AMC12 2011年B卷真题+英文详解

AMC2011B

Problem 1

What is

Problem 2

Josanna's test scores to date are , , , , and . Her goal is to raise her

test average at least points with her next test. What is the minimum test score she

would need to accomplish this goal?

Solution

Take the average of her current test scores, which is

This m eans that she wants her test average after the sixth test to be Let be the

score that Josanna receives on her sixth test. Thus, our equation is

Problem 3

LeRoy and Bernardo went on a week-long trip together and agreed to share the costs equally. Over the week, each of them paid for various joint expenses such as gasoline and car rental. At the end of the trip it turned out that LeRoy had paid

dollars and Bernardo had paid dollars, where . How m any dollars must

LeRoy give to Bernardo so that they share the costs equally?

Solution

The total amount of money that was spent during the trip was So each

person should pay if they were to share the costs equally. Because LeRoy has already paid dollars of his part, he still has to pay

Problem 4

In multiplying two positive integers and , Ron reversed the digits of the two-digit

number . His erroneous product was 161. What is the correct value of the product of and ?

Solution

Taking the prime factorization of reveals that it is equal to Therefore, the

only ways to represent as a product of two positive integers is and

Because neither nor is a two-digit number, we know that and are

and Because is a two-digit number, we know that a, with its two digits reversed,

gives Therefore, and Multiplying our two correct values of and

yields

Problem 5

Let be the second sm allest positive integer that is divisible by every positive

integer less than . What is the sum of the digits of ?

Solution

must be divisible by every positive integer less than , or

and . Each number that is divisible by each of these is is a multiple

of their least common multiple. , so each number

divisible by these is a multiple of . The smallest multiple of is clearly , so the second smallest multiple of is .

Therefore, the sum of the digits of is

Problem 6

Two tangents to a circle are drawn from a point . The points of contact and

divide the circle into arcs with lengths in the ratio . What is the degree m easure

of ?

Solution

In order to solve this problem, use of the tangent-tangent intersection theorem (Angle of intersection between two tangents dividing a circle into arc length A and arc length B = 1/2 (Arc A° - Arc B°).

In order to utilize this theorem, the degree measures of the arcs must be found. First, set A (Arc length A) equal to 3d, and B (Arc length B) equal to 2d.

Setting 3d+2d = 360° will find d = 72°, and so therefore Arc length A in degrees will equal 216° and arc length B will equal 144°.

Finally, simply plug the two arc lengths into the tangent-tangent intersection theorem, and the answer:

1/2 (216°-144°) = 1/2 (72°)

Problem 7

Let and be two-digit positive integers with m ean . What is the m aximum

value of the ratio ?

Solution

If and have a mean of , then and . To maximize

, we need to maximize and minimize . Since they are both two-digit positive integers, the maximum of is which gives . cannot be decreased because doing so would increase , so this gives the maximum

value of , which is

Problem 8

Keiko walks once around a track at exactly the sam e constant speed every day. The sides of the track are straight, and the ends are sem icircles. The track has width

meters, and it takes her seconds longer to walk around the outside edge of the

track than around the inside edge. What is Keiko's speed in meters per second?

Solution

To find Keiko's speed, all we need to find is the difference between the distance around the inside edge of the track and the distance around the outside edge of the track, and divide it by the difference in tim e it takes her for each distance. We are given the difference in tim e, so all we need to find is the difference between the distances.

The track is divided into lengths and curves. The lengths of the track will exhibit no difference in distance between the inside and outside edges, so we only need to concern ourselves with the curves.

The curves of the track are semicircles, but since there are two of them, we can consider both of the at the sam e time by treating them as a single circle. We need to find the difference in the circum ferences of the inside and outside edges of the circle.

The form ula for the circum ference of a circle is where is the radius

of the circle.

Let's define the circum ference of the inside circle as and the circum ference of the

outside circle as .

If the radius of the inside circle () is , then given the thickness of the track is 6

meters, the radius of the outside circle () is .

Using this, the difference in the circum ferences is:

is the difference between the inside and outside lengths of the track. Divided by the tim e differential, we get:

Problem 9

Two real numbers are selected independently and at random from the interval

. What is the probability that the product of those num bers is greater than zero?

Solution

For the product to be greater than zero, we m ust have either both numbers negative or both positive.

Both numbers are negative with a chance.

Both numbers are positive with a chance.

Therefore, the total probability is and we are done.

Problem 10

Rectangle has and . Point is chosen on side so

that . What is the degree m easure of ?

Solution

Since , hence . Therefore . Therefore

Problem 11

A frog located at , with both and integers, makes successive jumps of

length and always lands on points with integer coordinates. Suppose that the frog

starts at and ends at . What is the sm allest possible number of jumps the frog m akes?

