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.

triangulo

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

9 Comments so far

  1. Droper

    September 7th, 2006

    cool!, y remember my classes in pitagoras :D


  2. cachete

    September 7th, 2006

    demasiada geekness…


  3. E666

    September 8th, 2006

    what about triangle BCM perimeter’s?


  4. Jj

    September 8th, 2006

    What about BCM? Assumming you choose B or C instead of A, you will have M on AC or AB. Follow the same instructions and it should work.
    Or just rotate the vertex labels :P


  5. E666

    September 8th, 2006

    “…find a point M such that triangles ABM, ACM, and BCM have equal perimeters?…”

    what about triangle BCM perimeter’s!


  6. Orgen

    September 10th, 2006

    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.


  7. [...] It is always the same reason I fail in exams, I never read the whole thing, on my last post about Google’s problem I TOTALLY skipped the third triangle BCM, which my friend Enrique Llanos pointed out [...]


  8. Jj

    September 12th, 2006


  9. [...] It is always the same reason I fail in exams, I never read the whole thing, on my last post about Google’s problem I TOTALLY skipped the third triangle BCM, which my friend Enrique Llanos pointed out [...]


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