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Bulgarian Mathematical Competitions

Geometry Difficulty 6.1 National Olympiad Prove it Bulgaria

Problem:
Let $AA_{1}$ , $BB_{1}$ and $CC_{1}$ be the altitudes of an acute $\triangle ABC$ ( $A_{1} \in BC$ , $B_{1} \in CA$ and $C_{1} \in AB$ ). Denote by $O$ the circumcenter of $\triangle ABC$ , and by $H_{1}$ the orthocenter of $\triangle A_{1}B_{1}C_{1}$ . Prove that the midpoint of the segment $OH_{1}$ coincides with the incenter of the triangle with vertices at the midpoints of the sides of $\triangle A_{1}B_{1}C_{1}$ .

Solution

Solution:
Denote by $H$ the orthocenter of $\triangle ABC$ , and by $G_{1}$ the centroid of $\triangle A_{1}B_{1}C_{1}$ . Let $O_{1}$ be the midpoint of the segment $OH$ . It is well-known that $O_{1}$ is the circumcenter of $\triangle A_{1}B_{1}C_{1}$ , and $H$ is its incenter. Then $\overrightarrow{H_{1}G_{1}} = 2\, \overrightarrow{G_{1}O_{1}}$ .

Note that the dilation with center $G_{1}$ and ratio $-\frac{1}{2}$ maps $\triangle A_{1}B_{1}C_{1}$ into $\triangle A_{2}B_{2}C_{2}$ formed by the midpoints of the segments $B_{1}C_{1}$ , $A_{1}C_{1}$ and $A_{1}B_{1}$ . Hence the image of $H$ under this dilation is the incenter $I_{2}$ of $\triangle A_{2}B_{2}C_{2}$ . Since

$G_{1}$ is the centroid of $\triangle OHH_{1}$ , it follows that $I_{2}$ is the midpoint of the segment $OH_{1}$ .

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