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

Revision as of 11:27, 4 February 2016

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

$[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations pair P,Q,R,Pp,Qp,Rp; pair A,B; //Variable Definitions A=(-5, 0); B=(8, 0); P=(-2.828,1); Q=(0,2); R=(4.899,3); Pp=foot(P,A,B); Qp=foot(Q,A,B); Rp=foot(R,A,B); path PQR = P--Q--R--cycle; //Initial Diagram dot(P); dot(Q); dot(R); dot(Pp); dot(Qp); dot(Rp); draw(Circle(P, 1), linewidth(0.8)); draw(Circle(Q, 2), linewidth(0.8)); draw(Circle(R, 3), linewidth(0.8)); draw(A--B,Arrows); label("$P$",P,N); label("$Q$",Q,N); label("$R$",R,N); label("$P'$",Pp,S); label("$Q'$",Qp,S); label("$R'$",Rp,S); label("$l$",B,E); //Added lines draw(PQR); draw(P--Pp); draw(Q--Qp); draw(R--Rp); //Angle marks draw(rightanglemark(P,Pp,B)); draw(rightanglemark(Q,Qp,B)); draw(rightanglemark(R,Rp,B)); [/asy]$ 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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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Difference between revisions of "2016 AMC 10A Problems/Problem 21"

Revision as of 11:27, 4 February 2016

Solution

See Also

@@ Line 4: / Line 4: @@
 ==Solution==
+<asy>
+size(250);
+defaultpen(linewidth(0.4));
+//Variable Declarations
+pair P,Q,R,Pp,Qp,Rp;
+pair A,B;
+//Variable Definitions
+A=(-5, 0);
+B=(8, 0);
+P=(-2.828,1);
+Q=(0,2);
+R=(4.899,3);
+Pp=foot(P,A,B);
+Qp=foot(Q,A,B);
+Rp=foot(R,A,B);
+path PQR = P--Q--R--cycle;
+//Initial Diagram
+dot(P);
+dot(Q);
+dot(R);
+dot(Pp);
+dot(Qp);
+dot(Rp);
+draw(Circle(P, 1), linewidth(0.8));
+draw(Circle(Q, 2), linewidth(0.8));
+draw(Circle(R, 3), linewidth(0.8));
+draw(A--B,Arrows);
+label("$P$",P,N);
+label("$Q$",Q,N);
+label("$R$",R,N);
+label("$P'$",Pp,S);
+label("$Q'$",Qp,S);
+label("$R'$",Rp,S);
+label("$l$",B,E);
+//Added lines
+draw(PQR);
+draw(P--Pp);
+draw(Q--Qp);
+draw(R--Rp);
+//Angle marks
+draw(rightanglemark(P,Pp,B));
+draw(rightanglemark(Q,Qp,B));
+draw(rightanglemark(R,Rp,B));
+</asy>
 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>