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- AMC 美国数学竞赛 2001 AMC 10 试题及答案解析
- AMC 美国数学竞赛 2003 AMC 10A 试题及答案解析
- AMC 美国数学竞赛 2003 AMC 10B 试题及答案解析
- AMC 美国数学竞赛 2004 AMC 10B 试题及答案解析
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- AMC 美国数学竞赛 2003 AMC 10A 试题及答案解析
- AMC 美国数学竞赛 2004 AMC 10B 试题及答案解析
- AMC 美国数学竞赛 2000 AMC 10 试题及答案解析
- AMC 美国数学竞赛 2003 AMC 10B 试题及答案解析
- AMC 美国数学竞赛 2001 AMC 10 试题及答案解析

2002 AMC 10A

1、The ratio

is closest to which of the following numbers?

Solution

We factor answer is .

as

. As

, our

2、For the nonzero numbers

, , , define

. Find .

Solution . answer is then . Our

Alternate solution for the lazy: Without computing the answer exactly, we see that is the options are integers, the correct one is obviously . , , and . The sum , and as all

3、According to the standard convention for exponentiation,

. If the order in which the exponentiations are performed is changed, how many other values are possible?

Solution The best way to solve this problem is by simple brute force. It is convenient to drop the usual way how exponentiation is denoted, and to write the formula as , where denotes exponentiation. We are now examining all ways to add parentheses to this expression. There are 5 ways to do so: 1. 2. 3. 4. 5. We can note that . Therefore options 1 and

2 are equal, and options 3 and 4 are equal. Option 1 is the one given in the problem statement. Thus we only need to evaluate options 3 and 5.

Thus the only other result is

, and our answer is

.

4、For how many positive integers positive integer such that

does there exist at least one ? infinitely many

Solution Solution 1

For any answer is

we can pick .

, we get

, therefore the

Solution 2 Another solution, slightly similar to this first one would be using Simon's Favorite Factoring Trick.

Let

, then

This means that there are infinitely many numbers the inequality. So the answer is .

that can satisfy

5、Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.

Solution The outer circle has radius , and thus area . The little circles have area each; since there are 7, their total area is . Thus, our answer is .

6、 Cindy was asked by her teacher to subtract from a certain number and then divide the result by . Instead, she subtracted and then divided the result by , giving an answer of . What would her answer have been had she worked the problem correctly?

Solution

We work backwards; the number that Cindy started with is . Now, the correct result is answer is . . Our

7、If an arc of on circle has the same length as an arc of on circle , then the ratio of the area of circle to the area of circle is

Solution Let and be the radii of circles A and B, respectively.

It is well known that in a circle with radius r, a subtended arc opposite an angle of degrees has length . . The arc

Using that here, the arc of circle A has length of circle B has length

. We know that they are equal,

so , so we multiply through and simplify to get . As all circles are similar to one another, the ratio of the areas is just the square of the ratios of the radii, so our answer is .

8、 Betsy designed a flag using blue triangles, small white squares, and a red center square, as shown. Let be the total area of the blue the total area of the white squares, and the area of the triangles, red square. Which of the following is correct?

Solution

The blue that's touching the center red square makes up 8 triangles, or 4 squares. Each of the corners is 2 squares and each of the edges is 1, totaling 12 squares. There are 12 white squares, thus we have . 9、There are 3 numbers A, B, and C, such that and . What is the average of A, B, and C? More than 1 Solution Notice that we don't need to find what A, B, and C actually are, just their average. In other words, if we can find A+B+C, we will be done. Adding up the equations gives . Our answer is of all the . roots of so ,

and the average is 10 、 Compute the sum .

Solution Solution 1 We expand to get which is after combining like terms. Using the quadratic part of Vieta's Formulas, we find the sum of the roots is Solution 2 Combine terms to get , hence the roots are and , thus our answer is . .

11、Jamal wants to store computer files on floppy disks, each of which has a capacity of megabytes (MB). Three of his files require

MB of memory each, more require MB each, and the remaining require MB each. No file can be split between floppy disks. What is the minimal number of floppy disks that will hold all the files?

Solution A 0.8 MB file can either be on its own disk, or share it with a 0.4 MB. Clearly it is not worse to pick the second possibility. Thus we will have 3 disks, each with one 0.8 MB file and one 0.4 MB file. We are left with 12 files of 0.7 MB each, and 12 files of 0.4 MB each. Their total size is MB. The total capacity of 9 disks

is MB, hence we need at least 10 more disks. And we can easily verify that 10 disks are indeed enough: six of them will carry two 0.7 MB files each, and four will carry three 0.4 MB files each. Thus our answer is .

