Published September 7, 20202 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:

Initial diagram of the problem.

Rotate the diagram 60°60 \degree counterclockwise about AA:

Diagram after rotating about A.

At this point, we see that the triangle PCPPCP' has sidelengths aa, bb, and cc, so these sidelengths do indeed form a triangle as claimed. Next, rotate the original diagram 60°60\degree counterclockwise about CC:

Diagram after rotating about C.

And finally, about BB:

Diagram after rotating about B.

At this point, we have a large equilateral triangle with sidelength 22\ell, partitioned into twelve smaller triangles:[1]

  1. three triangles with sidelengths a,b,ca, b, c,
  2. two triangles with sidelengths a,b,a, b, \ell,
  3. two triangles with sidelengths a,c,a, c, \ell,
  4. two triangles with sidelengths b,c,b, c, \ell,
  5. and three equilateral triangles, with sidelengths a,b,ca, b, c respectively.

The triangles of type 2, 3, and 4 fit together neatly to form two equilateral triangles of sidelength \ell. We therefore have the following decomposition:

Decomposition of large triangle.

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 aa, bb, and cc. This combined area is proportional to a2+b2+c2a^2 + b^2 + c^2. Treating all points as vectors in R2\mathbb{R}^2, we obtain

a2+b2+c2=(AP)2+(BP)2+(CP)2=A2+B2+C2+3P22(A+B+C)P=(A2+B2+C23O2)+3(PO)2,\begin{aligned} &a^2 + b^2 + c^2 \\ &= (A-P)^2 + (B-P)^2 + (C - P)^2 \\ &= A^2 + B^2 + C^2 + 3P^2 - 2 \, (A + B + C) \cdot P \\ &= (A^2 + B^2 + C^2 - 3 O^2) + 3 \, (P - O)^2, \end{aligned}which is entirely determined by the length OPOP, as desired.


Footnotes

  1. [1]Strictly speaking, the decomposition works as shown if the original point lies within the triangle ABCABC. If not, then the proof still works using signed areas (but is a bit harder to follow).