The problem is just begging for vectors! We have $\bold{q}=\bold{p}+\overrightarrow{PQ}=\bold{b}+\overrightarrow{BP}+\overrightarrow{PQ}=\bold{b}+\bold{a}+\bold{c}$ so $Q$ is the orthocentre.

It has been some time since I last allowed myself to indulge in geometry. Please verify the following solution. Since $BOAP$ is a parallelogram and $BO=AO=R$, where $R$ is the circumradius of $\triangle ABC$, $BOAP$ is a rhombus of side length $R$. So $BP=AP=R$. Since $COPQ$ is a parallelogram with $CO=P$, we know that the opposite side $QP=R$. Then $QP=BP=AP=R$. Without loss of generality, suppose that $\angle B>\angle A$ (generality is not lost as the following arguments can be tailored for the case $A>B$ by replacing each $B$ with $A$, and each $A$ with $B$; the problem states that $AC\neq BC\iff \angle A\neq\angle B$, so that case needs not be considered). Note that $\angle APC,\angle BPC=\frac{\angle APB}{2}=\frac{\angle AOB}{2}=\angle C$. Also, $\angle BOC = 2\angle A\implies \angle POC = \angle C + 2\angle A\implies \angle OPQ = \angle B - \angle A$. We can now say that $\begin{cases}\angle BPQ = \angle C + \angle A - \angle B=\pi - 2\angle B\implies BQ= \sqrt{BP^2 + PQ^2-(BP)(QP)\cos \angle BQP}=2R\cos B\\ \angle APQ = \angle C + \angle B - \angle A=\pi-2\angle A\implies AQ= \sqrt{AP^2 + PQ^2-(AP)(QP)\cos \angle AQP}=2R\cos A\end{cases}$. Let $H$ be the orthocenter of $\triangle ABC$. It is well-known that $\begin{cases}BH = 2R\cos \angle B\\ AH =2R\cos \angle A\end{cases}$. There are up to two points that satisfy these distances from $A,B$; i.e. the interesections of the circle with center $B$ and radius $2R\cos \angle B$, and the circle with center $A$ and radius $2R\cos \angle A$. These two points are clearly the same point or on different sides of $AB$. Since $PB=PQ=PA=R$, $Q$ is inside $\triangle ABC$, and since $\triangle ABC$ is acute, $Q$ is the orthocenter.

Let $M$ be midpoint of $AB$. We have $CQ \perp AB$ since $CQ \parallel OP$ and clearly $OP$ is the perpendicular bisector of $AB$. Also $CQ = OP$ and $OP = 2OM = CH$ and therefore $CQ = CH$. Hence $Q=H$.

Clearly $OA=OB$, hence $P$ is on the perpendicular to $AB$ through its midpoint $M$, and $P$ is the symmetric of $O$ with respect to $AB$. Let $N$ be the midpoint of $CP$, clearly inside $ABC$ (otherwise $O$ would be outside $ABC$), because $ABC$ is acute. Note now that \[MN^2=\frac{ON^2+PN^2}{2}-\frac{OP^2}{4}=\frac{OC^2+OP^2}{4}-\frac{PC^2}{8}+\frac{PN^2}{2}-\frac{OP^2}{4}=\frac{OC^2}{4},\] hence $N$ is at a distance equal to $\frac{R}{2}$ from $M$, on the line halfway between the perpendicular bisector of $AB$ and the altitude from $C$ (by Thales' theorem and because $N$ is the midpoint of $PC$), and inside $ABC$. Hence $N$ is the nine-point center, and $Q$ the symmetric of $O$ with respect to it, thus the orthocenter.

This problem is just Servois' theorem itself.

semy wrote: This problem is just Servois' theorem itself. I've looked for this, but couldn't find it, where can I find a source which talks about this theorem?

This problem is actually trivial if you know that $2MO=CH$. Because of this $COPH$ is a parallelogram and so is $COPQ$, thus $H=Q$

Easy by complex numbers.....just set ($ABC$) as the unit circle and notice that $p=a+b$ and $q=a+b+c$ ..done.

