new years triangle

[Jim Buddenhagen]

Find a triangle with integer sides and such that the product of the
tangents of its half-angles is

8 2 2
(2) (5) (11)
--------------------
2 2 2 2
(3) (7) (13) (17) .

--Jim Buddenhagen (jb1...@daditz.sbc.com)
--
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[David Savitt]

Jim.Bud...@launchpad.unc.edu (Jim Buddenhagen) writes:

>Find a triangle with integer sides and such that the product of the
>tangents of its half-angles is

> 8 2 2
> (2) (5) (11)
> --------------------
> 2 2 2 2
> (3) (7) (13) (17) .

I haven't found a solution to the problem, but I've worked out what I think is
a neat relation regarding the desired product.

We know tan (.5x) = sqrt ((1-cos x)/(1+cos x)), and so if we find the prod.
of the squares of the tangents of the half-angles, we get

P^2 = ((1-cos x)*(1-cos y)*(1-cos z))/((1+cos x)*(1+cos y)*(1+cos z)) where
x,y,z are the angles of the triangle. If the triangle has sides of length
a, b, and c, then by the law of cosines,

cos x = (a^2+b^2-c^2)/(2ab), etc., and so

P^2=((2ab-a^2-b^2+c^2)*(2ac-a^2-c^2+b^2)*(2bc-b^2-c^2+a^2)) /
((2ab+a^2+b^2-c^2)*(2ac+a^2+c^2-b^2)*(2bc+b^2+c^2-a^2))
=((c^2-(a-b)^2)*(b^2-(a-c)^2)*(a^2-(b-c)^2))/
(((a+b)^2-c^2)*((a+c)^2-b^2)*((b+c)^2-a^2))
=((c+a-b)*(c-a+b)*(b+a-c)*(b-a+c)*(a+b-c)*(a-b+c)) /
((a+b+c)*(a+b-c)*(a+b+c)*(a+c-b)*(a+b+c)*(b+c-a))
=((a+b-c)*(a+c-b)*(b+c-a)) / (a+b+c)^3
=((a+b+c)*(a+b-c)*(a+c-b)*(b+c-a)) / (a+b+c)^4
=(2s)*(2s-2c)*(2s-2b)*(2s-2a) / (a+b+c)^4 where s=(a+b+c)/2
=16*(s*(s-a)*(s-b)*(s-c)) / (a+b+c)^4 .

But Heron's formula tells us that Area=Sqrt(s*(s-a)*(s-b)*(s-c)), and we know
that a+b+c = Perimeter. and thus

P^2 = 16 * Area^2 / Perimeter^4 => P = 4 * Area / Perimiter^2

This might be useful in solving the original problem.

Incidentally, one can see that this expression for P gives P independent of
triangle size, because multiplying the perim. by k multiplies the area by k^2.
So fixing the perimeter at 1, we know that area is maximized with an
equilateral triangle (with sides 1/3) whose area is (1/4)*(3^(-3/2)), and thus
the maximum for P is 3^-1.5 ~ .192. The greatest lower bound for P is, of
course, 0.

-Dave

--
------------------------------------------------------------------------------
Dave Savitt | 2nd year math/physics | AKA Little Dave | Go Canucks!
dsa...@unixg.ubc.ca | University of B.C. | AKA Goliath | Go Mounties?
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