# 2022 AIME II Problems/Problem 15

## 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. $[asy] import geometry; size(10cm); point O1=(0,0),O2=(15,0),B=9*dir(30); circle w1=circle(O1,9),w2=circle(O2,6),o=circle(O1,O2,B); point A=intersectionpoints(o,w2)[1],D=intersectionpoints(o,w2)[0],C=intersectionpoints(o,w1)[0]; filldraw(A--B--O1--C--D--O2--cycle,0.2*fuchsia+white,black); draw(w1); draw(w2); draw(O1--O2,dashed); draw(o); dot(O1); dot(O2); dot(A); dot(D); dot(C); dot(B); label("\omega_1",8*dir(110),SW); label("\omega_2",5*dir(70)+(15,0),SE); label("O_1",O1,W); label("O_2",O2,E); label("B",B,N+1/2*E); label("A",A,N+1/2*W); label("C",C,S+1/4*W); label("D",D,S+1/4*E); label("15",midpoint(O1--O2),N); label("16",midpoint(C--D),N); label("2",midpoint(A--B),S); label("\Omega",o.C+(o.r-1)*dir(270)); [/asy]$

## 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 $B'A'O_2O_1$ are congruent, so hexagons $ABO_1CDO_2$ and $A'B'O_1CDO_2$ have the same area. Furthermore, triangles $DO_2A'$ and $B'O_1C$ are congruent, so $A'D = B'C$ and quadrilateral $A'B'CD$ is an isosceles trapezoid. $[asy] import olympiad; size(180); defaultpen(linewidth(0.7)); pair Ap = dir(105), Bp = dir(75), O1 = dir(25), C = dir(320), D = dir(220), O2 = dir(175); draw(unitcircle^^Ap--Bp--O1--C--D--O2--cycle); label("A'",Ap,dir(origin--Ap)); label("B'",Bp,dir(origin--Bp)); label("O_1",O1,dir(origin--O1)); label("C",C,dir(origin--C)); label("D",D,dir(origin--D)); label("O_2",O2,dir(origin--O2)); draw(O2--O1,linetype("4 4")); draw(Ap--D^^Bp--C,linetype("2 2")); [/asy]$ Next, remark that $B'O_1 = DO_2$, so quadrilateral $B'O_1DO_2$ is also an isosceles trapezoid; in turn, $B'D = O_1O_2 = 15$, and similarly $A'C = 15$. Thus, Ptolmey's theorem on $A'B'CD$ yields $A'D\cdot B'C + 2\cdot 16 = 15^2$, whence $A'D = B'C = \sqrt{193}$. Let $\alpha = \angle A'B'D$. The Law of Cosines on triangle $A'B'D$ 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 $A’B’$ and $CD$ is $12$ (in fact, $\triangle A'B'D$ is a $9-12-15$ triangle with a $7-12-\sqrt{193}$ triangle removed), which implies the area of $A'B'CD$ is $\tfrac12\cdot 12\cdot(2+16) = 108$.

Now let $O_1C = O_2A' = r_1$ and $O_2D = O_1B' = r_2$; the tangency of circles $\omega_1$ and $\omega_2$ implies $r_1 + r_2 = 15$. Furthermore, angles $A'O_2D$ and $A'B'D$ are opposite angles in cyclic quadrilateral $B'A'O_2D$, which implies the measure of angle $A'O_2D$ is $180^\circ - \alpha$. Therefore, the Law of Cosines applied to triangle $\triangle A'O_2D$ 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 $A'O_2D$ 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)

## Solution 3

Let points $A'$ and $B'$ be the reflections of $A$ and $B,$ respectively, about the perpendicular bisector of $O_1 O_2.$ $$B'O_2 = BO_1 = O_1 P = O_1 C,$$ $$A'O_1 = AO_2 = O_2 P = O_2 D.$$ We establish the equality of the arcs and conclude that the corresponding chords are equal $$\overset{\Large\frown} {CO_1} + \overset{\Large\frown} {A'O_1} +\overset{\Large\frown} {A'B'} = \overset{\Large\frown} {B'O_2} +\overset{\Large\frown} {A'O_1} +\overset{\Large\frown} {A'B'} =\overset{\Large\frown} {B'O_2} +\overset{\Large\frown} {DO_2} +\overset{\Large\frown} {A'B'}$$ $$\implies A'D = B'C = O_1 O_2 = 15.$$ Similarly $A'C = B'D \implies \triangle A'CO_1 = \triangle B'DO_2.$

