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, Ptolemy'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

Additional Graph for Better Understanding (Rearranging of the Pizza Slices)

This is how the reflection mentioned above is actually done. The reflection is actually a reorganization of the red and blue triangles, creating a symmetric figure with a isosceles trapezoid without changing the area. Basically you rearrange them so that each side contains a red triangle above a blue one. Then you calculate the area of the trapezoid and the two congruent triangles beside.

~cassphe

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 $108 + 16 + 16 = \boxed{140}.$

vladimir.shelomovskii@gmail.com, vvsss

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:

eJMaSc

Solution 5 (Super bash)

$[asy] import geometry; size(10cm); point O1=(0,0),O2=(15,0),B=9*dir(30); circle w1=circle(O1,9),w2=circle(O2,6),o=circle(O1,O2,B); point A=intersectionpoints(o,w2)[1],D=intersectionpoints(o,w2)[0],C=intersectionpoints(o,w1)[0]; filldraw(A--B--O1--C--D--O2--cycle,0.2*fuchsia+white,black); draw(w1); draw(w2); draw(O1--O2,dashed); draw(o); dot(O1); dot(O2); dot(A); dot(D); dot(C); dot(B); label("$\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("$x$", midpoint(O1--B),N); label("$x$", midpoint(O1--C),E); label("$15-x$", midpoint(O2--A),W); label("$15-x$", midpoint(O2--D),E); label("$\Omega$",o.C+(o.r-1)*dir(270)); [/asy]$

Start by noting that the radius of $\omega_1$ and $\omega_2$ sum up to 15.

Let the radius of $\omega_1$ be $x$ , then the radius of $\omega_2$ will be $15-x$ . We can find the area of the hexagon as the sum of the area of the quadrilateral $[O_1BAO_2]$ and $[O_1O_2DC]$ , which is given by the Brahmagupta's Formula as: $\sqrt{(16-2)(16-15)(16-x)(1+x)} + \sqrt{(23-15)(23-16)(23-x)(8+x)}$ , we only need $x$ , and the condition is sufficient already. Note that the total number of diagonals and sides of the hexagon is $\frac{6 \times 5}{2} = 15$ , 2 pairs of them $O_1B, O_1C$ , and $O_2A, O_2D$ can all be represented by the same variable $x$ , and 3 of them ( $AB, O_1O_2, CD$ ) are given. Therefore, the total number of variables is $15-4+1-3=9$ . And we have $\frac{6 \times 5}{2} = 15$ equations by doing ptolemy on the quadrilaterals inside the hexagon, we can list some of them out in hope of bashing for a solution:

$2\times 15 + x(15-x) = BO_2\times AO_1$ $15\times 16 + x(15-x) = DO_1\times CO_2$ $AC\times x + 2x = BC\times AO_1$ $AC\times (15-x) + 16(15-x) = AD\times CO_2$ $BD\times x + 16x = BC\times DO_1$ $BD\times (15-x) + 2(15-x) = AD\times BO_2$

$2CO_2 + BC\times (15-x) = AC\times BO_2$ $2DO_1 + AD\times x = BD\times AO_1$ $BO_2\times x + CO_2\times x = 15BC$ $AD\times x + 16AO_1 = AC\times DO_1$ $BC\times (15-x) + 16BO_2 = BD\times CO_2$ $AO_1\times (15-x) + DO_1\times (15-x) = 15AD$

$AO_1\times CO_2 + AO_2\times x = 15AC$ $BC\times AD + 2\times 16 = AC\times BD$ $BO_2\times DO_1 + x(15-x) = 15BD$

15 equations, 9 variables, hopefully one can solve this.

The question is especially disgusting in the sense that there is not a lot of geometric properties or hypothesis that one can conduct. The constructions seemed hard even to USAJMO level and lacked motivation. The normal trig took way too much time and this solution took even more. The question is a straightforward and brutal one when using ptolemy, all the conditions given and inherent are used and it guarantees a solution.

~IDKHowtoaddsolution

Video Solution

https://youtu.be/TeAWS5H5bcc

~MathProblemSolvingSkills.com

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