I found this problem from the Math Challenges section of the 2002 *Pi in the Sky* Canadian math magazine for high school students to be truly astonishing.

“**Problem 4**. Inside of the square ABCD, take any point P. Prove that the perpendiculars from A on BP, from B on CP, from C on DP, and from D on AP are concurrent (i.e. they meet at one point).”

How could such a complicated arrangement produce such an amazing result? I didn’t know where to begin to try to prove it. My wandering path to discovery produced one of my most satisfying “aha!” moments.

See the Mysterious Doppelgänger Problem

**Update (12/27/2019)** I goofed. I had plotted the original figure incorrectly. (No figure was given in the *Pi in the Sky *statement of the problem.) Fortunately, the original solution idea still worked.