Orthic triangle is isosceles

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Orthic triangle is isosceles

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Source: Austrian Regional Competition for Advanced Students 2012, #4

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#1 h May 23, 2012, 12:30 PM • 2 Y

Y by Adventure10, Mango247

In a triangle $ABC$ , let $H_a$ , $H_b$ and $H_c$ denote the base points of the altitudes on the sides $BC$ , $CA$ and $AB$ , respectively.
Determine for which triangles $ABC$ two of the lengths $H_aH_b$ , $H_bH_c$ and $H_aH_c$ are equal.

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#2 h May 23, 2012, 1:05 PM • 1 Y

Y by Adventure10

This is little longish but this works:
$H_aC=b\cos C$
$H_bC=a\cos C$
Thus, $H_aH_b=\cos C\sqrt{a^2+b^2-2abcos C}$ (Use cosine rule in $\triangle CH_aH_b$ )
If $H_bH_c=H_aH_b$ , then, we have:
which gives us that $\angle A=\angle C.$
Peace.

Faustus.

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#3 h May 23, 2012, 2:47 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Martin N. wrote:

In a triangle $ABC$ , let $H_a$ , $H_b$ and $H_c$ denote the base points of the altitudes on the sides $BC$ , $CA$ and $AB$ , respectively.
Determine for which triangles $ABC$ two of the lengths $H_aH_b$ , $H_bH_c$ and $H_aH_c$ are equal.

EDIT: If $ABC$ is acute:
If $H_aH_b=H_aH_c$ , then their respective angles in the orthic triangle are equal, so $\angle B=\angle C$ . Obviously the converse is also true.

This post has been edited 1 time. Last edited by goodar2006, May 24, 2012, 8:50 AM

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#4 h May 24, 2012, 7:45 AM • 1 Y

Y by Adventure10

You are missing some cases! Note that any right-angled triangle and any triangle with angles $\{\alpha, \beta, \gamma\}=\{\varphi, 90^{\circ}+\varphi, 90^{\circ}-2\varphi\}$ for $\varphi\in{]0^{\circ}, 45^{\circ}[}$ satisfy the conditions.

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