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

(Added diagram)
Line 4: Line 4:
  
 
==Solution==  
 
==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>       
 
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>
 
<math>\break</math>

Revision as of 12:27, 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, $3\sqrt{2}+5\sqrt{6}=4\sqrt{2}+4\sqrt{6}+[PQR]$. $\newline$

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

See Also

2016 AMC 10A (ProblemsAnswer KeyResources)
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
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS