English Információ A lap Pontverseny Cikkek Hírek Fórum

Rendelje meg a KöMaL-t!

VersenyVizsga portál

Kísérletek.hu

Matematika oktatási portál

B. 4706. The sides of rectangle $\displaystyle ABCD$ are $\displaystyle AB= \frac{\sqrt{5}+1}2$ and $\displaystyle BC=1$. Let $\displaystyle E$ be the point in the interior of line segment $\displaystyle AB$ with $\displaystyle AE=1$. Show that $\displaystyle \angle ACE= 2\cdot \angle EDB$.

Suggested by Sz. Miklós, Herceghalom

(3 points)

Deadline expired on 11 May 2015.

Statistics on problem B. 4706.
 113 students sent a solution. 3 points: 86 students. 2 points: 8 students. 1 point: 17 students. 0 point: 2 students.

• Problems in Mathematics of KöMaL, April 2015

•  Támogatóink: Morgan Stanley