Given an acute triangle $ABC$ . It's altitudes $AA_1 , BB_1$ and $CC_1$ intersect at a point $H$ , the orthocenter of $\vartriangle ABC$. Let the lines $B_1C_1$ and $AA_1$ intersect at a point $K$, point $M$ be the midpoint of the segment $AH$. Prove that the circumscribed circle of $\vartriangle MKB_1$ touches the circumscribed circle of $\vartriangle ABC$ if and only if $BA1 = 3A1C$.
(Bondarenko Mykhailo)
Very nice problem
I dont know latex very much so i only write the steps
1-if BA(1)=3CA(1) then the circles touch each other:
This part is easy the point is the intersection (AH) and circumcircle of ABC(assume P)
Notice that if N is the midpoint BC then N,H,P are collinear and A(1) is the midpoint of MC.(its angle chasing)
2-if they touch each other then BA(1)=3CA(1)
The key point is that the tangent at B to (ABC) and the tangent at K at (KMB(1)) are parallel(angle chasing) so BK passes through the homothetic center of these circles so because they are tangent to eachother so if BK meet (ABC) at Q then we must have MKB(1)Q is cyclic now we can also show that if (ABC) and (AH) meet at R then
KB(1)QR is cyclic too(its angle chasing too! notice that RC(1)B(1) is similar to RBC)
So we must have Q and R are same because MKB(1)QR is cyclic and touches (ABC) and the rest is easy and only angle chasing