2024 SSMO Team Round Problems/Problem 15

Problem

In triangle inscribed in circle let be the midpoint of Denote as the intersection of with If and find the perimeter of triangle

Solution

We will use barycentric coordinates with as the reference triangle. Let Note that the circumcircle of can be represented as Since lies on cevian with we have for Substituting into the equation for the circumcircle of we have \begin{align*} a^2y_pz_p+b^2x_pz_p+c^2x_py_p&=0\implies\\ a^2y_p^2+b^2x_py_p+c^2x_py_p&=0\implies\\ a^2y_p &= -(b^2+c^2)x_p\implies\\ y_p&=-\frac{b^2+c^2}{a^2}x_p. \end{align*} From we have So, Now, we have So, \begin{align*} PB^2 &= \left|a^2y_{pb}z_{pb}+b^2x_{pb}z_{pb}+c^2x_{pb}y_{pb}\right|\\ &=\left|a^2\left(\frac{b^2+c^2-a^2}{2b^2+2c^2-a^2}\right)\left(-\frac{b^2+c^2}{2b^2+2c^2-a^2}\right)\\+b^2\left(\frac{a^2}{2b^2+2c^2-a^2}\right)\left(-\frac{b^2+c^2}{2b^2+2c^2-a^2}\right)\\+c^2\left(\frac{a^2}{2b^2+2c^2-a^2}\right)\left(\frac{b^2+c^2-a^2}{2b^2+2c^2-a^2}\right)\right|\\ &=\frac{|a^4b^2-2a^2b^4-2a^2b^2c^2|}{(2b^2+2c^2-a^2)^2}\\ &=\frac{a^2b^2(2b^2+2c^2-a^2)}{(2b^2+2c^2-a^2)^2}\\ &=\frac{a^2b^2}{2b^2+2c^2-a^2}\implies\\ PB&=\frac{ab}{\sqrt{2b^2+2c^2-a^2}} \end{align*} In the same manner, we have So, we have This means and Solving for and , we have Substituting, we get \begin{align*} \sqrt{2b^2+2c^2-a^2} &= 40\\ 2b^2+2c^2-a^2 &= 40^2\implies\\ 2\left(\left(\frac{9\cdot40}{a}\right)^2+\left(\frac{13\cdot40}{a}\right)^2\right)-a^2 &=40^2\implies\\ a^4+40^2a^2-2\left((9\cdot40)^2+(13\cdot40)^2\right) &= 0\implies\\ a^2 &= \frac{-40^2\pm\sqrt{40^4+8\left((9\cdot40)^2+(13\cdot40)^2\right)}}{2}\\ &=(20)\left(\pm\sqrt{40^2+8(9^2+13^2)}-40\right)\\ &=(20)\left(\pm60-40\right) = 400,-2000\implies\\ a&=20. \end{align*} So, and In conclusion, the perimeter of is

~SMO_Team