pounds, the mean weight of the rocks in the combined piles

pounds, and the mean weight of the rocks in the combined piles and ?

pounds. What is the greatest possible integer value for the mean in pounds of the rocks in the combined piles

Solution

Let pile have rocks, and so on. and can be expressed as and , we add the weight of . and subtract the weight of

The total weight of

To get the total weight of

Therefore, the mean of

and

is

, which is simplified to

.

We now need to eliminate

in the numerator. Since we know that , we have

Substituting,

, so the maximum value occurs when

. Since

must cancel to give an

integer, and the only fraction that satisfies both conditions is Plugging in, we get

Problem 17

A group of follows. The pirates agree to divide a treasure chest of gold coins among themselves as pirate to take a share takes of the coins that remain in the chest. The

number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the pirate receive?

Solution

The first pirate takes of the coins, leaving .

The second pirate takes

of the remaining coins, leaving

.

Continuing this pattern, the eleventh pirate must take first ten pirates have taken their share, which leaves this.

of the remaining coins after the . The twelfth pirate takes all of

Note that

All the 2s and 3s cancel out of

, leaving

in the denominator.

Problem 18

Six spheres of radius are positioned so that their centers are at the vertices of a regular hexagon of side length . The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?

Solution

Set up an isosceles triangle between the center of the 8th sphere and two opposite ends of the hexagon. Then set up another triangle between the point of tangency of the 7th and 8th spheres, and the points of tangency between the 7th sphere and 2 of the original spheres on opposite sides of the hexagon. Express each side length of the triangles in terms of r (the radius of sphere 8) and h (the height of the first triangle). You can then use Pythagorean Theorem to set up two equations for the two triangles, and find the values of h and r.

Problem 19

In points , and , and . Moreover and . A circle with center and radius ? intersects at have integer lengths. What is

Solution

Let . Let the circle intersect at and the diameter including intersect the circle again at So Obviously so we have three solution pairs for . By the Triangle Inequality, only yields a possible length of Therefore, the answer is D) 61. . . Use power of a point on point C to the circle centered at A. .

Problem 20

Let be the set or property that , . For , and ? , define to mean that either of elements of have the . How many ordered triples

Solution

Imagine 19 numbers are just 19 persons sitting evenly around a circle facing to the center. One may check that " implies that more than if and only if is one of the 9 persons on the left of . Therefore, " and cuts the circumference of , and and if ; each of them is

and only if is one of the 9 persons on the right of numbers sitting on it (inclusive).

into three arcs, each of which has no

We count the complement: where the cut generated by than persons in total. Suppose the number of persons on the longest arc is third person. Once the three places of put an arc of length

has ONE arc that has more

persons sitting on. Note that there can only be one such arc because there are only

. Then two places of

are

just chosen from the two end-points of the arc, and there are into them clockwise. Also, note that for any

possible places for the ways to choose

are chosen, there are three possible ways to , there are

. Therefore the total number of ways (of the complement) is

So the answer is

NOTE: this multiple choice problem can be done even faster -- after we realized the fact that each choice of the three places of and that each arc of length should be divisible by has corresponds to possible ways to put them in, equitable positions, it is evident that the answer from the five choices.

, which can only be

Problem 21

Consider the following intervals contains ? . Which of

Solution

Let and , and from the problem description,

We can reason out an approximation, by ignoring the

:

And a better approximation, by plugging in our first approximation for original definition for :

in our

And an even better approximation:

Continuing this pattern, obviously, will eventually terminate at original definition of .

, in other words our

However, at choices. So we take

, going further than

will not distinguish between our answer .

is nearly indistinguishable from and plug in.

Since

, we know

. This gives us our answer range:

Problem 22

A palindrome is a nonnegative integer number that reads the same forwards and backwards when written in base 10 with no leading zeros. A 6-digit palindrome random. What is the probability that is also a palindrome? is chosen uniformly at

Solution

Working backwards, we can multiply 5-digit palindromes palindrome: by , giving a 6-digit

Note that if

or

, then the symmetry will be broken by carried 1s for which and

Simply count the combinations of implies possible

(0 through 8), for each of which there are valid palindromes when

possible C, respectively. There are implies possible

(0 through 7), for each of which there are valid palindromes when

possible C, respectively. There are Following this pattern, the total is

6-digit palindromes are of the form there are So, the probability is combinations of

, and the first digit cannot be a zero, so

Problem 23

is a square of side length region bounded by region whose area is What is ? is rotated , where . Point is on such that . The square , sweeping out a . counterclockwise with center

, , and are positive integers and

Solution

We first note that diagonal is of length . It must be that divides the diagonal into two segments in the ratio by to . It is not difficult to visualize that when the

square is rotated, the initial and final squares overlap in a rectangular region of dimensions . The area of the overall region (of the initial and final squares) is therefore twice . the area of the original square minus the overlap, or

The area also includes circular segments. Two are quarter-circles centered at (the segment bounded by formed as and ) and (that bounded by and is the bottom-left vertex and

of radii ). Assuming

is the bottom-right one, it is clear that the third segment is overshoots the final square's left edge. to . Call the point of intersection and . This means (since it left the edge

swings out to the right of the original square [recall that the square is rotated

counterclockwise], while the fourth is formed when To find these areas, consider the perpendicular from . From the previous paragraph, it is clear that , and swings back inside edge

at a point unit above

unit below). The triangle of the circular sector is therefore an equilateral triangle of side length , and so the angle of the segment is clear that the situation is the same with point . Imagining the process in reverse, it is .

The area of the segments can be found by subtracting the area of the triangle from that of

the sector; it follows that the two quarter-segments have areas and both have area . . The other two segments

The total area is therefore

Since

,

, and

, the answer is

Problem 24

Three distinct segments are chosen at random among the segments whose end-points are the vertices of a regular 12-gon. What is the probability that the lengths of these three segments are the three side lengths of a triangle with positive area?

