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

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.

2017 AMC 12A Problem 16

Answer: B

Difficulty rating: 1840

Solution:

Problem 16 in Other Years