2019 Putnam Problem A2

On Saturday, December 7, 2019, I took the 2019 Putnam competition. The Putnam competition is the premier undergraduate math competition in the United States. It has two sessions of six problems each with 3 hours allotted for each session. The problems are all proof-based and are graded on a 0-10 scale, with 10 points being given only for complete proofs. This exam is infamous for having a median score of 0-1 points out of a total of 120. This was my first time taking the Putnam, so my goal was simply to get a positive score.

According to AoPS, discussion is now allowed on the 2019 Putnam.

Putnam Problem A2 concerns finding angles in a triangle so that the line between the incenter and the centroid is parallel to one of the sides. The question reads:

2019 Putnam Problem A2:
In the 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.

My solution for this problem was by far the least elegant of all of the solutions I wrote on the day of the competition. Here’s my solution:

Solution #1: Coordinate Bash Approach

WLOG, let AB = 1, let B be at (0,0), and let A be at (1,0). By the tangent addition formula, tan(beta) = (1/3+1/3)/(1-(1/3)*(1/3)) = 3/4. It follows that C has coordinates (4c,3c) for some real number c, and therefore G has coordinates (1+4c/3,c). Similarly, I has coordinates (3i,i) for some real number i. It is also known that I lies on the angle bisector of angle BAC. Thus, 2*tan^-1(i/(1-3i)) = tan^-1(3c/(1-4c)). Using the tangent addition formula, 3c/(1-4c) = (2i/(1-3i))/(1-(i/(1-3i))^2) = (2i(1-3i))/(1-6i+8i^2). In order for IG to be parallel to AB, we must have i = c. Setting i = c in this equation and solving, we get c = 1/4. It follows that C has coordinates (1,3/4), and therefore alpha = BAC = 90 degrees.

This was the initial solution I wrote during the competition. Luckily, while checking I noticed that c = 1/4 makes many of the expressions in the equation undefined. As a result, I added the note that this solution only proves that all values not equal to c = 1/4 result in bad triangles, and I showed that when c = 1/4, the property is satisfied.

Overall, this problem was relatively straightforward for a 2nd Problem on the Putnam, and it was actually the first problem I solved on the day of the competition. I really hope my coordinate bash is accepted, I am a bit nervous that my argument about c = 1/4 being the only solution will not be accepted.

Thanks for reading! See you tomorrow.