Mathematical and Physical Journal
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Problem B. 4818. (October 2016)

B. 4818. The intersection of diagonals \(\displaystyle AC\) and \(\displaystyle BD\) of a cyclic quadrilateral \(\displaystyle ABCD\) is \(\displaystyle M\). The angle bisectors of the angles \(\displaystyle \angle CAD\) and \(\displaystyle \angle ACB\) intersect the circumscribed circle of the cyclic quadrilateral \(\displaystyle ABCD\) at points \(\displaystyle E\) and \(\displaystyle F\), respectively. Prove that line \(\displaystyle EF\) is perpendicular to the angle bisector of the angle \(\displaystyle AMD\).

Proposed by B. Bíró, Eger

(5 pont)

Deadline expired on November 10, 2016.


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

Megoldás. Jelölje \(\displaystyle K_1\) az \(\displaystyle ADM\), \(\displaystyle K_2\) pedig a \(\displaystyle BCM\) háromszög beírt körének középpontját; ekkor a \(\displaystyle K_1K_2\) egyenes éppen az \(\displaystyle AMD\) szög felezője; ennek az \(\displaystyle AD\) és \(\displaystyle BC\) oldalakkal való metszéspontja \(\displaystyle G\) illetve \(\displaystyle H\). Az ábrán \(\displaystyle 2\alpha\)-val illetve \(\displaystyle 2\gamma\)-val jelölt szögek a kerületi szögek tétele miatt egyenlők egymással. Továbbá \(\displaystyle EK_1K_2\measuredangle = GK_1A\measuredangle = DK_1A\measuredangle /2 = (180^{\circ}-\alpha -\gamma)/2=BK_2C\measuredangle /2 = BK_2H\measuredangle = K_1K_2E\measuredangle\). Így \(\displaystyle K_1EK_2\) egyenlő szárú háromszög: \(\displaystyle EK_1=EK_2\). Ugyanígy kapjuk, hogy \(\displaystyle FK_1=FK_2\), azaz \(\displaystyle FK_1EK_2\) deltoid, aminek \(\displaystyle EF\) és \(\displaystyle K_1K_2\) átlói egymásra merőlegesek.


Statistics:

122 students sent a solution.
5 points:74 students.
4 points:30 students.
3 points:7 students.
2 points:6 students.
1 point:1 student.
0 point:4 students.

Problems in Mathematics of KöMaL, October 2016