Is not hard to prove with angle relations that $\Delta APQ$ is equlilateral. Now the nine point centre lies on the bisector of $\widehat{BAC}$. Because if $B',A'$ are the middle points of $AB,AC$ and $O'$ the nine point centre, the quadrilateral $O'B'A'C'$ is concyclic $( \widehat{B'O'C'}=120^0)$ ,So $H,O$ are symmetric points about the bisector of $\angle BAC$ , So $OP=HQ$

It's well known that in any triangle $\angle CAH=\angle BAO$. Also it's well known that $AH=AO$ iff $\angle BAC=60^\circ$. Also, APQ is equilateral. So triangles $AQH$ and $APO$ are congruent, and so $PO=QH$

This is a nice but recycled problem. See here: http://www.bmoc.maths.org/home/bmo2-2007.pdf .

Also in our national olympiad there was a problem from British Math Olympiad. We like very much Britain!

Bugi wrote: It's well known that in any triangle $\angle CAH=\angle BAO$. Also it's well known that $AH=AO$ iff $\angle BAC=60^\circ$. Possibly could you explain me why $\angle CAH=\angle BAO$ and if $\angle BAC=60^\circ \implies AH=AO$. Where can I find a proof , any appropriate link ?

To answer Obel1x's first question: $\angle CAH$ $= 90 - \angle C$ (since $AH$ is an altitude) $= 90 - \frac{1}{2} \angle AOB$ (since $\angle ACB$ is inscribed in the circle with center $O$) $=90- \frac{1}{2} (180- 2 \angle BAO)$ (since $\triangle AOB$ is isoceles since $OA=OB=R$) $=\angle BAO$ This is true in any triangle For the second question: Lemma: Let $X$ be the perpendicular from $O$ to $AB$. In any triangle, $CH=2OX$ Proof: Extend $AO$ to meet the circumcircle at $D$, then draw $BD$. $\angle ABD=90$ since it's inscribed in a semicircle, so $\triangle AOX \sim \triangle ADB$, and since $AB=2AX, BD=2OX$. Now we have $CH \parallel BD$, and $\angle DCB= \angle DAB= 90- \angle ADB= 90 - \angle C= \angle CBH$. Thus, $DC \parallel HB$, so $DCHB$ is a parallelogram, so $CH=DB=2OX$. Now for the question. Let $Y$ be the perpendicular from $H$ to $AC$. Then $\angle CHY=60$ so $\triangle CYH$ is a $30-60-90$ right triangle, so $HY= \frac{1}{2} CH= \frac{1}{2}(2OX)=OX$. Thus, since $\angle AYH= \angle AXO=90$ and $\angle HAY= \angle OAX$, $\triangle HYA \cong \triangle OXA$ so $AH=AO$.

thanks dgreenb801, your explanation was so needed.

ridgerso u kualifikove?

Denote the second intersection $S$ of the line $AI$ ($I$ is incenter) with the circumcircle $C(O,R)$ of $\triangle ABC$ . Since $m(\widehat {BOC})=m(\widehat {BHC})=m(\widehat{BIC})=120^{\circ}$ obtain that the points $O$ , $H$ , $I$ belong to the circle $C(S,R)$ and the quadrilateral $AOSH$ is a rhombus, $OH\perp AS$ a.s.o.

Anni wrote: ridgerso u kualifikove? Po, kete vit do ikim vetem 4 vete!

bravo vlla!mu be shume qejfi sinqerisht.ma dergo dhe njehere numrin se kam ndryshuar numer qe te pime ndonje kafe...

Anni wrote: bravo vlla!mu be shume qejfi sinqerisht.ma dergo dhe njehere numrin se kam ndryshuar numer qe te pime ndonje kafe... 0672051285 te enjten mbas ores 5 e gjysem jam ne tirane!

Please, english language ... There are and national communities on this site. Thank you.

Virgil Nicula wrote: Please, english language ... There are and national communities on this site. Thank you. He wrote personal things just for me, and nobody opened so far the requested Albanian Community!

ridgers wrote: Virgil Nicula wrote: Please, english language ... There are and national communities on this site. Thank you. He wrote personal things just for me, and nobody opened so far the requested Albanian Community! Exists and private messages, my dear ....