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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
2 years, 7 months ago

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is XY 26???? Nitish Deshpande · 2 years, 7 months ago

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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

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

\( \Rightarrow \sqrt{4a^{2}+4b^{2}}=XY=\sqrt{4× 169}=\boxed{26}\) Aneesh Kundu · 2 years, 7 months ago

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Length of XY is 26 Moumik Maitra · 2 years, 5 months ago

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XY = 26 KahYeen Lai · 2 years, 5 months ago

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XY=26 KahYeen Lai · 2 years, 5 months ago

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26 Unity 002 · 2 years, 5 months ago

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26 Sudharshan Sharma · 2 years, 6 months ago

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