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| − | ==Problem==
| + | #redirect [[2016 AMC 12A Problems/Problem 15]] |
| − | 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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| − | <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 1==
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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 Theorem]].
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| − | | |
| − | Let the center of the first circle of radius 1 be at <math>(0, 1)</math>.
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| − | | |
| − | Draw the trapezoid <math>PQQ'P'</math> and using the Pythagorean Theorem, 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 the Pythagorean Theorem, we get that <math>Q'R' = 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 Theorem!
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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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| − | | |
| − | ==Solution 3==
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| − | <math>PQ = 3</math> and <math>QR = 5</math> because they are the sum of two radii. <math>QQ' - PP' = 1</math> and <math>RR' - QQ' = 1</math>, the difference of the radii. Using pythagorean theorem, we find that <math>P'Q'</math> and <math>Q'R'</math> are <math>\sqrt{8}</math> and <math>\sqrt{24}</math>, <math>P'R' = \sqrt{8} + \sqrt{24}</math>.
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| − | Draw a perpendicular from <math>P</math> to line <math>RR'</math>, then we can use the Pythagorean theorem to find <math>PR</math>. <math>RR' - PP' = 2</math>. We get <cmath>PR^2 = (\sqrt{8} + \sqrt{24})^2 + 4 = 36 + 16\sqrt{3} \Rightarrow PR = \sqrt{36 + 16\sqrt{3}} = 2\sqrt{9 + 4\sqrt{3}}</cmath>
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| − | To make our calculations easier, let <math>\sqrt{9 + 4\sqrt{3}} = a</math>. The semi-perimeter of our triangle is <math>\frac{3 + 5 + 2a}{2} = 4 + a</math>. Symbolize the area of the triangle with <math>A</math>. Using Heron's formula, we have <cmath>A^2 = (4 + a)(4 + a - 2a)(4 + a - 3)(4 + a - 5) = (4 + a)(4 - a)(a + 1)(a - 1) = (16 - a^2)(a^2 - 1)</cmath> We can remove the outer root of a. <cmath>A^2 = (16 - 9 - 4\sqrt{3})(9 + 4\sqrt{3} - 1) = (7 - 4\sqrt{3})(8 + 4\sqrt{3}) = 8 - 4\sqrt{3} \rightarrow A = \sqrt{8 - 4\sqrt{3}}</cmath>
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| − | We solve the nested root. We want to turn <math>8 - 4\sqrt{3}</math> into the square of something. If we have <math>(a - b) ^ 2 = 8 - 4\sqrt{3}</math>, then we get <cmath>\begin{cases} a^2 + b^2 = 8 \\ ab = 2\sqrt{3} \end{cases}</cmath> Solving the system of equations, we get <math>a = \sqrt{6}</math> and <math>b = \sqrt{2}</math>. Alternatively, you can square all the possible solutions until you find one that is equal to <math>8 - 4\sqrt{3}</math>. <cmath>A = \sqrt{8 - 4\sqrt{3}} = \sqrt{(\sqrt{6} - \sqrt{2})^2} = \sqrt{6} - \sqrt{2} \rightarrow \fbox{D}</cmath>
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| − | ~ZericH
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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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