In triangle $ABC$, $M$ is the point of intersection of the medians, $O$ is the center of the inscribed circle, $A', B', C'$ are the touchpoints with the sides $BC, CA, AB$, respectively. Prove that if $CA'= AB$, then $OM$ and $AB$ are perpendicular. PS. There is a a typo
Problem
Source: 2003 Oral Moscow Geometry Olympiad grade 9 p4
Tags: geometry, incenter, Centroid, perpendicular
parmenides51
24.12.2023 18:08
There is a typo even in the source
If C' is touchpoint of incircle with side AB, then C'A < AB so it doesn't make sense the condition CA'=AB
any help is welcome to find the correct wording as noone likes incorrect problems
my thoughtsLooking backwards the condition $OM\perp AB \Leftrightarrow CA=CB$, so one possible fix of the typo is $C'A=C'B$
Om245
24.12.2023 20:45
Ok I guess problem statement is right We Claim that $M$ lie of $A'B'$ Let $N$ be midpoint of $AB$.and let $R$ be feet of perpendicular from $C$ to assume $CA>CB$(other cases like CA = CB are easy) We know that $C'O,A'B',CN$ concurrent at point let it be $X$ Now we have prove $X$ is $M$ which is equavalent to prove $RC' = 2 C'N$ Note from $CA' = AB$ we get $3c= a+b$ $C'N = c/2 - (s-b)$ = $ \frac{b-a}{2}$ Now $RC' = bcosA - (s-a) = b(\frac{b^2+c^2-a^2}{2bc}) + c-b$ Which should equal to $b-a$,So we have prove that $ \frac{b^2+c^2-a^2}{2c} + c-b = b-a$ $b^2+c^2-a^2 = 2c(5c-3a)$ $3c(b-a) + c^2 = 10c^2 - 6ca$ $3c(b+a) = 9c^2$ $b+a = 3c$ which is true hence $M$ lie on $C'O$ therefore $OM \perp AB$ Free to tell my calculations mistakes.