## 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(c2-12) (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×c2) (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, a2+b2=c2, so a2+b2=57600.

Suppose that a+b=529. Then b=529-a, and when we fill that in a2+b2=57600, we get a2+(529-a)2=57600, so a2-289×a+12960.5=0. This equation has no solutions if a must be integer.

Suppose that a+b=576. Then b=576-a, and when we fill that in a2+b2=57600, we get a2+(576-a)2=57600, so a2-336×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. Back to the puzzle
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