# 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
6 years, 1 month ago

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is XY 26????

- 6 years, 1 month ago

Yes .XY=26 is the correct answer.

- 2 years, 3 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

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

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

- 6 years, 1 month ago

26

- 5 years, 11 months ago

26

- 5 years, 11 months ago

XY=26

- 5 years, 11 months ago

XY = 26

- 5 years, 11 months ago

Length of XY is 26

- 5 years, 11 months ago