Given is an acute triangle ABC with circumcenter O. The point P on BC such that BP<BC2 and the point Q is on BC, such that CQ=BP. The line AO meets BC at D and N is the midpoint of AP. The circumcircle of (ODQ) meets (BOC) at E. The lines NO,OE meet BC at K,F. Show that AOKF is cyclic.
Problem
Source: Bulgarian Autumn Tournament 2023, 10.2
Tags: geometry
19.11.2023 14:02
Almost no one got this lol. Let AB intersect (ABC) again at G By menelaus on △APD ¯N−O−K we get ANNPPKKDPOOA=1 and thus KD=OD⋅PDDG Let FC=x by pop we have FC⋅FB=FQ⋅FB or x(x+BC)=(x+DC)(x+CQ) solving we get x=DC⋅CQPD and thus FD=FC+CD=DC⋅DBPD Finally by pop on D we get that AOKF is cyclic iff DO⋅DA=DK⋅DF or DO⋅DA=DO⋅PDDGDC⋅DBPD so its left to verify that DA⋅DG=DC⋅DB which is pop.
19.11.2023 18:31
Here is a completely synthetic solution, which relies on the idea to focus on triangle BRC, where AO∩(BOC)=R (this is also the main idea in IGO 2022 A3, and in fact, when P is the foot of the altitude from A, then E is the R-mixtilinear touchpoint for △BRC).
20.11.2023 00:11
Along with @Helixglich's approach, here is the other pedagogically best solution to this problem, found by a few contestants. To employ the midpoint N in midsegments, let AO intersect the circumcircle of ABC at T, so that NO is a midsegment in APT. Then PK∥OT, so ∠ATP=∠AON=∠DOK and to show this equals ∠AFD. it suffices to show that APTF is cyclic. By Power of a Point, we equivalently want PD⋅DF=AD⋅DT, but also from the circumcircle of ABC we have AD⋅DT=BD⋅DC, so it suffices to show PD⋅FD=BD⋅DC. On the other hand, FD=FE⋅FOFQ by Power of a Point on the cyclic ODQE and FE⋅FO=FC⋅FB from the circumcircle of BOC, so we obtain the relation FD⋅FQ=FB⋅FC. Now denote BP=CQ=x, FC=y, PD=z, DQ=t. Then we are given (x+y+t)(x+y)=(2x+y+z+t)y (equivalent to yz=x(x+t) after opening the brackets) and wish to show z(x+y+t)=(x+z)(x+t) (also equivalent to yz=x(x+t)), done.