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

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==Solution 2==  
 
==Solution 2==  
  
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Use the Shoelace Theorem.
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Use the [[Shoelace Theorem]].
  
 
Let the center of the first circle of radius 1 be at (0, 1).  
 
Let the center of the first circle of radius 1 be at (0, 1).  

Revision as of 15:20, 8 January 2018

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 1

[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 $[P'PQQ']+[Q'QRR']=[PQR]+[P'PRR']$ $\break$

$P'Q'=\sqrt{PQ^2-(QQ'-PP')^2}=\sqrt{9-1}=\sqrt{8}=2\sqrt{2}$. 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, $3\sqrt{2}+5\sqrt{6}=4\sqrt{2}+4\sqrt{6}+[PQR]$. $\newline$

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

Solution 2

Use the Shoelace Theorem.

Let the center of the first circle of radius 1 be at (0, 1).

Draw the trapezoid $PQQ'P'$ and using the Pythagorean Theorem, we get that $P'Q' = 2\sqrt{2}$ so the center of the second circle of radius 2 is at $(2\sqrt{2}, 2)$.

Draw the trapezoid $QRR'Q'$ and using the Pythagorean Theorem, we get that $Q'R' = 2\sqrt{6}$ so the center of the third circle of radius 3 is at $(2\sqrt{2}+2\sqrt{6}, 3)$.

Now, we may use the Shoelace Theorem!

$(0,1)$

$(2\sqrt{2}, 2)$

$(2\sqrt{2}+2\sqrt{6}, 3)$

$\frac{1}{2}|(2\sqrt{2}+4\sqrt{2}+4\sqrt{6})-(6\sqrt{2}+2\sqrt{2}+2\sqrt{6})|$

$= \sqrt{6}-\sqrt{2}$ $\fbox{D}$.

See Also

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
All AMC 10 Problems and Solutions
2016 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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 12 Problems and Solutions

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