2010 AMC 12A Problems and Solution 下载本文

Solution 2

We can use Principle of Inclusion-Exclusion to find the final volume of the cube. There are 3 \\

volumes, as the central

cubic inches. However, we can not just sum their

cube is included in each of these three cuts. To

get the correct result, we can take the sum of the volumes of the three cuts, and subtract the volume of the central cube twice. Hence the total volume of the cuts is

.

Therefore the volume of the rest of the cube is .

Solution 3

We can visualize the final figure and see a cubic frame. We can find the volume of the figure by adding up the volumes of the edges and corners. Each edge can be seen as a

box.

box, and each corner can be seen as a

.

Solution 4

First, you can find the volume, which is 27. Now, imagine there are three prisms of dimensions 2 x 2 x 3. Now subtract the prism volumes from 27. We have -9. From here we add two times 2^3, because we over-removed (LOL). This is 16 - 9 = 7 (A).

Problem 10

The first four terms of an arithmetic sequence are , , is the

term of this sequence?

, and

. What

Solution

and

are consecutive terms, so the common difference is

.

The common difference is . The first term is and the

term is

Problem 11

The solution of the equation What is ?

can be expressed in the form

.

Solution

This problem is quickly solved with knowledge of the laws of exponents and logarithms.

Since we are looking for the base of the logarithm, our answer is .

Problem 12

In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.

Brian: \ Chris: \ LeRoy: \

Mike: \ How many of these amphibians are frogs?

Solution Solution 1

We can begin by first looking at Chris and LeRoy.

Suppose Chris and LeRoy are the same species. If Chris is a toad, then what he says is true, so LeRoy is a frog. However, if LeRoy is a frog, then he is lying, but clearly Chris is not a frog, and we have a contradiction. The same applies if Chris is a frog.

Clearly, Chris and LeRoy are different species, and so we have exactly of the two of them.

frog out

Now suppose Mike is a toad. Then what he says is true because we already have toads. However, if Brian is a frog, then he is lying, yet his statement is true, a contradiction. If Brian is a toad, then what he says is true, but once again it conflicts with his statement, resulting in contradiction.

Therefore, Mike must be a frog. His statement must be false, which means that there is at most toad. Since either Chris or LeRoy is already a toad, Brain must be a frog. We can also verify that his statement is indeed false.

Both Mike and Brian are frogs, and one of either Chris or LeRoy is a frog, so we have frogs total.

Solution 2

Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog. As Mike is a frog, his statement is false, hence there is at most one toad.

As there is at most one toad, at least one of Chris and LeRoy is a frog. But then the other one tells the truth, and therefore is a toad. Hence we must have one toad and three frogs.

Problem 13

For how many integer values of do the graphs of intersect?

and

not

Solution

The image below shows the two curves for

.

. The blue curve is

,

which is clearly a circle with radius , and the red curve is a part of the curve

In the special case the blue curve is just the point , and as ,

this point is on the red curve as well, hence they intersect. The case

is symmetric to

: the blue curve remains the same and the red

.

curve is flipped according to the axis. Hence we just need to focus on

Clearly, on the red curve there will always be points arbitrarily far from the origin: for example, as approaches 0, approaches most .

At this point we can guess that on the red curve the point where closest to the origin, and skip the rest of this solution.

For an exact solution, fix and consider any point distance from the origin is minimize

on the red curve. Its

is always

. Hence the red curve intersects

the blue one if and only if it contains a point whose distance from the origin is at

. To minimize this distance, it is enough to

. By the Arithmetic Mean-Geometric Mean Inequality we get

, and that equality holds whenever

, i.e.,

that this value is at least

.

Now recall that the red curve intersects the blue one if and only if its closest point is at most from the origin. We just computed that the distance between the origin and the closest point on the red curve is integers such that

.

, hence the two curves are only disjoint for

values.

. Therefore, we want to find all positive

Clearly the only such integer is

and

. This is a total of

Solution 2:

From the graph shown above, we see that there is a specific point closest to the center of the circle. Using some logic, we realize that as long as said furthest point is not inside or on the graph of the circle. This should be enough to conclude that the hyperbola does not intersect the circle.

Therefore, for each value of k, we only need to check said value to determine intersection. Let said point, closest to the circle have coordinates

derived