A geometry problem by abhishek anand

Let $OX = a, OY = b$ such that:

$XN^{2} = OX^2 + ON^2 \Rightarrow 19^2 = a^2 + (\frac{b}{2})^2$ ; (i)

$YM^{2} = OY^2 + OM^2 \Rightarrow 22^2 = b^2 + (\frac{a}{2})^2$ . (ii)

If we multiply both (i) and (ii) through by $4$ and add together, then we arrive at:

$4(19^2+22^2) = 5(a^2+b^2) \Rightarrow \frac{4(845)}{5} = a^2 + b^2 \Rightarrow 4 \cdot 169 = 26^2 = a^2 + b^2 \Rightarrow XY^2 = OX^2 + OY^2$ (iii).

Hence per (iii), $\boxed{XY = 26}.$

We have ox=19, om=22 since m,n are the midpoints of legs ox and oy Easy to calculate, nm approximately = 13 (Pythagorean's theorem) => xy=26