Question

Please help

Let $ABC$ be an equilateral triangle. Let $A_1,B_1,C_1$ be interior points of $ABC$ such that $BA_1=A_1C$, $CB_1=B_1A$, $AC_1=C_1B$, and
$$\angle BA_1C+\angle CB_1A+\angle AC_1B=480^\circ$$Let $BC_1$ and $CB_1$ meet at $A_2,$ let $CA_1$ and $AC_1$ meet at $B_2,$ and let $AB_1$ and $BA_1$ meet at $C_2.$
Prove that if triangle $A_1B_1C_1$ is scalene, then the three circumcircles of triangles $AA_1A_2, BB_1B_2$
and $CC_1C_2$ all pass through two common points.

Answer 1

Oh hell, this is gonna be a minute.

Answer 2

Okay this is quite a bit man so be ready.


Angle: WLOG let $AB_1 > A_1B$, $BC_1 > B_1C$, $CA_1 > C_1A$. Denote all angles in degrees. Denote $\theta = \angle BAC = \angle CBA = \angle ACB$. It follows that $\angle BA_1C = 120 - \theta$, since $\angle B_1AC = \angle AC_1B = \theta$, and $\angle BA_1C$ must integrate up to $360 - 2\theta - 120 = 120 - 2\theta$. Then $\angle A_1CB_1 = \angle CB_1A = \theta - (120 - \theta) = 2\theta - 120$. Summing the angles, $\angle A_2B_1C_1 = 2\theta - 120 - \theta + 2\theta - 120 = 4\theta - 360$, so $\angle A_2C_1B_1 = 480 - (4\theta - 360) = 840 - 4\theta$. By Ceva's, $\prod \dfrac {\sin \angle \text{opposite}}{\sin \angle \text{adjacent}} = 1$, so we have: \[\dfrac {\sin (2\theta - 120)}{\sin \theta} \cdot \dfrac {\sin (2\theta - 120)}{\sin \theta} \cdot \dfrac {\sin (840 - 4\theta)}{\sin (120 - \theta)} = 1.\] Note that the variables are all quantified in degrees. Ergo $\sin \theta = \sin (120 - \theta) = \sin (840 - 4\theta)$, so $\theta = 60, 300, 420$. $\theta = 60$ engenders $A_1B_1C_1$ equilateral, so $\theta = 300, 420$. We get $\angle A_1CB_1 = 2\theta - 120 = 60, 180$ respectively. The former engenders $B_1$ and $C_1$ on the same side of $AB$, which facilely results in $\triangle A_1B_2C_1$ passing through $C$. The latter engenders $B_1$ and $C_1$ on antithesis sides of $AB$, so we get $\angle ACB_2 = 180 - 2\theta = 60, \angle A_1BC_2 = 420$. Utilizing the sine form of Ceva's, you should utilize a calculator to view the quantity $\dfrac {\sin \angle A_1BA_2}{\sin \angle A_2BC} \cdot \dfrac {\sin \angle C_1CB_2}{\sin \angle B_2BA} \cdot \dfrac {\sin \angle AC_1C_2}{\sin \angle C_2CA}$. The numerator of each fraction appears in another fraction as the denominator, and you can divide everything, in additament to any symmetry about the three angles you opt for. We will optate to cull the angles in $\triangle ABC$, so optate $\angle A, \angle B, \angle C$. Now, we calculate each quantity in turn: \[\dfrac {\sin \angle A_1BA_2}{\sin \angle A_2BC} = \dfrac {\sin 60}{\sin 2\theta - 60}\] \[\dfrac {\sin \angle C_1CB_2}{\sin \angle B_2BA} = \dfrac {\sin 2\theta - 120}{\sin 60}\] \[\dfrac {\sin \angle AC_1C_2}{\sin \angle C_2CA} = \dfrac {\sin (840-4\theta)}{\sin (120 - \theta)}.\] Multiplying all gives $\dfrac {\sin^2 60}{\sin (2\theta - 60)\sin (2\theta - 120)\sin (120 - \theta)\sin 60 \sin (840 - 4\theta)}$. Rescinding $\sin 60$'s, we get by standard identities that all this explodes into $\dfrac {3}{16}$, so $\dfrac {\sin \angle A_1BA_2}{\sin \angle A_2BC} \cdot \dfrac {\sin \angle C_1CB_2}{\sin \angle B_2BA} \cdot \dfrac {\sin \angle AC_1C_2}{\sin \angle C_2CA} = \dfrac {3}{16}$. Thus, by the trig form of Ceva's, $AA_1, BB_1, CC_1$ are concurrent, and by trig Ceva's on $\triangle A_1B_1C_1$, $AA_2, BB_2, CC_2$ are concurrent as well. By the radical axis theorem on $(AA_1CA_2), (BB_1AB_2), (CC_1BC_2)$, we deduce that $(AA_1A_2), (BB_1B_2), (CC_1C_2)$ are concurrent. Trivially, we visually perceive that $(AA_1A_2)$ passes through $A$ and the foot of the perpendicular from $C$ to $AB$, and symmetrically for the other two circles. Thus $(AA_1A_2), (BB_1B_2), (CC_1C_2)$ intersect at the circumcenter of $\triangle ABC$ and [tip = isogonal conjugate] This is a one liner by isogonal/homothety radical axis: $AA_2, BB_2, CC_2$ concur, as do their isagogals. Homothety sends the concurrency point to the Bevan point of $\triangle ABC$ (the circumcenter of $\triangle A_1B_1C_1$). Then, three circumcircles concur at this same point by radical axis.[/tip]. $\square$

Hope this helped you man, I spent half my life typin this for ya.