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, 2014

Geometry Difficulty 5.7 AIME, harder Prove it United States

Problem:

Let $ABC$ be a triangle with $AB = 13$ , $BC = 14$ , and $CA = 15$ . Let $\Gamma$ be the circumcircle of $ABC$ , let $O$ be its circumcenter, and let $M$ be the midpoint of minor arc $\widehat{BC}$ . Circle $\omega_1$ is internally tangent to $\Gamma$ at $A$ , and circle $\omega_2$ , centered at $M$ , is externally tangent to $\omega_1$ at a point $T$ . Ray $AT$ meets segment $BC$ at point $S$ , such that $BS - CS = 4/15$ . Find the radius of $\omega_2$ .

Solution

Solution:

Answer: $\quad \dfrac{1235}{108}$

Let $N$ be the midpoint of $BC$ . Notice that $BS - CS = \dfrac{4}{15}$ means that $NS = \dfrac{2}{15}$ . Let lines $MN$ and $AS$ meet at $P$ , and let $D$ be the foot of the altitude from $A$ to $BC$ . Then $BD = 5$ and $AD = 12$ , so $DN = 2$ and $DS = \dfrac{32}{15}$ . Thus $NP = AD \dfrac{SN}{SD} = 12 \cdot \dfrac{2/15}{32/15} = \dfrac{3}{4}$ .

Now $OB = R = \dfrac{abc}{4A} = \dfrac{(13)(14)(15)}{4 \cdot 84} = \dfrac{65}{8}$ , so $ON = \sqrt{OB^2 - BN^2} = \sqrt{\left(\dfrac{65}{8}\right)^2 - 7^2} = \dfrac{33}{8}$ . Thus $OP = \dfrac{27}{8}$ and $PM = OM - OP = \dfrac{19}{4}$ .

By Monge's theorem, the exsimilicenter of $\omega_1$ and $\Gamma$ (which is $A$ ), the insimilicenter of $\omega_1$ and $\omega_2$ (which is $T$ ), and the insimilicenter of $\omega_2$ and $\Gamma$ (call this $P'$ ) are collinear. But notice that this means $P' = OM \cap AT = P$ . From this we get
$\frac{\text{radius of } \omega_2}{R} = \frac{MP}{OP} = \frac{38}{27}$
Thus the radius of $\omega_2$ is $\dfrac{65}{8} \cdot \dfrac{38}{27} = \dfrac{1235}{108}$ .

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