Difference between revisions of "2022 AIME II Problems/Problem 15"

Revision as of 21:44, 18 February 2022

Problem

Two externally tangent circles $\omega_1$ and $\omega_2$ have centers $O_1$ and $O_2$ , respectively. A third circle $\Omega$ passing through $O_1$ and $O_2$ intersects $\omega_1$ at $B$ and $C$ and $\omega_2$ at $A$ and $D$ , as shown. Suppose that $AB = 2$ , $O_1O_2 = 15$ , $CD = 16$ , and $ABO_1CDO_2$ is a convex hexagon. Find the area of this hexagon. $[asy] import geometry; size(10cm); point O1=(0,0),O2=(15,0),B=9*dir(30); circle w1=circle(O1,9),w2=circle(O2,6),o=circle(O1,O2,B); point A=intersectionpoints(o,w2)[1],D=intersectionpoints(o,w2)[0],C=intersectionpoints(o,w1)[0]; filldraw(A--B--O1--C--D--O2--cycle,0.2*red+white,black); draw(w1); draw(w2); draw(O1--O2,dashed); draw(o); dot(O1); dot(O2); dot(A); dot(D); dot(C); dot(B); label("$\omega_1$",8*dir(110),SW); label("$\omega_2$",5*dir(70)+(15,0),SE); label("$O_1$",O1,W); label("$O_2$",O2,E); label("$B$",B,N+1/2*E); label("$A$",A,N+1/2*W); label("$C$",C,S+1/4*W); label("$D$",D,S+1/4*E); label("$15$",midpoint(O1--O2),N); label("$16$",midpoint(C--D),N); label("$2$",midpoint(A--B),S); label("$\Omega$",o.C+(o.r-1)*dir(270)); [/asy]$

Solution 1

First observe that $AO_2 = O_2D$ and $BO_1 = O_1C$ . Let points $A'$ and $B'$ be the reflections of $A$ and $B$ , respectively, about the perpendicular bisector of $\overline{O_1O_2}$ . Then quadrilaterals $ABO_1O_2$ and $A'B'O_2O_1$ are congruent, so hexagons $ABO_1CDO_2$ and $B'A'O_1CDO_2$ have the same area. Furthermore, triangles $DO_2B'$ and $A'O_1C$ are congruent, so $B'D = A'C$ and quadrilateral $B'A'CD$ is an isosceles trapezoid. $[asy] import olympiad; size(180); defaultpen(linewidth(0.7)); pair Bp = dir(105), Ap = dir(75), O1 = dir(25), C = dir(320), D = dir(220), O2 = dir(175); draw(unitcircle^^Bp--Ap--O1--C--D--O2--cycle); label("$B'$",Bp,dir(origin--Bp)); label("$A'$",Ap,dir(origin--Ap)); label("$O_1$",O1,dir(origin--O1)); label("$C$",C,dir(origin--C)); label("$D$",D,dir(origin--D)); label("$O_2$",O2,dir(origin--O2)); draw(O2--O1,linetype("4 4")); draw(Bp--D^^Ap--C,linetype("2 2")); [/asy]$ Next, remark that $A'O_1 = DO_2$ , so quadrilateral $A'O_1DO_2$ is also an isosceles trapezoid; in turn, $A'D = O_1O_2 = 15$ , and similarly $B'C = 15$ . Thus, Ptolmey's theorem on $B'A'CD$ yields $B'D\cdot A'C + 2\cdot 16 = 15^2$ , whence $B'D = A'C = \sqrt{193}$ . Let $\alpha = \angle B'A'D$ . The Law of Cosines on triangle $B'A'D$ yields $\cos\alpha = \frac{15^2 + 2^2 - (\sqrt{193})^2}{2\cdot 2\cdot 15} = \frac{36}{60} = \frac 35,$ and hence $\sin\alpha = \tfrac 45$ . Thus the distance between bases $AB$ and $CD$ is $12$ (in fact, $\triangle B'A'D$ is a $9-12-15$ triangle with a $7-12-\sqrt{193}$ triangle removed), which implies the area of $B'A'CD$ is $\tfrac12\cdot 12\cdot(2+16) = 108$ .

