Let $H$ be the orthocenter of $ABC$.Then it is well known that $AO$ and $AH$ are isogonal WRT $\angle BAC$,so $A$,$H$,$D$ are collinear. It is also well known that $BH=BE$,so $MD=MB=1/2BH=OP$. Let $MD$ intersect $AC$ at $K$. $\angle DKA=180-\angle KAD-\angle KDA=180-90+\gamma -(90-\angle MDB)=90+\gamma -(90-\angle MDB)=90+\gamma -(90-\angle CAD)=90+\gamma -\gamma =90 $. So $OP$ and $MD$ are parallel.So triangles $MDN$ and $NOP$ are congruent and conclusion follows.

Let $N =MP \cap OD$.We will prove $N$ is the midpoint of $OD$.As $AD$ is isogonal to $AO$ ,so $H$ lies on $AD$.By the well known property ,$D$ is the midpoint of $HE$ and $M$ is the midpoint of $BE$ So $MD \mid\mid BH $ Also $BH$ and $OP$ are parallel So $MD \mid\mid OP$.Now $MD=BM =\frac{1}{2}BH=\frac{1}{2} .2OP=OP$ So $\triangle MND$ and $\triangle NOP$ are congruent $ \Longrightarrow NO=OD$.Hence done.

Let $H$ be the orthocenter.A little angle chasing proves that $MD||OP$ Also since $\Delta EDB \sim \Delta CDA \Rightarrow \frac{MD}{DH}=\frac{MD}{DE}=\frac{DP}{DC}$ But $\Delta HCD \sim \Delta OAP \Rightarrow \frac{DP}{DC}=\frac{AP}{DC}=\frac{PO}{DH}\Rightarrow MD=OP \Rightarrow MDPO$ a parallelogram.Done!

Analitic solution: Let D(0,0),A(0,a),B(b,0),C(c,0).We find $O(\dfrac{b+c}{2},\dfrac{a^2+bc}{2a})$,and $E(0,\dfrac{bc}{a})$.So,$M(\dfrac{b}{2},\dfrac{bc}{2a}),N(\dfrac{b+c}{4},\dfrac{a^2+bc}{4a}),P(\dfrac{c}{2},\dfrac{a}{2})$.Because MN,MP has the same slope(equal with $\dfrac{a^2-bc}{a(c-b)}$),the conclusion follows

Hello, Can someone explain why $\frac{1}{2}BH=OP$?

It can be easily proved using trigonometry that $ \frac{1}{2}BH=OP= \frac{1}{2}\cot \beta $

The following analytical solution has some unexpected consequences, detailed in the sequel. Choose a coordinatesystem $xOy$ where $O(0,0), A(a,u), B(-b,v), C(b,v), A'(-a,-u)$. From $\angle BAD = \angle CAO$, i.e. $\angle BAE = \angle CAA'$, follows $E(a,-u)$ (thus $AD\perp BC$), hence $D(a,v)$. But then $M\left (\dfrac{a-b}{2},\dfrac{v-u}{2}\right )$, $N\left (\dfrac {a} {2},\dfrac{v}{2}\right )$, $P\left (\dfrac{a+b}{2},\dfrac{u+v}{2}\right )$. The slope of the line $MN$ is $\displaystyle \dfrac {\dfrac {v} {2} - \dfrac {v-u} {2}} {\dfrac {a} {2} - \dfrac {a-b} {2}} = \dfrac {u} {b}$, while the slope of the line $NP$ is $\displaystyle \dfrac {\dfrac {u+v} {2} - \dfrac {v} {2}} {\dfrac {a+b} {2} - \dfrac {a} {2}} = \dfrac {u} {b}$; as seen, they coincide, proving the colinearity. We can certify now a suspicion lingering from the very beginning, namely that $AB<AC$ is a useless restriction (not even the isosceles case produces any discomfort). Neither the requirement $\triangle ABC$ be acute-angled is instrumental (true, some degenerate cases may appear, but they only increase the wealth of the configurations; also, point $D$ may now be outside the side $BC$). It is strange the choice of the jury to abide by these irrelevant conditions !?!

trigisfun wrote: Hello, Can someone explain why $\frac{1}{2}BH=OP$? See this an important part of proving the existence of the euler line i.e $O,G,H$ are collinear. if $BO \cap \odot ABC =X$ First prove that $HAXC$ is a parallelogram and then you can see that $H ,P,X$ are collinear .Then by using similarity of $\triangle BHX$ and $OMX$ ,we get the required result.

mavropnevma wrote: We can certify now a suspicion lingering from the very beginning, namely that $AB<AC$ is a useless restriction (not even the isosceles case produces any discomfort). Neither the requirement $\triangle ABC$ be acute-angled is instrumental (true, some degenerate cases may appear, but they only increase the wealth of the configurations; also, point $D$ may now be outside the side $BC$). It is strange the choice of the jury to abide by these irrelevant conditions !?! In my opinion, it was done on purpose so that students won't lose time checking other configurations.