Solution

Suppose is the answer. We calculate . , , , Assume that the circumradius of the 12-gon is , and the 6 different lengths are , in increasing order. Then . So , , , , ,

.

Now, Consider the following inequalities: : Since . is greater than is greater than is greater than but less than but equal to . . .

. Then obviously any two segments with at least one them longer than have a sum greater than .

Therefore, all triples (in increasing order) that can't be the side lengths of a triangle are the following. Note that x-y-z means 1-1-3, 1-1-4, 1-1-5, 1-1-6, 1-2-4, 1-2-5, 1-2-6, 1-3-5, 1-3-6, 2-2-6 Note that there are segments of length segments of each length of . There are , , , , respectively, and :

segments in total.

In the above list there are triples of the type a-a-b without 6, triples of a-a-6 where a is not 6, triples of a-b-c without 6, and triples of a-b-6 where a, b are not 6. So,

So

.

Problem 25

Let such that

be defined by ?

. How many complex numbers are there are integers with

and both the real and the imaginary parts of

absolute value at most

Solution

Suppose integers where . . We look for with such that are

First, use the quadratic formula:

Generally, consider the imaginary part of a radical of a complex number: .

, where

. Now let , then , , .

Note that

if and only if , , , the equation

. The latter is true only when we take the

positive sign, and that or In other words, when region ); and when , and that When when pairs. So there are .

, or

. has unique solution in the such that integers

there is no solution. Therefore the number of

desired solution is the same as the number of ordered pairs

, there is no restriction on so there are , there are

pairs;

in total.

2013 AMC12-B Problem 1

On a particular January day, the high temperature in Lincoln, Nebraska, was In degrees, what was the low temperature in Lincoln that day? degrees . higher than the low temperature, and the average of the high and low temperatures was

Solution

Let be the low temperature. The high temperature is . Solving for , we get . The average is

Problem 2

Mr. Green measures his rectangular garden by walking two of the sides and finds that it is steps by steps. Each of Mr. Green’s steps is feet long. Mr. Green expects a half a pound of potatoes per square foot from his garden. How many pounds of potatoes does Mr. Green expect from his garden?

Solution

Since each step is feet, his garden is feet. Since he is expecting expected is by feet. Thus, the area of square

of a pound per square foot, the total amount of potatoes

Problem 3

When counting from to to , is the , is the number counted. When counting backwards from ? number counted. What is

Solution

Note that is equal to the number of integers between and , inclusive. Thus,

Problem 4

Ray's car averages miles per gallon of gasoline, and Tom's car averages miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline?

Solution

Let both Ray and Tom drive 40 miles. Ray's car would require Tom's car would require miles, on gallon of gas and

gallons of gas. They would have driven a total of gallons of gas, for a combined rate of

Problem 5

The average age of fifth-graders is . The average age of of their parents is . What is the average age of all of these parents and fifth-graders?

Solution

The sum of the ages of the fifth graders is is , while the sum of the ages of the parents , and given . . Therefore, the total sum of all their ages must be

people in total, their average age is

Problem 6

Real numbers and satisfy the equation . What is ?

Solution

If we complete the square after bringing the and to be and terms to the other side, we get and . Therefore, . Squares of real numbers are nonnegative, so we need both which only happens when

Problem 7

Jo and Blair take turns counting from to one more than the last number said by the other person. Jo starts by saying so on. What is the , so Blair follows by saying . Jo then says , and number said?

Solution

We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc. Thus, after nine "turns," numbers have been said. In the . Since we're . tenth turn, the eighth number will be the 53rd number said, as starting from 1 each time, the 53rd number said will be

Problem 8

Line point has equation at point . The area of and goes through . Line is . What is the slope of ? . Line has equation , and meets at and meets line has positive slope, goes through point

Solution

Line line has the equation will meet this line at point , and the area of slope, it will meet through to the right of when rearranged. Substituting for , which is point B. We call . Since , ,we find that the base and the . Because is . . has positive passes

altitude from A to the line connecting B and C,

, the height. The altitude has length

, and the point to the right of

and

, and thus has slope

Problem 9

What is the sum of the exponents of the prime factors of the square root of the largest perfect square that divides ?

Solution

Looking at the prime numbers under 12, we see that there are factors of 2, and factors of 5. All greater primes are represented once or not at all in factors of 3, , so

they cannot be part of the square. Since we are looking for a perfect square, the exponents on its prime factors must be even, so we can only use of the factors of . The prime factorization of the square is therefore halve the exponents, leaving . To find the square root of this, we . The sum of the exponents is

Problem 10

Alex has red tokens and blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token, and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges are possible. How many silver tokens will Alex have at the end?

Solution

We can approach this problem by assuming he goes to the red booth first. You start with and and at the end of the first booth, you will have and and . We now move to the blue booth, and working through each booth until we have none left, we will end up with: , and . So, the answer is

Problem 11

Two bees start at the same spot and fly at the same rate in the following directions. Bee travels foot north, then foot east, then foot upwards, and then continues to repeat this pattern. Bee travels foot south, then foot west, and then continues to repeat this pattern. feet away from each other? In what directions are the bees traveling when they are exactly east, north, north, up, up, west south west

south west

Solution

Let A and B begin at (0,0,0). In 6 steps, A will have done his route twice, ending up at (2,2,2), and B will have done his route three times, ending at (-3,-3,0). Their distance is We now move forward one step at a time until they are ten feet away: 7 steps: A moves north to (2,3,2), B moves south to (-3,-4,0), distance of moves west to (-4,-4,0), distance of Thus they reach 10 feet away when A is moving east and B is moving west, between moves 7 and 8. Thus the answer is 8 steps: A moves east to (3,3,2), B

Problem 12

Cities

,

,

,

, and

are connected by roads to

,

,

,

,

,

, and

.

