Let ABC, l and P be arbitrary triangle, line and point. A′,B′,C′ are reflections of A,B,C in point P. A″ is a point on B'C' such that AA''\parallel l. B'',C'' are defined similarly. Prove that A'',B'',C'' are collinear.
Problem
Source: Iranian National Olympiad (3rd Round) 2007
Tags: geometry, geometric transformation, reflection, parallelogram, geometry proposed
27.08.2007 12:42
Omid Hatami wrote: Let ABC, l and P be arbitrary triangle, line and point. A',B',C' are reflections of A,B,C in point P. A'' is a point on B'C' such that AA''\parallel l. B'',C'' are defined similarly. Prove that A'',B'',C'' are collinear. http://www.cut-the-knot.org/Curriculum/Geometry/Zaslavsky.shtml (interchange triangles ABC and A'B'C'). darij
02.07.2009 16:31
My solution: Let A_3,B_3,C_3 be points on BC,CA,AB respectively,such that A_2A_3 \| BC',B_2B_3 \| CA',C_2C_3 \| AB' Easy to see that AB' = A'B = C_2C_3,BC' = B'C = A_2A_3,CA' = C'A = B_2B_3.Then apply Thales's theorem and Menelaus's theorem we have: A_2,B_2,C_2 are collinear if and only if A_3,B_3,C_3 are collinear. Let d_1,d_2,d_3 be lines pass through A,B,C and parallel to A_2A_3,B_2B_3,C_2C_3 ,respectively.Then d_1,d_2,d_3 are concurrent at Q (the proof is simple,just take Q' = d_1 \cap d_2 then QB = CA' and QB \| CA' so QBA'C is parallelogram and from it we have CQ \| BA' \| C_2C_3. QAA_2A_3,QBB_2B_3,QCC_2C_3 are parallelograms so QA_3 \| QB_3 \| QC_3 \| l,then Q,A_3,B_3,C_3 are collinear and that complete the proof Image not found
07.11.2022 18:25
Trivial by moving point.Move the point A'' on B'C' then define B''and C'' such that AA''//BB''//CC'' deg(A'')=deg(B'')=deg(C'')=1 we want to show that A''-B''-C'' collinear so 4 case is enough.Take A'' C'',B'' ,AA''//BB' and \infty_{BC} .\blacksquare