Dear Mathlinkers, if we think to a Morley circle, we are rapidely done... http://jl.ayme.pagesperso-orange.fr/vol2.html vol. 2 Sincerely Jean-Louis

We have: $O_{1}XO_{2}=180-XCD-XDC=180-O_{1}AC-O_{2}AD=O_{1}AO_{2}=O_{1}BO_{2}$. DONE! All of them are angles!

∠CO1B=2∠CAB=∠BO2D

Join BC and BD. <O1BO2=<CBD (1). So, <O2BD=<O1BC. This implies that BXDC is cyclic. Using (1) we get <O1BO2=<O1XO2. Hence, BXO2O1 is cyclic.

\[\angle O_1BO_2 = \angle O_1AO_2 = 180^{\circ} - \angle O_1AC - \angle O_2AD = 180^{\circ} - \angle ACO_1 - \angle ADO_2 = \angle DXC=\angle O_1XO_2\]thus $O_1,O_2,B,X$ are concyclic.

Kezer wrote: The two circles $\Gamma_1$ and $\Gamma_2$ with the midpoints $O_1$ resp. $O_2$ intersect in the two distinct points $A$ and $B$. A line through $A$ meets $\Gamma_1$ in $C \neq A$ and $\Gamma_2$ in $D \neq A$. The lines $CO_1$ and $DO_2$ intersect in $X$. Prove that the four points $O_1,O_2,B$ and $X$ are concyclic. @OP are you sure that this is German TSTST 2016 - #1?

^Yes, I took part in the German TSTST and posted the problem when the ISL was released iirc.

Ridiculously easy

We have $$\angle O_1XO_2 = \pi - \angle DCX - \angle CDX = \angle CBD,$$and by spiral similarity $$\angle CBD \sim O_1BO_2,$$so we're done.

Why it seems as a first round problem? $$\angle O_1BO_2=\angle O_1AO_2=\pi-\angle XCD-\angle XDC=\angle O_1XO_2$$