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