How many different routes are there from

that use each road exactly once? (Such a

route will necessarily visit some cities more than once.)

Solution

Note that cities or and can be removed when counting paths because if a path goes in to or . So the diagram is as , there is only one possible path to take out of cities

follows: Now we proceed with casework. Remember that there are two ways to travel from to , to and to .: , if the path returns to , then the next path must go to . If the path goes to ,

Case 1

: From . There are

possibilities of the path

to

from

, then the path must continue with either possibilities. So, this case gives

or

. There are

different possibilities. . There are possibilities for

Case 2 this case.

: The path must continue with

Putting the two cases together gives

Problem 13

The internal angles of quadrilateral and are similar with form an arithmetic progression. Triangles and ? . Moreover, the angles

in each of these two triangles also form an arithmetic progression. In degrees, what is the largest possible sum of the two largest angles of

Solution

Since the angles of Quadrilateral ABCD form an arithmetic sequence, we can assign each angle with the value a, a+d, a+2d, and a+3d. Also, since these angles form an arithmetic progression, we can reason out that (a)+(a+3d)=(a+d)+(a+2d)=180. For the sake of simplicity, lets rename the angles of each similar triangle. Lets call Angle DBA and Angle DCB Angle 1. Also we rename Angle ADB and Angle CBD Angle 2. Finally we rename Angles BAD and BDC Angle 3. Now we can rename the four angles of Quadrilateral ABCD as Angle 2, Angle 1 + 2, Angle 3, and Angle 1 + 3. As for the similar triangles, whose Angles are equivalent, we can name them y, y+b, and y+2b. Therefore y+y+b+y+2b=180 and y+b=60. Because these 3 angles are each equal to one of the angles we named Angles 1, 2, and 3, we know that one of these three angles is equal to 60 degrees. Now we we use trial and error to find out which of these 3 angles has a value of 60. If we substitute 60 degrees into Angle 1. This would cause the angle values of ABCD to be Angle 2, 60+Angle 2, Angle 3, and 60 + Angle 3. Since these four angles add up to 360, then Angle 2 + Angle 3 = 120. If we list them in increasing value, we get Angle 2, Angle 3, 60 + Angle 2, 60+Angle 3. Note that this is the only sequence that works because the common

difference between each term cannot equal or exceed 45. So, this would give us the four angles 45, 75, 105, and 135. In this case, Angle 1, 2, and 3, the angles of both similar triangles, also form an arithmetic sequence with 45, 60, and 75, and the largest two angles of the quadrilateral add up to 240 which is an answer choice. If we apply the same reasoning to Angles 2 and 3, we would get the sum of the highest two angles as 220, which works but is lower than 240. Therefore, answer. is the correct

Problem 14

Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is possible value of ? . What is the smallest

Solution

Let the first two terms of the first sequence be sequence be and Note that this means that and and and the first two of the second . is ; . Computing the seventh term, we see that must have the same value modulo 8. To minimize, let . Thus, the smallest possible value of . To minimize, let . Thus,

one of them be 0; WLOG assume that since the sequences are non-decreasing .

Problem 15

The number is expressed in the form

,

where small as possible. What is

and ?

are positive integers and

is as

Solution

The prime factorization of minimize must equal factor of 2013 such that there must be a factor of well), so the answer is is is . To have a factor of in the numerator and to which is not a is , so as . Now we notice that there can be no prime

because this prime will not be represented in the less than in the denominator. It follows that (to minimize with

denominator, but will be represented in the numerator. The highest

. One possible way to express

Problem 16

Let be an equiangular convex pentagon of perimeter . The pairwise intersections of the lines that extend the sides of the pentagon determine a five-pointed star polygon. Let be the perimeter of this star. What is the difference between the maximum and the minimum possible values of ?

Solution

The five pointed star can be thought of as five triangles sitting on the five sides of the pentagon. Because the pentagon is equiangular, each of its angles has measure , and so the base angles of the aforementioned triangles (i.e., the angles adjacent to the pentagon) have measure equal, so the triangles must be isosceles. Let one of the sides of the pentagon have length (and the others ). Then, by . The base angles are

trigonometry, the non-base sides of the triangle sitting on that side of the pentagon each has length triangles to get , and so the two sides together have length . To find the (because the

perimeter of the star, we sum up the lengths of the non-base sides for each of the five

perimeter of the pentagon is ). The perimeter of the star is constant, so the difference between the maximum and minimum perimeters is .

Problem 17

Let and be real numbers such that

What is the difference between the maximum and minimum possible values of ?

Solution

. Now, by Cauchy-Schwarz, we have that we have that . Therefore,

. We then find the roots of that satisfy equality and find .

the difference of the roots. This gives the answer,

Problem 18

Barbara and Jenna play the following game, in which they take turns. A number of coins lie on a table. When it is Barbara’s turn, she must remove or coins, unless only one coin remains, in which case she loses her turn. What it is Jenna’s turn, she must remove or coins. A coin flip determines who goes first. Whoever removes the last coin wins the game. Assume both players use their best strategy. Who will win when the game starts with coins and when the game starts with Barbara will win with Jenna will win with Barbara will win with coins? coins. coins. coins.

coins and Jenna will win with

coins, and whoever goes first will win with coins, and whoever goes second will win with

Jenna will win with

coins, and Barbara will win with

coins.

