AMC 12A Problem and Solution

2017 AMC 12A

Problem15

Let

, using radian measure for the variable . In what

lie?

interval does the smallest positive value of for which

Solution

We must first get an idea of what Between 0 and 1,

looks like:

starts at and increases; clearly there is no zero here.

Between 1 and , either.

Between and 3, zero here either. Between 3 and , here either. Between and

starts at a positive number and increases to ; there is no zero here

starts at and increases to some negative number; there is no

starts at some negative number and increases to -2; there is no zero

, starts at -2 and increases

to

Intermediate Value Theorem. Therefore, the answer is

. There is a zero here by the

.

Problem16

In the figure below, semicircles with centers at and and with radii 2 and 1, respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter . The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. A circle centered at is drawn externally tangent to the two smaller semicircles and internally tangent to the largest semicircle. What is the radius of the circle centered at ?

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2017 AMC 12A

Solution 1

Connect the centers of the tangent circles! (call the center of the large circle )

Notice that we don't even need the circles anymore; thus, draw triangle cevian :

with

and use Stewart's Theorem:

From what we learned from the tangent circles, we have , , , ,

where is the radius of the circle centered at that we seek. Thus:

, and

,

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2017 AMC 12A

Solution 2

Like the solution above, connecting the centers of the circles results in triangle with cevian . The two triangles and share angle , which means we can use Law of Cosines to set up a system of 2 equations that solve for respectively:

(notice that the diameter of the largest

semicircle is 6, so its radius is 3 and

is 3 - r)

We can eliminate the extra variable of angle by multiplying the first equation by 3 and subtracting the second from it. Then, expand to find :

so =

,

Solution 3

Let be the center of the largest semicircle and be the radius of . We know that , , , , and . Notice that and are bounded by the same two parallel lines, so these triangles

have the same heights. Because the bases of these two triangles (that have the same heights) differ by a factor of 2, the area of must be twice that of , since the area of a triangle is

.

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2017 AMC 12A

Again, we don't need to look at the circle and the semicircles anymore; just focus on the triangles.

Let

equal to the area of

and

equal to the area of

and is

. Heron's

Formula states that the area of an triangle with sides where , or the semiperimeter, is The semiperimeter of to obtain

Using Heron's Formula again, find the area of

is

Use Heron's Formula

with sides

, , and

.

Now,

Solution 4

Let , the center of the large semicircle, to be at

, and to be at

.

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