# Difference between revisions of "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

## Solution 2

On the Spot STEM solves this problem here: https://www.youtube.com/watch?v=nJydO5CLuuI

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## See Also

 2019 AIME II (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

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

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