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Equilateral Triangle - Centroid/Incenter


Date: 09/27/2001 at 15:55:27
From: Casey Kachur
Subject: Equilateral triangles

Prove that a triangle is equilateral <--> its centroid equals its 
incenter.

Note: Requires an if and only if proof.

Hope you can help.


Date: 09/28/2001 at 05:10:14
From: Doctor Floor
Subject: Re: Equilateral triangles

Hi Casey,

Thanks for your question.

Let us start with a scalene triangle ABC with incenter I, radius of 
inscribed circle r, points of contact Ia, Ib and Ic, and sidelengths 
a = |BC|, b = |AC| and c = |AB| :


           C
          / \
         /   \
        /     \
      Ib       Ia
      /         \
     /     I     \
    /             \
   A-------Ic------B


First, we know that area(ABC) = area(AIB) + area(BIC) + area(CIA).

We can calculate area(AIB) = 0.5 * AB * IIc = 0.5cr. And in the same 
way we find area(BIC) = 0.5ar and area(CIA) = 0.5br.

That gives us that the ratio

  area(BIC) : area(CIA) : area(AIB) = a : b : c   [*]
  
Now we do the same for the centroid G in a scalene triangle, with 
midpoints Ma, Mb and Mc:

           C
          / \
         /   \
        /     \
      Mb       Ma
      /         \
     /     G     \
    /             \
   A-------Mc------B

We see that area(AMcC) = area (BMcC) because these triangles have the 
same altitude from C and have congruent bases AMc and McB. In the same 
way we see that area(AMcG) = area(BMcG). But then we can conclude that

  area(AMcC) - area (AmcG) = area(BMcC) - area (BMcG)

and thus

  area(AGC) = area(BGC).

In the same way we can find that area(AGC)=area(AGB).

So we have the ratio

 area(BGC) : area(AGC) : area(AGB) = 1 : 1 : 1  [**]

If the incenter and centroid coincide, then the two ratios [*] and 
[**] must be equal, so a:b:c = 1:1:1, which means that the sidelengths 
must be equal, and thus that the triangle must be equilateral.

The converse is also true, because there is only one point with a 
given triple of area ratios. So in an equilateral triangle the 
incenter gives ratio 1:1:1, and thus coincides with the centroid.

If you need more help, or have any follow-up questions, just write 
back.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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