In triangle $ABC$ the midpoints of sides $AC, BC$, vertex $C$ and the centroid lie on the same circle. Prove that this circle touches the circle passing through $A, B$ and the orthocenter of triangle $ABC$.
Problem
Source: Sharygin 2011 Final 10.1
Tags: geometry, orthocenter, Centroid, tangent circles
a_simple_guy
01.04.2019 05:04
Let $E,F$ be the midpoint of $AC$ and $BC$ . From the condition $E,F,O,C,G$ all lies on a circle tangent to $\odot(ABC)$ Let the tangent at $A$ to the circumcircle of $(ABC)$ intersects $BA$ at $J$.
$Claim$: $JC=JG$.
To show that we first proof that $G$ is the $C-HM$ point of $\triangle ABC$. Invert Around $AC$ with radius $\sqrt{\frac{AB.AC}{2}}$ followed by the reflextion of the $A$ angle bisector. Then $E \longleftarrow B$, $F \longleftarrow A$ and$O \longleftarrow CH \cap AB$. Hence $G$ sends to the intersection of the $C$ symmedian with the $BA$. Let it be $X$. Then from characterization 1.6 $G$ is the $C-HM$ point of $\triangle ABC$. It is well known that C appolonious circle passes through $AC-HM$ point. Hence $JC=JG$. By power of point $JC^2=JG^2=JC.JB$ and we are done
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zuss77
11.04.2019 13:59
It more simple than that. Official solution: Let $C'$ be a point symmetric to $C$ with respect to the middle of $AB$. Then $C' \in \odot (ABH)$ ($H$ - orthocenter of $\triangle ABC$) If $D, E$ are midpoints of $AC, BC$, $\triangle ABC'$ is homothetic to $\triangle EDC$ wrt $M$ - centroid of $\triangle ABC$ (with coefficient $-1/2$). Hence, the circumscribed circles of these triangles are touch each other at $M$.