Wednesday, December 26, 2007

Solution to the World's Hardest Easy Geometry Problem

A few weeks ago, someone posted this problem to a project mailing list at Google. For a few days afterward, every office I walked into had a white-board covered with elaborate color-coded triangles. By my rough estimate, various proofs that x = 20 probably cost Google on the order of a quarter of a million dollars in lost productivity over the next few days. It would obviously be Evil to allow this scourge to spread, so here's my solution.

First, we need some auxiliary lines and points. Draw a line through D parallel to AB, and let F be the intersection with BC. Now draw line AF. (It's just a mirror image of BD.) Let G be the intersection between AF and BD. Finally, draw line CG. It's easy to see that the line CG bisects the angle ACB, so ACG and GCB are both 10 degrees.

It's helpful to keep the endgame in mind while working toward it. Just from the fact that the angles of a triangle add up to 180 degrees, We know that angle AEB is 30 degrees. If we knew angle DEF, we could easily compute x: DEF - AEB. So, how to find angle DEF? First we need to establish three more-or-less unrelated equations between segment lengths:
  1. DF = FG. Angles ABG and BAG are both 60 degrees, so ABG is an equilateral triangle. Since angle DGF = AGB, angle DGF is also 60 degrees. By symmetry, angles FDG and DFG are equal, so they must be 60 degrees, too, and DFG must be equilateral. So the sides are equal.
  2. CF = AF. Since angles FCA and FAC are both 20 degrees, ACF is isosceles, which means that CF = AF.
  3. CE = AG. This is the trickiest one. It hinges on the fact that triangles ACG and ACE have equal sides. That's a little surprising, but true: both have angles 10, 20, and 150, and they have a common side AC. So they're congruent and have a common side, so they're equal. That means in particular that CE = AG.
Now subtract equation 3 from equation 2:
  • CF - CE = AF - AG
  • EF = FG
The rest is easy. By equation 1, FG = DF, so EF = DF, which means that triangle DEF is isosceles with DE as the base. We know that angle DFE is 80 degrees, so both DEF and EDF are 50 degrees. Finally, x = DEF - AEF = 50 - 30 = 20 degrees.


Derek said...

O MY GOD! Thanks sooo much! I was at my wits end trying to solve this!

Now I can get back to work...

Anonymous said...

is it bad that I'm a second year maths student at uni and this problem, despite the proof, is still confusing me :S

Anonymous said...

Couldn't get your tricky statement...

asd said...

not a good proof. easier one is at

Anonymous said...

solved it in 3mins just after listening (i'm 14)