Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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# Problem B. 4706. (April 2015)

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 pont)

Deadline expired on May 11, 2015.

### Statistics:

 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