AMC 美国数学竞赛 2002 AMC 10A 试题及答案解析

2002 AMC 10A

1、The ratio is closest to which of the following numbers?

Solution

We factor as . As , our

answer is .

2、For the nonzero numbers , , , define

.

Find .

Solution

. Our

answer is then .

Alternate solution for the lazy: Without computing the answer exactly,

we see that , , and . The sum

is , and as all the options are integers, the correct one is obviously .

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 does there exist at least one positive integer such that ?

infinitely many

Solution

Solution 1

For any we can pick , we get , therefore the

answer is .

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 that can satisfy

the inequality. So the answer is .

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 . Our

answer is .

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 .

Using that here, the arc of circle A has length . The arc

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

triangles, the total area of the white squares, and the area of the

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 so

and the average is . Our answer is .

10、Compute the sum of all the roots of

.

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 M

B 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 and

we find of an hour. Substituting t back in, we find . 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 . In this case, a and b are 40 and 60,

so our answer is , so .

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.

Then the time needed to get to work is .

We know that and . Summing these two

equations, we get: .

Substituting and dividing both sides by , we get ,

hence .

(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 and x is 12. Our answer is then

.

Solution 2

By Heron's formula, the area is , hence the shortest altitude's

length is .

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

Solution

Consider a general quadratic with the coefficient of being and the

roots being and . It can be factored as 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 , so .

Thus, our answer is .

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 ounces of coffee from cup to cup ,

getting:

?ounces of coffee in cup

?ounces of coffee and ounces of cream in cup

In the second step we pour ounce of coffee and ounces of cream from cup to cup , getting:

?ounces of coffee and ounces of cream in cup

?the rest in cup

Hence at the end we have ounces of liquid in cup , and out

of these 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 . Our answer is thus

.

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?

Solution

Part of what Spot can reach is of a circle with radius 2, which

gives him . He can also reach two parts of a unit circle, which

combines to give . The total area is then , which gives .

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 , there are at least two s.

As the range is and one of the numbers is , the largest one can be

at most .

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、A sit of tiles numbered 1 through 100 is modified repeatedly by the 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,

, so in the next step tiles

are removed. This gives , another perfect square.

Thus each two steps we cycle down a perfect square, and in

steps, we are left with tile, hence our answer is

.

23、Points and lie on a line, in that order, with and

. Point is not on the line, and . The perimeter

of is twice the perimeter of . Find .

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 is twice the perimeter of .

We have so

. Squaring both sides, we have

which nicely rearranges into

. 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 ways. Tina chooses two

distinct numbers in ways while Sergio chooses a number in

ways, so there are ways in all. Since , our

answer is .

25、In trapezoid with bases and , we have ,

, , and . The area of is

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 so that coincides with . We get

the following triangle:

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

, and therefore its altitude is

.

Now the area of the original trapezoid is

.

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关于公布绍兴县高一数学竞赛获奖名单的通知

与其到头来收拾残局,甚至做成蚀本生意,倒不如当时理智克制一些. 绍兴县教师发展中心 绍县教发[2011] 78号 关于公布绍兴县高一数学竞赛获奖名单的通知 各普通高中： 全县高一数学竞赛已经结束 现将获奖名单公布如下 希望获奖师生再接再历 为进一步提高数学学科的教学质量作出新的贡献 一、 A组（柯桥中学春季班、鲁迅中学柯桥校区春季班） 学校 班级 获奖学生 指导老师 获奖等级 鲁迅中学柯桥校区 高一(10) 徐世超 骆青 一等奖 鲁迅中学柯桥校区 高一(9) 周洁阳 田萌 一等奖 鲁迅中学柯桥校区 高一(9) 王灵微 田萌 一等奖 柯桥中学 高一(1) 王愿翔 陈冬良 一等奖 鲁迅中学柯桥校区 高一(10) 孙杰 骆青 一等奖 柯桥中学

高一(1) 虞楚尔 陈冬良 一等奖 柯桥中学 高一(1) 夏丹妮 陈冬良 一等奖 柯桥中学 高一(2) 李智涛 陈冬良 一等奖 鲁迅中学柯桥校区高一(9) 何建春 田萌 二等奖 柯桥中学 高一(1) 李佳群 陈冬良 二等奖 柯桥中学 高一(1) 洪哲瑜 陈冬良 二等奖 鲁迅中学柯桥校区高一(10) 屠刚亮 骆青 二等奖 柯桥中学 高一(2) 袁梦娣 陈冬良 二等奖 鲁迅中学柯桥校区高一(9) 赵梦娣 田萌 二等奖

鲁迅中学柯桥校区高一(10) 潘竹莹 骆青 二等奖 鲁迅中学柯桥校区高一(10) 阮烨玲 骆青 二等奖 柯桥中学 高一(2) 张露枫 陈冬良 二等奖 鲁迅中学柯桥校区高一(10) 郑浙秀 骆青 二等奖 柯桥中学 高一(1) 冯耀祺 陈冬良 二等奖 鲁迅中学柯桥校区高一(10) 徐迪青 骆青 二等奖 鲁迅中学柯桥校区高一(10) 秦王颖 骆青 三等奖 柯桥中学 高一(1) 金恒超 陈冬良 三等奖 柯桥中学 高一(1) 周洋 陈冬良

美国数学竞赛AMC题目及答案

2. is the value of friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $ to cover her portion of the total bill. What was the total bill is in the grade and weighs 106 pounds. His quadruplet sisters are tiny babies and weigh 5, 5, 6, and 8 pounds. Which is greater, the average (mean) weight of these five children or the median weight, and by how many pounds number in each box below is the product of the numbers in the two boxes that touch it in the row above. For example, . What is the missing number in the top row

and his mom stopped at a railroad crossing to let a train pass. As the train began to pass, Trey counted 6 cars in the first 10 seconds. It took the train 2 minutes and 45 seconds to clear the crossing at a constant speed. Which of the following was the most likely number of cars in the train fair coin is tossed 3 times. What is the probability of at least two consecutive heads Incredible Hulk can double the distance he jumps with each succeeding jump. If his first jump is 1 meter, the second jump is 2 meters, the third jump is 4 meters, and so on, then on which jump will he first be able to jump more than 1 kilometer is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594 11. Ted's grandfather used his treadmill on 3 days this week. He went 2 miles each day. On Monday he jogged at a speed of 5 miles per hour. He walked at the rate of 3 miles per hour on Wednesday and at 4 miles per hour on Friday. If Grandfather had always walked at 4 miles per hour, he would have spent less time on the treadmill. How many minutes less 12. At the 2013 Winnebago County Fair a vendor is offering a "fair special" on sandals. If you buy one pair of sandals at the regular price of $50, you get a second pair at a 40% discount, and a third pair at half the regular price. Javier took advantage of the "fair special" to buy three pairs of sandals. What percentage of the $150 regular price did he save