Solution

Since the frog always jumps in length and lands on a lattice point, the sum of its

coordinates m ust change either by (by jumping parallel to the x- or y-axis), or by

or (based off the 3-4-5 right triangle).

Because either , , or is always the change of the sum of the coordinates, the

sum of the coordinates will always change from odd to even or vice versa. Thus, it

is impossible for the frog to go from to in an even number of moves.

Therefore, the frog cannot reach in two m oves.

However, a path is possible in 3 m oves: from to to to . Thus, the answer is .

Problem 12

A dart board is a regular octagon divided into regions as shown below. Suppose that a dart thrown at the board is equally likely to land anywhere on the board. What is the probability that the dart lands within the center square?

Solution

Let's assume that the side length of the octagon is . The area of the center square is just . The triangles are all triangles,

with a side length ratio of . The area of each of the identical

triangles is , so the total area of all of the triangles is also . Now, we must find the area of all of the 4 identical rectangles.

One of the side lengths is and the other side length is , so the area of all of the rectangles is . The ratio of the area of the

square to the area of the octagon is . Cancelling from the fraction, the ratio becomes . Multiplying the numerator and the denominator each by will cancel out the radical, so the fraction is now

Problem 13

Brian writes down four integers whose sum is . The pairwise

positive differences of these num bers are and . What is the sum of the

possible values of ?

Solution

Assum e that results in the greatest

pairwise difference, and thus it is . This m eans . must be in

the set . The only way for 3 num bers in the set to add up to 9 is if they are

. , and then must be the rem aining two numbers which are and

. The ordering of must be either or .

Case 1

Case 2

The sum of the two w's is

Problem 14

A segm ent through the focus of a parabola with vertex is perpendicular to

and intersects the parabola in points and . What is ?

Solution

Name the directrix of the parabola . Define to be the distance between a

point and a line .

Now we remember the geom etric definition of a parabola: given any line (called

the directrix) and any point (called the focus), the parabola corresponding to the

given directrix and focus is the locus of the points that are equidistant from and

. Therefore . Let this distance be . Now note that , so

. Therefore . We now use the Pythagorean

Theorem on triangle ; . Similarly,

. We now use the Law of Cosines:

This shows that the answer is .

Problem 15

How many positive two-digit integers are factors of ?

Solution

From repeated application of difference of squares:

Aplying sum of cubes:

A quick check shows is prime. Thus, the only factors to be concerned about are

, since multiplying by will make any factor too large.

Multiply by or will give a two digit factor; itself will also work. The next

sm allest factor, , gives a three digit number. Thus, there are factors which are

multiples of .

Multiply by or will also give a two digit factor, as well as itself. Higher

numbers will not work, giving an additional factors.

Multiply by or for a two digit factor. There are no mare factors to check, as

all factors which include are already counted. Thus, there are an additional

factors.

Multiply by or for a two digit factor. All higher factors have been counted

already, so there are more factors.

Thus, the total number of factors is

Problem 16

Rhombus has side length and . Region consists of all points

inside of the rhombus that are closer to vertex than any of the other three

vertices. What is the area of ?

Solution

Suppose that is a point in the rhombus and let be the perpendicular

bisector of . Then if and only if is on the sam e side of as .

The line divides the plane into two half-planes; let be the half-plane

containing . Let us define similarly and . Then is equal to

. The region turns out to be an irregular pentagon. We can m ake it easier to find the area of this region by dividing it into four triangles:

Since and are

equilateral, contains , contains and , and contains . Then with and so

and has area .

Problem 17

Let , and

for integers . What is the sum of the digits of ?

Solution

Proof by induction that :

For

Assum e is true for n:

Therefore, if it is true for n, then it is true for n+1; since it is also true for n = 1, it is true for all positive integers n.

, which is the 2011-digit number 8888 (8889)

The sum of the digits is 8 times 2010 plus 9, or

Problem 18

A pyramid has a square base with side of length 1 and has lateral faces that are equilateral triangles. A cube is placed within the pyramid so that one face is on the base of the pyramid and its opposite face has all its edges on the lateral faces of the pyramid. What is the volume of this cube?

Solution

We can use the Pythagorean Theorem to split one of the triangular faces into two

30-60-90 triangles with side lengths and .

Next, take a cross-section of the pyram id, forming a triangle with the top of the triangle and the midpoints of two opposite sides of the square base.

This triangle is isosceles with a base of 1 and two sides of length .

The height of this triangle will equal the height of the pyram id. To find this height,

split the triangle into two right triangles, with sides and .

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The cube, touching all four triangular faces, will form a similar pyramid which sits on

top of the cube. If the cube has side length , the pyramid has side length .

Thus, the height of the cube plus the height of the sm aller pyramid equals the height of the larger pyramid.

.

side length of cube.

Problem 19

A lattice point in an -coordinate system is any point where both and are

integers. The graph of passes through no lattice point with

for all such that . What is the m aximum possible value of ?