12、Mr. Earl E. Bird leaves his house for work at exactly 8:00 A.M. every morning. When he averages miles per hour, he arrives at his workplace three minutes late. When he averages miles per hour, he arrives three minutes early. At what average speed, in miles per hour, should Mr. Bird drive to arrive at his workplace precisely on time?

Solution Solution 1 Let the time he needs to get there in be t and the distance he travels be d. From the given equations, we know that and

. Setting the two equal, we have we find of an hour. Substituting t back in, we find .

and . From

, we find that r, and our answer, is

Solution 2 Since either time he arrives at is 3 minutes from the desired time, the answer is merely the harmonic mean of 40 and 60. The harmonic mean of a and b is so our answer is Solution 3 A more general form of the argument in Solution 2, with proof: Let be the distance to work, and let be the correct average speed. . . Summing these two , so . In this case, a and b are 40 and 60, .

Then the time needed to get to work is We know that equations, we get: Substituting hence . and .

and dividing both sides by

, we get

,

(Note that this approach would work even if the time by which he is late was different from the time by which he is early in the other case - we would simply take a weighed sum in step two, and hence obtain a weighed harmonic mean in step three.)

13、Give a triangle with side lengths 15, 20, and 25, find the triangle's smallest height.

Solution Solution 1 This is a Pythagorean triple (a 3-4-5 actually) with legs 15 and 20. The area is then . Now, consider an altitude drawn to any

side. Since the area remains constant, the altitude and side to which it is drawn are inversely proportional. To get the smallest altitude, it must be drawn to the hypotenuse. Let the length be x; we have , so . Solution 2 By Heron's formula, the area is length is . are prime is , hence the shortest altitude's and x is 12. Our answer is then

14、Both roots of the quadratic equation numbers. The number of possible values of

Solution Consider a general quadratic with the coefficient of roots being and . It can be factored as being and the which is just

. Thus, the sum of the roots is the negative of the coefficient of and the product is the constant term. (In general, this leads to Vieta's Formulas). We now have that the sum of the two roots is while the product is . Since both roots are primes, one must be , otherwise the sum would be even. That means the other root is and the product must be . Hence, our answer is .

15、Using the digits 1, 2, 3, 4, 5, 6, 7, and 9, form 4 two-digit prime numbers, using each digit only once. What is the sum of the 4 prime numbers?

Solution

Only odd numbers can finish a two-digit prime number, and a two-digit number ending in 5 is divisible by 5 and thus composite, hence our answer is .

(Note that we did not need to actually construct the primes. If we had to, one way to match the tens and ones digits to form four primes is , , , and .)

16 、 Let ?

.

What

is

Solution Let . Since one of the sums involves a, b, c, and d, it makes sense to consider 4x. We have

. Rearranging, we have Thus, our answer is .

, so

.

17、 Sarah pours four ounces of coffee into an eight-ounce cup and four ounces of cream into a second cup of the same size. She then transfers half the coffee from the first cup to the second and, after stirring thoroughly, transfers half the liquid in the second cup back to the first. What fraction of the liquid in the first cup is now cream?

Solution We will simulate the process in steps. In the beginning, we have: ounces of coffee in cup ounces of cream in cup In the first step we pour getting: ounces of coffee from cup to cup ,

ounces of coffee in cup ounces of coffee and ounces of cream in cup In the second step we pour of cream from cup ounce of coffee and ounces

to cup , getting: ounces of cream in cup

ounces of coffee and the rest in cup Hence at the end we have of these

ounces of liquid in cup , and out .

ounces is cream. Thus the answer is

18、A cube is formed by gluing together 27 standard cubical dice. (On a standard die, the sum of the numbers on any pair of opposite faces is 7.) The smallest possible sum of all the numbers showing on the surface of the cube is

Solution In a 3x3x3 cube, there are 8 cubes with three faces showing, 12 with two faces showing and 6 with one face showing. The smallest sum with three faces showing is 1+2+3=6, with two faces showing is 1+2=3, and with one face showing is 1. Hence, the smallest possible sum is . 19、Spot's doghouse has a regular hexagonal base that measures one yard on each side. He is tethered to a vertex with a two-yard rope. What is the area, in square yards, of the region outside of the doghouse that Spot can reach? . Our answer is thus

Solution Part of what Spot can reach is gives him . He can also reach two of a circle with radius 2, which parts of a unit circle, which , which gives .

combines to give

. The total area is then

20、Points and lie, in that order, on , dividing it into five segments, each of length 1. Point is not on line . Point lies on , and point lies on . . The line segments and

are parallel. Find

Solution As is parallel to , angles FHD and FGA are congruent. Also, angle F is clearly congruent to itself. From SSS similarity, ; hence . Similarly, . Thus,

.

