[qimg]http://img703.imageshack.us/img703/1343/drawing2l.png[/qimg]
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If h*X = (pi*r2) then h = (pi*r2)/X
We deal with an isosceles triangle and a given circle if:
(r > 0 and r < ∞) and (X > r and X < ∞) and (h > 2r and h < ∞)
Well, at least you didn't get the constraints wrong, but you doronized them to the image and likeness of OM. The inscribed circle is a unite circle and that means its radius is equal to one. Sometimes I bemuse myself by inventing geometric arrangements where only one parameter is known, and so in this case it is the radius of the inscribed circle.
That X is a half of the base line
b of the triangle, coz the triangle is isosceles. So the constrains are
r = 1, b/2 > r, h > 2r
The expression "h < ∞" is a redundant addition to the works and its only purpose is to support the presence of the boolean operators as an opulent garnishment.
Since the area of a unit circle is equal to pi, your condition in the first line says
If h*X = pi then h = pi/X (where X = base/2)
which is formally correct, but useless to the purpose of finding the roots.
I think that there are only two theorems that are used to solve the problem altogether with a few basic laws of trigonometry. When all is done, the length of the altitude h which satisfies the given condition is given by the solution of this equation:
x
3 - 4*x*pi
2 + 8*pi
2 = 0
You can see below that there exist two real roots.
That means, the solution cannot be expressed in the exact format; it exists only as f(x) = 0 and that's the only way to draw the points onto the real line. The roots x = h are in the approximate format as
h
1 = 2.313757093139...
h
2 = 4.7982296547833...
So if you ever figure the geometry part, you can use the result to check if you got that right.
The next time OM gets tested through the "spaghetti triangles" -- the finite and the infinite cases.