Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.8 AIME, harder Prove it

Prove that the altitude from vertex $A$ of triangle $ABC$ is equal to the harmonic mean of the radii of the excircles opposite to sides $AB$ and $AC$ .

Let's prove that the altitude from vertex $A$ of triangle $ABC$ is equal to the harmonic mean of the radii of the excircles opposite to sides $AB$ and $AC$ .

Solution

Let the radii of the excircles opposite to sides $AC$ and $AB$ be denoted by $r_{b}$ and $r_{c}$ , respectively. Using the conventional notation,

$r_{b}=\frac{t}{s-b}, \quad r_{c}=\frac{t}{s-c}$

(The proof can be found, for example, on page 66 of I. Reiman's book "A geometria és határterületei" (Geometry and its Boundaries). Gondolat, Budapest, 1986).

The harmonic mean of $r_{b}$ and $r_{c}$ is:

$\frac{2}{\frac{1}{r_{b}}+\frac{1}{r_{c}}}=\frac{2 t}{s-b+s-c}=\frac{2 t}{a}=m_{a}$

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.