Let ABCD be a quadrilateral with a incircle ω. Let I be the center of ω, suppose that the lines AD and BC intersect at Q and the lines AB and CD intersect at P with B is in the segment AP and D is in the segment AQ. Let X and Y the incenters of △PBD and △QBD respectively. Let R be the intersection of PY and QX. Prove that the line IR is perpendicular to BD.
Problem
Source: Brazil Ibero TST 2020 #3
Tags: geometry, incenter, projective geometry, geometric transformation, inscribed quadrilateral
28.04.2022 17:09
Let IR∩PQ=S then note that its enough to show that the polar of S w.r.t. ω is BD which makes the problem purely projective so we make a projective transform that sends ABCD into a square which means that AC∩BD=I and AC⊥BD, now P is the infinity point of AB,CD and Q is the infinity point of BC,AD, X becomes the intersection of PI with the external angle bisector of ∠ABD and Y becomes the intersection of IQ with the external angle bisector of ∠BDA, now note that X,Y and P,Q are pairs of symetry w.r.t. AC so A,I,C,R,S are colinear, now that means S is sent to the infinity point of AC, but then that means I,S are inverses w.r.t. ω and since IS⊥BD we get that the polar of S w.r.t. ω is indeed BD thus we are done
30.01.2024 05:26
solution without transformations (ou solução namoral) Monge D'alembert on the three circles of the statement gives that S=XY∩BD lies on PQ. If T=IR∩PQ, since IT, PY and QX concur, we have (Q,P;T,S)=−1⇒T lies on AC since AT, PD and QB must concur. Let AB,BC,CD,DA touch ω on E,F,G,H respectively and let EG∩FH=U. Claim 1: U=BD∩AC (Newton's Theorem). Proof: Pascal on (GGFHHE) and (FFGEEH) gives that U lies on BD, and changing E,F,G,H by H,E,F,G respectively gives that U lies on AC. Claim 2: EF, GH, PQ and AC concur. Proof: See that △EGP and △FHQ are perspective to each other since EG∩FH=U, EP∩FQ=B, GP∩HQ=D are collinear, thus EF, GH and PQ concur by Desargues Theorem. Pascal on (FFEGGH) gives AC also concur in this point. Since T lies on the polar of B and D (i.e, EF and GH) wrt ω, BD is the polar of T by La Hire and then IR⊥BD as we wanted.