2023 SSMO Accuracy Round Problems/Problem 10

Problem

Let $\triangle ABC$ be a triangle such $AB = 13$ , $BC = 14$ , $CA = 15$ . Let the incircle of $\triangle ABC$ touch $BC$ at $D$ , $AC$ at $E$ , and $AB$ at $F$ . Let $\ell_A$ be the line through the midpoints of $AE$ and $EF$ . Define $\ell_B$ and $\ell_C$ similarily. Let the area of the star created by the union of $\triangle ABC$ and the triangle bound by $\ell_A$ , $\ell_B$ , and $\ell_C$ be $\frac{p}{q}$ for relatively prime $p$ and $q$ . Find $p + q$ .

Solution

First, note that $AE = AF = 7$ , $BD = BF = 6$ , $CE = CD = 8$ , and the incenter has radius $4$ with total area $84$ .

Let $M_a$ be the midpoint of $BC$ , and let $T_a$ be the intersection of $\ell_b$ and $\ell_c$ . Note that $T_a$ is the radical center of the circles centered at $B$ and $C$ with radius $0$ , and the incircle, so it lies on the perpendicular bisector of $BC$ .

We can find $M_aT_a$ using the fact that $T_a$ has equal power to $B$ and $C$ : $T_aB^2 = T_aC^2 = T_aI^2 - ID^2.$

Then, $M_aT_a^2 + 7^2 = (M_aT_a + 4)^2 + 1^2 - 4^2,$ which simplifies to $M_aT_a = 6$ .

Similarly, $M_bT_b^2 + \left(\frac{15}{2}\right)^2 = (M_bT_b + 4)^2 + \left(\frac{1}{2}\right)^2 - 4^2,$ so $M_bT_b = 7$ .

Finally, $M_cT_c^2 + \left(\frac{13}{2}\right)^2 = (M_cT_c + 4)^2 + \left(\frac{1}{2}\right)^2 - 4^2,$ giving $M_cT_c = \frac{21}{4}$ .

The total area is $84 + \frac{1}{2} \left(6 \cdot 7 + 7 \cdot \frac{15}{2} + \frac{21}{4} \cdot \frac{13}{2} \right) = \frac{2373}{16},$ so the final answer is $2373 + 16 = \boxed{2389}.$

~SMO_Team

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2023 SSMO Accuracy Round Problems/Problem 10

Problem

Solution