Actually the problem is kind of trivial. A quick look at the drawing shows that the definition of the points $ Y$ and $ Z$ are useless. What we really need to prove is that $ OF \perp FX$. For that we can use complex numbers: let $ a,b,c$ be the affixes of the triangle $ ABC$, $ a\overline a = b \overline b = c\overline c = 0$. Then $ h = a + b + c$, $ f = \frac { a + b + c - ab\overline c}{2}$, $ p = b + c - ab\overline c$. Then take $ X$ to be the intersection between the line $ BC$ and the perpendicular in $ F$ on $ OF$. We get the equations \begin{align*} x + \overline x bc & = b + c \\ x\overline f + \overline x f & = 2 |f|^2 . \end{align*} We get $ x = \frac { 2|f|^2 + \cfrac {b + c}{bc} f }{\overline f - \cfrac { f}{bc} }$, and now we just have to prove that $ h,x,p$ are collinear, namely $ \begin{array}{|ccc|}x & \overline x & 1 \\ h & \overline h & 1 \\ p & \overline p & 1 \end{array} = 0$, which is just an easy computation.

We really don't need complex numbers here. The following purely synthetic solution is quite natural. Denote by $ C^{\prime}$ the second intersection of $ CH$ with the circumcircle of triangle $ ABC$. Since $ F$ is the midpoint of the segment $ HC^{\prime}$, the quadrilateral $ AC^{\prime}PH$ is a parallelogram, and $ F$ is its center of symmetry. Since $ X$ lies on line $ PH$, its reflection $ X^{\prime}$ into $ F$ lies on $ AC^{\prime}$. Thus, according to the butterfly theorem applied in the cyclic quadrilateral $ AC^{\prime}BC$ with circumcenter $ O$, we have that $ OF \perp XF$, and therefore the points $ F$, $ M$, $ Y$, $ Z$ are concyclic.

pohoatza wrote: We really don't need complex numbers here. The following purely synthetic solution is quite natural. Well it took me around 2 minutes to find the above solution, and say another 10 minutes max to write it in the competition. I take 7 points at the Balkan in 12 minutes than the chance of not finding the synthetic solution in 20 or more

I have the pleasure to present you the synthetic solution I found during the competition.

I don't like complex numbers, but I like coordinates. Put origin in $ F$ and $ B(b,0), b<0$, $ A(a,0), a>0$ and $ C(0,1)$. Then $ H(0,-ab)$ and $ O(\frac {a+b}2, \frac {ab+1}2)$, $ P(-a,0)$ and we compute $ X(\frac {b(ab+1)}{1-b^2}, \frac{-b^2 - ab}{1-b^2})$. So, slope of $ FO$ is $ \frac {ab+1}{a+b}$ and the slope of $ FX$ is $ -\frac{a+b}{ab+1}$ so $ OF\bot FX$, that's it. Bye

Very nice Pohoata's and Nick Rapanos' proofs ! Indeed, the points $ M$ , $ Y$ , $ Z$ are useless . Here is a "cleaner" enunciation : Quote: Given a scalene acute triangle $ ABC$ with orthocenter $ H$ , circumcenter $ O$ and $ AC > BC$ . Denote : $ F\in AB\ ,\ CF\perp AB$ ; the reflection $ P$ of $ A$ w.r.t. $ F$ ; $ X\in HP\cap BC$ . Prove that $ FO\perp FX$ . Virgil Nicula wrote: Lemma. Let $ ABC$ be a triangle for which exists a point $ D\in (BC)$ so that $ AD\perp BC$ . Consider an interior point $ X$ of the triangle $ ABD$ and an interior point $ Y$ of the triangle $ ACD$ so that $ \widehat {XAB}\equiv\widehat {XBA}\equiv\widehat {YAD}\equiv\widehat {YCD}$ . Then $ DX\perp DY\ \Longleftrightarrow\ AB = AC$ .

Remark. If we'll apply the upper lemma to the $ C$ - isosceles triangle $ CAP$ , i.e. $ CA = CP$ and the points $ O$ , $ X$ then we'll obtain the required property $ FO\perp FX$ .

Just observe that reflection of $ X$ wrt $ CF$ is isogonaly conjugate to $ O$ wrt triangle $ AFC$ so $ FO \perp FX$.

i cant find what is difficult with this problem as we see that it can be done only after few trigonometric computations....