Whoever goes first will win with coins.

coins, and whoever goes second will win with

Solution

We spit into 2 cases: 2013 coins, and 2014 coins. Notice that when there are coins left, whoever moves first loses, as they must leave an amount of coins the other person can take. If Jenna goes first, she can take coins. Then, whenever Barbara takes coins, Jenna will take the amount that makes the total coins taken in that round . (For instance, if Barbara takes coins, Jenna will take ). Eventually, since it will be Barbara's move with coins remaining, so she , it will be Barbara's will lose. If Barbara goes first, on each round, Jenna will take the amount of coins that makes the total coins taken on that round . Since last coin. Therefore, Jenna will win with coins. coins, and she wins as . After every move move with coins remaining, so she will have two take coins, allowing Jenna to take the

If Jenna moves first, she will take coin, leaving shown above. If Barbara moves first, she can take coins, leaving round . Since

by Jenna, Barbara will then take the number of coins that makes the total taken in that , it will be Jenna's turn with coins left, so Barbara will win. In this case, whoever moves first wins. Based on this, the answer is

Problem 19

In triangle segments , , , and , , and , respectively, such that can be written as ? . Distinct points , , where and , , and , and are relatively lie on

. The length of segment prime positive integers. What is

Solution

Since

, quadrilateral . In addition, since

is cyclic. It follows that , triangles . By Ptolemy, we have and are

similar. It follows that . Cancelling

, the rest is easy. We obtain

Problem 20

For and , points are the vertices of a trapezoid. What is ?

Solution

Let be (not respectively). Then we have four points , and a pair of lines each connecting two points must be parallel (as we are dealing with a trapezoid). WLOG, take the line connecting the first two points and the line connecting the last two points to be parallel, so that or . to , we have so the sum of the largest and the smallest is equal to . : so that the above equation , ,

Now, we must find how to match up has a solution. On the interval and the sum of the other two, namely,

Now, we perform some algebraic manipulation to find

Solve the quadratic to find .

, so that

Problem 21

Consider the set of 30 parabolas defined as follows: all parabolas have as focus the point (0,0) and the directrix lines have the form and with a and b integers such that . No three of these parabolas have a

common point. How many points in the plane are on two of these parabolas?

Solution

Being on two parabolae means having the same distance from the common focus and both directrices. In particular, you have to be on an angle bisector of the directrices, and clearly on the same "side" of the directrices as the focus. So it's easy to see there are at most two solutions per pair of parabolae. Convexity and continuity imply exactly two solutions unless the directrices are parallel and on the same side of the focus.

So out of

possible intersection points, only solutions.

fail to exist. This leaves

Problem 22

Let equation integer. What is ? and be integers. Suppose that the product of the solutions for of the is the smallest possible

Solution

Rearranging logs, the original equation becomes

By Vieta's Theorem, the sum of the possible values of

is

. But the sum of the possible values of is the logarithm of the product of the possible values of possible values of is equal to . . Since . , we can check that . Thus the product of the

It remains to minimize the integer value of and work. Thus the answer is

Problem 23

Bernardo chooses a three-digit positive integer and writes both its base-5 and base-6 . For , in order, representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer example, if the sum the same as those of . For how many choices of ? are the two rightmost digits of , Bernardo writes the numbers 10,444 and 3,245, and LeRoy obtains

Solution

First, we can examine the units digits of the number base 5 and base 6 and eliminate some possibilities. Say that also that Substituting these equations into the question and setting the units digits of 2N and S equal to each other, it can be seen that , Therefore, can be written as and can be written as , and , so , ,

Keep in mind that can be digits of

, five choices; Also, we have already found which .

will add up into the units digits of

Now, examine the tens digit, and take it representation.

by using and

and

to find the tens digit (units

digits can be disregarded because

will always work) Then we see that to find the last two digits in the base and Both of those must add up to

( Now, since units digit. Say that

) will always work if works, then we can treat as a units digit is now the

instead of a tens digit in the respective bases and decrease the mods so that

(m is between 0-6, n is 0-4 because of constraints on x) Then

and this simplifies to

From inspection, when

This gives you choices for

, and choices for

, so the answer is

Problem 24

Let addition be a triangle where with on . Let is the midpoint of . What is , and ? is the angle bisector of and the bisector . In be the intersection of the median

is equilateral with

Solution

Let and . From the conditions, let's deduct some convenient conditions that seems sufficient to solve the problem.

is the midpoint of side This implies that

. . Given that angle is degrees and angle is

degrees, we can use the area formula to get

So,

.....(1)

is angle bisector. In the triangle , one has , therefore , so ....(3) .....(2) ,

Furthermore, triangle therefore

is similar to triangle

By (2) and (3) and the subtraction law of ratios, we get

Therefore

, or

. So , we get

.

Finally, using the law of cosine for triangle

Problem 25

Let be the set of polynomials of the form where are integers and

has distinct roots of the form ?

with

and integers. How many polynomials are in

Solution

If we factor into irreducible polynomials (in factorization and degree at most (since the product equal ; for compute , let , for negative . that can be repeated are and multiply ), each factor with has exponent in the come in conjugate pairs with with constant term , . It's easy to

). Clearly we want the product of constant terms of these polynomials to be the number of permitted , , , , and obviously

Note that by the distinctness condition, the only constant terms those with by and at the end. , i.e. and . Also, the

s don't affect the product, so

we can simply count the number of polynomials with no constant terms of

We do casework on the (unique) even constant term For convenience, let (so only using be the number of ways to get a product of ) and recall

in our product. without using

; then our final answer will be . It's easy to compute ,

,

, ,

, , so we get

2014 AMC12-A Problem 1

What is

Solution

We have Making the denominators equal gives

Finally, simplifying gives

Problem 2

At the theater children get in for half price. The price for adult tickets and child tickets is . How much would adult tickets and child tickets cost?

Solution

Suppose is the price of an adult ticket. The price of a child ticket would be .

Plug in for 8 adult tickets and 6 child tickets.

Problem 3

Walking down Jane Street, Ralph passed four houses in a row, each painted a different color. He passed the orange house before the red house, and he passed the blue house before the yellow house. The blue house was not next to the yellow house. How many orderings of the colored houses are possible?

Solution

Attack this problem with very simple casework. The only possible locations for the yellow house Case 1: is the rd house and the last house. is the rd house.

