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 = √(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 = √(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.

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