# Pre-RMO 2014/15

Let $XOY$ be a triangle with $\angle XOY = 90^\circ$. Let $M$ and $N$ be the midpoints of legs $OX$ and $OY$, respectively. Suppose that $XN = 19$ and $YM = 22$. What is $XY$?

This note is part of the set Pre-RMO 2014 Note by Pranshu Gaba
5 years, 1 month ago

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Length of XY is 26

- 4 years, 11 months ago

XY = 26

- 4 years, 11 months ago

XY=26

- 4 years, 11 months ago

26

- 4 years, 11 months ago

26

- 4 years, 11 months ago

Let $XM=MO=a$ and $ON=NY=b$

By applying PT, we get

$4a^{2}+b^{2}=19^{2} \rightarrow Eq.1$

$a^{2}+4b^{2}=22^{2} \rightarrow Eq.2$

Eq.1+Eq.2

$a^{2}+b^{2}=\frac{19^{2}+22^{2}}{5}=13^{2}$

$\Rightarrow \sqrt{4a^{2}+4b^{2}}=XY=\sqrt{4× 169}=\boxed{26}$

- 5 years, 1 month ago

is XY 26????

- 5 years, 1 month ago

Yes .XY=26 is the correct answer.

- 1 year, 3 months ago