2017 AMC 12A Problem 16

Question

In the figure below, semicircles with centers at

A

and

B

and with radii

2

and

1,

respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter

\overline{JK}.

The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. A circle centered at

P

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

P?

\dfrac{3}{4}

\dfrac{6}{7}

\dfrac{1}{2}\sqrt3

\dfrac{5}{8}\sqrt2

\dfrac{11}{12}

Answer: B

Concepts:tangent circles Pythagorean Theorem coordinate geometry

Difficulty rating: 1840

Small hint:

Let

C

be the center of the large semicircle; then

PA=2+r,

PB=1+r,

PC=3-r

Big hint:

Drop a perpendicular from

P

to

\overline{JK}

and set the three expressions for its squared height equal

Solution:

The large semicircle has radius

3

and center

C,

the midpoint of

\overline{JK}.

Placing

J

at the origin,

A=2,

B=5,

C=3,

K=6

along the base. Let

r

be the radius of the circle at

P.

By tangency,

PA=2+r,

PB=1+r,

and

PC=3-r.

Dropping a perpendicular from

P

to the base at horizontal position

3+x

with height

h,

the Pythagorean theorem gives

\begin{aligned} h^2 &=(2+r)^2-(1+x)^2 \\ &=(3-r)^2-x^2 \\ &=(1+r)^2-(2-x)^2. \end{aligned}

Equating the first expression with the middle one gives

5r-x=3,

while equating the last expression with the middle one gives

2r+x=3.

Adding yields

7r=6,

so

r=\dfrac{6}{7}.

Thus, the correct answer is B.