title: Google's triangle problem
link: http://jj.isgeek.net/2006/09/googles-triangle-problem/
author: Jj
description: 
post_id: 195
date: 2006/09/06 22:47:07
created_gmt: 2006/09/07 03:47:07
comment_status: open
slug: googles-triangle-problem

I was reading through [Google's aptitude tests](http://googlesystem.blogspot.com/2005/12/google-labs-aptitude-test.html), and among the questions, this one stuck on my head... Later as I walked to the bus station came up with the answer.

![triangulo](https://bm.isgeek.net/4b/dd/07ed3fb6a49d95fb64bb457b52fccd650324d83d7eddbcfd87d73f06de09.png)

> 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](http://mathworld.wolfram.com/HeronsFormula.html) :P.

![Google's triangle](https://bm.isgeek.net/9c/c0/27890d5c1b502cfd482fc554fb35e1573a9c1ac87a4185a1b632def133cc.jpg)

## Comments

**[Droper](#2030 "2006-09-07 01:05:25"):** cool!, y remember my classes in pitagoras :D

**[cachete](#2035 "2006-09-07 21:22:52"):** demasiada geekness...

**[E666](#2037 "2006-09-08 12:06:59"):** what about triangle BCM perimeter's?

**[Jj](#2040 "2006-09-08 15:21:53"):** Or just rotate the vertex labels :P

**[E666](#2042 "2006-09-08 18:50:01"):** what about triangle BCM perimeterâ€™s!

**[Orgen](#2104 "2006-09-10 23:38:12"):** 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.

**[Jj](#2127 "2006-09-12 00:48:53"):** The correct answer can be found at http://jj.isgeek.net/2006/09/12/google%e2%80%99s-triangle-problem-correction/

