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Difference between revisions of "2024 SSMO Team Round Problems/Problem 15"

(Created page with "==Problem== In triangle <math>ABC</math> inscribed in circle <math>\omega,</math> let <math>M</math> be the midpoint of <math>BC.</math> Denote <math>P</math> as the intersec...")
 
 
Line 4: Line 4:
  
 
==Solution==
 
==Solution==
 +
We will use barycentric coordinates with <math>ABC</math> as the reference triangle. Let <math>A = (1:0:0),B = (0:1:0),C(0:0:1).</math> Note that the circumcircle of <math>ABC</math> can be represented as <math>a^2yx+b^2xz+c^2xy = 0.</math> Since <math>P</math> lies on cevian <math>AM,</math> with <math>M = \left(0:\frac{1}{2}:\frac{1}{2}\right),</math> we have <math>P = (x_p:y_p:z_p),</math> for <math>y_p=z_p.</math> Substituting into the equation for the circumcircle of <math>ABC,</math> 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 <math>x_p+y_p+z_p=1,</math> we have <cmath>x_p+\left(-\frac{(b^2+c^2)}{a^2}\right)x_p+\left(-\frac{(b^2+c^2)}{a^2}\right)x_p = 1\implies x_p = \frac{a^2}{2b^2+2c^2-a^2}.</cmath> So, <math>P = \left(-\frac{a^2}{2b^2+2c^2-a^2}:\frac{b^2+c^2}{2b^2+2c^2-a^2}:\frac{b^2+c^2}{2b^2+2c^2-a^2}\right).</math> Now, we have <math>\overrightarrow{PB} = (x_{pb}:y_{pb}:z_{pb}) = \left(\frac{a^2}{2b^2+2c^2-a^2}:\frac{b^2+c^2-a^2}{2b^2+2c^2-a^2}:-\frac{b^2+c^2}{2b^2+2c^2-a^2}\right).</math> 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 <cmath>PC = \frac{ac}{\sqrt{2b^2+2c^2-a^2}}.</cmath> So, we have <cmath>AM = 20\implies\frac{\sqrt{2b^2+2c^2-a^2}}{2} = 20\implies\sqrt{2b^2+2c^2-a^2} = 40.</cmath> This means <cmath>ab = PB\sqrt{2b^2+2c^2-a^2} = 9\cdot40</cmath> and <cmath>ac = PC\sqrt{2b^2+2c^2-a^2}=13\cdot40.</cmath> Solving for <math>b</math> and <math>c</math>, we have <math>b = \frac{9\cdot40}{a},c = \frac{13\cdot40}{a}.</math> 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, <math>b = \frac{9\cdot40}{a} = \frac{9\cdot40}{20} = 18</math> and <math>c = \frac{13\cdot40}{20} = 26.</math> In conclusion, the perimeter of <math>ABC</math> is <math>20+18+26 = \boxed{64}.</math>
 +
 +
~SMO_Team

Latest revision as of 14:42, 10 September 2025

Problem

In triangle $ABC$ inscribed in circle $\omega,$ let $M$ be the midpoint of $BC.$ Denote $P$ as the intersection of $AM$ with $\omega.$ If $BP = 9, CP = 13,$ and $AM = 20,$ find the perimeter of triangle $ABC.$

Solution

We will use barycentric coordinates with $ABC$ as the reference triangle. Let $A = (1:0:0),B = (0:1:0),C(0:0:1).$ Note that the circumcircle of $ABC$ can be represented as $a^2yx+b^2xz+c^2xy = 0.$ Since $P$ lies on cevian $AM,$ with $M = \left(0:\frac{1}{2}:\frac{1}{2}\right),$ we have $P = (x_p:y_p:z_p),$ for $y_p=z_p.$ Substituting into the equation for the circumcircle of $ABC,$ 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 $x_p+y_p+z_p=1,$ we have \[x_p+\left(-\frac{(b^2+c^2)}{a^2}\right)x_p+\left(-\frac{(b^2+c^2)}{a^2}\right)x_p = 1\implies x_p = \frac{a^2}{2b^2+2c^2-a^2}.\] So, $P = \left(-\frac{a^2}{2b^2+2c^2-a^2}:\frac{b^2+c^2}{2b^2+2c^2-a^2}:\frac{b^2+c^2}{2b^2+2c^2-a^2}\right).$ Now, we have $\overrightarrow{PB} = (x_{pb}:y_{pb}:z_{pb}) = \left(\frac{a^2}{2b^2+2c^2-a^2}:\frac{b^2+c^2-a^2}{2b^2+2c^2-a^2}:-\frac{b^2+c^2}{2b^2+2c^2-a^2}\right).$ 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 \[PC = \frac{ac}{\sqrt{2b^2+2c^2-a^2}}.\] So, we have \[AM = 20\implies\frac{\sqrt{2b^2+2c^2-a^2}}{2} = 20\implies\sqrt{2b^2+2c^2-a^2} = 40.\] This means \[ab = PB\sqrt{2b^2+2c^2-a^2} = 9\cdot40\] and \[ac = PC\sqrt{2b^2+2c^2-a^2}=13\cdot40.\] Solving for $b$ and $c$, we have $b = \frac{9\cdot40}{a},c = \frac{13\cdot40}{a}.$ 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, $b = \frac{9\cdot40}{a} = \frac{9\cdot40}{20} = 18$ and $c = \frac{13\cdot40}{20} = 26.$ In conclusion, the perimeter of $ABC$ is $20+18+26 = \boxed{64}.$

~SMO_Team

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