orthocenter criterion, distances related (2017 Metropolitan Olympiad MIPT 11.4)

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orthocenter criterion, distances related (2017 Metropolitan Olympiad MIPT 11.4)

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#1 h Mar 11, 2021, 8:12 PM

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Point $H$ is chosen inside an acute-angled triangle $ABC$ . It is known that the product of the distance from point $H$ to any of the vertices of triangle $ABC$ with the distance from point $H$ to the opposite side of triangle $ABC$ to this vertex, is the same for each of the vertices. Prove that $H$ is the intersection point of the altitudes of the triangle $ABC$ .

This post has been edited 1 time. Last edited by parmenides51, Mar 11, 2021, 8:55 PM
Reason: improved translation

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#2 h Mar 11, 2021, 8:32 PM • 1 Y

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I don't understand the wording. What does "the product the distance from point $H$ to any of the vertices of triangle $ABC$ " mean?

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#3 h Mar 11, 2021, 8:47 PM • 1 Y

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It's worded kinda weirdly (I think because of translation stuff, also conveying that meaning concisely in english is already hard enough)

It essentially means that if $D,E,F$ are the projections of $H$ onto $BC,AC,AB$ respectively, then we have $HA\cdot HD=HB\cdot HE=HC\cdot HF$ .

The converse is easy to prove with orthocenter reflections power of point.

This post has been edited 1 time. Last edited by natmath, Mar 11, 2021, 8:50 PM

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#4 h Mar 12, 2021, 2:48 AM

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From $\frac{HA}{HC} = \frac{HF}{FD}$ , then $\triangle HAF \sim \triangle HCD$ . Then, $\angle AHF=\angle CHD=\alpha$ . Similary, $\angle BHF = \angle CHE=\beta$ and $\angle AHE = \angle BHD=\gamma$ . Then $\alpha + \beta +\gamma=180^\circ \Longrightarrow C,H,F$ collinear. And also $A,H,D$ collinear and $B,H,E$ collinear. Then, $H$ is orthocenter of $\triangle ABC$ .

This post has been edited 4 times. Last edited by Wildabandon, Mar 12, 2021, 3:00 AM

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