2009 AMC 12A Problems(答案)

2009 AMC 12A Problems

Problem 1

Kim's flight took off from Newark at 10:34 AM and landed in Miami at 1:18 PM. Both cities are in the same time zone. If her flight took hours and minutes, with

, what is ?

Solution

There is hour and minutes between 10:34 AM and noon; and there

is hour and minutes between noon and 1:18 PM. Hence the flight took hours

and minutes, and .

Problem 2

Which of the following is equal to ?

Solution

We compute:

Problem 3

What number is one third of the way from to ?

Solution

Solution 1

We can rewrite the two given fractions as and . (We multiplied all numerators

and denominators by .)

Now it is obvious that the interval between them is divided into three parts by the

fractions and .

Solution 2

The number we seek can be obtained as a weighted average of the two endpoints, where the closer one has weight and the further one . We compute:

Problem 4

Four coins are picked out of a piggy bank that contains a collection of pennies, nickels, dimes, and quarters. Which of the following could not be the total value of the four coins, in cents?

Solution

As all five options are divisible by , we may not use any pennies. (This is because

a penny is the only coin that is not divisible by , and if we used between and

pennies, the sum would not be divisible by .)

Hence the smallest coin we can use is a nickel, and thus the smallest amount we can

get is . Therefore the option that is not reachable is .

We can verify that we can indeed get the other ones:

?

?

?

?

Problem 5

One dimension of a cube is increased by , another is decreased by , and the third

is left unchanged. The volume of the new rectangular solid is less than that of the

cube. What was the volume of the cube?

Solution

Let the original cube have edge length . Then its volume is . The new box has

dimensions , , and , hence its volume is .

The difference between the two volumes is . As we are given that the difference is

, we have , and the volume of the original cube was .

Problem 6

Suppose that and . Which of the following is equal to for

every pair of integers ?

Solution

We have .

Problem 7

The first three terms of an arithmetic sequence are , , and

respectively. The th term of the sequence is . What is ?

Solution

As this is an arithmetic sequence, the difference must be constant:

. This solves to . The first

three terms then are , , and . In general, the -th term is . Solving

we get .

Problem 8

Four congruent rectangles are placed as shown. The area of the outer square is

times that of the inner square. What is the ratio of the length of the longer side of each rectangle to the length of its shorter side?

Solution

The area of the outer square is times that of the inner square. Therefore the side

of the outer square is times that of the inner square.

Then the shorter side of the rectangle is of the side of the outer square, and the

longer side of the rectangle is of the side of the outer square, hence their ratio

is .

Problem 9

Suppose that and . What is ?

Solution

Solution 1

As , we have .

To compute , set in the first formula. We get

.

Solution 2

Combining the two formulas, we know that .

We can rearrange the right hand side to .

Comparing coefficients we have , , and . From

the second equation we get , and then from the third we get . Hence

.

Problem 10

In quadrilateral , , , , , and is an

integer. What is ?

Solution

By the triangle inequality we have , and also

, hence .

We got that , and as we know that is an integer, we must have

.

Problem 11

The figures , , , and shown are the first in a sequence of figures. For

, is constructed from by surrounding it with a square and placing one

more diamond on each side of the new square than had on each side of its

outside square. For example, figure has diamonds. How many diamonds are

there in figure ?

Solution

Solution 1

Color the diamond layers alternately blue and red, starting from the outside. You'll get the following pattern:

In the figure , the blue diamonds form a square, and the red diamonds

form a square. Hence the total number of diamonds in is

.

Solution 2

When constructing from , we add new diamonds. Let be the

number of diamonds in . We now know that and

.

Hence we get:

Problem 12

How many positive integers less than are times the sum of their digits?

Solution

Solution 1

The sum of the digits is at most . Therefore the number is at most . Out of the numbers to the one with the largest sum of digits is ,

and the sum is . Hence the sum of digits will be at most .

Also, each number with this property is divisible by , therefore it is divisible by ,

and thus also its sum of digits is divisible by .

We only have six possibilities left for the sum of the digits: , , , , , and .

These lead to the integers , , , , , and . But for the sum of digits

is , which is not , therefore is not a solution. Similarly we can throw

away , , , and , and we are left with just solution: the number .

Solution 2

We can write each integer between and inclusive as

where and . The sum of digits of this number is

, hence we get the equation . This

simplifies to . Clearly for there are no solutions, hence

and we get the equation . This obviously has only one valid solution

, hence the only solution is the number .

Problem 13

A ship sails miles in a straight line from to , turns through an angle between

and , and then sails another miles to . Let be measured in miles.

Which of the following intervals contains ?

Solution

Answering the question

To answer the question we are asked, it is enough to compute for two different

angles, preferably for both extremes (and degrees).

Alternately, it is enough to compute for one of the extreme angles. In case it falls inside one of the given intervals, we are done. In case it falls on the boundary

between two options, we also have to argue whether our is the minimal or the

maximal possible value of .

Below we show a complete solution in which we also show that all possible values of

do indeed lie in the given interval.

Complete solution

Let be the point the ship would reach if it turned , and the point it would

reach if it turned . Obviously, is the furthest possible point from , and is

the closest possible point to .

Hence the interval of possible values for is .

We can find and as follows:

Let and be the feet of the heights from and onto . The angles in the

triangle are , , and , hence .

Similarly, the angles in the triangle are , , and , hence

and .

Hence we get:

Therefore for any valid the value is surely in the interval .

