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# 2003 AMC12B(美国数学竞赛)

Which of the following is the same as

Solution

Al gets the disease algebritis and must take one green pill and one pink pill each day for two weeks. A green pill costs 1 dollar more than a pink pill, and Al's pills cost a

total of 546 dollars for the two weeks. How much does one green pill cost?

Solution

Rose fills each of the rectangular regions of her rectangular flower bed with a

different type of flower. The lengths, in feet, of the rectangular regions in her flower bed are as shown in the figure. She plants one flower per square foot in each region. Asters cost \$1 each, begonias \$1.50 each, cannas \$2 each, dahlias \$2.50 each, and Easter lilies \$3 each. What is the least possible cost, in dollars, for her garden?

Solution

Moe uses a mower to cut his rectangular 90-foot by 150-foot lawn. The swath he cuts is 28 inches wide, but he overlaps each cut by 4 inches to make sure that no grass is missed. he walks at the rate of 5000 feet per hour while pushing the mower.

Which of the following is closest to the number of hours it will take Moe to mow his lawn?

Solution

Many television screens are rectangles that are measured by the length of their diagonals. The ratio of the horizontal length to the height in a standard television screen is 4 : 3. The horizontal length of a "27-inch" television screen is closest, in

inches, to which of the following?

Solution

The second and fourth terms of a geometric sequence are 2 and 6. Which of the following is a possible first term?

Solution

Penniless Pete's piggy bank has no pennies in it, but it has 100 coins, all

nickels,dimes, and quarters, whose total value is \$8.35. It does not necessarily contain coins of all three types. What is the difference between the largest and

smallest number of dimes that could be in the bank? Solution

Let denote the sum of the digits of the positive integer

. For and

Solution For how many two-digit values example,

of

is Let be a linear function for which What is

Solution

Several figures can be made by attaching two equilateral triangles to the regular pentagon ABCDE in two of the five positions shown. How many non-congruent figures can be constructed in this way?

Solution

Cassandra sets her watch to the correct time at noon. At the actual time of 1:00 PM, she notices that her watch reads 12:57 and 36 seconds. Assuming that her watch loses time at a constant rate, what will be the actual time when her watch first reads

10:00 PM?

Solution

What is the largest integer that is a divisor

of for all positive even integers ?

Solution

An ice cream cone consists of a sphere

of vanilla ice cream and a right circular cone that has the same diameter as the sphere. If the ice cream melts, it will exactly fill the cone. Assume that the melted ice cream occupies of the volume of the

frozen ice cream. What is the ratio of the cone’s height to its radius

?

Solution

In rectangle ABCD, AB = 5 and BC = 3. Points F and G are on CD so that DF = 1 and

GC = 2. Lines AF and BG intersect at E. Find the area

of

.

Solution

A regular octagon

of the rectangle ? has an area of one square unit. What is the area

Solution

Three semicircles of radius 1 are constructed on diameter AB of a semicircle of radius 2. The centers of the small semicircles divide AB into four line segments of equal length, as shown. What is the area of the shaded region that lies within the

large semicircle but outside the smaller semicircles? Solution

If and

, what is

?

Solution

Let be the set of permutations

of the sequence

for which the first

term is not . A permutation is chosen randomly from . The probabilitythat the second term is , in lowest terms, is . What is

Solution ?

Part of the graph of is shown. What is ?

Solution

An object moves cm in a straight line

from to , turns at an angle , measured in radians and chosen at random from the interval straight line to . What is the probability that Solution , and moves cm in a ?

Let be a

rhombus with and . Let

to be a point on and, , and let and be the feet of the perpendiculars from

respectively. Which of the following is closest to the minimum possible value of ?

Solution

The number of -intercepts on the graph of

interval

is closest to Solution in the

Positive integers

of equations

and and

are chosen so that , and the system

has exactly one solution. What is the minimum value of

?

Solution

Three points are chosen randomly and independently on a circle. What is the probability that all three pairwise distance between the points are less than the radius of the circle?

