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Wednesday, July 2, 2014

Centroids and area

There is a really surprising fact about centroids, one I had not known about until this year. The area of a triangle is equal to 1.5 times the distance of centroid to a side times the length of the side. It is not too hard to prove this result.



As I showed before, the area of a triangle is 0.5 (a*d - b*c) where u = (a, b) and v = (c, d) are two vectors that are the triangle's sides. To prove the really surprising fact about the centroid, all that must be shown is that 1.5·||gv||·||v||=0.5·(ad-bc). The vector v is the side, so we need its magnitude; the perpendicular vector is adjusted by the scale factor g so that it is the proper distance from the side v to the centroid.

Finding g takes a little more work. Again as shown before, the vector w to the centroid is (u+v)/3, that is (a+c, b+d)/3. Using the Pythagorean theorem, it is true that ||w||2=||gv||2+||fv||2. One also knows from vector addition that fv=w+gv. This allows a substitution for fv: fv=<a+c,b+d>/3+g<d, -c>. Therefore, substituting into the Pythagorean theorem yields:

||w||2=||gv||2+||fv||2(a+c)29+(b+d)29=g2·(c2+d2)+||<a+c,b+d>3+g<d, -c>||20=2g2·(c2+d2)+2g(a+c)·d3+-2g(b+d)c3

This sets up showing that g = 0 (the trivial solution) or that:

g=(a+c)·d-(b+d)c3(c2+d2)=ad-bc3(c2+d2)

Therefore, our proof is nearly done. Going back to the original statement to prove,

1.5·||gv||·||v||=0.5·(ad-bc)1.5·g·(c2+d2)=0.5·(ad-bc)1.5·ad-bc3(c2+d2)·(c2+d2)=0.5·(ad-bc)

And our work is done.

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