Putnam 2019, Problem A-2

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 \(\alpha\).

Solution:
**Solution 1.** 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 $\triangle ABC$. 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 $EB = r/\tan(\beta/2) = 3r$. 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 $\alpha = \frac{\pi}{2}$.

**Remark.** One can obtain a similar solution by fixing a coordinate system with $B$ at the origin and $A$ on the positive $x$-axis. Since $\tan \frac{\beta}{2} = \frac{1}{3}$, we may assume without loss of generality that $I = (3,1)$. Then $C$ lies on the intersection of the line $y = 3$ (because $CD = 3r$ as above) with the line $y = \frac{3}{4}x$ (because $\tan \beta = \frac{3}{4}$ as above), forcing $C = (4,3)$ and so forth.

**Solution 2.** Let $a,b,c$ be the lengths of $BC,CA,AB$, respectively. Let $r$, $s$, and $K$ denote the inradius, semiperimeter, and area of $\triangle ABC$. By Heron's Formula,
$$r^2 s^2 = K^2 = s(s-a)(s-b)(s-c).$$

If $IG$ is parallel to $AB$, then
$$\frac{1}{2}rc = \text{area}(\triangle ABI) = \text{area}(\triangle ABG) = \frac{1}{3}K = \frac{1}{3}rs$$
and so $c = \frac{a+b}{2}$. Since $s = \frac{3(a+b)}{4}$ and $s-c = \frac{a+b}{4}$, we have $3r^2 = (s-a)(s-b)$. Let $E$ be the point at which the incircle meets $AB$; then $s-b = EB = r/\tan(\beta/2)$ and $s-a = EA = r/\tan(\alpha/2)$. It follows that $\tan(\alpha/2)\tan(\beta/2) = \frac{1}{3}$ and so $\tan(\alpha/2) = 1$. This implies that $\alpha = \frac{\pi}{2}$.

**Remark.** The equality $c = \frac{a+b}{2}$ can also be derived from the vector representations
$$G = \frac{A+B+C}{3}, \quad I = \frac{aA + bB + cC}{a+b+c}.$$

**Solution 3.** (by Catalin Zara) It is straightforward to check that a right triangle with $AC = 3,AB = 4,BC = 5$ works. For example, in a coordinate system with $A = (0,0),B = (4,0),C = (0,3)$, we have
$$G = \left(\frac{4}{3},1\right), \quad I = (1,1)$$
and for $D = (1,0)$,
$$\tan \frac{\beta}{2} = \frac{ID}{BD} = \frac{1}{3}.$$