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May 14th, 2015, 09:35 AM  #1 
Senior Member Joined: Jan 2015 From: USA Posts: 107 Thanks: 2  Ratio of Areas inside a very interesting triangle
Does somebody know how to solve this problem? I don't even know how to start... Last edited by skipjack; May 14th, 2015 at 06:10 PM. 
May 14th, 2015, 10:50 AM  #2 
Senior Member Joined: Mar 2011 From: Chicago, IL Posts: 214 Thanks: 77 
I am strongly recommending to make the picture, it will help you to understand. (1) The ∆ABH and ∆AHM have the same base AH. The height of the ∆ABH BP is twice greater then the height of the ∆ABH that we draw from the vertex M because of AM=MC. So, the ratio of the areas is 2:1. (2) BP:PC=1 because the ∆ABC is isosceles. 
May 14th, 2015, 06:08 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,617 Thanks: 2072 
A diagram would help, as you seem to be using an incorrect diagram. I got 4:1 and 2:1.

May 14th, 2015, 07:25 PM  #4 
Senior Member Joined: Apr 2014 From: Europa Posts: 584 Thanks: 177  We have: $\displaystyle \Delta ABM,\ m(\hat{A})=90^0,\ AH \perp BM\ (H \in BM),\ AB = 2 AM .$ We'll see soon enough the triangles ABH and MAH. These two triangles are similar, with ratio $\displaystyle \dfrac{2}{1}.$ What can we say about their areas ? (The answer is simple, if we just draw a geometrical figure) Last edited by skipjack; May 15th, 2015 at 04:00 AM. Reason: to correct typo 
May 15th, 2015, 04:01 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,617 Thanks: 2072 
There's a standard theorem that gives the ratio of the areas.


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