HMMT February

Geometry Difficulty 5.1 AIME, harder Find the answer United States

Problem:

Regular hexagon $ABCDEF$ has side length $2$ . Circle $\omega$ lies inside the hexagon and is tangent to segments $\overline{AB}$ and $\overline{AF}$ . There exist two perpendicular lines tangent to $\omega$ that pass through $C$ and $E$ , respectively. Given that these two lines do not intersect on line $AD$ , compute the radius of $\omega$ .

Proposed by: Karthik Venkata Vedula

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:

Let $O$ be the center of $\omega$ , and let the two tangent lines intersect at $P$ . Note that $O$ lies on the external angle bisector of $\angle CPE$ because the tangents are symmetric about line $PO$ . Additionally, $O$ lies on the perpendicular bisector of $CE$ by symmetry. By Fact 5, $COPE$ is cyclic and $\angle COE = 90^{\circ}$ . To finish, observe that $\angle COD = 45^{\circ}$ . Dropping the altitude $CH$ down to $AD$ gives $OH = CH = \sqrt{3}$ . So, $AO = AH - OH = 3 - \sqrt{3}$ . The desired answer is then $\frac{\sqrt{3}}{2} \cdot AO = \left[\frac{3\sqrt{3} - 3}{2}\right]$ .

Solution 2: Another way to get $\angle COE = 90^{\circ}$ is as follows.
Let $\omega$ meet the tangents from $C$ and $E$ at $Q$ and $R$ , respectively. Observe $OC = OE$ (as $O$ lies on the perpendicular bisector of $CE$ ) and $OQ = OR$ , so $\triangle OCQ \cong \triangle OER$ . Then $\angle COE = \angle QOR = 90^{\circ}$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

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