Also was Romania JBMO TST 1 Problem 1. Solution: $O_1$ is circumcenter of $ABC$;then,since the circumcircles of $ABC$ and $HBC$ are symmetrical wrt $BC$,we have that $O$ is the reflection of $O_1$ over $BC$;it is also obvious that $AD=\frac{1}{2}AH=O_1M=OM$ as well as $AD||OM$,Q.E.D.

2015 year?

IstekOlympiadTeam wrote: 2015 year? yes.

It was actually taken from Junior Balkan MO shortlist 2014. Solution which uses simple trigonometric manipulations is also an option.

Complex numbers: Let each lowercase letter denote its corresponding point in the plane. $d=a+\frac{b+c}{2}, m=\frac{b+c}{2}$, and it is well known that $o=b+c$. Indeed $a+o=d+m=a+b+c$, so we're done.

sorry from where you know that o=b+c and another question if a+o=d+m can you say that damo is parallelogram please can you explain

By theorem $OM$ = $AD$ = $DH$. Since $AH$ $//$ $OM$ and $AD$ = $OM$ $\longrightarrow$ $AMOD$ is a parallelogram.

$OM=\frac{a}{2\tan{A}}$ and $AH=\frac{a}{\tan{A}}$ so $AD=OM$ but $AD\parallel OM$ the conclusion follows

Dear Mathlinkers, just a simple application of the Carnot's formula... AH = 2.OM where M is the midpoint of BC and O the circumcenter of ABC... Sincerely Jean-Louis

what's e Carnot's formula?

Dear Mathlinkers, because this formula was found by Carnot around 1800... Sincerely Jean-Louis

We know $DH||OM $ and, by angle chasing and trig, $DH=OM=\frac {BC}{2\tan A} \implies OMDH \text { is parallelogram}$

Let $l_1$ denote the Euler line , by similarity $AH=2\times{OM}$ ,since $AM\cap{l_1}=G$ where $G$ is the centroid and $HG=2\times{OG}$ . Note that $\angle AKM = \angle OMC = 90^{\circ}\Rightarrow{AK\parallel{OM}}\Rightarrow{AH\parallel{OM}}$ Since $AD=OM$ and $AD\parallel{OM}$ $DAMO$ is a parallelogram . $\blacksquare$

Attachments: