Jj's web stream

Google's triangle problem

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.

Google's triangle


Droper: cool!, y remember my classes in pitagoras :D

cachete: demasiada geekness...

E666: what about triangle BCM perimeter's?

Jj: Or just rotate the vertex labels :P

E666: what about triangle BCM perimeter’s!

Orgen: Cierto. El problema pide que los perímetros de los tres triángulos formados sean iguales y, en esta solución, el perímetro del triángulo BCM (2*BC) no es igual al de los triángulos ABM y ACM.

Jj: The correct answer can be found at http://jj.isgeek.net/2006/09/12/google%e2%80%99s-triangle-problem-correction/