Now let $O_1C = O_2B' = r_1$ and $O_2D = O_1A' = r_2$ ; the tangency of circles $\omega_1$ and $\omega_2$ implies $r_1 + r_2 = 15$ . Furthermore, angles $B'O_2D$ and $B'A'D$ are opposite angles in cyclic quadrilateral $A'B'O_2D$ , which implies the measure of angle $B'O_2D$ is $180^\circ - \alpha$ . Therefore, the Law of Cosines applied to triangle $\triangle B'O_2D$ yields $\begin{align*} 193 &= r_1^2 + r_2^2 - 2r_1r_2(-\tfrac 35) = (r_1^2 + 2r_1r_2 + r_2^2) - \tfrac45r_1r_2\\ &= (r_1+r_2)^2 - \tfrac45 r_1r_2 = 225 - \tfrac45r_1r_2. \end{align*}$

Thus $r_1r_2 = 40$ , and so the area of triangle $B'O_2D$ is $\tfrac12r_1r_2\sin\alpha = 16$ .

Thus, the area of hexagon $ABO_{1}CDO_{2}$ is $108 + 2\cdot 16 = \boxed{140}$ .

~djmathman

Solution 2

Denote by $O$ the center of $\Omega$ . Denote by $r$ the radius of $\Omega$ .

We have $O_1$ , $O_2$ , $A$ , $B$ , $C$ , $D$ are all on circle $\Omega$ .

Denote $\angle O_1 O O_2 = 2 \theta$ . Denote $\angle O_1 O B = \alpha$ . Denote $\angle O_2 O A = \beta$ .

Because $B$ and $C$ are on circles $\omega_1$ and $\Omega$ , $BC$ is a perpendicular bisector of $O_1 O$ . Hence, $\angle O_1 O C = \alpha$ .

Because $A$ and $D$ are on circles $\omega_2$ and $\Omega$ , $AD$ is a perpendicular bisector of $O_2 O$ . Hence, $\angle O_2 O D = \beta$ .

In $\triangle O O_1 O_2$ , $O_1 O_2 = 2 r \sin \theta .$

Hence, $2 r \sin \theta = 15 .$

In $\triangle O AB$ , $\begin{align*} AB & = 2 r \sin \frac{2 \theta - \alpha - \beta}{2} \\ & = 2 r \sin \theta \cos \frac{\alpha + \beta}{2} - 2 r \cos \theta \sin \frac{\alpha + \beta}{2} \\ & = 15 \cos \frac{\alpha + \beta}{2} - 2 r \cos \theta \sin \frac{\alpha + \beta}{2} . \end{align*}$

Hence, $15 \cos \frac{\alpha + \beta}{2} - 2 r \cos \theta \sin \frac{\alpha + \beta}{2} = 2 . \hspace{1cm} (1)$

In $\triangle O CD$ , $\begin{align*} CD & = 2 r \sin \frac{360^\circ - 2 \theta - \alpha - \beta}{2} \\ & = 2 r \sin \left( \theta + \frac{\alpha + \beta}{2} \right) \\ & = 2 r \sin \theta \cos \frac{\alpha + \beta}{2} + 2 r \cos \theta \sin \frac{\alpha + \beta}{2} \\ & = 15 \cos \frac{\alpha + \beta}{2} + 2 r \cos \theta \sin \frac{\alpha + \beta}{2} . \end{align*}$

Hence, $15 \cos \frac{\alpha + \beta}{2} + 2 r \cos \theta \sin \frac{\alpha + \beta}{2} = 16 . \hspace{1cm} (2)$

Taking $\frac{(1) + (2)}{30}$ , we get $\cos \frac{\alpha + \beta}{2} = \frac{3}{5}$ . Thus, $\sin \frac{\alpha + \beta}{2} = \frac{4}{5}$ .

Taking these into (1), we get $2 r \cos \theta = \frac{35}{4}$ . Hence, $\begin{align*} 2 r & = \sqrt{ \left( 2 r \sin \theta \right)^2 + \left( 2 r \cos \theta \right)^2} \\ & = \frac{5}{4} \sqrt{193} . \end{align*}$

Hence, $\cos \theta = \frac{7}{\sqrt{193}}$ .

In $\triangle O O_1 B$ , $O_1 B = 2 r \sin \frac{\alpha}{2} .$

In $\triangle O O_2 A$ , by applying the law of sines, we get $O_2 A = 2 r \sin \frac{\beta}{2} .$

Because circles $\omega_1$ and $\omega_2$ are externally tangent, $B$ is on circle $\omega_1$ , $A$ is on circle $\omega_2$ , $\begin{align*} O_1 O_2 & = O_1 B + O_2 A \\ & = 2 r \sin \frac{\alpha}{2} + 2 r \sin \frac{\beta}{2} \\ & = 2 r \left( \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} \right) . \end{align*}$

Thus, $\sin \frac{\alpha}{2} + \sin \frac{\beta}{2} = \frac{12}{\sqrt{193}}$ .

