Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.4 AIME, harder Prove it United States

Problem:

Let $XYZ$ be a triangle with $\angle XYZ = 40^{\circ}$ and $\angle YZX = 60^{\circ}$ . A circle $\Gamma$ , centered at the point $I$ , lies inside triangle $XYZ$ and is tangent to all three sides of the triangle. Let $A$ be the point of tangency of $\Gamma$ with $YZ$ , and let ray $\overrightarrow{XI}$ intersect side $YZ$ at $B$ . Determine the measure of $\angle AIB$ .

Solution

Solution:

Answer: $10^{\circ}$

Let $D$ be the foot of the perpendicular from $X$ to $YZ$ . Since $I$ is the incenter and $A$ the point of tangency, $IA \perp YZ$ , so

$AI \parallel XD \Rightarrow \angle AIB = \angle DXB$

Since $I$ is the incenter,
$\angle BXZ = \frac{1}{2} \angle YXZ = \frac{1}{2}\left(180^{\circ} - 40^{\circ} - 60^{\circ}\right) = 40^{\circ}.$

Consequently, we get that
$\angle AIB = \angle DXB = \angle ZX B - \angle ZXD = 40^{\circ} - (90^{\circ} - 60^{\circ}) = 10^{\circ}$

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