Maths Olympiad Prep

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, 2014

Geometry Difficulty 5.0 AIME Prove it United States

Problem:

Let ABCABC be a triangle with AB=AC=5AB = AC = 5 and BC=6BC = 6. Denote by ω\omega the circumcircle of ABCABC. We draw a circle Ω\Omega which is externally tangent to ω\omega as well as to the lines ABAB and ACAC (such a circle is called an AA-mixtilinear excircle). Find the radius of Ω\Omega.

Solution

Solution:

Let MM be the midpoint of BCBC. Let DD be the point diametrically opposite AA on the circumcircle, and let the AA-mixtilinear excircle be tangent to lines ABAB and ACAC at XX and YY. Let OO be the center of the AA-mixtilinear excircle.

Notice that AOXABM\triangle AOX \sim \triangle ABM. If we let xx be the desired radius, we have
x+ADx=53. \frac{x + AD}{x} = \frac{5}{3}.
We can compute AD5=54\frac{AD}{5} = \frac{5}{4} since ADBABM\triangle ADB \sim \triangle ABM, we derive AD=254AD = \frac{25}{4}. From here it follows that x=758x = \frac{75}{8}.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.