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Difference between revisions of "2006 Romanian NMO Problems/Grade 9/Problem 2"
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==Problem==
 
==Problem==
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Let <math>\displaystyle ABC</math> and <math>\displaystyle DBC</math> be isosceles triangle with the base <math>\displaystyle BC</math>. We know that <math>\displaystyle \angle ABD = \frac{\pi}{2}</math>. Let <math>\displaystyle M</math> be the midpoint of <math>\displaystyle BC</math>. The points <math>\displaystyle E,F,P</math> are chosen such that <math>\displaystyle E \in (AB)</math>, <math>\displaystyle P \in (MC)</math>, <math>\displaystyle C \in (AF)</math>, and <math>\displaystyle \angle BDE = \angle ADP = \angle CDF</math>. Prove that <math>\displaystyle P</math> is the midpoint of <math>\displaystyle EF</math> and <math>\displaystyle DP \perp EF</math>.
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Let <math>\displaystyle ABC</math> and <math>\displaystyle DBC</math> be [[isosceles triangle]]s with the base <math>\displaystyle BC</math>. We know that <math>\displaystyle \angle ABD = \frac{\pi}{2}</math>. Let <math>\displaystyle M</math> be the [[midpoint]] of <math>\displaystyle BC</math>. The points <math>\displaystyle E,F,P</math> are chosen such that <math>\displaystyle E \in (AB)</math>, <math>\displaystyle P \in (MC)</math>, <math>\displaystyle C \in (AF)</math>, and <math>\displaystyle \angle BDE = \angle ADP = \angle CDF</math>. Prove that <math>\displaystyle P</math> is the midpoint of <math>\displaystyle EF</math> and <math>\displaystyle DP \perp EF</math>.
 
==Solution==
 
==Solution==
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{{solution}}
 
==See also==
 
==See also==
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*[[2006 Romanian NMO Problems/Grade 9/Problem 1 | Previous problem]]
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*[[2006 Romanian NMO Problems/Grade 9/Problem 3 | Next problem]]
 
*[[2006 Romanian NMO Problems]]
 
*[[2006 Romanian NMO Problems]]
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]

Revision as of 15:01, 13 October 2006

Problem

Let $\displaystyle ABC$ and $\displaystyle DBC$ be isosceles triangles with the base $\displaystyle BC$. We know that $\displaystyle \angle ABD = \frac{\pi}{2}$. Let $\displaystyle M$ be the midpoint of $\displaystyle BC$. The points $\displaystyle E,F,P$ are chosen such that $\displaystyle E \in (AB)$, $\displaystyle P \in (MC)$, $\displaystyle C \in (AF)$, and $\displaystyle \angle BDE = \angle ADP = \angle CDF$. Prove that $\displaystyle P$ is the midpoint of $\displaystyle EF$ and $\displaystyle DP \perp EF$.

Solution

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See also

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