Published September 7, 2020｜2 minute read
Update 2020-09-08: This solution has been added to Kiran Kedlaya’s solution page.
I've found an elegant geometry-only proof of Putnam 2003 B5, which, to the best of my knowledge, hasn't yet been discovered.
We begin with a diagram of the problem:
Rotate the diagram counterclockwise about :
At this point, we see that the triangle has sidelengths , , and , so these sidelengths do indeed form a triangle as claimed. Next, rotate the original diagram counterclockwise about :
And finally, about :
At this point, we have a large equilateral triangle with sidelength , partitioned into twelve smaller triangles:
The triangles of type 2, 3, and 4 fit together neatly to form two equilateral triangles of sidelength . We therefore have the following decomposition:
In particular, the area of the triangle we’re looking for is entirely determined by the combined area of the three equilateral triangles with sidelengths , , and . This combined area is proportional to . Treating all points as vectors in , we obtain
which is entirely determined by the length , as desired.