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/* end of picture */ [/asy][/asy] Lemma: It is well known that the reflection of the orthocentre $H$ through the midpoint of $BC$ and $AC$ is the antipode of $A, B$ respectively. If we let primes denote the antipodes of points, we have $BA'CH$ (Bach hehe) and $AB'CH$ are parallelograms. Part 1: Through dilations with ratio $2$ about $E$ and $A$, we take $MD \to BH$ and $OP \to A'C$. But, $BA'CH$ is a parallogram, so $BH \parallel A'C \implies MD \parallel OP$. Part 2: Through dilations with ratio $2$ about $E$ and $H$, we take $OM \to E'B$ and $DP \to EB'$. But $EB'BE'$ is a rectangle (points and their antipodes), so $E'B \parallel EB' \implies OM \parallel DP$. Therefore, $OMDP$ is a parallelogram, so the $MP$ goes through the midpoint of $OD$, or $M, N, P$ are collinear, as desired $\blacksquare$.

We have $ACEB$ is cyclic quadriteral and dioganals perpendicular - $AE\bot BC.$ So, $N$ - center of the eight points circle.

useful lemma: Given cyclic quadrilateral $ABCD$ with perpendicular diagonals. Let $AC\cap BD=E$ and foot of perpendicular from $E$ to the side $AB$ be $X_{1}$ and that perpendicular line meets $CD$ at $Y_{1}$. Define $X_{2}, Y_{2}, X_{3}, Y_{3}, X_{4}, Y_{4}$. Then these 8 points are concyclic.

liimr wrote: useful lemma: Given cyclic quadrilateral $ABCD$ with perpendicular diagonals. Let $AC\cap BD=E$ and foot of perpendicular to the side $AB$ be $X_{1}$ and meets $CD$ at $Y_{1}$. Define $X_{2}, Y_{2}, X_{3}, Y_{3}, X_{4}, Y_{4}$. Then these 8 points are concyclic. From Where the foot of the perpendicular is drawn to $AB$?

BBAI wrote: From Where the foot of the perpendicular is drawn to $AB$? I edited my post I think now it is understandable

liimr can you prove the lemma?

It's just angle chasing , and maybe Brahmagupta's lemma can be use here.

Math-lover123 wrote: liimr can you prove the lemma? proof is just angle chasing. Firstly prove that $X_{1}, X_{2}, X_{3}, X_{4}$ are concyclic then try to prove $X_{1}, X_{2}, X_{3}, Y_{3}$ are concyclic then others will follow directly.

Not too hard.

it's obvious to see that , $AD$ is perpendicular to $BC$ and $E$ is the reflection of $H$ w.r.t. $BC$. now , WLOG , assume that , the circumcircle of $ABC$ is the unit circle centred at origin. then $h=a+b+c$ . also , we get , $d=(a+b+c-bc/a)/2$. so , $e=-bc/a$ . hence , $p=\frac{a+c}{2}$. $n=[a+b+c-bc/a]/4$ $m=b(a-c)/2$ . then it is easy to see that $\mu_{NP}=(b-a-c-bc/a)/(1/b-a/bc-1/a-1/c)$ and , $\mu_{PM}=\frac{a+c-b+bc/a}{a/bc+1/a+1/c-1/b}$ . hence $M,P,N$ are collinear.

Author of this problem is Stefan Lozanovski (Стефан Лозановски).

$\triangle BDE\sim\triangle ADC$ and being right-angled means the median $DP$ of $\triangle ADC$ is altitude in $\triangle BDE$, so $DP\parallel OM$ ($OM$ is perpendicular bisector of $BE$), and similarly the median $DM$ is altitude of $\triangle ADC$, hence $DM\parallel OP$ ( the latter being perpendicular bisector of $AC$), hence $OMDP$ is parallelogram, done. Best regards, sunken rock

It's well known that the orthocenter is the isogonal conjugate of the circumcenter; therefore, we know that $AD \perp BC$. Then we know that $BM=MD = ME$, and $\angle BCA = \angle BEA = \angle MDE$. Extend $ME$ to intersect $AC$ at $P'$. Therefore, $\angle ADP' = \angle BCA = 90^{\circ} - \angle EAC$, so we see that $\angle DP'A = 90^{\circ}$, so $DM || OP$. Let $AO$ intersect $\omega$ at $Q$. Then, we know that $CQ = BE$, so $OP = MD$. Therefore, $MDPO$ is a parallelogram, so the diagonals bisect each other, and we are done.

Triangles POC and DEC are similar, using PC:CD=PO:DE and AC:BE=CD:DE. Since AC=2PC and BE=2ME=2MD we get MD=PO. N is midpoint of DO so DN=NO. We will prove angle MND=PNO, respectively triangles MDN and NOP are congruent. Easy to prove that angle MDN=PON. And we have MD=PO , DN=NO and angle MDN=PON, so triangles MDN and NOP are congruent.

