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Solution to: Angled Triangle

Name the sides of the triangle a, b, and c, where c = 240 is the hypotenuse. For the sought values of a and b, the following must hold:

The triangle has the minimal perimeter when a = 1 and b = √(c² - 1²) = √(240² - 1²) (approximately 240). The perimeter in that case is approximately 480.

The triangle has the maximal perimeter when a and b are equal: a = b = √(½ × c²) = √(½ × 240²) (approximately 169.7). The perimeter in that case is approximately 579.4.

The only squares of integers that lie between 480 and 579.4 are 484, 529, and 576.

Now we know that a + b + c equals 484, 529, or 576. In addition, a² + b² = c², so a² + b² = 240² = 57600.

Suppose that a + b + c = 484. Then b = 244 - a, and if we substitute that into a² + b² = 57600, we get a² + (244 - a)² = 57600, so a² - 244 × a + 968 = 0. This equation has no solution if a must be an integer.

Suppose that a + b + c = 529. Then b = 289 - a, and if we substitute that into a² + b² = 57600, we get a² + (289 - a)² = 57600, so a² - 289 × a + 12960.5 = 0. This equation has no solution if a must be an integer.

Suppose that a + b + c = 576. Then b = 336 - a, and if we substitute that into a² + b² = 57600, we get a² + (336 - a)² = 57600, so a² - 336 × a + 27648 = 0. This equation has solutions a = 144 (b = 192) and a = 192 (b = 144).

For a = 144 and b = 192, the area of the triangle is ½ × 144 × 192 = 13824 = 24³. Therefore, the area is the cube of an integer.

Therefore, the sides of the triangle are a = 144, b = 192, and c = 240.


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