Let $ABC$ be a triangle with $AB < AC < BC$ and $\Gamma$ its circumcircle. Let $\omega_1$ be the circle with center $B$ and radius $AC$ and $\omega_2$ the circle with center $C$ and radius $AB$. The circles $\omega_1$ and $\omega_2$ intersect at point $E$ such that $A$ and $E$ are on opposite sides of the line $BC$. The circles $\Gamma$ and $\omega_1$ intersect at point $F$ and the circles $\Gamma$ and $\omega_2$ intersect at point $G$ such that the points $F$ and $G$ are on the same side as $E$ in relation to the line $BC$. With $K$ being the point such that $AK$ is a diameter of $\Gamma$, prove that $K$ is circumcenter of triangle $EFG$.
Problem
Source: Brazil Cono Sur TST 2024 - T1/P2
Tags: geometry
Vivouaf
31.10.2024 15:02
First, notice that $E$ is such that $ABEC$ is a parallelogram. Now, $K$ is the orthocenter of $BEC$, because reflection with respect to the midpoint of $BC$ maps $E$ to $A$ and $K$ to the orthocenter of $ABC$, implying the result. Therefore with a bit of angle chase we get $\angle BFC = \angle BKC = \angle CEB$, meaning $BFE$ is isoscele, and as $BK \perp CE$, $KFE$ is isoscele too. Similarly, $KGE$ is isoscele, thus obtaining $KE=KF=KG$ and concluding that $K$ is the circumcneter of $EFG$.
moony_
31.10.2024 17:29
the weirdest problem formulation i have ever seen... lmao
Ianis
31.10.2024 19:58
First, it is clear that the points $D$ such that $ABCD$ is an isosceles trapezoid with $AD\parallel BC$ and $E'$ such that $ABE'C$ is a parallelogram belong to both $\omega _1$ and $\omega _2$, so they are the points where $\omega _1$ intersects $\omega _2$. Since $D$ lies on the same side of $A$ wrt $BC$ and $E'$ lies on the opposite side of $A$ wrt $BC$, we get that $E=E'$. Similarly, $F\neq C$ is the point on $\Gamma$ such that $AB\parallel CF$ and $G\neq B$ is the point on $\Gamma$ such that $CA\parallel BG$ (because none of them can be $D$).
Now proceed with the solution. Use complex numbers with $\Gamma$ as the unit circle. Then $e=b+c-a$, $f=\frac{ab}{c}$, $g=\frac{ca}{b}$ and $k=-a$. Hence $e-k=b+c$, $f-k=\frac{ab}{c}+a=\frac{a}{c}(b+c)$ and $g-k=\frac{ca}{b}+a=\frac{a}{b}(b+c)$, so $|e-k|=|f-k|=|g-k|$, because they are all equal to $|b+c|$.