Problem 14

A triangle has vertices , , and , and the line divides the

triangle into two triangles of equal area. What is the sum of all possible values of ?

Solution

Let's label the three points as , , and .

Clearly, whenever the line intersects the inside of the triangle, it will intersect the side . Let be the point of intersection.

The triangles and have the same height, which is the distance between

the point and the line . Hence they have equal areas if and only if is the

midpoint of .

The midpoint of the segment has coordinates . This point lies on

the line if and only if . This simplifies to

. This is a quadratic equation with roots and .

Both roots represent valid solutions, and their sum is .

For illustration, below are pictures of the situation for , ,

, and .

Problem 15

For what value of is ?

Note: here .

Solution

Obviously, even powers of are real and odd powers of are imaginary. Hence the

real part of the sum is , and the imaginary part is

.

Let's take a look at the real part first. We have , hence the real part

simplifies to . If there were an odd number of terms, we

could pair them as follows: , hence the result would be negative. As we need the real part to be , we must have an even number of

terms. If we have an even number of terms, we can pair them as

. Each parenthesis is equal to , thus there are of

them, and the last value used is . This happens for and . As

is not present as an option, we may conclude that the answer is .

In a complete solution, we should now verify which of and will give

us the correct imaginary part.

We can rewrite the imaginary part as follows:

. We need to

obtain . Once again we can repeat the same reasoning: If the number of terms were even, the left hand side would be negative, thus the

number of terms is odd. The left hand side can then be rewritten as

. We need parentheses, therefore the last value

used is . This happens when or , and we are done.

Problem 16

A circle with center is tangent to the positive and -axes and externally tangent

to the circle centered at with radius . What is the sum of all possible radii of

the circle with center ?

Solution

Let be the radius of our circle. For it to be tangent to the positive and axes, we

must have . For the circle to be externally tangent to the circle centered

at with radius , the distance between and must be exactly .

By the Pythagorean theorem the distance between and is

, hence we get the equation .

Simplifying, we obtain . By Vieta's formulas the sum of the two

roots of this equation is .

(We should actually solve for to verify that there are two distinct positive roots. In

this case we get .)

Problem 17

Let and be two different

infinite geometric series of positive numbers with the same first term. The sum of the first series is , and the sum of the second series is . What is ?

Solution

Using the formula for the sum of a geometric series we get that the sums of the

given two sequences are and .

Hence we have and . This can be rewritten as

.

As we are given that and are distinct, these must be precisely the two roots of

the equation .

Using Vieta's formulas we get that the sum of these two roots is .

Alternate Solution

Using the formula for the sum of a geometric series we get that the sums of the

given two sequences are and .

Hence we have and . This can be rewritten as

.

Which can be further rewritten as . Rearranging the equation we

get . Expressing this as a difference of squares we get

.

Dividing by like terms we finally get as desired.

Problem 18

For , let , where there are zeros between the and the .

Let be the number of factors of in the prime factorization of . What is the

maximum value of ?

Solution

The number can be written as .

For we have . The first value in the parentheses is odd, the second one is even, hence their sum is odd and we have

.

For we have . For the value in the

parentheses is odd, hence .

This leaves the case . We have . The value is

obviously even. And as , we have , and therefore

. Hence the largest power of that divides is , and

this gives us the desired maximum of the function : .

Problem 19

Andrea inscribed a circle inside a regular pentagon, circumscribed a circle around the pentagon, and calculated the area of the region between the two circles. Bethany did the same with a regular heptagon (7 sides). The areas of the two regions were and , respectively. Each polygon had a side length of . Which of

the following is true?

Solution

In any regular polygon with side length , consider the isosceles triangle formed by

the center of the polygon and two consecutive vertices and . We are given

that . Obviously , where is the radius of the circumcircle.

Let be the midpoint of . Then , and , where is the

radius of the incircle.

Applying the Pythagorean theorem on the triangle , we get that .

Then the area between the circumcircle and the incircle can be computed as

.

Hence , , and therefore .

Problem 20

Convex quadrilateral has and . Diagonals and

intersect at , , and and have equal areas. What is

?

1 Problem

2 Solution

2.1

Solution 1

2.2

Solution 2

3 See also

Solution

Solution 1

Let denote the area of triangle . , so

. Since triangles

and share a base, they also have the same height and thus and

with a ratio of . .

Solution 2

Using the sine area formula on triangles and , as ,

we see that

Since , triangles and are similar. Their ratio is

. Since , we must have , .

Problem 21

Let , where , , and are complex numbers. Suppose

that

What is the number of nonreal zeros of ?

Solutions

Solution 1

From the three zeroes, we have

.

Then

.

Let's do each factor case by case:

: Clearly, all the fourth roots are going to be complex.

?: The real roots are , and there are two complex roots.

?: The real roots are , and there are two complex roots.

So the answer is .

Problem 22

A regular octahedron has side length . A plane parallel to two of its opposite faces

cuts the octahedron into the two congruent solids. The polygon formed by the

intersection of the plane and the octahedron has area , where , , and are

positive integers, and are relatively prime, and is not divisible by the square of

any prime. What is ?

Solution

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file]

If the plane divides the octahedron into two congruent solids, it goes through the center of the octahedron. As it is parallel to two opposite faces (colored above in green), it passes through the midpoints of the edges connecting the corresponding vertices of the faces. The distance between the center and any of the midpoints, as

well as the distance between any consecutive midpoints, is found to be (by midline and so forth). Thus, the intersection of the plane and the octahedron is a

regular hexagon, and the answer is , and

.

Problem 23

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