Let $ABC$ be a triangle, and let $P$ be a distinct point on the plane. Moreover, let $A'B'C'$ be a homothety of $ABC$ with ratio $2$ and center $P$, and let $O$ and $O'$ be the circumcenters of $ABC$ and $A'B'C'$, respectively. The circumcircles of $AB'C'$, $A'BC'$, and $A'B'C$ meet at points $X$, $Y$, and $Z$, different from $A'$, $B'$, and $C'$. In a similar way, the circumcircles of $A'BC$, $AB'C$, and $ABC'$ meet at $X'$, $Y'$, and $Z'$, different from $A$, $B$, $C$. Let $W$ and $W'$ be the circumcenters of $XYZ$ and $X'Y'Z'$, respectively. Prove that $OW$ is parallel to $O'W'$. Proposed by Mateus Thimóteo, Brazil.
Problem
Source: 1st International Mathematical Olympic Revenge
Tags: IMOR, geometry
pinetree1
22.07.2017 21:15
This must be a nightmare to draw for anyone using paper and pencil. Here's a nice diagram: [asy][asy] size(350); defaultpen(fontsize(10pt)); pair P, A, B, C, AA, BB, CC, X, Y, Z, XX, YY, ZZ, W, WW, O, OO; A = dir(120); B = dir(220); C = dir(340); P = (-3,0); AA = 2*A-P; BB = 2*B-P; CC = 2*C-P; O = circumcenter(A, B, C); OO = circumcenter(AA, BB, CC); draw(A--B--C--cycle, orange+linewidth(1.2)); draw(AA--BB--CC--cycle, orange+linewidth(1.2)); draw(P--AA^^P--BB^^P--CC, red); path w1, w2, w3, g1, g2, g3; w1 = circumcircle(A, BB, CC); w2 = circumcircle(AA, B, CC); w3 = circumcircle(AA, BB, C); g1 = circumcircle(AA, B, C); g2 = circumcircle(A, BB, C); g3 = circumcircle(A, B, CC); draw(w1^^w2^^w3, heavygreen+dashed); draw(g1^^g2^^g3, blue+dashed); X = IP(w2, w3, 0); Y = IP(w1, w3, 1); Z = IP(w1, w2, 0); XX = IP(g2, g3, 1); YY = IP(g1, g3, 0); ZZ = IP(g1, g2, 0); W = circumcenter(X, Y, Z); WW = circumcenter(XX, YY, ZZ); draw(O--W, magenta); draw(OO--WW, magenta); draw(X--Y--Z--cycle, purple); draw(XX--YY--ZZ--cycle, purple); clip(Circle((O+W)/2, 4.5)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(270)); dot("$A'$", AA, dir(70)); dot("$B'$", BB, dir(215)); dot("$C'$", CC, dir(CC)); dot("$P$", P, dir(P)); dot("$X$", X, dir(30)); dot("$Y$", Y, dir(315)); dot("$Z$", Z, dir(270)); dot("$X'$", XX, dir(300)); dot("$Y'$", YY, dir(YY)); dot("$Z'$", ZZ, dir(140)); dot("$O$", O, dir(90)); dot("$O'$", OO, dir(45)); dot("$W$", W, dir(270)); dot("$W'$", WW, dir(180)); [/asy][/asy]
User335559
23.07.2017 02:48
pinetree1 wrote: This must be a nightmare to draw for anyone using paper and pencil. Here's a nice diagram: [asy][asy] size(350); defaultpen(fontsize(10pt)); pair P, A, B, C, AA, BB, CC, X, Y, Z, XX, YY, ZZ, W, WW, O, OO; A = dir(120); B = dir(220); C = dir(340); P = (-3,0); AA = 2*A-P; BB = 2*B-P; CC = 2*C-P; O = circumcenter(A, B, C); OO = circumcenter(AA, BB, CC); draw(A--B--C--cycle, orange+linewidth(1.2)); draw(AA--BB--CC--cycle, orange+linewidth(1.2)); draw(P--AA^^P--BB^^P--CC, red); path w1, w2, w3, g1, g2, g3; w1 = circumcircle(A, BB, CC); w2 = circumcircle(AA, B, CC); w3 = circumcircle(AA, BB, C); g1 = circumcircle(AA, B, C); g2 = circumcircle(A, BB, C); g3 = circumcircle(A, B, CC); draw(w1^^w2^^w3, heavygreen+dashed); draw(g1^^g2^^g3, blue+dashed); X = IP(w2, w3, 0); Y = IP(w1, w3, 1); Z = IP(w1, w2, 0); XX = IP(g2, g3, 1); YY = IP(g1, g3, 0); ZZ = IP(g1, g2, 0); W = circumcenter(X, Y, Z); WW = circumcenter(XX, YY, ZZ); draw(O--W, magenta); draw(OO--WW, magenta); draw(X--Y--Z--cycle, purple); draw(XX--YY--ZZ--cycle, purple); clip(Circle((O+W)/2, 4.5)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(270)); dot("$A'$", AA, dir(70)); dot("$B'$", BB, dir(215)); dot("$C'$", CC, dir(CC)); dot("$P$", P, dir(P)); dot("$X$", X, dir(30)); dot("$Y$", Y, dir(315)); dot("$Z$", Z, dir(270)); dot("$X'$", XX, dir(300)); dot("$Y'$", YY, dir(YY)); dot("$Z'$", ZZ, dir(140)); dot("$O$", O, dir(90)); dot("$O'$", OO, dir(45)); dot("$W$", W, dir(270)); dot("$W'$", WW, dir(180)); [/asy][/asy] I don't understand how could someone solve this? It seems terrible with computer diagram, but with man-hand? You probably wouldn't see anything on the figure, so problem couldn't be solved.
stronto
23.07.2017 02:51
That's the point... now the contestants can get their sweet revenge for problem #3
zuss77
19.05.2018 12:28
You have to be a real masochist to tackle such a messy problem. Meanwhile, another nice picture.
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