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# 2004 AMC12A(美国数学竞赛)

Alicia earns dollars per hour, of which

is deducted to pay local taxes. How

many cents per hour of Alicia's wages are used to pay local taxes?

Solution

On the AMC 12, each correct answer is worth points, each incorrect answer is

worth points, and each problem left unanswered is worth

leaves

of the points. If Charlyn problems unanswered, how many of the remaining problems

?

must she answer correctly in order to score at least

Solution

For how many ordered pairs of positive integers

Solution is

? Bertha has daughters and no sons. Some of her daughters have daughters, and the rest have none. Bertha has a total of daughters and granddaughters, and no great-granddaughters. How many of Bertha's daughters and grand-daughters have no children?

Solution

The graph of the line is shown. Which of the following is true?

Solution

Let

and ,

, , ,

. Which of the following is the largest?

Solution

A game is played with tokens according to the following rules. In each round, the player with the most tokens gives one token to each of the other players and also places one token into a discard pile. The game ends when some player runs out of tokens. Players , and start with

many rounds will there be in the game?

Solution , and tokens, respectively. How

In the overlapping triangles

side

and of ,

and

and and intersect at

?

and

are right angles,

sharing common , , ,

. What is the difference between the areas

Solution

A company sells peanut butter in cylindrical jars. Marketing research suggests that using wider jars would increase sales. If the diameter of the jars is increased by without altering the volume, by what percent must the height be decreased?

Solution

The sum of consecutive integers is . What is their median?

Solution

The average value of all the pennies, nickels, dimes, and quarters in Paula's purse is

cents. If she had one more quarter, the average value would be cents. How

many dimes does she have in her purse? Solution

Let

and and

and

. Points

and are on the line , ? intersect at . What is the length of

Solution

Let

be the set of points

be in the coordinate plane, where each of

and

may

, , or

. How many distinct lines pass through at least two members of ?

Solution

A sequence of three real numbers forms an arithmetic progression with a first term of . If is added to the second term and is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term in the geometric progression?

Solution

Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run

They next meet after Sally has run meters. meters past their first meeting point. Each

girl runs at a constant speed. What is the length of the track in meters?

Solution

The set of all real numbers for which

is defined is . What is the value of

?

Solution

Let be a function with the following properties:

, and

, for any positive integer .

What is the value of ?

Solution

Square has side length

. A semicircle with diameter is constructed inside the square, and the tangent to the semicircle from

intersects side at

. What is the length of ?

Solution

Circles

circle and

are externally tangent to each other, and internally tangent to . Circles and

are congruent. Circle has radius and passes through . What is the radius of circle

? the center of

Solution

Select numbers and between and independently and at random, and

let be their sum. Let and be the results when and , respectively, are

? rounded to the nearest integer. What is the probability that

Solution

If , what is the value of

?

Solution

Three mutually tangent spheres of radius rest on a horizontal plane. A sphere of

radius rests on them. What is the distance from the plane to the top of the larger

sphere?

Solution

A polynomial

has real coefficients with

zeroes , and with distinct complex and real, , and

Which of the following quantities can be a nonzero number?

Solution

A plane contains points and

with

radius in the plane that cover . Let be the union of all disks of . What is the area of ?

Solution

For each integer

product , let denote the base- number , where . The can be expressed as and are positive integers

?

and is as small as possible. What is the value of

Solution

20 dollars is the same as 2000 cents, and is

cents .

of 2000 Problem 2

She gets points for the problems she didn't answer. She must

get problems right to score at least 100

Problem 3

Every integer value of leads to an integer solution for Since

must be positive,

Also,

Since

must be positive,

This leaves

equation solutions to the values for y, which mean there are

Problem 4

Since Bertha has 6 daughters, Bertha has

none have daughters. Of Bertha's daughters,

so do not have daughters. granddaughters, of which have daughters,

Therefore, of Bertha's daughters and granddaughters,

daughters

OR .

do not have

Draw a tree diagram and see that the answer can be found in the sum of 6 + 6 granddaughters, 5 + 5 daughters, and 4 more daughters Problem 5

It looks like it has a slope of

and is shifted

up.

Problem 6

After comparison,

is the largest. Problem 7 Look at a set of 3 rounds, where the players have

, , and tokens. Each of the players will gain two tokens from the others and give away 3 tokens, so overall, each player will lose 1 token.

Therefore, after 12 sets of 3 rounds, or 36 rounds, the players will have 3, 2, and 1 tokens, repectively. After 1 more round, player will give away his last 3 tokens and the game will stop .

Problem 8

Solution 1 Since

we find that and , . By

alternate interior angles and AA~,

, with side length ratio . Their heights also have

the same ratio, and since the two heights add up to , we have

that areas, and . . Subtracting the Solution 2

Let

that represent the area of figure . Note and . .

Problem 9

When the diameter is increased by , it is increased by , so the area of the base is increased by .

To keep the volume the same, the height must be

which is a reduction of the original height,

Problem 10

The median of a sequence is the middle number of the sequence when the sequence is arranged in order. Since the integers are consecutive, the median is also the mean, so the median is

.