Solution

Alternatively, notice that each term in the numerator is of a term in the

denominator, so the quotient has to be . Problem 6

Call the first term by

for and

and the common ratio . Then the nth term is given and and . Substituting 2 and 6 . Dividing the first equation

. Therefore, respectively gives

, so by the second gives

gives Substituting this into the first equation and rationalizing the denominator

. Dividing by

gives

. The negative value corresponds to answer choice Problem 8

Let and be the digits of ,

Clearly can only be or and only and are possible to have two digits sum to.

If

If

The total number of solutions is sums to , there are 3 different solutions : sums to , there are 7 different solutions:

Problem 9

Solution

Since

is a linear function with slope ,

Problem 10 Place the first triangle. Now, we can place the second triangle either adjacent to the first, or with one side between them, for a total of Problem 13

Let

be the common radius of the sphere and the cone, and be the cone’s height. Then

Thus . Problem 16 Each small semicircle shares a radius with an adjacent circle. Therefore, the radii to the points of intersection will create equilateral triangles. Draw these triangles, and then bisect the angles on the sides and complete the incomplete triangles so that the unshaded region is broken into two types of pieces: circular segments from a sector of a circle with radius , and equilateral triangles with side length .

There are equilateral triangles and circular segments. One equilateral triangle

has area area , and one segment has .

So the shaded region is the area of the large semicircle minus the area of the unshaded parts, which

is . Problem 17

Since

givesHence

It is not difficult to find .

.

Summing Problem 18

Suppose Since and ,

, so there are

of that are divisible by . values

Problem 19

There are

choices for the first element of , and for each of these choices there

are ways to arrange the remaining elements. If the second element must be , then there are only choices for the first element and ways to arrange the

remaining elements. Hence the answer is

and.

,

Problem 20

Solution 1

SinceIt follows that

. Also, , so

. Solution 2

Two of the roots of Then are , and we let the third one be .

Notice that , so .

Problem 21

By the Law of

Cosines, It follows that , and the probability is .

Problem 22

Let then and

and intersect at . Since is a rhombus,

, so is , so

are perpendicular bisectors. Thus arectangle. Since the diagonals of a rectangle are of equal length,

we want to minimize

. It follows that we want Finding the area in two different ways,.

Problem 23

Solution

The function

we want

are

on has roots in the form of

, so

solutions for

on this interval for all integers . Therefore,

. There Problem 24

Step 1: Finding some promising bound

Does the system have a solution where

For such a solution we would have

hence , which solves to

,

, there will always ? , . If we want to avoid this solution, we need to have hence

be one solution , hence such that . In other words, if .

Step 2: Showing one solution

We will now find out whether there is a for which (and some ) the system has only one solution. We already know of one such solution, so we need to make sure that no other solution appears.

Obviously, there are three more theoretically possible solutions: one in in , and one in

. The first case solves to

, and the third to , one

, the

. We second to need to make sure that the following three conditions hold:

1.

2.

3.

.

Let

1.

2. and

. We then have:

3.

Hence for solution

, and any valid the system has exactly one

.

Step 3: Proving the optimality of our solution

We will now show that for that the system always has a solution such . This will mean that the system has at least two solutions, and thus the

is optimal.

, we have ,

, we solution with 1. As we are looking for a hence . To make sure that the value falls outside

, or need to make it larger than , thus

equivalently

2. The condition we just derived,

as

becomes

outside

inequality , then as . , can be rewritten , which . Thus to make sure that the second value falls , we need to make it larger than . The simplifies to .

3. To avoid the last solution, we must have

to .

, which simplifies The last two inequalities contradict each other, thus there are no satisfy both of them. that would Conclusion

We just showed that whenever

solutions: one with We also showed that for exactly one solution. Hence the optimal value of is .

and one with , the system has at least two different .

for which the system has there are some

Problem 25

First: One can choose the first point anywhere on the circle. Secondly: The Next point must lie within of degrees possible, chance.

degrees of both. This ranges from degrees arc

degrees (if degrees of arc on either side, a total

The last point must lie within to sit on (if the first two are degrees apart) and a probability, to

they are negligibly apart) and a chance. As the second point moves

from degrees away to the first point, the probability changes linearly (every degree it moves, adds one degree to where the third could be), the probabilities at each end can be averaged to find .

Therefore the total probability is or