The incircle of a triangle ABC touches the sides BC,CA,AB at points D,E,F, respectively. Let X be a point on the incircle, different from the points D,E,F. The lines XD and EF,XE and FD,XF and DE meet at points J,K,L, respectively. Let further M,N,P be points on the sides BC,CA,AB, respectively, such that the lines AM,BN,CP are concurrent. Prove that the lines JM,KN and LP are concurrent. Dinu Serbanescu
Problem
Source: Romanian TST 2011
Tags: geometry, incenter, search, geometry proposed
09.04.2012 22:51
WLOG assume that J is between E,F. △JKL is obviously autopolar WRT the incircle (I) of △ABC and A∈LK, because its polar EF WRT (I) passes through J. Analogously, B∈LJ and C∈JK. Since KL does not cut (I), then project KL to infinity and (I) into a circle, whose center is then J. So △DEF becomes isosceles right at D and BCEF is a rectangle. Let Y≡AM∩BN∩CP, T≡KN∩LP and M′≡TJ∩BC. Thus, it suffices to show that M≡M′. Let TP,TN cut BC at U,V. Then isosceles right △TUV and △JBC are homothetic through M′ ⟹ ¯M′B¯M′C=¯M′U¯M′V=¯BU¯CV=¯PB¯NC⟹YM′∥PB∥NC⟹M≡M′.
10.04.2012 20:49
Lemma:- In a triangle △ABC, P is a point inside it.Let △PaPbPc be the cevian triangle of P.Consider a point Q on the circumference of ⊙PaPbPc.Let X=PbPc∩PaQ.Analogously define Y,Z.Then AX,BY,CZ are concurrent. Proof:- Let us consider a projective transformation that takes the circle ⊙PaPbPc to the circle P′aP′bP′c and the tri-liner polar of P to infinity(It can be proven that there exists such a projective transformation).Let A′B′C′ be the image of ABC.Clearly P′ becomes the centroid of △A′B′C′.Now the result of the lemma is trivially true for the medial triangle.Hence the lemma is proven. Main Proof:- Now we just need to prove that poles of JM,NK,LP wrt the incircle are collinear.Note that J is the pole of KL.Similarly K is the pole of JL and L is the pole of JK.Now clearly polars of M,N,P pass through D,E,F respectively.Since AM,BN,CP are concurrent, clearly △A1B1C1 is perspective with △DEF where △A1B1C1 is the triangle formed by the polars of M,N,P.So △DEF is a cevian triangle of △A1B1C1.Now using the above lemma we conclude that JM,NK,LP are concurrent.
07.11.2014 03:37
My solution: Let I be the incenter of △ABC . From Cevian nest theorem we get AJ,BK,CL are concurrent at Z . Since KL is the polar of J WRT (I) , so K,A,L are collinear . Similarly, we can prove B∈LJ and C∈JK . Since △ABC is the Z -cevian triangle of △JKL , so from Cevian nest theorem we get JM,KN,LP are concurrent . Q.E.D
07.11.2014 04:52
What's cevian-nest theorem?
07.11.2014 06:38
Dear Mathlinkers, see for example http://jl.ayme.pagesperso-orange.fr/ vol. 3 the cevian nests theorem or on search on mathlinks Sincerely Jean-Louis
07.11.2014 15:19
cevian nest theorem, twice