# 2019 AIME II Problems/Problem 11

## Problem

Triangle $ABC$ has side lengths $AB=7, BC=8,$ and $CA=9.$ Circle $\omega_1$ passes through $B$ and is tangent to line $AC$ at $A.$ Circle $\omega_2$ passes through $C$ and is tangent to line $AB$ at $A.$ Let $K$ be the intersection of circles $\omega_1$ and $\omega_2$ not equal to $A.$ Then $AK=\tfrac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

## Solution 1

$[asy] unitsize(20); pair B = (0,0); pair A = (2,sqrt(45)); pair C = (8,0); draw(circumcircle(A,B,(-17/8,0)),rgb(.7,.7,.7)); draw(circumcircle(A,C,(49/8,0)),rgb(.7,.7,.7)); draw(B--A--C--cycle); label("A",A,dir(105)); label("B",B,dir(-135)); label("C",C,dir(-75)); dot((2.68,2.25)); label("K",(2.68,2.25),dir(-150)); label("\omega_1",(-6,1)); label("\omega_2",(14,6)); label("7",(A+B)/2,dir(140)); label("8",(B+C)/2,dir(-90)); label("9",(A+C)/2,dir(60)); [/asy]$ -Diagram by Brendanb4321

Note that from the tangency condition that the supplement of $\angle CAB$ with respects to lines $AB$ and $AC$ are equal to $\angle AKB$ and $\angle AKC$, respectively, so from tangent-chord, $$\angle AKC=\angle AKB=180^{\circ}-\angle BAC$$ Also note that $\angle ABK=\angle KAC$, so $\triangle AKB\sim \triangle CKA$. Using similarity ratios, we can easily find $$AK^2=BK*KC$$ However, since $AB=7$ and $CA=9$, we can use similarity ratios to get $$BK=\frac{7}{9}AK, CK=\frac{9}{7}AK$$

• Now we use Law of Cosines on $\triangle AKB$: From reverse Law of Cosines, $\cos{\angle BAC}=\frac{11}{21}\implies \cos{(180^{\circ}-\angle BAC)}=-\frac{11}{21}$. This gives us $$AK^2+\frac{49}{81}AK^2+\frac{22}{27}AK^2=49$$ $$\implies \frac{196}{81}AK^2=49$$ $$AK=\frac{9}{2}$$ so our answer is $9+2=\boxed{011}$.

-franchester

The motivation for using the Law of Cosines ("LoC") is after finding the similar triangles it's hard to figure out what to do with $BK$ and $CK$ yet we know $BC$ which somehow has to help us solve the problem--a common theme in solving geometry problems is figuring out how to use what you haven't used yet. We know all three sides of some triangle though, and we're dealign with angles (that's how we found similarity), so why not try the Law of Cosines? This is to help with motivation--the solution is franchester's and I learned about using LoC from reading his solution (I only solved half the problem and got struck). To anyone in the future reading this, math is beautiful. -First

## Solution 2 (Inversion)

Consider an inversion with center $A$ and radius $r=AK$. Then, we have $AB\cdot AB^*=AK^2$, or $AB^*=\frac{AK^2}{7}$. Similarly, $AC^*=\frac{AK^2}{9}$. Notice that $AB^*KC^*$ is a parallelogram, since $\omega_1$ and $\omega_2$ are tangent to $AC$ and $AB$, respectively. Thus, $AC^*=B^*K$. Now, we get that $$\cos(\angle AB^*K)=-\cos(180-\angle BAC)=-\frac{11}{21}$$ so by Law of Cosines on $\triangle AB^*K$ we have $$(AK)^2=(AB^*)2+(B^*K)^2-2\cdot AB^*\cdot B^*K \cdot \cos(\angle AB^*K)$$ $$\Rightarrow (AK)^2=\frac{AK^4}{49}+\frac{AK^4}{81}-2\cdot \frac{AK^2}{7}\frac{AK^2}{9}\frac{-11}{21}$$ $$\Rightarrow 1=\frac{AK^2}{49}+\frac{AK^2}{81}+\frac{11AK^2}{63\cdot21}$$ $$\Rightarrow AK=\frac{9}{2}$$ Then, our answer is $9+2=\boxed{11}$. -brianzjk

## Solution 3 (Death By Trig Bash)

14. Let the centers of the circles be $O_{1}$ and $O_{2}$ where the $O_{1}$ has the side length $7$ contained in the circle. Now let $\angle BAC =x.$ This implies $\angle AO_{1}B = \angle AO_{2}C = 2x$ by the angle by by tangent. Then we also know that $\angle O_{1}AB = \angle O_{1}BA = \angle O_{2}AC = \angle O_{2}CA = 90^{\circ}-x.$ Now we first find $\cos x.$ We use law of cosines on $\bigtriangleup ABC$ to obtain $64 = 81 + 48 - 2 \cdot 9 \cdot 7 \cdot \cos{x} \implies \cos{x} =\frac{11}{21} \implies \sin{x} =\frac{8\sqrt{5}}{21}.$ Then applying law of sines on $\bigtriangleup AO_{1}B$ we obtain $\frac{7}{\sin{2x}} =\frac{OB_{1}}{\sin{90^{\circ}-x}} \implies\frac{7}{2\sin{x}\cos{x}} =\frac{OB_{1}}{\cos{x}} \implies OB_{1} = O_{1}A=\frac{147}{16\sqrt{5}}.$ Using similar logic we obtain $OA_{1} =\frac{189}{16\sqrt{5}}.$ Now we know that $\angle O_{1}AO_{2}=180^{\circ}-x.$ Thus using law of cosines on $\bigtriangleup O_{1}AO_{2}$ yields $O_{1}O_{2} =\sqrt{\left(\frac{147}{16\sqrt{5}}\right)^2+\left(\frac{189}{16\sqrt{5}}\right)^2-2\:\cdot \left(\frac{147}{16\sqrt{5}}\right)\cdot \frac{189}{16\sqrt{5}}\cdot -\frac{11}{21}}.$ While this does look daunting we can write the above expression as $\sqrt{\left(\frac{189+147}{16\sqrt{5}}\right)^2 - 2\cdot \left(\frac{147}{16\sqrt{5}}\right)\cdot \frac{189}{16\sqrt{5}}\cdot \frac{10}{21}} =\sqrt{\left(\frac{168}{8\sqrt{5}}\right)^2 - \left(\frac{7 \cdot 189 \cdot 5}{8 \sqrt{5} \cdot 8\sqrt{5}}\right)}.$ Then factoring yields $\sqrt{\frac{21^2(8^2-15)}{(8\sqrt{5})^2}} =\frac{147}{8\sqrt{5}}.$ The area $[O_{1}AO_{2}] =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot \sin(180^{\circ}-x) =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot\frac{8\sqrt{5}}{21}.$ Now $AK$ is twice the length of the altitude of $\bigtriangleup O_{1}AO_{2}$ so we let the altitude be $h$ and we have $\frac{1}{2} \cdot h \cdot\frac{147\sqrt{5}}{8\sqrt{5}} =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot\frac{8\sqrt{5}}{21} \implies h =\frac{9}{4}.$ Thus our desired length is $\frac{9}{2} \implies n+n = \boxed{11}.$