Difference between revisions of "2006 Romanian NMO Problems/Grade 7/Problem 3"

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b) <math>CH=DE</math>.
 
b) <math>CH=DE</math>.
 
==Solution==
 
==Solution==
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{{solution}}
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==See also==
 
==See also==
 
*[[2006 Romanian NMO Problems]]
 
*[[2006 Romanian NMO Problems]]
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]

Revision as of 07:33, 27 August 2008

Problem

In the acute-angle triangle $ABC$ we have $\angle ACB = 45^\circ$. The points $A_1$ and $B_1$ are the feet of the altitudes from $A$ and $B$, and $H$ is the orthocenter of the triangle. We consider the points $D$ and $E$ on the segments $AA_1$ and $BC$ such that $A_1D = A_1E = A_1B_1$. Prove that

a) $A_1B_1 = \sqrt{ \frac{A_1B^2+A_1C^2}{2} }$;

b) $CH=DE$.

Solution

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See also