KöMaL - Középiskolai Matematikai és Fizikai Lapok
Sign In
Sign Up
 Magyar
Information
Contest
Journal
Articles

 

Problem A. 431. (September 2007)

A. 431. Triangle ABC is not isosceles. The incenter is I, the excenter is O. The incircle touches the sides BC, CA, AB at points D, E, F, respectively. Lines FD and AC intersect at P, lines DE and AB intersect at Q. The midpoints of segments EP amd FQ are M and N, respectively. Prove that MN and OI are perpendicular.

(Chinese Mathematical Olympiad, 2007)

(5 pont)

Deadline expired on 15 October 2007.


Solution. We prove that MN is the radical line of the incircle and the circumcircle.

Let AE=AF=x, BD=BF=y and CD=CE=z. Denote the signed distance QA by u; let u be positive if the direction of QA matches the direction of AB. From Menelaus' theorem

 \frac{AQ}{QB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = -1,

 \frac{-u}{u+x+y} \cdot \frac{y}{z} \cdot \frac{z}{x} = -1,

 u = \frac{x(x+y)}{y-x}.

Then QF = QA+AF = u+x = \frac{2xy}{y-x}, NF = \frac12QF = \frac{xy}{y-x}, NA = NF - AF = \frac{xy}{y-x} - x = \frac{y^2}{y-x} and NB = NF +FB = \frac{xy}{y-x} + y = \frac{x^2}{y-x}. Since

 NF^2 = NA \cdot NB = \frac{x^2y^2}{(y-x)^2},

the powers of point N with respect to the incircle and the circumcircle match, so N lies on the radical line.

It can be obtained similarly that M lies on the radical line of the two circles. So MN itself is the radical line which is perpendicular to the central line OI.


Statistics:

10 students sent a solution.
5 points:Blázsik Zoltán, Gombor Tamás, Korándi Dániel, Lovász László Miklós, Nagy 235 János, Nagy 314 Dániel, Szűcs Gergely, Tomon István, Tuan Nhat Le, Wolosz János.

Our web pages are supported by:   Ericsson   Cognex   Emberi Erőforrás Támogatáskezelő   Emberi Erőforrások Minisztériuma   Nemzeti Tehetség Program    
MTA Energiatudományi Kutatóközpont   MTA Wigner Fizikai Kutatóközpont     Nemzeti
Kulturális Alap   ELTE   Morgan Stanley