Problem B. 5005. (January 2019)

B. 5005. The feet of the altitudes of an acute-angled triangle \(\displaystyle ABC\) drawn to the sides \(\displaystyle BC\), \(\displaystyle CA\), \(\displaystyle AB\) are \(\displaystyle D\), \(\displaystyle E\), \(\displaystyle F\), respectively. The orthocentre of triangle \(\displaystyle ABC\) is \(\displaystyle M\). Let \(\displaystyle k_1\) denote the circle of diameter \(\displaystyle AB\), and let \(\displaystyle k_2\) be the circumscribed circle of triangle \(\displaystyle DEM\). Let \(\displaystyle P\) be an interior point of the arc \(\displaystyle EM\) of \(\displaystyle k_2\) that does not contain point \(\displaystyle D\). Let line \(\displaystyle DP\) intersect circle \(\displaystyle k_1\) again at point \(\displaystyle Q\), and let the midpoint of line segment \(\displaystyle PQ\) be \(\displaystyle R\). Show that lines \(\displaystyle AQ\), \(\displaystyle MP\), \(\displaystyle FR\) are concurrent.

Proposed by B. Bíró, Eger

(6 pont)

Deadline expired on February 11, 2019.

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

Megoldás. Legyen \(\displaystyle X=MP\cap AQ\). Azt kell igazolnunk, hogy az \(\displaystyle X,F,R\) pontok egy egyenesen vannak, avagy (irányított, modulo \(\displaystyle 180^\circ\) szögekkel) \(\displaystyle MXR\measuredangle = MXF\measuredangle\).

Legyen \(\displaystyle ABC\measuredangle=\beta\). Az \(\displaystyle ABDQ\) húrnégyszögben \(\displaystyle XQP\measuredangle=AQD\measuredangle=ABD\measuredangle=\beta\), a \(\displaystyle CPMD\) húrnégyszögben \(\displaystyle QPX\measuredangle=DPM\measuredangle=DCM\measuredangle=90^\circ-\beta\), tehát \(\displaystyle PXQ\measuredangle=90^\circ\); az \(\displaystyle X\) ponton átmegy az \(\displaystyle AFME\) és az \(\displaystyle R\) középpontú \(\displaystyle PQE\) kör.

Az \(\displaystyle RPX\) egyenlő szárú háromszögből, a \(\displaystyle CPMD\) húrnégyszögből, az \(\displaystyle DCM\measuredangle\) és \(\displaystyle MAF\measuredangle\) merőleges szárú szögekből, majd az \(\displaystyle AFMX\) húrnégyszögből látjuk, hogy

\(\displaystyle MXR\measuredangle = RPX\measuredangle = DPM\measuredangle = DCM\measuredangle = MAF\measuredangle = MXF\measuredangle. \)

Ezzel az állítást igazoltuk.

Statistics:

28 students sent a solution.
6 points:	Balogh Zsófia, Baski Bence, Beke Csongor, Csaplár Viktor, Dobák Dániel, Fekete Richárd, Füredi Erik Benjámin, Geretovszky Anna, Győrffy Ágoston, Hámori Janka, Hegedűs Dániel, Hervay Bence, Jánosik Áron, Kerekes Anna, Kovács 129 Tamás, Mátravölgyi Bence, Nagy Nándor, Nguyen Bich Diep, Pooya Esmaeil Akhoondy, Rareș Polenciuc, Szabó 417 Dávid, Tiderenczl Dániel, Tóth Ábel, Várkonyi Zsombor, Velich Nóra, Weisz Máté.
3 points:	1 student.
0 point:	1 student.

Problems in Mathematics of KöMaL, January 2019