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 $AK=m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution

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

2019 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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2019 AIME II Problems/Problem 11

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