The vertex $C$ of equilateral triangles $ABC$ and $CDE$ lies on the segment $AE$, and the vertices $B$ and $D$ lie on the same side with respect to this segment. The circumcircles of these triangles centered at $O_1$ and $O_2$ meet for the second time at point $F$. The lines $O_1O_2$ and $AD$ meet at point $K$. Prove that $AK = BF$.
Problem
Source: Sharygin 2018
Tags: geometry
04.04.2018 22:24
I complex bashed this. (The fact that $A, K, F, D$ are collinear simplified stuff) If anyone is interested in a synthetic solution. I noticed that: Let $X$ be the around intersection of $CK$ and the circle donating $B$. Then $BX\parallel AK$ and $\triangle AKX$ is equilateral but I couldn't prove it though.
04.04.2018 22:27
Ptolemy's is enough! (after noticing that $\triangle KFC $ is equilateral)
05.04.2018 09:39
Solution: [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11.12, xmax = 13.28, ymin = -6.38, ymax = 7.1; /* image dimensions */ pair B = (-3.,0.), C = (3.,0.), A = (0.,5.196152422706633), D = (1.8216540881398724,-2.040954988232818), F = (1.6399929863610558,-1.3192477692155296), O_1 = (0.,1.7320508075688779), O_2 = (3.,-1.360636658821879), K = (1.1774944111539294,0.5181767385285706); draw(B--C--A--cycle, linewidth(1.)); draw((4.178345911860127,-2.0409549882328184)--C--D--cycle, linewidth(1.)); /* draw figures */ draw(B--C, linewidth(1.)); draw(C--A, linewidth(1.)); draw(A--B, linewidth(1.)); draw((4.178345911860127,-2.0409549882328184)--C, linewidth(1)); draw(C--D, linewidth(1.)); draw(D--(4.178345911860127,-2.0409549882328184), linewidth(1)); draw(circle(O_1, 3.4641016151377553), linewidth(1)); draw(circle(O_2, 1.360636658821879), linewidth(1)); draw(A--D, linewidth(1.)); draw(B--(4.178345911860127,-2.0409549882328184), linewidth(1.)); draw(O_2--O_1, linewidth(1.)); draw(K--C, linewidth(1.)); draw(F--C, linewidth(1.)); /* dots and labels */ dot(B,dotstyle); label("$B$", (-3.36,0.04), NE * labelscalefactor); dot(C,dotstyle); label("$C$", (3.24,0.06), NE * labelscalefactor); dot(A,dotstyle); label("$A$", (0.08,5.4), NE * labelscalefactor); dot((4.178345911860127,-2.0409549882328184),dotstyle); label("$E$", (4.32,-2.08), NE * labelscalefactor); dot(D,dotstyle); label("$D$", (1.42,-2.28), NE * labelscalefactor); dot(F,linewidth(4.pt) + dotstyle); label("$F$", (1.38,-1.76), NE * labelscalefactor); dot(O_1,linewidth(4.pt) + dotstyle); label("$O_1$", (-0.3,1.34), NE * labelscalefactor); dot(O_2,linewidth(4.pt) + dotstyle); label("$O_2$", (2.7,-1.68), NE * labelscalefactor); dot(K,linewidth(4.pt) + dotstyle); label("$K$", (1.26,0.68), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Claim 1: $F = AD \cap BE$. Proof: $\angle BFE = \angle BFC + \angle CFE = (180^{\circ}- 60^{\circ}) + \angle CDE = 180^{\circ}$. So, $F$ lies on $BE$. Similarly $\angle AFD=\angle AFC+\angle CFD=\angle ABC+(180^{\circ}-60^{\circ})=180^{\circ}$. So $F$ lies on $AD$ which gives us that $F$ is the intersection of $AD$ and $BE$. Claim 2: $\triangle CFK$ is equilateral. Proof: As $CF$ is the radical axis of circles centered at $O_1$ and $O_2$, $O_1O_2$ bisects $CF$ perpendicularly. We know that $K$ lies on $O_1O_2$. So, $KF = KC$. We have $\angle KFC = \angle AFC = \angle ABC=60^{\circ}$. So, $KFC$ is an equilateral triangle. Main Problem: Consider triangles $AKC$ and $BFC$. We have $AC = BC$, $KC = FC$. We also have, \[\angle BCF =\angle BAF=\angle BAC-\angle CAF= 60^{\circ} -\angle CAF =\angle FKC - \angle CAF =\angle ACK.\]So, $\triangle BCF \cong \triangle ACK$ which gives us that $BF = AK$. Proved.
05.04.2018 14:23
Tumon2001 wrote: Ptolemy's is enough! (after noticing that $\triangle KFC $ is equilateral) Congruency is more than enough! (my solution is the same as above)
05.04.2018 16:07
Spiral similarity does it instantly. Along with similarity.