I am sorry, but I must to say that this problem is "painted" clumsy. Here is "disgusting paint" : if we suppress the points $ M$ , $ Y$ , $ Z$ and replace the its conclusion with only "Prove that $ FX\perp FO$ " then obtain a classical problem which appears and in the book "Harmonic division" of V. N. & C.P. , where it is solved with the butterfly property (see the upper Pohoata's proof) of the point $ F\in AB\cap CH$ in the circumcircle of $ \triangle ABC$ . Best regards, Virgil Nicula P.S.1. I have a curiosity : who is the author of the problem 1 from BMO 2008 ? P.S.2. My upper proof is an entertaiment for to find a metrical proof - my great pleasure !

This problem is equivalent to 1996 shortlist problem.

Hmm this problem appears on the Blue MOP homework... This problem can be solved directly by two applications of trig ceva.

Another approach: $ \angle OCA = \angle OAC = m,\angle OCF = n$ $ \implies$ $ \angle HCB = m = \angle ABH$ $ CF\perp AP,AF = FP$ $ \implies$ $ \angle HPA = m,\angle BCP = n$ $ \angle XFB = \alpha,\angle HFZ = \beta$ let we use in $ \triangle AFC,\triangle PFC$ trig ceva $ \implies$ $ \alpha = \beta$ $ \implies$ $ FZ\perp FX$ $ \implies$ $ FMYZ$ concylic

Dear Mathlinkers, this problem has a relation with the butterfly theorem... see also http://perso.orange.fr/jl.ayme vol. 7 Butterfly... p. 55 Sincerely Jean-Louis

Umm... if I'm not mistaken and if my computations aren't either, then $OM\parallel FX \Leftrightarrow \dfrac{\tan{A}\cdot \cos{C}}{\sin{(A-B)}}=1$, and I think one can find angles $A,B,C$ to satisfy the condition and hence the point $Y$ (might) not exist in some situations..

My solution: Let $ X' $ be the reflection of $ X $ in $ CF $ . Easy to see $ X' \in AH $ . Since $ \angle X'CA=\angle PCX=\angle FCO $ , so we get $ X', O $ are isogonal conjugate of $ \triangle CAF $ , hence $ \angle YFZ=\angle XFO=90^{\circ}-\angle OFC+\angle AFX'=90^{\circ}=\angle YMZ $ . ie. $ F, M, Y, Z $ are concyclic Q.E.D

Let $B'$ be the antipode of $B$ wrt $O$, and let $N$ be the midpoint of $AB$. It suffices to prove that $OF \perp FX$. It is well-known that $CH = 2ON = B'A$. Also, $PB=2AF-AB=2(AF-AN)=2NF$. From $\angle XCH=\angle HAB = \angle XPB$ and $\angle CXH = \angle PXB$, we have $\triangle CXH \sim \triangle PXB$. Now note that $\frac{FD}{DX}=\frac{LX}{DX}=\frac{CX \sin \angle XCH}{XP \sin \angle XPB}=\frac{CX}{XP}=\frac{CH}{PB}=\frac{2ON}{2NF}=\frac{ON}{NF}$, so $\triangle ONF \sim \triangle FDX$. Therefore, we have $OF \perp FX$, as desired $\blacksquare$

We need $ \measuredangle XFO=90$ which is equivalent to $\measuredangle BFX = \measuredangle CFO $ or $\measuredangle XFH = \measuredangle OFA $. applying the sine law in triangles $OFA $, $OFC $, $XFB$, $XFH$ we get : $$ \frac {sin\measuredangle BFX }{sin\measuredangle XFH} =\frac {sin\measuredangle CFO}{sin\measuredangle OFA}$$and we know that $$\measuredangle BFX + \measuredangle XFH =\measuredangle OFA + \measuredangle CFO =90 $$hence $\measuredangle BFX = \measuredangle CFO $ and $\measuredangle XFH = \measuredangle OFA $ and we are done.

Why does $FO$ being perpendicular to $FX$ imply that $F, M, Y, Z$ are concyclic?

orthocentre wrote: Why does $FO$ being perpendicular to $FX$ imply that $F, M, Y, Z$ are concyclic? Because OM is already perpendicular to AC.