High School Olympiads Forum 9

by franzliszt, Jul 23, 2024, 8:23 PM

High School Olympiads Forum 9 wrote:
Let $ABC$ be an acute triangle with circumcircle $\Gamma$ and intouch triangle $DEF$. A circle is drawn tangent to $\overline{BC}$ at $D$ and to minor arc $BC$ of $\Gamma$ at $X$. Define $Y$ and $Z$ similarly. Prove that lines $\overline{DX}$, $\overline{EY}$, $\overline{FZ}$ are concurrent.
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Proof. Let $D'$ be the midpoint of $\widehat{BAC}$ containing $A$ and define $E',F'$ similarly. By Shooting Lemma, $XD\cap \Gamma=D'$. In Barycentric Coordinates with reference triangle $ABC$ with $A=(1,0,0),B=(0,1,0),C=(0,0,1)$ we have $D=(0:s-c:s-b)$ and $D'=(a^2:-b^2+bc:-c^2+bc)$ where we computed the coordinates of $D'=\frac{I_B+I_C}2$ using the midpoint formula on normalized $I_B=(a:-b:c),I_C=(a:b:-c)$. We can compute $E',F'$ similarly. We claim that the concurrency point is $X_{55} =( a^2(s-a) : b^2(s-b) : c^2(s-c))$.


Verify that \begin{align*}\begin{vmatrix}0&s-c&s-b\\a^2(s-a) & b^2(s-b) & c^2(s-c)\\  a^2&-b^2+bc&-c^2+bc\end{vmatrix}&=\frac{a^2}{4}\begin{vmatrix}0&a+b-c&a-b+c\\-a+b+c & b^2(a-b+c) & c^2(a+b-c)\\  1&-b^2+bc&-c^2+bc\end{vmatrix}\\&=0+(a+b-c)[c^2(a+b-c)-(-a+b+c)(-c^2+bc)]+(a-b+c)[(-a+b+c)(-b^2+bc)-b^2(a-b+c)]\\&=0\end{align*}where we simplified the last part with a little help.

We can use the exact same process to show that $X_{55}$ lies on $EE'$ and $FF'$ completing the proof.

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franz never dissapoints

by K124659, Mar 27, 2025, 12:15 AM

Every Barycentric Coordinates

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