21、The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection is

Solution As the unique mode is As the range is at most . , there are at least two s.

and one of the numbers is

, the largest one can be

If the largest one is , then the smallest one is , and thus the mean is strictly larger than , which is a contradiction. If the largest one is , then the smallest one is . This means that we already know four of the values: , , , . Since the mean of all the numbers is , their sum must be . Thus the sum of the missing four numbers is . But if is the smallest number, then the sum of the missing numbers must be at least , which is again a contradiction. If the largest number is , we can easily find the solution .

. Hence, our answer is

Note The solution for is, in fact, unique. As the median must be , this means that both the and the number, when ordered by size, must be s. This gives the partial solution . For the

mean to be each missing variable must be replaced by the smallest allowed value. 22、 sit of tiles numbered 1 through 100 is modified repeatedly by the A following operation: remove all tiles numbered with a perfect square, and renumber the remaining tiles consecutively starting with 1. How many times must the operation be performed to reduce the number of tiles in the set to one?

Solution Solution 1 The pattern is quite simple to see after listing a couple of terms.

Solution 2 Given tiles, a step removes tiles, leaving tiles behind. Now, tiles ,

, so in the next step are removed. This gives another perfect square.

Thus each two steps we cycle down a perfect square, and in steps, we are left with . 23、Points and lie on a line, in that order, with and . Point is not on the line, and . The perimeter is twice the perimeter of . Find . of tile, hence our answer is

Solution First, we draw an altitude to BC from E.Let it intersect at M. As triangle BEC is isosceles, we immediately get MB=MC=6, so the altitude is 8. Now, let . Using the Pythagorean Theorem on triangle EMA, we find . From symmetry,

as well. Now, we use the fact that the perimeter of We is twice the perimeter of have . Squaring which both nicely sides, we . so have into

rearranges .

. Hence, AB is 9 so our answer is

24 、 Tina randomly selects two distinct numbers from the set and Sergio randomly selects a number from the set . The probability that Sergio's number is larger than the sum of the two numbers chosen by Tina is

Solution This is not too bad using casework. Tina gets a sum of 3: This happens in only one way (1,2) and Sergio can choose a number from 4 to 10, inclusive. There are 7 ways that Sergio gets a desirable number here. Tina gets a sum of 4: This once again happens in only one way (1,3). Sergio can choose a number from 5 to 10, so 6 ways here. Tina gets a sum of 5: This can happen in two ways (1,4) and (2,3). Sergio can choose a number from 6 to 10, so 2*5=10 ways here. Tina gets a sum of 6: Two ways here (1,5) and (2,4). Sergio can choose a number from 7 to 10, so 2*4=8 here. Tina gets a sum of 7: Two ways here (2,5) and (3,4). Sergio can choose from 8 to 10, so 2*3=6 ways here. Tina gets a sum of 8: Only one way possible (3,5). Sergio chooses 9 or 10, so 2 ways here. Tina gets a sum of 9: Only one way (4,5). Sergio must choose 10, so 1 way. In all, there are distinct numbers in ways, so there are answer is . , we have is , ways. Tina chooses two ways while Sergio chooses a number in ways in all. Since , our

25、In trapezoid with bases and , , and . The area of

Solution Solution 1 It shouldn't be hard to use trigonometry to bash this and find the height, but there is a much easier way. Extend and to meet at point :

Since

we

have

,

with

the

ratio

of

proportionality being . Thus So the sides of are , which we recognize to be a right triangle. Therefore (we could simplify some of the calculation using that the ratio of areas is equal to the ratio of the sides squared),

Solution 2 Draw altitudes from points and :

Translate the triangle the following triangle:

so that

coincides with

. We get

The length of in this triangle is equal to the length of the original , minus the length of . Thus .

Therefore

is a well-known

right triangle. Its area is is

, and therefore its altitude . Now the area of the . original trapezoid

is