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 
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