Now, we compute $\sin \alpha$ and $\sin \beta$ .

Recall $\cos \frac{\alpha + \beta}{2} = \frac{3}{5}$ and $\sin \frac{\alpha + \beta}{2} = \frac{4}{5}$ . Thus, $e^{i \frac{\alpha}{2}} e^{i \frac{\beta}{2}} = e^{i \frac{\alpha + \beta}{2}} = \frac{3}{5} + i \frac{4}{5}$ .

We also have $\begin{align*} \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} & = \frac{1}{2i} \left( e^{i \frac{\alpha}{2}} - e^{-i \frac{\alpha}{2}} + e^{i \frac{\beta}{2}} - e^{-i \frac{\beta}{2}} \right) \\ & = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \frac{\alpha + \beta}{2}} } \right) \left( e^{i \frac{\alpha}{2}} + e^{i \frac{\beta}{2}} \right) . \end{align*}$

Thus, $\begin{align*} \sin \alpha + \sin \beta & = \frac{1}{2i} \left( e^{i \alpha} - e^{-i \alpha} + e^{i \beta} - e^{-i \beta} \right) \\ & = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \left( \alpha + \beta \right)}} \right) \left( e^{i \alpha} + e^{i \beta} \right) \\ & = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \left( \alpha + \beta \right)}} \right) \left( \left( e^{i \frac{\alpha}{2}} + e^{i \frac{\beta}{2}} \right)^2 - 2 e^{i \frac{\alpha + \beta}{2}} \right) \\ & = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \left( \alpha + \beta \right)}} \right) \left( \left( \frac{2 i \left( \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} \right)}{1 - \frac{1}{e^{i \frac{\alpha + \beta}{2}} }} \right)^2 - 2 e^{i \frac{\alpha + \beta}{2}} \right) \\ & = - \frac{1}{i} \left( e^{i \frac{\alpha + \beta}{2}} - e^{-i \frac{\alpha + \beta}{2}} \right) \left( \frac{2 \left( \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} \right)^2} {e^{i \frac{\alpha + \beta}{2}} + e^{-i \frac{\alpha + \beta}{2}} - 2 } + 1 \right) \\ & = \frac{167 \cdot 8}{193 \cdot 5 } . \end{align*}$

Therefore, $\begin{align*} {\rm Area} \ ABO_1CDO_2 & = {\rm Area} \ \triangle O_3 AB + {\rm Area} \ \triangle O_3 BO_1 + {\rm Area} \ \triangle O_3 O_1 C \\ & \quad + {\rm Area} \ \triangle O_3 C D + {\rm Area} \ \triangle O_3 D O_2 + {\rm Area} \ \triangle O_3 O_2 A \\ & = \frac{1}{2} r^2 \left( \sin \left( 2 \theta - \alpha - \beta \right) + \sin \alpha + \sin \alpha + \sin \left( 360^\circ - 2 \theta - \alpha - \beta \right) + \sin \beta + \sin \beta \right) \\ & = \frac{1}{2} r^2 \left( \sin \left( 2 \theta - \alpha - \beta \right) - \sin \left( 2 \theta + \alpha + \beta \right) + 2 \sin \alpha + 2 \sin \beta \right) \\ & = r^2 \left( - \cos 2 \theta \sin \left( \alpha + \beta \right) + \sin \alpha + \sin \beta \right) \\ & = r^2 \left( \left( 1 - 2 \cos^2 \theta \right) 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha + \beta}{2} + \sin \alpha + \sin \beta \right) \\ & = \boxed{\textbf{(140) }} . \end{align*}$

~Steven Chen (www.professorchenedu.com)

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Difference between revisions of "2022 AIME II Problems/Problem 15"

