Solution to:
Angled Triangle
Let the sides of the triangle be a, b, and c, with c=240 being the hypotenuse.
The triangle has the minimal circumference when a=1 and b=sqrt(c^{2}1^{2}) (approximately 240.0).
The circumference in that case is approximately 480.0.
The triangle has the maximal circumference when a and b are equal: a=b=sqrt(^{1}/_{2}×c^{2})
(approximately 169.7).
The circumference in that case is approximately 579.4.
The only two squares of whole numbers that lie in the interval [480.0, 579.4] are 529 and 576.
Now we know that a+b=529 or a+b=576. In addition, a^{2}+b^{2}=c^{2},
so a^{2}+b^{2}=57600.
Suppose that a+b=529.
Then b=529a, and when we fill that in a^{2}+b^{2}=57600, we get a^{2}+(529a)^{2}=57600,
so a^{2}289×a+12960.5=0.
This equation has no solutions if a must be integer.
Suppose that a+b=576. Then b=576a, and when we fill that in a^{2}+b^{2}=57600,
we get a^{2}+(576a)^{2}=57600, so a^{2}336×a+27648=0.
This equation has solutions a=192 (b=144) and a=144 (b=192).
Therefore, the sides of the triangle are a=144, b=192, and c=240.
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