I've been thinking about this ever since high school & never got a particularly good answer. Here's some of my thinking, though. It took me a while to figure out how & why the perimeter got into the equation. If you "unroll" the triangle by breaking it into three triangles, where the bases of the triangles are the original 3 sides, and the altitudes of the three triangles are all equal to the radius of the incenter, then the area is the sum of the 3 smaller triangles, and yet this area is _independent_ of everything but the perimeter and the radius of the incenter: A = A_1 + A_2 + A_3 = p*r, where r = radius of the incircle The next question is how to find the incenter. If we label the triangle in the usual way (a is the side opposite to angle/vertex A, etc.), and if we choose A as the origin, and if we let b,c be _vectors_ (not just lengths), then incenter = (|c|b+|b|c)/(2s), [s = (|a|+|b|+|c|)/2] = |b||c|/(2s) * (b/|b| + c/|c|) (Sorry that this isn't symmetrical.) --- Coxeter (Intro to Geometry, 2nd Ed. 1961, 1969, Wiley) has a cool theorem: (Area)^2 = r * r_a * r_b * r_c, where r_a, r_b, r_c are the "exradii" (radii of the excircles tangent to the three sides). Also (Coxeter) r = r_a + r_b + r_c - 4*R, where R = circumradius All of these appear to be intimately connected to the Descartes circle theorem, recently [Amer. Math. Monthly 109 (2002), 338-361.] extended to the complex numbers. At 05:38 PM 4/25/2006, Gareth McCaughan wrote:
As all munsters know, area^2 = s(s-a)(s-b)(s-c) = xyz(x+y+z). Is there a really good proof of this? The best one I've been able to cook up[1] says: let t,u,v = tan A/2 etc.; then xt=r (the inradius) etc., so t:u:v = 1/x:1/y:1/z, and (tangent-sum formula, angle-sum of triangle) uv+vt+tu=1, so t = 1/x / sqrt(1/yz+1/zx+1/xy), so r^2 = 1/(1/yz+1/zx+1/xy) = xyz/(x+y+z) and as rs=area we're done. But this feels a bit too complicated. Is there perhaps a neat proof in 4 or 6 dimensions that looks at the cartesian product of the triangle (or some tetrahedron with the triangle as base) with itself?
[1] Presumably it's been known for centuries.
-- g