Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2016 AMC 12A Problems/Problem 15"

(Removed redirect to 2016 AMC 10A Problems/Problem 21)
(Tag: Removed redirect)
(Added Pi Academy video. Removed the second video because the video was deleted. Added a subheading for Richard's solution.)
 
(28 intermediate revisions by 9 users not shown)
Line 2: Line 2:
 
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>?
 
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>?
  
−
<math>\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{6}/3\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{6}/2</math>
+
<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==  
 
==Solution 1==  
 +
 
<asy>
 
<asy>
 
size(250);
 
size(250);
Line 52: Line 53:
 
draw(rightanglemark(R,Rp,B));
 
draw(rightanglemark(R,Rp,B));
 
</asy>
 
</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> Since we want <math>[PQR],</math> we use the latter method, so we have <math>[P'PQQ']+[Q'QRR']=[PQR]+[P'PRR'].</math>       
 
<math>\break</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>
+
<math>P'Q'=\sqrt{PQ^2-(QQ'-PP')^2}=\sqrt{3^2-1^2}=\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>.
+
Plugging into first equation, the two sums of areas, <math>3\sqrt{2}+5\sqrt{6}=4\sqrt{2}+4\sqrt{6}+[PQR],</math> so therefore <math>[PQR]=\boxed{\textbf{(D) }\sqrt{6}-\sqrt{2}.}</math>
  
 
==Solution 2==  
 
==Solution 2==  
Line 71: Line 71:
 
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>.
 
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!
+
Now, we may use the [[Shoelace Theorem]]!
  
 
<math>(0,1)</math>
 
<math>(0,1)</math>
Line 81: Line 81:
 
<math>\frac{1}{2}|(2\sqrt{2}+4\sqrt{2}+4\sqrt{6})-(6\sqrt{2}+2\sqrt{2}+2\sqrt{6})|</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>.
+
<math>=\boxed{\textbf{(D) }\sqrt{6}-\sqrt{2}.}</math>
  
 
==Solution 3==
 
==Solution 3==
Line 90: Line 90:
 
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>
 
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>
  
−
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>
+
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> <math>\newline A = \sqrt{8 - 4\sqrt{3}} = \sqrt{(\sqrt{6} - \sqrt{2})^2} = \boxed{\textbf{(D) }\sqrt{6}-\sqrt{2}.}</math>
 
~ZericH
 
~ZericH
  
Line 175: Line 175:
 
==Solution 5 (Heron’s)==
 
==Solution 5 (Heron’s)==
  
−
We can use the Pythagorean theorem to find that the lengths are <math>3,5,2\sqrt3</math>. If we apply Heron’s, we know that it must be the sum (or difference) of two or more square roots, by instinct. This means that <math>\boxed{\textbf{(D)} \sqrt{6}-\sqrt{2}}</math> is the answer.
+
We can use the Pythagorean theorem to find that the lengths are <math>3,5,\sqrt{36+16\sqrt{3}}</math>. If we apply Heron’s, we know that it must be the sum (or difference) of two or more square roots, by instinct. This means that <math>\boxed{\textbf{(D)} \sqrt{6}-\sqrt{2}}</math> is the answer.
−
 
 
  
 
==Solution 6 (Educated Guess)==
 
==Solution 6 (Educated Guess)==
  
 
Like Solution 1, we can use the Pythagorean theorem to find <math>P'Q'</math> and <math>Q'R'</math>, which are <math>2\sqrt2</math> and <math>2\sqrt6</math> respectively. Since the only answer choice that has <math>\sqrt2</math> and <math>\sqrt6</math> is <math>D</math>, we can make an educated guess that <math>\boxed{\textbf{(D)} \sqrt{6}-\sqrt{2}}</math> is the answer.
 
