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Question Number 201150 by Mingma last updated on 30/Nov/23

	Answered by mr W last updated on 02/Dec/23

	$a = s i d e l e n g t h o f s q u a r e$ ${(\frac{a^{2} + x^{2} - 15^{2}}{2 a x})}^{2} + {(\frac{a^{2} + x^{2} - 20^{2}}{2 a x})}^{2} = 1$ $a^{4} + x^{4} - (15^{2} + 20^{2}) (a^{2} + x^{2}) + \frac{15^{4} + 20^{4}}{2} = 0$ $x^{4} - (15^{2} + 20^{2}) x^{2} + a^{4} - (15^{2} + 20^{2}) a^{2} + \frac{15^{4} + 20^{4}}{2} = 0$ $\frac{{(15^{2} + 20^{2})}^{2}}{4} ⩾ x^{4} - (15^{2} + 20^{2}) x^{2} + \frac{15^{4} + 20^{4}}{2}$ $x^{4} - (15^{2} + 20^{2}) x^{2} + \frac{{(20^{2} - 15^{2})}^{2}}{4} ⩽ 0$ $\frac{{(20 - 15)}^{2}}{2} ⩽ x^{2} ⩽ \frac{{(20 + 15)}^{2}}{2}$ $\Rightarrow \frac{20 - 15}{\sqrt{2}} ⩽ x ⩽ \frac{20 + 15}{\sqrt{2}}$ $x^{2} + y^{2} = 15^{2} + 20^{2} = 625$ $f o r x, y \in N t h e r e a r e 2 p o s s i b i l i t i e s :$ $(x, y) = (15, 20) o r (7, 24)$ $a^{2} = \frac{1}{2} [625 \pm \sqrt{4 x^{2} (625 - x^{2}) - 30625}]$ $\Rightarrow a^{2} = \frac{625 \pm 25 \sqrt{527}}{2} o r \frac{625 \pm 7 \sqrt{1679}}{2}$ $i f w e o n l y c o n s i d e r t h e c a s e s t h a t$ $t h e p o i n t l i e s i n s i d e t h e s q u a r e, t h e n$ $t h e r e a r e t w o s o l u t i o n s :$ $(x, y) = (15, 20) a n d t h e a r e a o f t h e$ $s q u a r e i s \frac{625 + 25 \sqrt{527}}{2} o r$ $(x, y) = (7, 24) a n d t h e a r e a o f t h e$ $s q u a r e i s \frac{625 + 7 \sqrt{1679}}{2} .$
	Commented by mr W last updated on 02/Dec/23

	Commented by mr W last updated on 02/Dec/23

	Commented by Mingma last updated on 05/Dec/23
	very elegant solution!
	Answered by ajfour last updated on 02/Dec/23

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