2006 Romanian NMO Problems/Grade 7/Problem 3


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) $CH=DE$.


a) Note that quadrilateral $ABA_1B_1$ is cyclic, because $\angle{AB_1B}=\angle{AA_1B}=90^{\circ}$. Thus $\angle{B_1A_1B}=180-\angle{B_1AB}$ and $\angle{B_1A_1C}=\angle{B_1AB}$. Similarly $\angle{A_1B_1C}=\angle{ABC}$. Therefore $\triangle{A_1B_1C} \sim \triangle{ABC}$ and $\dfrac{A_1B_1}{AB}=\dfrac{B_1C}{BC}$. However $\triangle{BB_1C}$ is a $45-45-90$ triangle, so $\dfrac{B_1C}{BC}=\dfrac{1}{\sqrt{2}}$ and $AB=A_1B_1\sqrt{2}$. By Pythagorean theorem, $AB=\sqrt{A_1B^2+AA_1^2}$. However $AA_1=A_1C$, so $AB=\sqrt{A_1B^2+A_1C^2}$, and thus $A_1B_1\sqrt{2}=\sqrt{A_1B^2+A_1C^2}$, or $A_1B_1=\sqrt{\dfrac{A_1B^2+A_1C^2}{2}}$.

b) $DE=\sqrt{A_1E^2+A_1D^2}$, $A_1E=A_1D=A_1B_1$, $DE=\sqrt{2A_1B_1^2}=A_1B_1\sqrt{2}=\sqrt{A_1B^2+A_1C^2}=\sqrt{A_1H^2+A_1C^2}=CH$.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See also