Question

Circles with centers $P, Q$ and $R,$ having radii $1, 2$ and $3,$ respectively, lie on the same side of line $l$ and are tangent to $l$ at $P', Q'$ and $R',$ respectively, with $Q'$ between $P'$ and $R'.$ The circle with center $Q$ is externally tangent to each of the other two circles. What is the area of $\triangle PQR?$

$0$

$\sqrt{\dfrac{2}{3}}$

$1$

$\sqrt{6}-\sqrt{2}$

$\sqrt{\dfrac{3}{2}}$

Answer: D

Concepts:tangent circles Pythagorean Theorem coordinate geometry

Difficulty rating: 1970

Solution:

Put the tangent line on the $x$ -axis and take $P=(0,1).$ Because $PQ=1+2=3$ and the centers differ in height by $1,$ the horizontal distance from $P$ to $Q$ is $\sqrt{3^2-1^2}=2\sqrt2.$ Similarly, $QR=2+3=5$ and its centers also differ in height by $1,$ so their horizontal distance is $2\sqrt6.$

Thus we may use $\begin{aligned} P&=(0,1),\qquad Q=(2\sqrt2,2), \\ R&=(2\sqrt2+2\sqrt6,3). \end{aligned}$ The coordinate-area formula gives $\begin{aligned} [PQR] &=\frac12\left|4\sqrt2-(2\sqrt2+2\sqrt6)\right| \\ &=\sqrt6-\sqrt2. \end{aligned}$

Thus, the correct answer is D .

2016 AMC 10A Problem 21

Answer: D

Difficulty rating: 1970

Solution:

Problem 21 in Other Years