Revision as of 21:44, 18 February 2022

Contents

Problem

Solution 1

Solution 2

See Also

@@ Line 33: / Line 33: @@
 </asy>
-==Solution==
+==Solution 1==
 First observe that <math>AO_2 = O_2D</math> and <math>BO_1 = O_1C</math>. Let points <math>A'</math> and <math>B'</math> be the reflections of <math>A</math> and <math>B</math>, respectively, about the perpendicular bisector of <math>\overline{O_1O_2}</math>. Then quadrilaterals <math>ABO_1O_2</math> and <math>A'B'O_2O_1</math> are congruent, so hexagons <math>ABO_1CDO_2</math> and <math>B'A'O_1CDO_2</math> have the same area. Furthermore, triangles <math>DO_2B'</math> and <math>A'O_1C</math> are congruent, so <math>B'D = A'C</math> and quadrilateral <math>B'A'CD</math> is an isosceles trapezoid.
 <asy>
@@ Line 64: / Line 64: @@
 ~djmathman
+==Solution 2==
+Denote by <math>O</math> the center of <math>\Omega</math>.
+Denote by <math>r</math> the radius of <math>\Omega</math>.
+We have <math>O_1</math>, <math>O_2</math>, <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math> are all on circle <math>\Omega</math>.
+Denote <math>\angle O_1 O O_2 = 2 \theta</math>.
+Denote <math>\angle O_1 O B = \alpha</math>.
+Denote <math>\angle O_2 O A = \beta</math>.
+Because <math>B</math> and <math>C</math> are on circles <math>\omega_1</math> and <math>\Omega</math>, <math>BC</math> is a perpendicular bisector of <math>O_1 O</math>. Hence, <math>\angle O_1 O C = \alpha</math>.
+Because <math>A</math> and <math>D</math> are on circles <math>\omega_2</math> and <math>\Omega</math>, <math>AD</math> is a perpendicular bisector of <math>O_2 O</math>. Hence, <math>\angle O_2 O D = \beta</math>.
+In <math>\triangle O O_1 O_2</math>,
+<cmath>
+\[
+O_1 O_2 = 2 r \sin \theta .
+\]
+</cmath>
+Hence,
+<cmath>
+\[
+r \sin \theta = 15 .
+\]
+</cmath>
+In <math>\triangle O AB</math>,
+<cmath>
+\begin{align*}
+AB & = 2 r \sin \frac{2 \theta - \alpha - \beta}{2} \\
+& = 2 r \sin \theta \cos \frac{\alpha + \beta}{2}
+- 2 r \cos \theta \sin \frac{\alpha + \beta}{2} \\
+& = 15 \cos \frac{\alpha + \beta}{2}
+- 2 r \cos \theta \sin \frac{\alpha + \beta}{2} .
+\end{align*}
+</cmath>
+Hence,
+<cmath>
+\[
+\cos \frac{\alpha + \beta}{2}
+- 2 r \cos \theta \sin \frac{\alpha + \beta}{2}
+= 2 . \hspace{1cm} (1)
+\]
+</cmath>
+In <math>\triangle O CD</math>,
+<cmath>
+\begin{align*}
+CD & = 2 r \sin \frac{360^\circ - 2 \theta - \alpha - \beta}{2} \\
+& = 2 r \sin \left( \theta + \frac{\alpha + \beta}{2} \right) \\
+& = 2 r \sin \theta \cos \frac{\alpha + \beta}{2}
++ 2 r \cos \theta \sin \frac{\alpha + \beta}{2} \\
+& = 15 \cos \frac{\alpha + \beta}{2}
++ 2 r \cos \theta \sin \frac{\alpha + \beta}{2} .
+\end{align*}
+</cmath>
+Hence,
+<cmath>
+\[
+\cos \frac{\alpha + \beta}{2}
++ 2 r \cos \theta \sin \frac{\alpha + \beta}{2}
+= 16 . \hspace{1cm} (2)
+\]
+</cmath>
+Taking <math>\frac{(1) + (2)}{30}</math>, we get <math>\cos \frac{\alpha + \beta}{2} = \frac{3}{5}</math>.
+Thus, <math>\sin \frac{\alpha + \beta}{2} = \frac{4}{5}</math>.
+Taking these into (1), we get <math>2 r \cos \theta = \frac{35}{4}</math>.
+Hence,
+<cmath>
+\begin{align*}
+r & = \sqrt{ \left( 2 r \sin \theta \right)^2 + \left( 2 r \cos \theta \right)^2} \\
+& = \frac{5}{4} \sqrt{193} .
+\end{align*}
+</cmath>
+Hence, <math>\cos \theta = \frac{7}{\sqrt{193}}</math>.
+In <math>\triangle O O_1 B</math>,
+<cmath>
+\[
+O_1 B = 2 r \sin \frac{\alpha}{2} .
+\]
+</cmath>
+In <math>\triangle O O_2 A</math>, by applying the law of sines, we get