Like Solution 1, we can use the Pythagorean theorem to find <math>P'Q'</math> and <math>Q'R'</math>, which are <math>2\sqrt2</math> and <math>2\sqrt6</math> respectively. Since the only answer choice that has <math>\sqrt2</math> and <math>\sqrt6</math> is <math>D</math>, we can make an educated guess that <math>\boxed{\textbf{(D)} \sqrt{6}-\sqrt{2}}</math> is the answer.
 +
 +
 +
==Solution 7 (Quickest and Easiest Using Trapezoids)==
 +
 +
The fastest way to find the area of <math>[PQR]</math> is by breaking it into shapes with easy-to-calculate areas. We can see that the area of the whole shape <math>[PQRR'P']</math> is made up of trapezoids <math>[PP'QQ']</math> and <math>[Q'QRR']</math>. So, <math>[PQR]=[PP'QQ']+[Q'QRR']-[PRR'P']</math>.
 +
 +
We already know the lengths of <math>PP'</math> and <math>QQ'</math> being <math>1</math> and <math>2</math> respectively (they are the radiuses of the circles). The "height" of the trapezoid is <math>P'Q'=\sqrt{3^2-1^2}=\sqrt{8}=2\sqrt{2}</math> which can be found from the Pythagorean Theorem. So, the area of trapezoid <math>[PP'QQ']=\frac{1+2}2*{(2\sqrt{2})}=3\sqrt{2}</math>. Similarly, we can find the area of trapezoid <math>[Q'QRR']</math>. <math>Q'R'=5\sqrt{6}</math>, so <math>[Q'QRR']=\frac{2+3}2*{(2\sqrt{6})}=5\sqrt{2}</math>.
 +
 +
Now we can put the pieces together. <math>[PQR]=[PP'QQ']+[Q'QRR']-[PRR'P']=3\sqrt{2}+5\sqrt{6}-[PRR'P']=</math>. The area of <math>PRR'P'</math> is the long rectangle <math>[PR"R'P]'+ [PRR"]=4\sqrt{2}+4\sqrt{6}</math>. <math>R"</math> is the point on <math>RR'</math> in which <math>PR"</math> is perpendicular to <math>RR'</math>.
 +
 +
So, <math>[PQR]=3\sqrt{2}+5\sqrt{6}-(4\sqrt{2}+4\sqrt{6})=\boxed{\textbf{(D) }\sqrt{6}-\sqrt{2}.}</math>
 +
 +
~hwan
 +
 +
== Video Solution 1 by Pi Academy ==
 +
https://youtu.be/tMVGfNSwLq0?si=TL8gtFQr3xC1Hg1R
 +
 +
~ Pi Academy
 +
 +
== Video Solution 2 ==
 +
https://www.youtube.com/watch?v=UanfIBpDTh8&ab_channel=ArtofProblemSolving
 +
~by the official Art of Problem Solving channel on YouTube
  
 
==See Also==
 
==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}}
 
{{AMC12 box|year=2016|ab=A|num-b=14|num-a=16}}
−
{{AMC10 box|year=2016|ab=A|num-b=20|num-a=22}}
 
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:56, 8 October 2024

Problem

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 $[PQR]+[P'PRR'].$ Since we want $[PQR],$ we use the latter method, so we have $[P'PQQ']+[Q'QRR']=[PQR]+[P'PRR'].$ $\break$

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

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!

$(0,1)$

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

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

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

$=\boxed{\textbf{(D) }\sqrt{6}-\sqrt{2}.}$

Solution 3

$PQ = 3$ and $QR = 5$ because they are the sum of two radii. $QQ' - PP' = 1$ and $RR' - QQ' = 1$, the difference of the radii. Using pythagorean theorem, we find that $P'Q'$ and $Q'R'$ are $\sqrt{8}$ and $\sqrt{24}$, $P'R' = \sqrt{8} + \sqrt{24}$.

Draw a perpendicular from $P$ to line $RR'$, then we can use the Pythagorean theorem to find $PR$. $RR' - PP' = 2$. We get \[PR^2 = (\sqrt{8} + \sqrt{24})^2 + 4 = 36 + 16\sqrt{3} \Rightarrow PR = \sqrt{36 + 16\sqrt{3}} = 2\sqrt{9 + 4\sqrt{3}}\]

To make our calculations easier, let $\sqrt{9 + 4\sqrt{3}} = a$. The semi-perimeter of our triangle is $\frac{3 + 5 + 2a}{2} = 4 + a$. Symbolize the area of the triangle with $A$. Using Heron's formula, we have \[A^2 = (4 + a)(4 + a - 2a)(4 + a - 3)(4 + a - 5) = (4 + a)(4 - a)(a + 1)(a - 1) = (16 - a^2)(a^2 - 1)\] We can remove the outer root of a. \[A^2 = (16 - 9 - 4\sqrt{3})(9 + 4\sqrt{3} - 1) = (7 - 4\sqrt{3})(8 + 4\sqrt{3}) = 8 - 4\sqrt{3} \rightarrow A = \sqrt{8 - 4\sqrt{3}}\]

We solve the nested root. We want to turn $8 - 4\sqrt{3}$ into the square of something. If we have $(a - b) ^ 2 = 8 - 4\sqrt{3}$, then we get \[\begin{cases} a^2 + b^2 = 8 \\ ab = 2\sqrt{3} \end{cases}\] Solving the system of equations, we get $a = \sqrt{6}$ and $b = \sqrt{2}$. Alternatively, you can square all the possible solutions until you find one that is equal to $8 - 4\sqrt{3}.$ $\newline A = \sqrt{8 - 4\sqrt{3}} = \sqrt{(\sqrt{6} - \sqrt{2})^2} = \boxed{\textbf{(D) }\sqrt{6}-\sqrt{2}.}$ ~ZericH

