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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>?
| + | #redirect [[2016 AMC 12A Problems/Problem 15]] |
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| − | <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>
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| − | | |
| − | ==Solution==
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| − | <asy>
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| − | size(250);
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| − | defaultpen(linewidth(0.4));
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| − | //Variable Declarations
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| − | pair P,Q,R,Pp,Qp,Rp;
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| − | pair A,B;
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| − | | |
| − | //Variable Definitions
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| − | A=(-5, 0);
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| − | B=(8, 0);
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| − | P=(-2.828,1);
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| − | Q=(0,2);
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| − | R=(4.899,3);
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| − | Pp=foot(P,A,B);
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| − | Qp=foot(Q,A,B);
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| − | Rp=foot(R,A,B);
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| − | path PQR = P--Q--R--cycle;
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| − | //Initial Diagram
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| − | dot(P);
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| − | dot(Q);
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| − | dot(R);
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| − | dot(Pp);
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| − | dot(Qp);
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| − | dot(Rp);
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| − | draw(Circle(P, 1), linewidth(0.8));
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| − | draw(Circle(Q, 2), linewidth(0.8));
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| − | draw(Circle(R, 3), linewidth(0.8));
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| − | draw(A--B,Arrows);
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| − | label("$P$",P,N);
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| − | label("$Q$",Q,N);
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| − | label("$R$",R,N);
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| − | label("$P'$",Pp,S);
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| − | label("$Q'$",Qp,S);
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| − | label("$R'$",Rp,S);
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| − | label("$l$",B,E);
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| − | | |
| − | //Added lines
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| − | draw(PQR);
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| − | draw(P--Pp);
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| − | draw(Q--Qp);
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| − | draw(R--Rp);
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| − | | |
| − | //Angle marks
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| − | draw(rightanglemark(P,Pp,B));
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| − | draw(rightanglemark(Q,Qp,B));
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| − | draw(rightanglemark(R,Rp,B));
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| − | </asy>
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| − | 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>
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| − | <math>\break</math>
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| − | <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>
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| − | 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>
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| − | <math>[PQR]=\sqrt{6}-\sqrt{2}\rightarrow \fbox{D}</math>.
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| − | | |
| − | ==Solution 2==
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| − | Use the Shoelace Thm!
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| − | Let the center of the first circle of radius 1 be at (0, 1).
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| − | Draw the trapezoid <math>PQQ'P'</math> and using Pythagorean Thm., we get that <math>P'Q' = 2\sqrt{2}</math> so the center of the second circle of radius 2 is at <math>(2\sqrt{2}, 2)</math>.
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| − | Draw the trapezoid <math>QRR'Q'</math> and using Pythagorean Thm., we get that <math>Q'R' = 2\sqrt{2} + 2\sqrt{6}</math> so the center of the third circle of radius 3 is at <math>(2\sqrt{2}+2\sqrt{6}, 3)</math>.
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| − | Now, we may use the Shoelace Thm!
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| − | <math>(0,1)</math>
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| − | <math>(2\sqrt{2}, 2)</math>
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| − | <math>(2\sqrt{2}+2\sqrt{6}, 3)</math>
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| − | <math>\frac{1}{2}|(2\sqrt{2}+4\sqrt{2}+4\sqrt{6})-(6\sqrt{2}+2\sqrt{2}+2\sqrt{6})|</math>
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| − | <math>= \sqrt{6}-\sqrt{2}</math> <math>\fbox{D}</math>.
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| − | ==See Also==
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| − | {{AMC10 box|year=2016|ab=A|num-b=20|num-a=22}}
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| − | {{AMC12 box|year=2016|ab=A|num-b=14|num-a=16}}
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| − | {{MAA Notice}}
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