Let $AB$ and $CD$ be two perpendicular diameters of a circle with centre $O$. Consider a point $M$ on the diameter $AB$, different from $A$ and $B$. The line $CM$ cuts the circle again at $N$. The tangent at $N$ to the circle and the perpendicular at $M$ to $AM$ intersect at $P$. Show that $OP = CM$.
Problem
Source: Pan African MO 2006 Q1
Tags: geometry unsolved, geometry
sunken rock
02.05.2013 12:01
See that the tangents at $N, D$ concur at $P$!
Best regards, sunken rock
srinidhi321
30.05.2013 21:45
$OPNM$ is cyclic. so,$\angle OPM=\angle ONM=\angle OCM$ $\Rightarrow OPMC\quad $ is a parallelogram. So,$OP=CM$
Itticantdomath
14.04.2016 17:51
srinidhi321 wrote: $OPNM$ is cyclic. so,$\angle OPM=\angle ONM=\angle OCM$ $\Rightarrow OPMC\quad $ is a parallelogram. So,$OP=CM$ $OPNM$ is not cyclic, $ODNM$ is cyclic.
JoelBinu
11.06.2017 15:52
Hi, How did you reach the conclusion that $ODNM$ is cyclic. Thanks
ACGNmath
02.11.2017 17:34
JoelBinu wrote: Hi, How did you reach the conclusion that $ODNM$ is cyclic. Thanks $\angle DNC = 90^{\circ}$, thus $\angle DNM = 90^{\circ} = \angle DOM$, which implies that $ODNM$ is cyclic
ACGNmath
02.11.2017 17:41
Itticantdomath wrote: srinidhi321 wrote: $OPNM$ is cyclic. so,$\angle OPM=\angle ONM=\angle OCM$ $\Rightarrow OPMC\quad $ is a parallelogram. So,$OP=CM$ $OPNM$ is not cyclic, $ODNM$ is cyclic. Actually, $OPNM$ is cyclic too ($\angle ONP = 90^{\circ} = \angle OMP$)