Difference between revisions of "2016 AMC 10A Problems/Problem 21"

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Circles with centers <math>P, Q</math> and <math>R</math>, having radii <math>1, 2</math> and <math>3</math>, respectively, lie on the same side of line <math>l</math> and are tangent to <math>l</math> at <math>P', Q'</math> and <math>R'</math>, respectively, with <math>Q'</math> between <math>P'</math> and <math>R'</math>. The circle with center <math>Q</math> is externally tangent to each of the other two circles. What is the area of triangle <math>PQR</math>?
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#redirect [[2016 AMC 12A Problems/Problem 15]]
 
 
<math>\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}}</math>
 
 
 
==Solution== [edit]
 
 
 
Notice that we can find <math>[P'PQRR']</math> in two different ways: <math>[P'PQQ']+[Q'QRR']</math> and <math>[PQR]+[P'PRR']</math>, so <math>[P'PQQ']+[Q'QRR']=[PQR]+[P'PRR']</math>     
 
<math>\break</math>
 
 
 
Thus, these are equal. <math>P'Q'=\sqrt{PQ^2-(QQ'-PP')^2}=\sqrt{9-1}=\sqrt{8}=2\sqrt{2}</math>. Additionally, <math>Q'R'=\sqrt{QR^2-(RR'-QQ')^2}=\sqrt{5^2-1^2}=\sqrt{24}=2\sqrt{6}</math>. Therefore, <math>[P'PQQ']=\frac{P'P+Q'Q}{2}*2\sqrt{2}=\frac{1+2}{2}*2\sqrt{2}=3\sqrt{2}</math>. Similarly, <math>[Q'QRR']=5\sqrt6</math>. We can calculate <math>[P'PRR']</math> easily because <math>P'R'=P'Q'+Q'R'=2\sqrt{2}+2\sqrt{6}</math>. <math>[P'PRR']=4\sqrt{2}+4\sqrt{6}</math>. <math>\newline</math>
 
 
 
Plugging into first equation, the two sums of areas, <math>3\sqrt{2}+5\sqrt{6}=4\sqrt{2}+4\sqrt{6}+[PQR]</math>. <math>\newline</math>
 
 
 
<math>[PQR]=\sqrt{6}-\sqrt{2}\rightarrow \fbox{D}</math>.
 
 
 
==See Also==
 
{{AMC10 box|year=2016|ab=A|num-b=20|num-a=22}}
 
{{MAA Notice}}
 

Latest revision as of 15:33, 26 January 2021

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