Solution

Answer: (B)

It is very easy to see that the in the graph does not impact whether it passes through lattice.

We need to make sure that cannot be in the form of for , otherwise the graph passes through lattice point at . We only

need to worry about very close to , , will be the only case we need to worry about and we want the minimum of those, clearly for

, the smallest is , so answer is (B)

Problem 20

Triangle has , and . The points , and

are the midpoints of , and respectively. Let be the intersection

of the circum circles of and . What is ?

Solution 1

Answer: (C)

Let us also consider the circum circle of .

Note that if we draw the perpendicular bisector of each side, we will have the

circum center of which is , Also, since .

is cyclic, similarly, and are also cyclic. With this, we know

that the circum circles of , and all intercept at , so is

.

The question now becom es calculate the sum of distance from each vertices to the circum center.

We can do it will coordinate geometry, note that because of

being circum center.

Let , , ,

Then is on the line and also the line with slope and passes through

.

So

and

Solution 2

Consider an additional circum circle on . After drawing the diagram, it is

noticed that each triangle has side values: , , . Thus they are congruent, and their respective circum circles are. By inspection, we see that , , and

are the circum diameters, and so they are congruent. Therefore, the solution can be found by calculating one of these circum diam eters and multiplying it by a factor of . We can find the circumradius quite easily with the form ula

, s.t. and R is the circumradius.

Since :

After a few algebraic m anipulations:

.

Problem 21

The arithm etic m ean of two distinct positive integers and is a two-digit integer. The geom etric m ean of and is obtained by reversing the digits of the arithm etic

mean. What is ?

Solution

Answer: (D)

for som e ,.

Note that in order for x-y to be integer, has to be for some perfect

square . Since is at m ost , or

If , , if , . In AMC, we are done. Otherwise,

we need to show that is impossible.

-> , or or and , ,

respectively. And since , , , but there is no integer solution for , .

Problem 22

Let be a triangle with sides , and . For , if

and , and are the points of tangency of the incircle of to the sides

, and , respectively, then is a triangle with side lengths ,

and , if it exists. What is the perim eter of the last triangle in the sequence ?

Solution

Answer: (D)

Let , , and

Then , and

Then , ,

Hence:

Note that and for , I claim that it is true for all , assum e for induction that it is true for som e , then

Furthermore, the average for the sides is decreased by a factor of 2 each tim e.

So is a triangle with side length , ,

and the perimeter of such is

Now we need to find what fails the triangle inequality. So we need to find the last

such that

For , perimeter is

Problem 23

A bug travels in the coordinate plane, moving only along the lines that are parallel

to the -axis or -axis. Let and . Consider all possible paths of the bug from to of length at m ost . How m any points with integer

coordinates lie on at least one of these paths?

Solution

Answer: (C)

If a point satisfy the property that ,

then it is in the desire range because is the shortest path from

to , and is the shortest path from to

If , then satisfy the property. there are

lattice points here.

else let (and for it is symmetrical,

,

So for , there are lattice points,

for , there are lattice points,

etc.

For , there are lattice points.

Hence, there are a total of lattice points.

Problem 24

Let . What is the minimum perimeter am ong all the -sided polygons in the complex plane whose vertices are precisely

the zeros of ?

Solution

Answer: (B)

First of all, we need to find all such that

So or

or

Now we have a solution at if we look at them in polar coordinate, further m ore, the 8-gon is symm etric (it is an equilateral octagon) . So we only need to find the side length of one and multiply by .

So answer distance from to

Side length

Hence, answer is .

Problem 25

For every and integers with odd, denote by the integer closest to . For

every odd integer , let be the probability that

for an integer randomly chosen from the interval . What is the

minimum possible value of over the odd integers in the interval

?

Solution

Answer:

First of all, you have to realize that

if

then

So, we can consider what happen in and it will repeat. Also since range of

is to , it is always a multiple of . So we can just consider for .

LET be the fractional part function

This is an AMC exam, so use the given choic es wisely. With the given choices, and

the previous explanation, we only need to consider , , , .

For , . 3 of the that should consider lands in here.

For , , then we need

else for , , then we need

For ,

So, for the condition to be true, . ( , no worry for the

rounding to be )

, so this is always true.

For , , so we want , or

For k = 67,

For k = 69,

etc.

We can clearly see that for this case, has the m inimu m, which is .

Also, .

So for AMC purpose, answer is (D).

Now, let's say we are not given any answer, we need to consider .

I claim that

If got round down, then all satisfy the condition along with

because if and , so m ust

and for , it is the sam e as .

, which m akes

If got round up, then all satisfy the condition along with

because if and

Case 1)

->

Case 2)

->

and for , since is odd,

->->, and is prime so or ,

which is not in this set , which m akes

Now the only case without rounding, . It must be true.

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