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

m (Solution 2)
(Redirected page to 2016 AMC 12A Problems/Problem 15)
(Tag: New redirect)
 
(12 intermediate revisions by 9 users not shown)
Line 1: Line 1:
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]]
 
 
<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>
 
 
 
==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 <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>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>
 
 
 
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>
 
 
 
<math>[PQR]=\sqrt{6}-\sqrt{2}\rightarrow \fbox{D}</math>.
 
 
 
==Solution 2==
 
 
 
Use the [[Shoelace Theorem]].
 
 
 
Let the center of the first circle of radius 1 be at <math>(0, 1)</math>.
 
 
 
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>.
 
 
 
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>.
 
 
 
Now, we may use the Shoelace Theorem!
 
 
 
<math>(0,1)</math>
 
 
 
<math>(2\sqrt{2}, 2)</math>
 
 
 
<math>(2\sqrt{2}+2\sqrt{6}, 3)</math>
 
 
 
<math>\frac{1}{2}|(2\sqrt{2}+4\sqrt{2}+4\sqrt{6})-(6\sqrt{2}+2\sqrt{2}+2\sqrt{6})|</math>
 
 
 
<math>= \sqrt{6}-\sqrt{2}</math>  <math>\fbox{D}</math>.
 
 
 
==See Also==
 
{{AMC10 box|year=2016|ab=A|num-b=20|num-a=22}}
 
{{AMC12 box|year=2016|ab=A|num-b=14|num-a=16}}
 
{{MAA Notice}}
 

Latest revision as of 15:33, 26 January 2021

Invalid username
Login to AoPS