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# 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
3 years ago

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is XY 26????

- 3 years 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

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

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

- 3 years ago

Length of XY is 26

- 2 years, 10 months ago

XY = 26

- 2 years, 10 months ago

XY=26

- 2 years, 10 months ago

26

- 2 years, 10 months ago

26

- 2 years, 10 months ago

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