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2011AMC12B详细答案

2011AMC12B答案

1. Add up the numbers in each fraction to get , which equals . Doing the subtraction yields

2. Take the average of her current test scores, which is

This means 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

3. 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

4. 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

5. 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

6. 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°)

7. 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

8. 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 time it takes her for each distance. We are given the difference in time, 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 same time by treating them as a single circle. We need to find the difference in the circumferences of the inside and outside edges of the circle.

The formula for the circumference of a circle is where is the radius of the circle.

Let's define the circumference of the inside circle as and the circumference 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 circumferences is:

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

9. For the product to be greater than zero, we must 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.

10.

Since , hence . Therefore . Therefore

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

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 moves.

However, a path is possible in 3 moves: from to to to .

Thus, the answer is .

12. 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

13. Assume that results in the greatest pairwise difference, and

thus it is . This means . must be in the set . The only way for 3 numbers in

the set to add up to 9 is if they are . , and then must be the remaining two numbers which

are and . The ordering of must be either or .

Case 1

Case 2

The sum of the two w's is

14. Name the directrix of the parabola . Define to be the distance between a point and a line .

Now we remember the geometric 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 .

15. From repeated application of difference of squares:

2412633

21)6597=21)7,

???????

2-1=(+

21)(2+1)(2+1)(2-1),2412122

2-1=(+(+5133

Aplying sum of cubes:

1248412

+=+-++=?

21(21)(221),2117241.

????, since A quick check shows 241 is prime. Thus, the only factors to be concerned about are 23571317 multiplying by 241 will make any factor too large.

Multiply 17 by 3 or 5 will give a two digit factor; 17 itself will also work. The next smallest factor, 7, gives a three digit number. Thus, there are 3 factors which are multiples of 17.

Multiply 13 by 3、5 or 7 will also give a two digit factor, as well as 13 itself. Higher numbers will not work, giving an additional 4 factors.

Multiply 7 by 3、5 or 23for a two digit factor. There are no mare factors to check, as all factors which include 13 are already counted. Thus, there are an additional 3 factors.

Multiply 5 by 3 or 23for a two digit factor. All higher factors have been counted already, so there are more factors.

Thus, the total number of factors is 3+4+3+2=12.D

16. Suppose that is a point in the rhombus and let be the perpendicular bisector of .

Then if and only if is on the same 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

make 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 .

17.

Proof by induction that :

For

Assume 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

18. We can use the Pythagorean Theorem to split one of the triangular faces into two 30-60-90 triangles with side lengths

，1 and .

Next, take a cross-section of the pyramid, 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 pyramid. To find this height, split the triangle into two right triangles, with sides

12，22

and .

Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and activated. (click here to install Java now )

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

x , the pyramid has side length

Thus, the height of the cube plus the height of the smaller pyramid equals the height of the larger pyramid.

，

，side length of cube.

19. It is very easy to see that the +2 in the graph does not impact whether it passes through lattice.

We need to make sure that

m cannot be in the form of

for , otherwise the graph y mx =

passes through lattice point at x b =. We only need to worry about

a b very close to 12, 121m m ++, 12m m

+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) 20. Answer: (C) Let us also consider the circumcircle of ADF ?.

Note that if we draw the perpendicular bisector of each side, we will have the circumcenter of ABC ? which is p, Also, since 0

90m ADP m AFP ∠=∠=. ADPF is cyclic, similarly, BDPE and CEPF are also cyclic. With this, we know that the circumcircles of ADF ?, BDE ?and CEF

?all intercept at P, so P is X.

The question now becomes calculate the sum of distance from each vertices to the circumcenter. We can do it will coordinate geometry, note that XA=XB=XC because of X being circumcenter. Let

00(5,12),(0,0),(14,0),(,)A B C X x y ====,

Then X is on the line 7x = and also the line with slope 5

and passes through (2.5,6).

and .

Solution 2

Consider an additional circumcircle on ADF ?. After drawing the diagram, it is noticed that each triangle has side values:7,

152, 132

. Thus they are congruent, and their respective circumcircles are. By inspection, we see that XA,XB, and XC are the circumdiameters, and so they are congruent. Therefore, the solution can be found by calculating one of these circumdiameters and multiplying it by a factor of 3. We can find the circumradius quite easily with the formula

, s.t.

and R is

the circumradius. Since :

After a few algebraic manipulations:

21. Answer: (D)

for some ,.

Note that in order for x-y to be integer, has to be for some perfect square . Since is at

most , 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 , .

22. Answer: (D) Let , , and

Then , and

Then , ,

Hence:

Note that and for , I claim that it is true for all , assume for induction that it is

true for some , then

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

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

23. 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.

24.Answer: (B) First of all, we need to find all such that

So or

Now we have a solution at if we look at them in polar coordinate, further more, the 8-gon is symmetric (it is a regular 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 .

25. Answer:

First of all, you have to realize that

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 choices 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 minimum , 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 must

and for , it is the same as .

, which makes

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 makes

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