Let $ABC$ be an acute-angled triangle and let $E$ and $F$ be the feet of the altitudes from $B$ and $C$ to the sides $AC$ and $AB$ respectively. Suppose $AD$ is the diameter of the circle $ABC$. Let $M$ be the midpoint of $BC$. Let $K$ be the imsimilicenter of the incircles of the triangles $BMF$ and $CME$. Prove that the points $K, M, D$ are collinear. Proposed by Bilegdembrel Bat-Amgalan.
Problem
Source: Mongolian Mathematical Olympiad 2024 P5
Tags: geometric transformation, geometry
thdnder
12.04.2024 16:58
Quidditch wrote: I drew this but it wasn’t true? I think there is a typo, in the actual contest, it was incircles of $BMF$ and $CME$, not circumcircles.
Iveela
12.04.2024 17:01
Quidditch wrote: I drew this but it wasn’t true? My bad, it should be fixed now.
math_comb01
12.04.2024 17:09
Reposted, there is also a obvious projective soln btw
NO_SQUARES
12.04.2024 17:15
Also this problem has a generalization: Let $ABCD$ be a cyclic quadrilateral, $O$ be center of $(ABCD)$, $X=AC \cap BD$, $K$ be a insimilicenter of the incircles of the triangles $AOB$ and $COD$. Then points $O,K,X$ are collinear. P.S. Maybe, the same solution as in post above works, but I didn't checked it...
Quidditch
12.04.2024 17:41
math_comb01 wrote: Reposted, there is also a obvious projective soln btw Here it is. Add the exsimilicenter, say $T$, of those two circles. It obviously lies on $BC$. Let the incenters be $I_B,I_C$. Projecting from $M$ to $BE$: $$-1=(T,K;I_B,I_C)=(MK\cap BE,B;MI_B\cap BH,MI_C\cap BE).$$However, $MI_B\parallel CF$ and it bisects $BF$, so it also bisects $BH$. Also, $MI_C\parallel BE$ so they meet at $\infty$. Thus, we can deduce that $MK\cap BE$ is actually $H$ and it’s well-known that $M,H,D$ are collinear, so we’re done.