AoPS - Problem

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Tags: concurrent, geometry, circles

parmenides51

17.08.2020 00:49

Three circles are constructed for the triangle $ABC$ : the circle ${{w} _ {A}}$ passes through the vertices $B$ and $C$ and intersects the sides $AB$ and $AC$ at points ${{A} _ {1}}$ and ${{A} _ {2}}$ respectively, the circle ${{w} _ {B}}$ passes through the vertices $A$ and $C$ and intersects the sides $BA$ and $BC$ at the points ${{B} _ {1}}$ and ${{B} _ {2}}$ , ${{w} _ {C}}$ passes through the vertices $A$ and $B$ and intersects the sides $CA$ and $CB$ at the points ${{C} _ {1}}$ and ${{C} _ {2}}$ . Let ${{A} _ {1}} {{A} _ {2}} \cap {{B} _ {1}} {{B} _ {2}} = {C} '$ , ${{A} _ {1}} {{A} _ {2}} \cap {{C} _ {1}} {{C} _ {2}} = {B} '$ ta ${ {B} _ {1}} {{B} _ {2}} \cap {{C} _ {1}} {{C} _ {2}} = {A} '$ 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)

PROA200

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 $\omega_a$ to $\omega_a'$ , the result holding for $\omega_a'$ , $\omega_b$ , $\omega_c$ is necessary and sufficient. (Here we're dropping a perpendicular from $A$ onto $A_1A_2$ and noting that by antiparallel lines, all lines $A_1A_2$ are parallel. So this perpendicular is fixed as well.) By perturbing $\omega_b$ and $\omega_c$ as necessary, it suffices to show the result for one choice of $\omega_a$ , $\omega_b$ , $\omega_c$ . However, the case where $\omega_a$ is the circle with diameter $BC$ (choose $\omega_b$ , $\omega_c$ similarly) is well-known; the lines concur at the circumcenter. (This is also a property of isogonal conjugates.)