Putnam

Geometry Difficulty 7.6 National Olympiad, round 2 Find the answer

In the triangle $\triangle ABC$ , let $G$ be the centroid, and let $I$ be the center of the inscribed circle. Let $\alpha$ and $\beta$ be the angles at the vertices $A$ and $B$ , respectively. Suppose that the segment $IG$ is parallel to $AB$ and that $\beta = 2 \tan^{-1} (1/3)$ . Find $\alpha$ .

Your answer

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

Solution

Let $M$ and $D$ denote the midpoint of $AB$ and the foot of the altitude from $C$ to $AB$ , respectively, and let $r$ be the inradius of $\bigtriangleup ABC$ . Since $C,G,M$ are collinear with $CM = 3GM$ , the distance from $C$ to line $AB$ is $3$ times the distance from $G$ to $AB$ , and the latter is $r$ since $IG \parallel AB$ ; hence the altitude $CD$ has length $3r$ . By the double angle formula for tangent, $\frac{CD}{DB} = \tan\beta = \frac{3}{4}$ , and so $DB = 4r$ . Let $E$ be the point where the incircle meets $AB$ ; then $EB = r/\tan(\frac{\beta}{2}) = 3r$ . It follows that $ED = r$ , whence the incircle is tangent to the altitude $CD$ . This implies that $D=A$ , $ABC$ is a right triangle, and $\alpha = \frac{\pi}{2}$ .

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