Solution 4

[asy] // Initial Pen Sizing size(250); defaultpen(linewidth(0.4)); defaultpen(fontsize(10pt)); // Variable Declarations pair P,Q,R,Pp,Qp,Rp,X,Y,Z,A,B; // Variable Definitions A=(-5, 0); B=(8, 0); P=(-2.828,1); Q=(0,2); R=(4.899,3); X=(0,1); Y=(4.899,1); Z=(4.899,2); 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); dot(X); dot(Y); dot(Z); draw(Circle(P, 1), linewidth(0.3)); draw(Circle(Q, 2), linewidth(0.3)); draw(Circle(R, 3), linewidth(0.3)); 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); label("$X$",X,NE); label("$Y$",Y,E); label("$Z$",Z,E); //Added lines filldraw(PQR,gray(0.8)); draw(P--Pp,linetype("8 8")); draw(Q--Qp,linetype("8 8")); draw(R--Rp,linetype("8 8")); draw(P--Y,linetype("8 8")); draw(Q--Z,linetype("8 8")); //Angle marks draw(rightanglemark(R,Y,P)); draw(rightanglemark(Q,X,P)); draw(rightanglemark(R,Z,Q)); //Length labeling label("$2\sqrt{2}$",P--X,fontsize(8pt)); label("$2\sqrt{6}$",X--Y,fontsize(8pt)); label("$2\sqrt{6}$",Q--Z,fontsize(8pt)); label("$1$",R--Z,E,fontsize(8pt)); label("$1$",Z--Y,E,fontsize(8pt)); label("$3$",(-2.828,1.3)--Q,W,fontsize(8pt)); label("$5$",Q--R,N,fontsize(8pt)); [/asy] The above diagram can be achieved relatively simply using basic knowledge of the Pythagorean theorem and the fact that the radius from the center to the point of tangency is perpendicular to the tangent line. From there, observe that $[PQRY]$ can be calculated in two ways: $[\triangle PQX] + [QZYX] + [\triangle QRZ]$ and $[\triangle PRY] + [\triangle PQR]$. Solving, we get: \begin{align*} [\triangle PQR] &= [PQRY] - [\triangle PRY] \\ &= [\triangle PQX] + [QZYX] + [\triangle QRZ] - [\triangle PRY] \\ &= \sqrt{2} + 2\sqrt{6} + \sqrt{6}- 2\sqrt{2}-2\sqrt{6} \\ &= \boxed{\textbf{(D)} \sqrt{6}-\sqrt{2}} \end{align*}

- ColtsFan10, diagram partially borrowed from Solution 1

Solution 5 (Heron’s)

We can use the Pythagorean theorem to find that the lengths are $3,5,\sqrt{36+16\sqrt{3}}$. If we apply Heron’s, we know that it must be the sum (or difference) of two or more square roots, by instinct. This means that $\boxed{\textbf{(D)} \sqrt{6}-\sqrt{2}}$ is the answer.

Solution 6 (Educated Guess)

Like Solution 1, we can use the Pythagorean theorem to find $P'Q'$ and $Q'R'$, which are $2\sqrt2$ and $2\sqrt6$ respectively. Since the only answer choice that has $\sqrt2$ and $\sqrt6$ is $D$, we can make an educated guess that $\boxed{\textbf{(D)} \sqrt{6}-\sqrt{2}}$ is the answer.


Solution 7 (Quickest and Easiest Using Trapezoids)

The fastest way to find the area of $[PQR]$ is by breaking it into shapes with easy-to-calculate areas. We can see that the area of the whole shape $[PQRR'P']$ is made up of trapezoids $[PP'QQ']$ and $[Q'QRR']$. So, $[PQR]=[PP'QQ']+[Q'QRR']-[PRR'P']$.

We already know the lengths of $PP'$ and $QQ'$ being $1$ and $2$ respectively (they are the radiuses of the circles). The "height" of the trapezoid is $P'Q'=\sqrt{3^2-1^2}=\sqrt{8}=2\sqrt{2}$ which can be found from the Pythagorean Theorem. So, the area of trapezoid $[PP'QQ']=\frac{1+2}2*{(2\sqrt{2})}=3\sqrt{2}$. Similarly, we can find the area of trapezoid $[Q'QRR']$. $Q'R'=5\sqrt{6}$, so $[Q'QRR']=\frac{2+3}2*{(2\sqrt{6})}=5\sqrt{2}$.

Now we can put the pieces together. $[PQR]=[PP'QQ']+[Q'QRR']-[PRR'P']=3\sqrt{2}+5\sqrt{6}-[PRR'P']=$. The area of $PRR'P'$ is the long rectangle $[PR"R'P]'+ [PRR"]=4\sqrt{2}+4\sqrt{6}$. $R"$ is the point on $RR'$ in which $PR"$ is perpendicular to $RR'$.

So, $[PQR]=3\sqrt{2}+5\sqrt{6}-(4\sqrt{2}+4\sqrt{6})=\boxed{\textbf{(D) }\sqrt{6}-\sqrt{2}.}$

~hwan

Video Solution 1 by Pi Academy

https://youtu.be/tMVGfNSwLq0?si=TL8gtFQr3xC1Hg1R

~ Pi Academy

Video Solution 2

https://www.youtube.com/watch?v=UanfIBpDTh8&ab_channel=ArtofProblemSolving ~by the official Art of Problem Solving channel on YouTube

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_15&oldid=229322"