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Difference between revisions of "2023 SSMO Accuracy Round Problems/Problem 10"

(Created page with "==Problem== Let <math>\triangle ABC</math> be a triangle such <math>AB = 13</math>, <math>BC = 14</math>, <math>CA = 15</math>. Let the incircle of <math>\triangle ABC</math>...")
 
 
Line 3: Line 3:
  
 
==Solution==
 
==Solution==
 +
First, note that <math>AE = AF = 7</math>, <math>BD = BF = 6</math>, <math>CE = CD = 8</math>, and the incenter has radius <math>4</math> with total area <math>84</math>.
 +
 +
Let <math>M_a</math> be the midpoint of <math>BC</math>, and let <math>T_a</math> be the intersection of <math>\ell_b</math> and <math>\ell_c</math>. Note that <math>T_a</math> is the radical center of the circles centered at <math>B</math> and <math>C</math> with radius <math>0</math>, and the incircle, so it lies on the perpendicular bisector of <math>BC</math>.
 +
 +
We can find <math>M_aT_a</math> using the fact that <math>T_a</math> has equal power to <math>B</math> and <math>C</math>:
 +
<cmath>T_aB^2 = T_aC^2 = T_aI^2 - ID^2.</cmath>
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 +
Then,
 +
<cmath>M_aT_a^2 + 7^2 = (M_aT_a + 4)^2 + 1^2 - 4^2,</cmath>
 +
which simplifies to <math>M_aT_a = 6</math>.
 +
 +
Similarly,
 +
<cmath>M_bT_b^2 + \left(\frac{15}{2}\right)^2 = (M_bT_b + 4)^2 + \left(\frac{1}{2}\right)^2 - 4^2,</cmath>
 +
so <math>M_bT_b = 7</math>.
 +
 +
Finally,
 +
<cmath>M_cT_c^2 + \left(\frac{13}{2}\right)^2 = (M_cT_c + 4)^2 + \left(\frac{1}{2}\right)^2 - 4^2,</cmath>
 +
giving <math>M_cT_c = \frac{21}{4}</math>.
 +
 +
The total area is
 +
<cmath>84 + \frac{1}{2} \left(6 \cdot 7 + 7 \cdot \frac{15}{2} + \frac{21}{4} \cdot \frac{13}{2} \right) = \frac{2373}{16},</cmath>
 +
so the final answer is
 +
<cmath>2373 + 16 = \boxed{2389}.</cmath>
 +
 +
~SMO_Team

Latest revision as of 21:13, 9 September 2025

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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