2016 AMC 10A Problems/Problem 21

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 $PQR$ ?

$\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}$

==Solution== [edit]

Notice that we can find $[P'PQRR']$ in two different ways: $[P'PQQ']+[Q'QRR']$ and $[PQR]+[P'PRR']$ , so $[P'PQQ']+[Q'QRR']=[PQR]+[P'PRR']$ $\break$

Thus, these are equal. $P'Q'=\sqrt{PQ^2-(QQ'-PP')^2}=\sqrt{9-1}=\sqrt{8}=2\sqrt{2}$ . Additionally, $Q'R'=\sqrt{QR^2-(RR'-QQ')^2}=\sqrt{5^2-1^2}=\sqrt{24}=2\sqrt{6}$ . Therefore, $[P'PQQ']=\frac{P'P+Q'Q}{2}*2\sqrt{2}=\frac{1+2}{2}*2\sqrt{2}=3\sqrt{2}$ . Similarly, $[Q'QRR']=5\sqrt6$ . We can calculate $[P'PRR']$ easily because $P'R'=P'Q'+Q'R'=2\sqrt{2}+2\sqrt{6}$ . $[P'PRR']=4\sqrt{2}+4\sqrt{6}$ . $\newline$

Plugging into first equation, the two sums of areas, $3\sqrt{2}+5\sqrt{6}=4\sqrt{2}+4\sqrt{6}+[PQR]$ . $\newline$

$[PQR]=\sqrt{6}-\sqrt{2}\rightarrow \fbox{D}$ .

2016 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2016 AMC 10A Problems/Problem 21

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