Circle $\omega$ touches the side $AC$ of the equilateral triangle $ABC$ at point $D$, and its circumcircle at the point $E$ lying on the arc $\overarc{BC}$. Prove that with segments $AD$, $BE$ and $CD$, you can form a triangle, in which the difference of two of its angles is $60^{\circ}$.
Problem
Source: 239 2019 J3
Tags: geometry
rstenetbg
25.05.2024 21:04
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -13.047398257790807, xmax = 22.561421602255844, ymin = -11.888441912408274, ymax = 2.6750090225405216; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); pen zzttqq = rgb(0.6,0.2,0); draw((-2.94,-5.81)--(5.16,-5.89)--(1.1792820323027575,1.1648057706539539)--cycle, linewidth(1)); draw(arc((-2.94,-5.81),0.4172908577349217,-41.93567999937778,-0.5658658432466473)--(-2.94,-5.81)--cycle, linewidth(1) + qqwuqq); draw(arc((-2.94,-5.81),0.4172908577349217,-0.5658658432466349,40.80394831288451)--(-2.94,-5.81)--cycle, linewidth(1) + qqwuqq); draw(arc((1.1792820323027575,1.1648057706539539),0.4172908577349217,-60.565865843246655,-19.19605168711551)--(1.1792820323027575,1.1648057706539539)--cycle, linewidth(1) + qqwuqq); draw(arc((-2.94,-5.81),0.4172908577349217,-0.5658658432466349,40.80394831288451)--(-2.94,-5.81)--cycle, linewidth(1) + qqwuqq); draw(arc((1.1792820323027575,1.1648057706539539),0.4172908577349217,-101.93567999937783,-60.565865843246655)--(1.1792820323027575,1.1648057706539539)--cycle, linewidth(1) + qqwuqq); draw(arc((4.001208354045201,0.18232509319929963),0.4172908577349217,-139.19605168711547,-120.56586584324667)--(4.001208354045201,0.18232509319929963)--cycle, linewidth(1) + blue); draw(arc((5.16,-5.89),0.4172908577349217,179.43413415675337,198.06432000062216)--(5.16,-5.89)--cycle, linewidth(1) + blue); draw((-0.7171913340232638,-7.8069121366182594)--(0.29532751687364045,-3.0169511135216607)--(4.001208354045201,0.18232509319929963)--cycle, linewidth(1) + zzttqq); draw((-0.7171913340232638,-7.8069121366182594)--(1.8982859844352098,-5.857785540587015)--(5.16,-5.89)--cycle, linewidth(1) + zzttqq); /* draw figures */ draw((-2.94,-5.81)--(5.16,-5.89), linewidth(1)); draw((5.16,-5.89)--(1.1792820323027575,1.1648057706539539), linewidth(1)); draw((1.1792820323027575,1.1648057706539539)--(-2.94,-5.81), linewidth(1)); draw(circle((1.133094010767585,-3.511731409782015), 4.676765263869177), linewidth(1)); draw((1.1792820323027575,1.1648057706539539)--(4.001208354045201,0.18232509319929963), linewidth(1)); draw((-2.94,-5.81)--(-0.7171913340232638,-7.8069121366182594), linewidth(1)); draw((1.8982859844352098,-5.857785540587015)--(-0.7171913340232638,-7.8069121366182594), linewidth(1)); draw((-0.7171913340232638,-7.8069121366182594)--(5.16,-5.89), linewidth(1)); draw((-2.94,-5.81)--(4.001208354045201,0.18232509319929963), linewidth(1)); draw((1.0869059892324173,-8.188268590217984)--(4.001208354045201,0.18232509319929963), linewidth(1)); draw((-0.7171913340232638,-7.8069121366182594)--(4.001208354045201,0.18232509319929963), linewidth(1)); draw((1.1792820323027575,1.1648057706539539)--(-0.7171913340232638,-7.8069121366182594), linewidth(1)); draw((4.001208354045201,0.18232509319929963)--(5.16,-5.89), linewidth(1)); draw((5.16,-5.89)--(1.0869059892324173,-8.188268590217984), linewidth(1)); draw((-0.7171913340232638,-7.8069121366182