The only possible arrangement is Case 2: is the last house.

There are two possible ways: and so our answer is

Problem 4

Suppose that will cows give gallons of milk in days. At this rate, how many gallons of milk cows give in days?

Solution 1

We need to multiply by , or for the new cows and for the new time, so the answer is .

(Solution by ahaanomegas)

Solution 2

We see that the the amount of cows is inversely proportional to the amount of days and directly proportional to the gallons of milk. So our constant is Let be the answer to the question. We have .

Problem 5

On an algebra quiz, of the students scored points, scored points, scored points, and the rest scored points. What is the difference between the mean and median

score of the students' scores on this quiz?

Solution

Without loss of generality, let there be took the test. We have students score points and students score points. students(the least whole number possible) who points, students score points, students score

The median can be obtained by eliminating members from each group. The median is points. The mean is equal to the total number of points divided by the number of people, which gives Thus, the difference between the median and the mean is equal to

Problem 6

The difference between a two-digit number and the number obtained by reversing its digits is times the sum of the digits of either number. What is the sum of the two digit number and its reverse?

Solution

Let the two digits be This yields and

and . Then, because . Then,

, or

.

Problem 7

The first three terms of a geometric progression are term? , , and . What is the fourth

Solution

The terms are will be , , and , which are equivalent to . , , and . So the next term , so the answer is

Problem 8

A customer who intends to purchase an appliance has three coupons, only one of which may be used: Coupon 1: Coupon 2: Coupon 3: off the listed price if the listed price is at least dollars off the listed price if the listed price is at least off the amount by which the listed price exceeds

For which of the following listed prices will coupon offer a greater price reduction than either coupon or coupon ?

Solution

Let the listed price be . Since all the answer choices are above , we can assume . Thus the prices after coupons will be as follows: Coupon 1: Coupon 2:

Coupon 3: For coupon to give a better price reduction than the other coupons, we must have and From the first inequality, . From the second inequality, . The only answer choice that satisfies these constraints is .

Problem 9

Five positive consecutive integers starting with consecutive integers that start with ? have average . What is the average of

Solution 1

Let is . Our list is . Our average is with an average of . which means that the answer is . . Our next set starting with

Therefore, we notice that

Solution 2

We are given that We are asked to find the average of the 5 consecutive integers starting from By substitution, this is Thus, the answer is in terms of .

Problem 10

Three congruent isosceles triangles are constructed with their bases on the sides of an equilateral triangle of side length . The sum of the areas of the three isosceles triangles is the same as the area of the equilateral triangle. What is the length of one of the two congruent sides of one of the isosceles triangles?

Solution

Since the total area of each congruent isosceles triangle is the same, the area of each is the total area of the equilateral triangle of side length 1, or each can be defined as x = . Likewise, the area of x , or = .A

with base equaling 1, meaning that

side length of the isosceles triangle is the hypotenuse with legs

and

. Using the

Pythagorean Theorem, the side length is

, or

.

Problem 11

David drives from his home to the airport to catch a flight. He drives speed by miles in the first minutes early. hour, but realizes that he will be hour late if he continues at this speed. He increases his miles per hour for the rest of the way to the airport and arrives How many miles is the airport from his home?

Solution 1 (Algebra)

Note that he drives at arrives. Let be the distance still needed to travel after the first hour. We have that comes from hour late decreased to hours early. , miles per hour after the first hour and continues doing so until he

where the

Simplifying gives Now, we must add an extra

, or

. miles.

miles traveled in the first hour, giving a total of

Solution 2 (Answer Choices)

Instead of spending time thinking about how one can set up an equation to solve the problem, one can simply start checking the answer choices. Quickly checking, we know that neither choice or choice . Since . work, but does. We can verify as follows. After hour at hours at . But is hours less than , our answer is , David has miles left. This then takes him

Problem 12

Two circles intersect at points smaller circle? and . The minor arcs measure on one circle and on the other circle. What is the ratio of the area of the larger circle to the area of the

Solution 1

Let the radius of the larger and smaller circles be centers be and and , respectively. Also, let their Draw the belongs

, respectively. Then the ratio we need to find is and

radii from the centers of the circles to to the larger circle, and the and is drawn, it has length

. We can easily conclude that the

degree arc belongs to the smaller circle. Therefore, . Note that is equilateral, so when chord AB :

. Now, applying the Law of Cosines on

Problem 13

A fancy bed and breakfast inn has rooms, each with a distinctive color-coded decor. One day friends arrive to spend the night. There are no other guests that night. The friends can

room in any combination they wish, but with no more than friends per room. In how many ways can the innkeeper assign the guests to the rooms?

Solution

We can discern three cases. Case 1: Each room houses one guest. In this case, we have guests to choose for the first room, for the second, ..., for a total of assignments.

Case 2: Three rooms house one guest; one houses two. We have three rooms with guest, and

ways to choose the

to choose the remaining one with . There are

ways to place guests in the first three rooms, with the last two residing in the two-person room, for a total of ways.

Case 3: Two rooms house two guests; one houses one. We have rooms with two people, and

to choose the two

to choose one remaining room for one person. Then there for the two in the first two-person room. The last ways. .

are choices for the lonely person, and

two will stay in the other two-room, so there are In total, there are assignments, or

Problem 14

Let be three integers such that is an arithmetic progression and ? is a geometric progression. What is the smallest possible value of

Solution

We have have and thus ,

, so

. Since

is geometric, . Since , we can't . Since , or .

. Then our arithmetic progression is . The smallest possible value of is

Problem 15

A five-digit palindrome is a positive integer with respective digits Let , where . is non-zero. be the sum of all five-digit palindromes. What is the sum of the digits of

Solution

For each digit s contribute there are are (since there are (ways of choosing and ) palindromes. So the to the sum. For each digit ) palindromes. So the s contribute to the sum. Similarly, for each palindromes, so the contributes there to the sum.

