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May 14th, 2015, 09:35 AM   #1
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Ratio of Areas inside a very interesting triangle

Does somebody know how to solve this problem? I don't even know how to start...
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Last edited by skipjack; May 14th, 2015 at 06:10 PM.
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May 14th, 2015, 10:50 AM   #2
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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.
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May 14th, 2015, 06:08 PM   #3
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A diagram would help, as you seem to be using an incorrect diagram. I got 4:1 and 2:1.
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May 14th, 2015, 07:25 PM   #4
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Quote:
Originally Posted by matisolla View Post
I don't even know how to start...

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)
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Last edited by skipjack; May 15th, 2015 at 04:00 AM. Reason: to correct typo
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May 15th, 2015, 04:01 AM   #5
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There's a standard theorem that gives the ratio of the areas.
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