Problem 11

Solution 1

Let the total value (in cents) of the coins Paula has originally be , and the number of coins she has be .

Then and .Substituting yields

. It is easy to see now that Paula has 3

quarters, 1 nickel, so she has dimes.

Solution 2

If the new coin was worth 20 cents, adding it would not change the mean at all. The additional 5 cents raise the mean by 1, thus the new number of coins must be 5. Therefore initially there were 4 coins worth a total of cents. As in the previous solution, we conclude that the only way to get 80 cents using 4 coins is 25+25+25+5.

Problem 12

Problem 13

Solution 1 Let's count them by cases:

? Case 1

: The line is horizontal or vertical, clearly .

? Case 2: The line has slope , with through and additional ones one unit

above or below those. These total .

? Case 3: The only remaining lines pass through two points, a vertex and a non-vertex

point on the opposite side. Thus we have each vertex pairing up with two points on the two opposites sides, giving lines.

These add up to .

Solution 2 There are ways to pick two points, but we've clearly overcounted all of the lines which pass through three points. In fact, each line which passes through

three points will have been counted times, so we have to subtract for

each of these lines. Quick counting yields horizontal, vertical, and

diagonal

lines, so the answer is distinct lines.

Problem 14

Let be the common difference. Then are the terms of the

geometric progression. Since the middle term is the geometric mean of the other two terms,

, and the third term . The smallest possible value occurs when

is .

Problem 15

Solution 1 Call the length of the race track . When they meet at the first meeting point, Brenda has run meters, while Sally has run meters. By the second

meters. meeting point, Sally has run meters, while Brenda has run

Since they run at a constant speed, we can set up

a proportion:

that . . Cross-multiplying, we get

Solution 2 The total distance the girls run between the start and the first meeting is one half of

the track length.

The total distance they run between the two meetings is the track length.

As the girls run at constant speeds, the interval between the meetings is twice as long as the interval between the start and the first meeting.

Thus between the meetings Brenda will run length of the track is meters meters. Therefore the

Problem 16

We know that the domain of

So

have

Then

and , where is a constant, is .

. By the definition of logarithms

, we then .

.

Problem 17

Problem 18

Solution 1

Let the point of tangency be . By the

The on and yields . Thus .

Hence .

Solution 2 Clearly,

. Thus, the sides of

and are in arithmetic progression. Thus it is

to the triangle

since , .

Problem 19

Solution 1

Note that

since is the center of the larger circle of radius . Using

, the Pythagorean Theorem on

Now using the on ,

Substituting ,

Solution 2

We can apply .

The four circles have curvatures , and .

We have

Simplifying, we get

Problem 20

Solution 1

:

1.

. The probability that

and is . Notice that the sum ranges from to

with a symmetric distribution across ,

and we want . Thus the chance is . 2.

now is

.

3.

4. . The probability that and is

, but automatically. Hence the chance

, which makes

. This is the same as the previous case. . We recognize that this is equivalent to the first case.

Our answer is .

Solution 2

Use areas to deal with this continuous probability problem. Set up a unit square with values of on x-axis and on y-axis.

If

if then this will work because . Similarly then this will work because in order for this to happen, and are

making , and . Each of these triangles in each greater than

the unit square has area of 1/8.

The only case left is when . Then each of and must be 1 and 0, in any order. These cut off squares of area 1/2 from the upper left and lower right corners of the unit square.

Then the area producing the desired result is 3/4. Since the area of the unit square is 1, the probability is

Problem 21

This is an infinite geometric series, which sums

to

. Using the formula .

Problem 22

By the

, we have

the heights up, we get . Adding Problem 23

We have to evaluate the answer choices and use process of elimination:

? : We are given that

then . , so . If one of the roots is zero,

? : By , we know that is the sum of all of the roots of .

Since that is real, , and , so . ?

of : All of the coefficients are real. For sake of contradiction suppose none are zero. Then for each complex root , its

product is equal to zero. is also a root. So the roots should pair up, but we have an odd number of imaginary roots! This gives us the contradiction, and therefore the

? : We are given that . Since the coefficients are real, it follows

that if a root is complex, its conjugate is also a root; and the sum of the imaginary parts of complex conjugates is zero. Hence the RHS is zero.

There is, however, no reason to believe that quantity is

should be zero (in fact, that ).

, and there is no evidence that is a root of

Problem 24

As the red circles move about segment , they cover the area we are looking for. On the left side, the circle must move around pivoted on . On the right side, the circle must move pivoted on However, at the top and bottom, the circle must lie on both A and B, giving us our upper and lower bounds.

This egg-like shape is .

The area of the region can be found by dividing it into several sectors, namely

Problem 25

This is an infinite geometric series

with common ratio

term , and initial

so

.

Alternatively, we could have used the algebraic manipulation for repeating decimals,

Telescoping,

Some factors cancel, (after all,

isn't one of the answer choices)

Since the only factor in the numerator that goes into Therefore the answer is . is , is minimized.