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%207.pdf p. 54... Sincerely Jean-Louis

Let $H$ be the orthocenter of $ABC$. Let us prove $COPH$ is a parallelogram. Claim 1: $HP = OC$ Notice that $BP=OA=OC=OB=PA$. Thus $O$ and $P$ are reflections of each other over line $AB$. Realizing the circle centered at $P$ passing through $A$ (say, $\omega)$ has the same radius as the circle centered at $P$ passing through $A$ (say, $\omega')$. The conclusion follows $\omega$ and $\omega'$ are reflections of each other about $AB$. Recall, the reflection of $H$ about $AB$ lies on $\omega$ so, $H \in \omega'$. Hence, $OH = OA = OB$. Proving the claim. Claim 2: $CH \parallel OP$ Clearly, $O$ and $P$ lie on the perpendicular bisector of $AB$ so, $OP \perp AB$. Clearly, $CH \perp AB$ so $CH \parallel OP$. Proving the claim. Claim 1 and Claim 2 are sufficient to ensure $COPH$ is a parallelogram. So, we are done.

Well Just use complex numbers Let $a,b,c$ lie on the unit circle . I don't like vectors (Alone) , so we use the fact that the diagonals of a parallelogram bisect each other to get $p=a+b$ and $q=a+b+c$, Thus $q$ is the orthocenter . Moral of the story , Don't be sticky with Synthetic solutions. Like complex numbers finish this in a second or two.

We have $OA=OB=OC$ because $O$ is the circumcenter of $ABC$, so the paralellograms imply that $PA=PB=PQ$, hence $P$ is the circumcenter of $ABQ$, this gives$$\angle AQB=180^\circ -\frac{\angle BPA}{2}=180^\circ -\frac{\angle AOB}{2}=180^\circ -\angle ACB.$$Observe that $OA=OB$ and $PA=PB$ imply that $OP\perp AB$, so $CQ\perp AB$. Hence $Q$ is the orthocenter of $ABC$.

βλέπε σχήμα:

$$Q=P+C=A+B+C$$

Clarly $P$ is the reflection of $O$ wrt $AB.$ So $Q$ is a point in the $C-altitude$ such that $OP=CQ.$ Clarly the orthocenter satisfies this lenght condition, and thus we are donne.

So we use complex numbers... Set $(ABC)$ as the unit circle. We got $p+0=a+b$ so $p=a+b$. We got $c+p=0+q$ so $q=c+p=a+b+c$ which means $Q$ is the orthocenter.

Complex is actually quite immediate but we give a synthetic solution anyways. First, we show $Q$ lies on $\overline{CH}$. Claim : The line $\overline{CQ}$ is infact the $C$-altitude of $\triangle ABC$. We note that, \begin{align*} \measuredangle QCO &= \measuredangle POC \\ &= \measuredangle POB + \measuredangle BOC\\ &= \measuredangle OPA + \measuredangle BOC \\ &= \measuredangle AOP + \measuredangle BOC\\ &= 90 + \measuredangle OAB + \measuredangle BOC \\ &= \measuredangle ACB + 2\measuredangle BAC \end{align*}Also, we can verify that \begin{align*} \measuredangle HCO & = \measuredangle BCA + 2 \measuredangle ACO\\ &= \measuredangle BCA + \measuredangle AOC\\ &= \measuredangle BCA + 2 \measuredangle ABC \\ &= \measuredangle ACB + 2 \measuredangle BAC \end{align*}from which it is clear that $\measuredangle QCO = \measuredangle HCO$ which implies that $Q$ lies on $\overline{CH}$ as claimed. Now, it is well known that (and easy to prove) if $M_C$ is the midpoint of $AB$ and $N_C$ is the midpoint of $CH$, then $CN_CM_CO$ is a parallelogram. First note that since in a parallelogram, diagonals bisect each other, $PO$ passes through the midpoint of $AB$ (which is $M_C$). Thus, \[QC = OP = 2OM_C = 2CN_C=CH\]But combining this with the previous fact, concludes that indeed $Q$ is the orthocenter of $\triangle ABC$ and we are done.

Clearly the point $Q$ is unique. Let $H$ be the orthocentre of $\Delta ABC$. Our goal is to prove that $COPH$ is a parallelogram. Observe that if $F$ is the midpoint of $AB$, then $OF \perp AB \implies P \text{ is the reflection of } O \text{about line} AB$. Now clearly $OP \parallel CH$, and also $OP = 2R\cos C = CH$ where $R$ is the circumradius of $\Delta ABC$, from standard trigonometry results, so we are done.