Problem K. 693. (March 2021)

K. 693. Quadrilateral \(\displaystyle ABCD\) has an inscribed circle centred at \(\displaystyle O\). Show that the sum of \(\displaystyle \angle DOC\) and \(\displaystyle \angle BOA\) is \(\displaystyle 180^{\circ}\).

(6 pont)

Deadline expired on April 12, 2021.

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

Megoldás. Ha a középpontból rendre az \(\displaystyle OT\), \(\displaystyle OR\), \(\displaystyle OS\) és \(\displaystyle OQ\) merőlegeseket állítjuk az \(\displaystyle AB\), \(\displaystyle BC\), \(\displaystyle CD\) és \(\displaystyle DA\) oldalakra, akkor a szimmetria miatt az alábbi szögek egyenlők lesznek: \(\displaystyle TOB \sphericalangle= BOR \sphericalangle\), \(\displaystyle ROC \sphericalangle = COS \sphericalangle\), \(\displaystyle SOD \sphericalangle = DOQ \sphericalangle\), \(\displaystyle QOA \sphericalangle = AOT \sphericalangle\). Ezen nyolc szög összege \(\displaystyle 360^{\circ}\). Mivel \(\displaystyle DOC \sphericalangle=DOS\sphericalangle+COS\sphericalangle\) és \(\displaystyle BOA \sphericalangle= TOB \sphericalangle + AOT\sphericalangle\), ezért \(\displaystyle DOC\sphericalangle+BOA\sphericalangle=DOS\sphericalangle+COS\sphericalangle+TOB \sphericalangle + AOT\sphericalangle=\frac{360^{\circ}}{2}=180^{\circ}\), mert minden szögpárból az egyik szerepel az összegben.

Statistics:

78 students sent a solution.
6 points:	62 students.
4 points:	1 student.
2 points:	1 student.
1 point:	2 students.
0 point:	10 students.
Unfair, not evaluated:	1 solutions.
Not shown because of missing birth date or parental permission:	1 solutions.

Problems in Mathematics of KöMaL, March 2021