It just so happens that so the sum of the digits of the sum is , or .

Problem 16

The product have a sum of , where the second factor has . What is ? digits, is an integer whose digits

Solution

Note that for

,

, which has a digit sum of . Since we are given that said number has a digit sum of ,

we have

Problem 17

A rectangular box contains a sphere of radius and eight smaller spheres of radius ? . The smaller spheres are each tangent to three sides of the box, and the larger sphere is tangent to each of the smaller spheres. What is

Solution

Let be the point in the same plane as the centers of the top spheres equidistant from said be the analogous point for the bottom spheres, and let and be the midpoint of and the midpoint of the large sphere. Let be the points at which line centers. Let

intersects the top of the box and the bottom, respectively. Let be the center of any of the top spheres (you choose!). We have , so . Similarly, . and , and are clearly equal

to the radius of the small spheres, . Thus the total height is , or

Problem 18

The domain of the function , where and are relatively prime positive integers. What is is an interval of length ?

Solution

For simplicity, let The domain of is , so , and . Thus, . .

Since

we have . Finally, since ,

. Since

, we have

.

The length of the

interval is

and the answer is

Problem 19

There are exactly distinct rational numbers . What is such that ? and has at least one integer solution for

Solution

Factor the quadratic into where and . is our integer solution. Then, when , excluding

which takes rational values between . This leads to an answer of

Problem 20

In , , , and . Points and lie on ? and respectively. What is the minimum possible value of

Solution

Let be the reflection of across , and let be the reflection of across . Then it . is well-known that the quantity case, see Heron's Shortest Path Problem.) As . Furthermore, reflection, so Cosines is minimized when it is equal to lies on both and , we have

(Proving this is a simple application of the triangle inequality; for an example of a simpler by the nature of the . Therefore by the Law of

Problem 21

For every real number , let denote the greatest integer not exceeding The set of all numbers such that , and let and is

a union of disjoint intervals. What is the sum of the lengths of those intervals?

Solution

Let for some integer . Then we can rewrite as . In order for this to be less than or equal to , we need . Combining this with the fact that gives that , and so the length of the interval is

. We want the sum of all possible intervals such that the inequality holds true; since all of these intervals must be disjoint, we can sum from that the desired sum is to to get

Problem 22

The number such that

is between and

and

. How many pairs of integers

are there

Solution

Between any two consecutive powers of 5 there are either 2 or 3 powers of 2 (because ). Consider the intervals number of intervals with 3 powers of 2. From the given that 2013 powers of 2. Let we have the system is . , we know that these 867 intervals together have of them have 2 powers of 2 and of them have 3 powers of 2. Thus , so the answer from which we get . We want the

Problem 23

The fraction where is the length of the period of the ?

repeating decimal expansion. What is the sum

Solution

So, the answer is or .

Problem 24

Let many values of is ? , and for , let . For how

Solution

1. Draw the graph of by dividing the domain into three parts. 2. Look at the recursive rule. Take absolute of the previous function and down by 1 to get the next function. 3. Count the x intercepts of the each function and find the pattern. The pattern turns out to be extra solution after answer is thus . solutions,for x interval:[1,99], the function gain only one

because there is no summit on the graph any more, and the

Problem 25

The parabola many points has focus and goes through the points and . For how ? with integer coefficients is it true that

Solution

The parabola is symmetric through directrix is the line through and , and the common distance is , so the . That's the line Using the which

point-line distance formula, the parabola is the locus rearranges to Let , . Put . to obtain

and accordingly we find by solving the system that and .

One can show that the values of . For . this is

that make

an integer pair are precisely odd integers , so values work and the answer is

2014 AMC-12B Problem 1

Leah has coins, all of which are pennies and nickels. If she had one more nickel than she has now, then she would have the same number of pennies and nickels. In cents, how much are Leah's coins worth?

Solution

She has so of pennies and and cents. nickels, where . If she had nickels then , . So she has 6 nickels and 7 pennies, which clearly have a value

Problem 2

Orvin went to the store with just enough money to buy balloons. When he arrived he discovered that the store had a special sale on balloons: buy balloon at the regular price and get a second at could buy? off the regular price. What is the greatest number of balloons Orvin

Solution

If every balloon costs dollars, then Orvin has dollars. For every balloon he buys for dollars to buy a bundle

dollars, he can buy another for of balloons. With

dollars. This means it costs him

dollars, he can buy

sets of two balloons, so the total

number of balloons he can buy is

Problem 3

Randy drove the first third of his trip on a gravel road, the next

miles on pavement, and

the remaining one-fifth on a dirt road. In miles, how long was Randy's trip?

Solution

If the first and last legs of his trip account for accounts for the entire trip equal and of his trip, then the middle leg miles. Letting the length of

ths of his trip. This is equal to , we have

Problem 4

Susie pays for muffins and bananas. Calvin spends twice as much paying for muffins and bananas. A muffin is how many times as expensive as a banana?

Solution

Let stand for the cost of a muffin, and let stand for the value of a banana. We we need , the ratio of the price of the muffins to that of the bananas. We have to find

Problem 5

Doug constructs a square window using equal-size panes of glass, as shown. The ratio of the height to width for each pane is , and the borders around and between the panes are inches wide. In inches, what is the side length of the square window?

Solution

Let the height of the panes equal have , and let the width of the panes equal . Now notice that the total width of the borders equals , and the total height of the borders is . We

Now, the total side length of the window equals

Problem 6

Ed and Ann both have lemonade with their lunch. Ed orders the regular size. Ann gets the large lemonade, which is 50% more than the regular. After both consume of their drinks,

Ann gives Ed a third of what she has left, and 2 additional ounces. When they finish their lemonades they realize that they both drank the same amount. How many ounces of lemonade did they drink together?

