Anti-Pythagoras

Relevant wiki: Triangle Inequality

The perimeter of any square, where the side length is $s$ , is always $4s$ . Therefore, we can restate the problem using an inequality:

If $a, b, \text{and}~c$ are sides of a triangle and $c$ is the longest of the sides, then for what cases is $4a+4b>4c$ ?

If we divide both sides of the inequality by 4, then we get $a+b>c$ . Most (if not all) geometry students should know this inequality as the Triangle Inequality Theorem, which states that the sum of any two sides of any triangle is always greater than the third side.

Since $a+b>c$ is always true, then $4a+4b>4c$ is $\boxed{\text{always true}}$ .

The sum of the perimeters of squares $a$ and $b$ would be $P_{ab} = 4a + 4b$ . From the law of cosines, $c = \sqrt{a^2 + b^2 - 2ab \cos C}$ , so the perimeter of square $c$ would be $P_c = 4\sqrt{a^2 + b^2 - 2ab \cos C}$ . We were given that c is the longest side, so $60° < C < 180°$ , which means $\frac{1}{2} < \cos C < -1$ , so $P_c < 4\sqrt{a^2 + b^2 - 2ab(-1)}$ which simplifies to $P_c < 4a + 4b$ . Therefore, $P_{ab} = 4a + 4b > P_c$ , so the sum of the perimeters of squares $a$ and $b$ is always greater than the perimeter of square $c$ .