Ptolemy's theorem on $A'CDB'$ yields $$B'D \cdot A'C + A'B' \cdot CD = A'D \cdot B'C \implies$$ $$B'D^2 + 2 \cdot 16 = 15^2 \implies B'D = A'C = \sqrt{193}.$$ The area of the trapezoid $A'CDB'$ is equal to the area of an isosceles triangle with sides $A'D = B'C = 15$ and $A'B' + CD = 18.$

The height of this triangle is $\sqrt{15^2-9^2} = 12.$ The area of $A'CDB'$ is $108.$

$$\sin \angle B'CD = \frac{12}{15} = \frac{4}{5},$$ $$\angle B'CD + \angle B'O_2 D = 180^o \implies \sin \angle B'O_2 D = \frac{4}{5}.$$

Denote $\angle B'O_2 D = 2\alpha.$ $\angle B'O_2 D > \frac{\pi}{2},$ hence $\cos \angle B'O_2 D = \cos 2\alpha = -\frac{3}{5}.$ $$\tan \alpha =\frac { \sin 2 \alpha}{1+\cos 2 \alpha} = \frac {4/5}{1 - 3/5}=2.$$

Semiperimeter of $\triangle B'O_2 D$ is $s = \frac {15 + \sqrt{193}}{2}.$

The distance from the vertex $O_2$ to the tangent points of the inscribed circle of the triangle $B'O_2 D$ is equal $s – B'D = \frac{15 – \sqrt{193}}{2}.$

The radius of the inscribed circle is $r = (s – B'D) \tan \alpha.$

The area of triangle $B'O_2 D$ is $[B'O_2 D] = sr = s (s – B'D) \tan \alpha = \frac {15^2 – 193}{2} = 16.$

The hexagon $ABO_1 CDO_2$ has the same area as hexagon $B'A'O_1 CDO_2.$

The area of hexagon $B'A'O_1 CDO_2$ is equal to the sum of the area of the trapezoid $A'CDB'$ and the areas of two equal triangles $B'O_2 D$ and $A'O_1 C,$ so the area of the hexagon $ABO_1 CDO_2$ is $$108 + 16 + 16 = \boxed{140}.$$

## Solution 4

Let circle $O_1$'s radius be $r$, then the radius of circle $O_2$ is $15-r$. Based on Brahmagupta's Formula,

the hexagon's Area $= \sqrt{14(1)(16-r)(1+r)} + \sqrt{7(8)(23-r)(8+r)}$.

Now we only need to find the $r$.

Connect $O_1$ and $A$ , $O_1$ and $D$ , and let $X$ be the point of intersection between $O_1D$ and circle $O_2$ , based on the " 2 Non-Congruent Triangles of 'SSA' Scenario " , we can immediately see $O_1X = O_1A$ and therefore get an equation from the "Power of A Point Theorem:

$(O_1A)(O_1D) = r(15+15-r) = 15r + r(15-r)$ (1).

Similarly,

$(O_2B)(O_2C) = (15-r)(15+r) = 15(15-r) + r(15-r)$ (2).

We can also get two other equations about these 4 segments from Ptolemy's Theorem:

$(O_1A)(O_2B) = 30 + r(15-r)$ (3)

$(O_1D)(O_2C) = 240 + r(15-r)$ (4)

Multiply equations (1) and (2), and equations (3) and (4) respectively, we will get a very simple and nice equation of $r$:

$2(15^2)r(15-r) = 7200 + 270r(15-r)$,

then:

$r(15-r) = 40$.

This result is good enough for us to find the hexagon's area, which:

     $= \sqrt{14(1)(16-r)(1+r)} + \sqrt{7(8)(23-r)(8+r)}$
$= \sqrt{14(1)(1+15-r)(1+r)} + \sqrt{7(8)(8+15-r)(8+r)}$
$= \sqrt{14(1)(1+15+40)} + \sqrt{7(8)(64+8(15)+40)}$
$= 28 + 112 = \boxed{\textbf{140}}$.


eJMaSc

## Video Solution

~MathProblemSolvingSkills.com