Solution

Let the size of Ed's drink equal After both consume

ounces, and let the size of Ann's drink equal and

ounces.

of their drinks, Ed and Ann have ounces to Ed.

ounces of their drinks

remaining. Ann gives away

In the end, Ed drank everything in his original lemonade plus what Ann gave him, and Ann drank everything in her original lemonade minus what she gave Ed. Thus we have The total amount the two of them drank is simply

Problem 7

For how many positive integers is also a positive integer?

Solutions

We know that or else will be negative, resulting in a negative fraction. We also know that or else the fraction's denominator will exceed its numerator making the fraction unable to equal a positive integer value. Substituting all values from to gives us integer values for . Counting them up, we have possible values for .

Problem 8

In the addition shown below are possible for ? , , , and are distinct digits. How many different values

Solution

From the first column, we see fourth column, we see that sums less than and letting equals because it yields a single digit answer. From the and therefore . We know that , can take is equal to the number of possible

. Therefore, the number of values

that can be formed by adding two distinct natural numbers. Letting , we have

Problem 9

Convex quadrilateral has , , , , and , as shown. What is the area of the quadrilateral?

Note that by the pythagorean theorem, because and

. Also note that

is a right angle

is a right triangle. The area of the quadrilateral is the sum of the areas of which is equal to

Problem 10

Danica drove her new car on a trip for a whole number of hours, averaging 55 miles per hour. At the beginning of the trip, 3-digit number with miles. What is and . miles was displayed on the odometer, where is a . At the end of the trip, the odometer showed

Solution

We know that the number of miles she drove is divisible by , so be apart. Because we know that possible values for we have and and and must either be the and must . Therefore, equal or differ by . We can quickly conclude that the former is impossible, so and are and , respectively. Because ,

, we find that the only

Problem 11

A list of 11 positive integers has a mean of 10, a median of 9, and a unique mode of 8. What is the largest possible value of an integer in the list?

Solution

We start off with the fact that the median is , so we must have listed in ascending order. Note that the integers do not have to be distinct. Since the mode is , we have to have at least occurrences of in the list. If there are occurrences of in the list, we will have . In this case, since is the in unique mode, the rest of the integers have to be distinct. So we minimize ,

order to maximize . If we let the list be .

, then

Next, consider the case where there are occurrences of in the list. Now, we can have two occurrences of another integer in the list. We try same process as above, we get . As this is the highest choice in the list, we know this is our answer. Therefore, the answer is . Following the

Problem 12

A set have? consists of triangles whose sides have integer lengths less than 5, and no two are congruent or similar. What is the largest number of elements that can elements of

Solution

Define to be the set of all integral triples . Now we enumerate the elements of such that : , , and

It should be clear that duplicate of or .

is simply is

minus the larger "duplicates" (e.g.

is a larger

). Since

and the number of higher duplicates is , the answer is

Problem 13

Real numbers side lengths and are chosen with and or such that no triangles with positive area has and . What is the smallest possible value of ?

Solution

Notice that . Using the triangle inequality, we find

In order for us the find the lowest possible value for , we try to create two degenerate triangles where the sum of the smallest two sides equals the largest side. Thus we get and Substituting, we get Solving for using the

quadratic equation, we get

Problem 14

A rectangular box has a total surface area of 94 square inches. The sum of the lengths of all its edges is 48 inches. What is the sum of the lengths in inches of all of its interior diagonals?

Solution

Let the side lengths of the rectangular box be and . From the information we get

The sum of all the lengths of the box's interior diagonals is

Squaring the first expression, we get:

Hence

Problem 15

When factor of ? , the number is an integer. What is the largest power of 2 that is a

Solution

Let's write out the sum. Our sum is equal to Raising to the power of this quantity eliminates the natural logarithm, which leaves us with This product has powers of in the powers of in the factor, and powers of in the factor, factor. This adds up to

powers of two which divide into our quantity, so our answer is

Problem 16

Let be a cubic polynomial with ? , , and . What is

Solution

Let and and for . Plugging in for , we obtain the following equations: Adding these two equations together, we get in for , we find that Multiplying the third equation by and adding gives us our desired result, so If we plug in , we find , and plugging in

Problem 17

Let be the parabola with equation ? and let . There are real numbers if and only if . and such that the line through What is with slope does not intersect

Solution (Algebra Based)

Let . Equating them:

For there to be no solutions, the discriminant must be less than zero:

. So for where and are the roots of . and their sum by

Vieta's formulas is

Problem 18

The numbers , , , , , are to be arranged in a circle. An arrangement is true that for every from to consecutively on the circle that sum to

if it is not

one can find a subset of the numbers that appear . Arrangements that differ only by a rotation or a

reflection are considered the same. How many different bad arrangements are there?

Solution

We see that there are there are only out all cases. total ways to arrange the numbers. However, we can always rotate these numbers so that, for example, the number 1 is always at the top of the circle. Thus, ways under rotation, which is not difficult to list out. We systematically list

Now, we must examine if they satisfy the conditions. We can see that by choosing one number at a time, we can always obtain subsets with sums 1, 2, 3, 4, and 5. By choosing the full circle, we can obtain 15. By choosing everything except for 1, 2, 3, 4, and 5, we can obtain subsets with sums of 10, 11, 12, 13, and 14. This means that we now only need to check for 6, 7, 8, and 9. However, once we have found a set summing to 6, we can choose everything else and obtain a set summing to 9, and similarly for 7 and 8. Thus, we only need to check each case for weather or not we can obtain 6 or 7. We find that there are only 4 arrangements that satisfy these conditions. However, each of these is a reflection of another. We divide by 2 for these reflections to obtain a final answer of .

Problem 19

A sphere is inscribed in a truncated right circular cone as shown. The volume of the truncated cone is twice that of the sphere. What is the ratio of the radius of the bottom base

of the truncated cone to the radius of the top base of the truncated cone?

