Problem K. 695. (September 2021)

K. 695. A point \(\displaystyle P\) is selected on side \(\displaystyle BC\) of a square sheet of paper \(\displaystyle ABCD\). The sheet is folded along the line \(\displaystyle AP\) so that point \(\displaystyle B\) should lie equidistant from vertices \(\displaystyle C\) and \(\displaystyle D\). The new position of point \(\displaystyle B\) is denoted by \(\displaystyle B'\). Determine the measure of angle \(\displaystyle CB'D\).

(5 pont)

Deadline expired on October 11, 2021.

Sorry, the solution is available only in Hungarian. Google translation

Megoldás. A hajtás miatt az \(\displaystyle ABP\) és \(\displaystyle AB'P\) háromszögek egybevágók, így \(\displaystyle AB' = AB\). Mivel \(\displaystyle B'C = B'D\), így \(\displaystyle B'\) rajta van a \(\displaystyle CD\) oldal felezőmerőlegesén, ami egybeesik az \(\displaystyle AB\) oldal felezőmerőlegesével, így \(\displaystyle AB' = BB'\), tehát az \(\displaystyle ABB'\) háromszög egyenlő oldalú, minden szöge \(\displaystyle 60^\circ\)-os. \(\displaystyle B'AD\sphericalangle = BAD\sphericalangle-BAB'\sphericalangle=90^{\circ}-60^{\circ}=30^\circ\), a \(\displaystyle B'AD\) háromszögben (\(\displaystyle AB =)AB' = AD\), így \(\displaystyle ADB'\sphericalangle = AB'D\sphericalangle =\frac{180^{\circ}-B'AD\sphericalangle}{2}=75^\circ\). Hasonlóan \(\displaystyle BB'C\sphericalangle = 75^\circ\), így \(\displaystyle CB'D \sphericalangle= 360^{\circ}-AB'D\sphericalangle-AB'B\sphericalangle-BB'C\sphericalangle=360^\circ- 75^\circ- 60^\circ- 75^\circ= 150^\circ\).

Statistics:

97 students sent a solution.
5 points:	60 students.
4 points:	7 students.
3 points:	2 students.
2 points:	14 students.
1 point:	7 students.
0 point:	2 students.
Unfair, not evaluated:	1 solutions.
Not shown because of missing birth date or parental permission:	4 solutions.

Problems in Mathematics of KöMaL, September 2021