What does JOM stand for?

Joy of Mathematics/Junior Olympiad of Mathematics, it's a contest created by seniors for juniors in the Malaysian IMO training camp

notice that $A, I, O$ are collinear and $AO \perp DC$. since $\angle ADI = \angle ACI = \angle BCI$, $DBIC$ is cyclic. then $J, O, M$ are on the simson line of triangle $BCD$ wrt point $I$, finish

$ADC$ is isosceles with $DC$ as its base. $AI$ is a bisector and $AO$ is a median. It follows that $AO \equiv AI$ so $A$, $I$, $O$ are collinear. Also for the same reason $AO \perp DC$. Notice $\Delta ADI \equiv \Delta AIC$ Thus $\angle ADI = \angle ACI$. But $\angle ACI = \angle BCI$ so $DCIB$ is cyclic. Consider the circumscribed circle of $\Delta BCD$ $I$ is the incenter of $\Delta ABC$ so $IM \perp AD$ and $IJ \perp BC$. We mentioned before that $AO \perp DC$ Apply the simson line theorem for point $I$ on $\Delta BDC$ and notice that $M$, $J$, $O$ are indeed collinear. Here is the diagram for those curious.

Attachments:

Since $\angle BIC=\dfrac{\pi+\angle BAC}{2}=\pi-\angle{ADC}$ implies $BICD$ cyclic. Then $M$, $J$ and $O$ are concurret by Simson line.

ehuseyinyigit wrote: Since $\angle BIC=\dfrac{\pi+\angle BAC}{2}=\pi-\angle{ADC}$ implies $BICD$ cyclic. Then $M$, $J$ and $O$ are concurret by Simson line. wow! solution!

Solution: The key idea is that triangles $AMI$ and $AOC$ are similar, where $I$ is the incentre of triangle $ABC$. From here we have $\angle AMO= \angle AIC = \pi/2 +B/2$, and $\angle AMJ= \pi - \angle BMJ = \pi - \pi/2 +B/2$, so these angles are equal and we're done.