Equilateral triangles $ACB'$ and $BDC'$ are drawn on the diagonals of a convex quadrilateral $ABCD$ so that $B$ and $B'$ are on the same side of $AC$, and $C$ and $C'$ are on the same sides of $BD$. Find $\angle BAD + \angle CDA$ if $B'C' = AB+CD$.
Problem
Source: International Zhautykov Olympiad 2012 - D2 - P2
Tags: complex numbers, geometry unsolved, geometry
IZO
31.01.2012 17:27
Draw an equilateral triangle $AMD$ so that $M$ and $C$ are on the same side of $AD$. Since $\angle DAC=MAB'$, it follows that $\triangle MAB'=\triangle DAC$. Similarly, $\angle MDC'=\angle ADB$ implies that $\triangle MDC'=\triangle ADB$. In particular, $MB'=CD$ and $MC'=AB$. Therefore $B'C'=AB+CD=MC'+MB'$, which implies that $M\in [B'C']$. Now, it is easy to see that $\angle BAD+\angle CDA=\angle C'MD+\angle B'MA=180^o-\angle AMD=120^o$.
125907
07.02.2012 16:14
nice solution!
anonymouslonely
07.02.2012 17:16
a solution with complex numbers! let be $ w $ such that $ w^{2}+w+1=0 $. we have $ c'= -dw^{2}-bw $ and $ b'=-aw-cw^{2} $ so the conditon is: $ |b-a+w(d-c)|=|b-a|+|d-c| $. then $ \frac{b-a}{w(d-c)} $ is a positive real number. then the angle between $ AB,DC $ will $ arg w $ so $ 120 $ then the sum of the angles $ BAD,CDA $ will be $ 120 $.
horizon
09.02.2012 18:40
my solution $|Z_{B'}-Z_{C'}|=|(\frac{1}{2}+\frac{\sqrt{3}}{2}i(Z_{A}-Z_{B})+(\frac{1}{2}-\frac{\sqrt{3}}{2}i)(Z_{c}-Z_{D})=|Z_{A}-Z_{B}|+|Z_{C}-Z_{D}|$ so ${Z_{1}=(\frac{1}{2}+\frac{\sqrt{3}}{2}i)(Z_{A}-Z_b})$ and $Z_{2}=(\frac{1}{2}-\frac{\sqrt{3}}{2}i)(Z_{C}-Z_{D})$ have the same direction so the sum of angles will be $120$
Ardest__
13.03.2012 17:56
Why in this year IZHO was easy?
62861
19.05.2016 04:19
Let the $60^{\circ}$ rotation about $C$ which takes $A$ to $B'$ take $B$ to $P$. Since $CB=CP$ and $\angle BCP=60^{\circ}$, $\Delta BCP$ is equilateral, so $BC=BP$. This implies that there exists a $60^{\circ}$ rotation about $B$ taking $\Delta BDC$ to $\Delta BC'P$.
Thus, $B'C'=AB+CD=B'P+P'C\geq B'C'$, so $B'P+P'C=B'C'$ and $P$ lies on segment $B'C'$. Then, we have $\angle ABC+\angle DCB=\angle B'PC+\angle C'PB=\angle B'PC'+\angle BPC=240^{\circ}$, so $\angle BAD+\angle CDA=360^{\circ}-240^{\circ}=120^{\circ}$.
EugenyVassilievitch
16.03.2023 18:58
I have a nice solution using sin rule. Let $AC \cap BD = E, BC' \cap CB' = F, C'B \cap AB' = X, B'C \cap DC' = Y$. Obviously, using pairs of angles each equal to $60^\circ$, we have three circumscribed quadrilaterals: $AXBE, DYCE, B'C'YX$. Let $\angle AEB = \angle DEC = \varphi$, therefore, $\angle BXB' = \varphi$. Then: $\frac{AB}{\sin \varphi} = \frac{XE}{\sin 120^\circ}$, $\frac{CD}{\sin \varphi} = \frac{YE}{\sin 120^\circ}$, $\frac{B'C'}{\sin \varphi} = \frac{XY}{\sin 60^\circ}.$ Equality $B'C' = AB + CD$ leads to: $\frac{XY}{\sin 60^\circ}=\frac{B'C'}{\sin \varphi} = \frac{AB + CD}{\sin \varphi} = \frac{XE + YE}{\sin 120^\circ}.$ In other words, $XY=XE+YE$, so points $X, Y, E$ are collinear. Then goes simple angle-chasing: $\angle BAD+\angle CDA=(\angle BAE+\angle CDE)+(\angle EAD+\angle EDA)=(\angle BXE+\angle CYE)+(180^\circ-\angle AED)=(180^\circ-\angle BFC)+(180^\circ-\angle BEC)=360^\circ-\angle BFC-\angle BEC=120^\circ.$
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