[math-fun] Hero[n]'s formula
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
--- Gareth McCaughan <gareth.mccaughan@pobox.com> 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
Adjoin two congruent triangles along their c edge to obtain a parallogram of area 2A = a b sinC. (2A)^2 = a^2 b^2 (1 - (cosC)^2). Using the law of cosines, A^2 can be expressed as a polynomial P(a,b,c) of degree 4. P vanishes when c=a+b, etc. So P(a,b,c) = (-a+b+c)(a-b+c)(a+b-c)Q(a,b,c). Q is of degree 1, and since P is symmetric, so is Q. Then Q is a multiple of a+b+c. The fixed constant can be found by consdering e.g. an equilateral triangle. Gene __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com
Here is a heuristic, not a proof. But is very suggestive, and may have influenced people two thousand or more years ago. Note that one doesn't need to think about imaginary numbers --- merely that the factors vanish as the triangle degenerates. Leave out the square roots until the dimension has got too big and needs reducing. R. What is a triangle? a + b > c, etc. Might expect area to contain factors sqrt(b+c-a) sqrt(c+a-b) sqrt(b+c-a) To get dimension up to 2 we need another factor, one that is symmetric in a,b,c. sqrt(a+b+c) -- and we're home within a constant, which an easy case --- say 3,4,5 --- shows to be 1/4. Note also: Conway's `extraversion' gives equal weight to the 4 quantities s s-a s-b s-c On Wed, 26 Apr 2006, Eugene Salamin wrote:
--- Gareth McCaughan <gareth.mccaughan@pobox.com> 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
Adjoin two congruent triangles along their c edge to obtain a parallogram of area 2A = a b sinC. (2A)^2 = a^2 b^2 (1 - (cosC)^2). Using the law of cosines, A^2 can be expressed as a polynomial P(a,b,c) of degree 4. P vanishes when c=a+b, etc. So
P(a,b,c) = (-a+b+c)(a-b+c)(a+b-c)Q(a,b,c).
Q is of degree 1, and since P is symmetric, so is Q. Then Q is a multiple of a+b+c. The fixed constant can be found by consdering e.g. an equilateral triangle.
Gene
On Wednesday 26 April 2006 16:22, Eugene Salamin wrote:
Adjoin two congruent triangles along their c edge to obtain a parallogram of area 2A = a b sinC. (2A)^2 = a^2 b^2 (1 - (cosC)^2). Using the law of cosines, A^2 can be expressed as a polynomial P(a,b,c) of degree 4. P vanishes when c=a+b, etc. So
P(a,b,c) = (-a+b+c)(a-b+c)(a+b-c)Q(a,b,c).
Q is of degree 1, and since P is symmetric, so is Q. Then Q is a multiple of a+b+c. The fixed constant can be found by consdering e.g. an equilateral triangle.
Entertaining! -- g
I did a web search and found a dialogue between Peter Doyle & John Conway, where they discussed this particular problem. http://www.math.dartmouth.edu/~doyle/docs/heron/heron.txt 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
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
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.)
It's equal to (aA+bB+cC)/(a+b+c) where a,b,c are lengths and A,B,C are vertices. I expect there's a not-too-painful proof of Hero[n] that goes via barycentric coordinates... -- g
Relationship between the 3x3 magic square of squares problem and the Heron's formula for tetrahedrons: http://www.mathpages.com/home/kmath427.htm Christian.
participants (5)
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Christian Boyer -
Eugene Salamin -
Gareth McCaughan -
Henry Baker -
Richard Guy