Solution

First, we draw the vertical cross-section passing through the middle of the frustum. let the top base equal 2 and the bottom base to be equal to 2r

then using the Pythagorean theorem we have: equivalent to: solving for s we get: sphere and we know we get: we get so we have: dividing by equivalent to we get so subtracting so we can solve for using

which is from both sides

next we can find the area of the frustum and of the

using

which is

Problem 20

For how many positive integers is ?

Solution

The domain of the LHS implies that Begin from the left hand side

Hence, we have integers from 41 to 49 and 51 to 59. There are

integers.

Problem 21

In the figure, is a square of side length . The rectangles and are

congruent. What is

?

Solution

Draw the attitude from to and call the foot and . . Then . Consider . It is the and hypotenuse of both right triangles , so we must have , and we know

Notice that all four triangles in this picture are similar and thus we have . This means is the midpoint of . So , along with all other similar triangles in the picture, is a 30-60-90 triangle, and we have

and subsequently , which gives answer is .

. This means , so the

Problem 22

In a small pond there are eleven lily pads in a row labeled 0 through 10. A frog is sitting on pad 1. When the frog is on pad and to pad , , it will jump to pad with probability

with probability

. Each jump is independent of the previous

jumps. If the frog reaches pad 0 it will be eaten by a patiently waiting snake. If the frog reaches pad 10 it will exit the pond, never to return. What is the probability that the frog will escape without being eaten by the snake?

Solution

A long, but straightforward bash: Define to be the probability that the frog survives starting from pad N.

Then note that by symmetry,

, since the probabilities of the frog moving

subsequently in either direction from pad 5 are equal.

We therefore seek to rewrite

in terms of

, using the fact that

as said in the problem.

Hence

Returning to our original equation:

Returning to our original equation:

Cleaing up the coefficients, we have:

Hence,

Problem 23

The number 2017 is prime. Let by 2017? . What is the remainder when is divided

Solution

Note that . We have for

Therefore is simply an alternating series of triangular numbers that goes like this: After finding the first few sums of the series, it becomes apparent that Obviously, category, so our desired value is

This

and falls in the second

Problem 24

Let and and be a pentagon inscribed in a circle such that . The sum of the lengths of all diagonals of are relatively prime positive integers. What is ? , is equal to , where ,

Solution

Note that Then we have: and are isosceles trapezoids. They must be cyclic quadrilaterals, , , and . so we can apply Ptolemy's Theorem. Let

According to the first equation,

. Plugging this into the third equation results in . Substituting into the

. The only positive root of this cubic is first and second equations gives and

and thus the sum of all diagonals is .

. Our answer is therefore

Problem 25

What is the sum of all positive real solutions to the equation

Solution

Rewrite have For the first case, multiple of

as

. Now let Notice that either only when : and ,

, and let and or when and

. We .

is an integer. only when

is an even . This , in

, and since

is an odd divisor of For the case where

gives us these possible values for , so order for to equal for , so

, where m is odd.

must also be an odd multiple of

must be odd. We can quickly see that dividing an even and . Therefore, the sum of all our

number by an odd number will never yield an odd number, so there are no possible values , and therefore no cases where is

possible values for

AMC 12 Key Answers

2010 A

1. C 2. A 3. E 4. D 5. C 6. E 7. C 6. D 8. C 9. A 10. A 11. C 12. D 13. C 14. B 15. D 16. B 17. E 18. D 19. A 20. C 21. A 22. A 23. A 24. B 25. C 21. B 21.A 21.D 21.E 21.A 21.A 21.C 21.A 21.C 16. E 17. D 18. C 19. E 20. E 16.C 17.D 18.D 19.C 20.C 16.C 17.B 18.A 19.B 20.C 16.E 17.B 18.A 19.B 20.B 16.B 17.C 18.B 19.A 20.D 16.E 17.D 18.B 19.D 20.B 16.A 17.D 18.B 19.B 20.A 16.D 17.A 18.C 19.E 20.D 16.E 17.E 18.B 19.E 20.B 11. E 12. D 13. C 14. B 15. D 11.C 12.D 13.B 14.E 15.A 11.B 12.A 13.B 14.D 15.D 11.B 12.D 13.D 14.E 15.A 11.C 12.E 13.D 14.A 15.C 11.C 12.A 13.B 14.B 15.D 11.A 12.D 13.D 14.C 15.B 11.C 12.D 13.B 14.C 15.B 11.E 12.B 13.C 14.D 15.C 7. C 8. B 9. E 10. B 7. B 8. C 9. B 10.B 7. B 8. A 9. D 10.E 7. C 8. C 9. A 10.D 7.E 8.A 9.B 10.B 7. C 8. D 9. C 10.D 7.E 8.B 9.C 10.E 7.A 8.C 9.B 10.B 7.D 8.C 9.B 10.D 6. A 6 .C 6. D 6.A 6. B 6.B 6.D 6.D 2. A 3. E 4. B 5. D 2. E 3. E 4. C 5. C 2. E 3. C 4. E 5. A 2. D 3. D 4. C 5. D 2.E 3.D 4.B 5.A 2. C 3. E 4. C 5. B 2.A 3.D 4.B 5.C 2.B 3.B 4.A 5.C 2.C 3.E 4.B 5.A

2011 B

1. C

2012 B

1. C

2013 B

1.C

2014 B

1.C

A

1. D

A

1. E

A

1. E

A

1.C

B

1.C

22. D 23. A 24. C 25. D

22.C 23.C 24.C 25.D

22.D 23.C 24.B 25.D

22.C 23.C 24.C 25.C

22.E 23.B 24.D 25.B

22.E 23.C 24.E 25.A

22.A 23.E 24.A 25.B

22.B 23.B 24.C 25.B

22.C 23.C 24.D 25.D