Problem A. 799. (April 2021)

A. 799. For a given quadrilateral \(\displaystyle A_1A_2B_1B_2\) point \(\displaystyle P\) is called phenomenal, if line segments \(\displaystyle A_1A_2\) and \(\displaystyle B_1B_2\) subtend the same angle at point \(\displaystyle P\) (i.e. triangles \(\displaystyle PA_1A_2\) and \(\displaystyle PB_1B_2\) which can be also also degenerate have equal inner angles at point \(\displaystyle P\) disregarding orientation).

Three non-collinear points, \(\displaystyle A_1\), \(\displaystyle A_2\) and \(\displaystyle B_1\) are given on the plane. Prove that it is possible to find a disc on the plane such that for every point \(\displaystyle B_2\) on the disc quadrilateral \(\displaystyle A_1A_2B_1B_2\) is convex for which it is possible to construct seven distinct phenomenal points only using a right ruler.

With a right ruler the following two steps are allowed:

\(\displaystyle i)\) given two points it is possible to draw the straight line connecting them;

\(\displaystyle ii)\) given a point and a straight line, it is possible to draw the straight line passing through the given point which is perpendicular to the given line.

Proposed by Á. Bán-Szabó, Budapest

(7 pont)

Deadline expired on May 10, 2021.

Statistics:

3 students sent a solution.
7 points:	Bán-Szabó Áron.
4 points:	1 student.
0 point:	1 student.

Problems in Mathematics of KöMaL, April 2021