594)--(0.29532751687364045,-3.0169511135216607), linewidth(1) + zzttqq); draw((0.29532751687364045,-3.0169511135216607)--(4.001208354045201,0.18232509319929963), linewidth(1) + zzttqq); draw((4.001208354045201,0.18232509319929963)--(-0.7171913340232638,-7.8069121366182594), linewidth(1) + zzttqq); draw((-0.7171913340232638,-7.8069121366182594)--(1.8982859844352098,-5.857785540587015), linewidth(1) + zzttqq); draw((1.8982859844352098,-5.857785540587015)--(5.16,-5.89), linewidth(1) + zzttqq); draw((5.16,-5.89)--(-0.7171913340232638,-7.8069121366182594), linewidth(1) + zzttqq); /* dots and labels */ dot((-2.94,-5.81),dotstyle); label("$A$", (-3.268882491535808,-6.046369904119358), NE * labelscalefactor); dot((5.16,-5.89),dotstyle); label("$C$", (5.257760701514425,-6.157647466182004), NE * labelscalefactor); dot((1.1792820323027575,1.1648057706539539),dotstyle); label("$B$", (0.9874842573603934,1.3118588872731085), NE * labelscalefactor); dot((1.0869059892324173,-8.188268590217984),linewidth(4pt) + dotstyle); label("$M$", (0.9457551715869011,-8.536205355271061), NE * labelscalefactor); dot((1.8982859844352098,-5.857785540587015),dotstyle); label("$D$", (2.0446210969555283,-5.768175998962743), NE * labelscalefactor); dot((4.001208354045201,0.18232509319929963),linewidth(4pt) + dotstyle); label("$E$", (4.061526909340984,0.2964511334514638), NE * labelscalefactor); dot((-0.7171913340232638,-7.8069121366182594),linewidth(4pt) + dotstyle); label("$F$", (-1.0155118597672308,-8.11891449753614), NE * labelscalefactor); dot((0.29532751687364045,-3.0169511135216607),linewidth(4pt) + dotstyle); label("$X$", (-0.13920105852389525,-3.0557854236857467), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] It is well-known that if $M$ is the midpoint of small arc ${AC}$, then $M\in DE$ (Shooting lemma). Let $F$ be a point on the small arc $AC$ such that $AF=BE$. Hence, we get that $CF=CE$ and thus $$\angle EBC =\angle CBF =\angle EAC =\angle CAF=\theta.$$Also, note that $BM=2R$ is a diameter and $MC=R$. Let $BF\cap AE = X$. Claim 1: $\triangle FXE\sim \triangle FDC.$ Proof: Clearly, we have that $\angle XEF = \angle AEF = \angle DCF.$ This means that it suffices to show that $$\frac{DC}{CF}=\frac{XE}{EF}\iff \frac{DC}{CE}=\frac{XE}{EF}.$$Note that $\triangle MDC\sim \triangle MCE,$ thus $$\frac{DC}{CE}=\frac{MC}{ME}=\frac{R}{2R\sin(30^{\circ}+\theta)}=\frac{1}{2\sin(30^{\circ}+\theta)}.$$ By construction, we have that $BEFA$ is an isosceles trapezoid, so $\angle XAB=\angle XBA = \angle XFE = 60^{\circ}-\theta.$ Hence, $\angle FXE = 60^{\circ}+2\theta$ and therefore $$\frac{XE}{EF}=\frac{\sin(60^{\circ}-\theta)}{\sin(60^{\circ}+2\theta)}=\frac{\cos(30^{\circ}+\theta)}{2\sin(30^{\circ}+\theta)\cos(30^{\circ}+\theta)}=\frac{1}{2\sin(30^{\circ}+\theta)},$$as desired. Since $\triangle FXE$ is isosceles, we obtain that $\triangle FDC$ is also isosceles. In particular, $DC=DF.$ Thus, with segments $AD,DC,BE$ we can construct $\triangle ADF$. Now to finish the problem, just observe that $$60^{\circ}+2\theta = \angle FDC = \angle DAF+\angle DFA=\theta+\angle DFA.$$Therefore, $\angle DFA=\theta+60^{\circ}=\angle DAF+60^{\circ}$, so we may conclude. Remark: At first glance, this problem seemed to be quite annoying, but it actually has a really cute solution.