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Harvard-MIT November Tournament

Geometry Difficulty 5.1 AIME, harder Prove it United States

Problem:

Let $ABC$ be an equilateral triangle with $AB = 3$ . Circle $\omega$ with diameter $1$ is drawn inside the triangle such that it is tangent to sides $AB$ and $AC$ . Let $P$ be a point on $\omega$ and $Q$ be a point on segment $BC$ . Find the minimum possible length of the segment $PQ$ .

Solution

Solution:

The minimum possible length is $\frac{3\sqrt{3} - 3}{2}$ .

Let $P$ , $Q$ be the points which minimize the distance. We see that we want both to lie on the altitude from $A$ to $BC$ . Hence, $Q$ is the foot of the altitude from $A$ to $BC$ and $AQ = \frac{3\sqrt{3}}{2}$ .

Let $O$ , which must also lie on this line, be the center of $\omega$ , and let $D$ be the point of tangency between $\omega$ and $AC$ . Then, since $OD = \frac{1}{2}$ , we have $AO = 2OD = 1$ because $\angle OAD = 30^{\circ}$ , and $OP = \frac{1}{2}$ .

Consequently,

$PQ = AQ - AO - OP = \frac{3\sqrt{3} - 3}{2}$

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