+<cmath>
+\[
+O_2 A = 2 r \sin \frac{\beta}{2} .
+\]
+</cmath>
+Because circles <math>\omega_1</math> and <math>\omega_2</math> are externally tangent, <math>B</math> is on circle <math>\omega_1</math>, <math>A</math> is on circle <math>\omega_2</math>,
+<cmath>
+\begin{align*}
+O_1 O_2 & = O_1 B + O_2 A \\
+& = 2 r \sin \frac{\alpha}{2} + 2 r \sin \frac{\beta}{2} \\
+& = 2 r \left( \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} \right) .
+\end{align*}
+</cmath>
+Thus, <math>\sin \frac{\alpha}{2} + \sin \frac{\beta}{2} = \frac{12}{\sqrt{193}}</math>.
+Now, we compute <math>\sin \alpha</math> and <math>\sin \beta</math>.
+Recall <math>\cos \frac{\alpha + \beta}{2} = \frac{3}{5}</math> and <math>\sin \frac{\alpha + \beta}{2} = \frac{4}{5}</math>.
+Thus, <math>e^{i \frac{\alpha}{2}} e^{i \frac{\beta}{2}} = e^{i \frac{\alpha + \beta}{2}} =  \frac{3}{5} + i \frac{4}{5}</math>.
+We also have
+<cmath>
+\begin{align*}
+\sin \frac{\alpha}{2} + \sin \frac{\beta}{2}
+& = \frac{1}{2i} \left( e^{i \frac{\alpha}{2}} - e^{-i \frac{\alpha}{2}}
++ e^{i \frac{\beta}{2}} - e^{-i \frac{\beta}{2}}  \right) \\
+& = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \frac{\alpha + \beta}{2}} } \right)
+\left( e^{i \frac{\alpha}{2}} + e^{i \frac{\beta}{2}} \right)
+.
+\end{align*}
+</cmath>
+Thus,
+<cmath>
+\begin{align*}
+\sin \alpha + \sin \beta
+& = \frac{1}{2i} \left( e^{i \alpha} - e^{-i \alpha}
++ e^{i \beta} - e^{-i \beta}  \right) \\
+& = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \left( \alpha + \beta \right)}} \right)
+\left( e^{i \alpha} + e^{i \beta} \right) \\
+& = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \left( \alpha + \beta \right)}}  \right)
+\left(
+\left( e^{i \frac{\alpha}{2}} + e^{i \frac{\beta}{2}} \right)^2
+- 2 e^{i \frac{\alpha + \beta}{2}}
+\right) \\
+& = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \left( \alpha + \beta \right)}}  \right)
+\left(
+\left(  \frac{2 i \left( \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} \right)}{1 - \frac{1}{e^{i \frac{\alpha + \beta}{2}} }} \right)^2
+- 2 e^{i \frac{\alpha + \beta}{2}}
+\right) \\
+& = - \frac{1}{i} \left( e^{i \frac{\alpha + \beta}{2}} - e^{-i \frac{\alpha + \beta}{2}} \right)
+\left(
+\frac{2 \left( \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} \right)^2}
+{e^{i \frac{\alpha + \beta}{2}} + e^{-i \frac{\alpha + \beta}{2}} - 2 }
++ 1
+\right) \\
+& = \frac{167 \cdot 8}{193 \cdot 5 } .
+\end{align*}
+</cmath>
+Therefore,
+<cmath>
+\begin{align*}
+{\rm Area} \ ABO_1CDO_2
+& = {\rm Area} \ \triangle O_3 AB + {\rm Area} \ \triangle O_3  BO_1
++ {\rm Area} \ \triangle O_3 O_1 C \\
+& \quad + {\rm Area} \ \triangle O_3 C D
++ {\rm Area} \ \triangle O_3 D O_2 + {\rm Area} \ \triangle O_3 O_2 A \\
+& = \frac{1}{2} r^2 \left(
+\sin \left( 2 \theta - \alpha - \beta \right) + \sin \alpha + \sin \alpha
++ \sin \left( 360^\circ - 2 \theta - \alpha - \beta \right)
++ \sin \beta + \sin \beta \right) \\
+& = \frac{1}{2} r^2 \left(
+\sin \left( 2 \theta - \alpha - \beta \right)
+- \sin \left( 2 \theta + \alpha + \beta \right)
++ 2 \sin \alpha + 2 \sin \beta
+\right) \\
+& = r^2 \left(
+- \cos 2 \theta \sin \left( \alpha + \beta \right)
++ \sin \alpha + \sin \beta
+\right) \\
+& = r^2 \left( \left( 1 - 2 \cos^2 \theta \right) 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha + \beta}{2}
++ \sin \alpha + \sin \beta \right) \\
+& = \boxed{\textbf{(140) }} .
+\end{align*}
+</cmath>
+~Steven Chen (www.professorchenedu.com)
 ==See Also==