Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.2 National Olympiad Prove it Soviet Union

Problem:
Given point $O$ inside the acute-angled triangle $ABC$ , and point $O'$ inside the acute-angled triangle $A'B'C'$ . $D$ , $E$ , $F$ are the feet of the perpendiculars from $O$ to $BC$ , $CA$ , $AB$ respectively, and $D'$ , $E'$ , $F'$ are the feet of the perpendiculars from $O'$ to $B'C'$ , $C'A'$ , $A'B'$ respectively. $OD$ is parallel to $O'A'$ , $OE$ is parallel to $O'B'$ and $OF$ is parallel to $O'C'$ . Also $OD - O'A' = OE - O'B' = OF - O'C'$ . Prove that $O'D'$ is parallel to $OA$ , $O'E'$ to $OB$ and $O'F'$ to $OC$ , and that $O'D' \cdot OA = O'E' \cdot OB = O'F' \cdot OC$ .

Solution

Solution:

Let $\Gamma$ be the circumcircle of $DEF$ . Let $OD$ , $OE$ , $OF$ meet it again at $A''$ , $B''$ , $C''$ respectively. Then the figure $O'A'B'C'$ must be similar to $OA''B''C''$ . So to prove that $OD$ is parallel to $O'A'$ , we have to prove that $AO$ is perpendicular to $B''C''$ .

So $AO$ meets $B''C''$ at $D''$ . Now since $\square AFC'' = 90^{0}$ and $\square AD''C'' = 90^{0}$ , both $F$ and $D'$ lie on the circle diameter $AC''$ . Hence $AO - OD'' = OF - OC''$ . Similarly, $BO$ meets $C''A''$ at $E''$ , and $CO$ meets $A''B''$ at $F''$ , and $BO - OE'' = OD - OA''$ and $CO - OF'' = OE - OB''$ . Hence $OD'' \cdot OA = OE'' \cdot OB = OF'' \cdot OC$ . So using the similarity, $OD' - OA = OE' - OB = OF' - OC$ .

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