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/
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