I was reading through Google's aptitude tests, and among the questions, this one stuck on my head... Later as I walked to the bus station came up with the answer.
Given a triangle ABC, how would you use only a compass and straight edge to find a point M such that triangles ABM, ACM, and BCM have equal perimeters? (Assume that ABC is constructed so that a solution does exist.)
Once you realize that the line AM is the same for both, then the problem is only to divide ABC's permieter in 2 equal parts. This is how it works..

Choose either AB or AC, I chose AB, which is the longest in this case, it works the same way if you choose the shortest.

With your compass draw the arc with radius AB

Prolong AC until it reaches the arc with radius AB (point B'), this extra segment is how much it takes to make AC the same length as AB

Again, with your compass, draw an arc over BC with radius CB', This point is N, so now AC + CN = AB

Using your compass one last time, plot M at the midpoint of NB

With your straightedge, draw AM

Triangles ABM and ACM should have the same perimeter :)
Then I also figured that you could come with some interesting relationship if you had the patience to solve using Heron's formula :P.
Comments
#2030" title="20060907 01:05:25">Droper: cool!, y remember my classes in pitagoras :D
#2035" title="20060907 21:22:52">cachete: demasiada geekness...
#2037" title="20060908 12:06:59">E666: what about triangle BCM perimeter's?
#2040" title="20060908 15:21:53">Jj: Or just rotate the vertex labels :P
#2042" title="20060908 18:50:01">E666: what about triangle BCM perimeterâ€™s!
#2104" title="20060910 23:38:12">Orgen: Cierto. El problema pide que los perÃmetros de los tres triÃ¡ngulos formados sean iguales y, en esta soluciÃ³n, el perÃmetro del triÃ¡ngulo BCM (2*BC) no es igual al de los triÃ¡ngulos ABM y ACM.
#2127" title="20060912 00:48:53">Jj: The correct answer can be found at http://jj.isgeek.net/2006/09/12/google%e2%80%99striangleproblemcorrection/