Three circles are constructed for the triangle ABC: the circle wA passes through the vertices B and C and intersects the sides AB and AC at points A1 and A2 respectively, the circle wB passes through the vertices A and C and intersects the sides BA and BC at the points B1 and B2, wC passes through the vertices A and B and intersects the sides CA and CB at the points C1 and C2. Let A1A2∩B1B2=C′, A1A2∩C1C2=B′ ta B1B2∩C1C2=A′ is Prove that the perpendiculars, which are omitted from the points A′,B′,C′ to the lines BC, CA and AB respectively intersect at one point. (Rudenko Alexander)
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Tags: concurrent, geometry, circles
17.08.2020 04:18
[asy][asy] unitsize(0.3cm); pair A, B, C, A1, A2, B1, B2, C1, C2, A3, A4; A=(5,12); B=(0,0); C=(13,0); draw(A--B--C--cycle); path circle1= circle(circumcenter((4,7),B,C),circumradius((4,7),B,C)); path circle4= circle(circumcenter((4,8),B,C),circumradius((4,8),B,C)); path circle2= circle(circumcenter((9,-2),A,C),circumradius((9,-2),A,C)); path circle3= circle(circumcenter((9,0),B,A),circumradius((9,0),B,A)); //draw(circle1); //draw(circle2); //draw(circle3); A1=intersectionpoint(circle1, A--B); A2=intersectionpoint(circle1, A--C); B1=intersectionpoint(circle2, B--C); B2=intersectionpoint(circle2, B--A); C1=intersectionpoint(circle3, C--A); C2=intersectionpoint(circle3, C--B); A3=intersectionpoint(circle4, A--B); A4=intersectionpoint(circle4, A--C); dot(A1); dot(A2); draw(A1--A2); dot(A3); dot(A4); //draw(A3--A4); dot(B1); dot(B2); draw(B1--B2); dot(C1); dot(C2); draw(C1--C2); pair Ap=extension(B1,B2,C1,C2); pair Bp=extension(A1,A2,C1,C2); pair Cp=extension(B1,B2,A1,A2); draw(Ap--Bp--Cp--cycle); pair Aq=extension(B1,B2,C1,C2); pair Bq=extension(A3,A4,C1,C2); pair Cq=extension(B1,B2,A3,A4); draw(Aq--Bq--Cq--cycle,dotted); draw(A--foot(A,Cq,Bq),dotted); label("A",A,N); label("B",B,WSW); label("C",C,ESE); [/asy][/asy] By a well-known fact about orthologic triangles, it suffices to have the perpendiculars from A onto B′C′, B onto A′C′, and C onto A′B′ concur. Note that if we perturb ωa to ω′a, the result holding for ω′a, ωb, ωc is necessary and sufficient. (Here we're dropping a perpendicular from A onto A1A2 and noting that by antiparallel lines, all lines A1A2 are parallel. So this perpendicular is fixed as well.) By perturbing ωb and ωc as necessary, it suffices to show the result for one choice of ωa, ωb, ωc. However, the case where ωa is the circle with diameter BC (choose ωb, ωc similarly) is well-known; the lines concur at the circumcenter. (This is also a property of isogonal conjugates.)