Using some simple trig bash, we find that OPDM is a ||gm. If PM meets OD in N', then N'D = N'O. This implies that N and N' coincides and hence the result follows.

Very challenging and beautiful problem! Solution with v_Enhance, msinghal, linpaws, and mathcool2009. By $\sqrt{bc}$ inversion, $AD$, and $AO$ map to each other, so $AD$ passes through the reflection of $A$ over $BC$, hence $D$ is the foot of the $A$-altitude. Consider a homothety centered at $D$ with factor two, so that $M$ maps to $M'$, $P$ maps to $P'$, and $N$ to $O$. Also, let $Q,R$ be the reflections of $D$ over the midpoints of $AB$, $CE$, so that $QP'RM'$ is a rectangle. Since its center is the intersection of the perpendicular bisectors of $P'R$ and $RM'$, which are precisely the perpendicular bisectors of $AE$ and $BC$, which meet at $O$, $O$ is on the diagonal $P'M'$. $\blacksquare$

Let $H$ be the orthocenter of $ABC$. From Brahmagupta's theorem, $MD$ is perpendicular to $AC$, which is perpendicular to $OP$. Also we have $MD = 0.5 BE = 0.5 BH = OP$, so $MDOP$ is a parallelogram, and thus we are done.

Straight-forward with complex numbers. Set $(ABC)$ as unit circle. $m=\frac{b+(\frac{-bc}a)}2$ $p=\frac{a+c}2$ $d=\frac{1}2 \left(a+b+c-bc\overline{a} \right)$ Midpoint of $MP$ is $\frac{1}2 \left (\frac{b+(\frac{-bc}a)}2+\frac{a+c}2 \right)=\frac{1}2 \left(\frac{1}2(a+b+c-bc\overline{a}) \right)=\frac{o+d}2=n$ Hence, $N$ is midpoint of $MP \implies M,N,P$ are collinear, as desired. $\blacksquare$

(A non-guessing solution using Complex numbers,isogonal conjugates) The orthocenter $H$ is the isogonal conjugate of the circumcenter $O$, combining this with the given angle condition we have $AD \perp BC$.Working in the complex plane we set $|A|=|B|=|C|=|E|=1$. Since $AE \perp BC$ we have $ae+bc=0$ so $e=\frac{-bc}{a}$. $d=\frac{1}{2} (a+b+c-\frac{bc}{a})$, $n=\frac{1}{4}(a+b+c-\frac{bc}{a})$, $m=\frac{1}{2}(b-\frac{bc}{a})$ $p=\frac{1}{2}(a+c)$ $P,M,N$ are collinear if $\frac{p-n}{p-m}$ is a real number. $\frac{p-n}{p-m}=\frac{\frac{1}{2}(a+c)-\frac{1}{4}(a+b+c-\frac{bc}{a})}{\frac{1}{2}(a+c)-\frac{1}{2}(b-\frac{bc}{a})}=\frac{\frac{1}{4}(a+c-b-e)}{\frac{1}{2}(a+c-b-e)}=\frac{1}{2}$ So $M,N$ and $P$ are collinear.

this problem becom very easy in the moment when you realized a known confiraguration. That MDPO is a parallelogram.

It is well known that the isogonal of $AC$ is the $A$-Altitude of $\triangle ABC \implies AD$ is the $A$-Altitude. Let $(ABC)$ be the unit circle. $$h = a + b + c$$$$d = \frac{a + b + c - \frac{bc}{a}}{2}$$Also since $E$ is the reflection of $H$ across $BC \in (ABC)$ $$\implies d = \frac{h + e}{2} \implies e = 2d - h$$$$p = \frac{a+c}{2}$$$$n = \frac{d + o}{2} = \boxed{\frac{d}{2}}$$$$m = \frac{b + e}{2} = \frac{b - h + 2d}{2} = d - \frac{a+c}{2} = d - p$$$$\implies m + p = d$$$$\implies \frac{m+p}{2} = \frac d2 = n$$Hence $N$ is the midpoint of $PM$ and we are done.

@mathur and @liimr in posts 13 and 14 respectively, I read the 8 point theorem and I found it really impressive! But when it comes to the accual problem, I couldnt combine these two.... It is clear that the ABEC has perpendicular diameters and N is the midpoint of BE and P the midpoint of AC, but I cant keep applying this theorem. Can anybody helps me with that???

It is well known that $N$ is the center of the eight point circle of quadrilateral $ABCD$ with diameter $MP,$ and the conclusion follows. Note: This problem is also killed by the so-called parallelogram trick(which is in author's book).

Note that: $AD$ is altitude from $A$ to $BC$ So: $MO \perp BE$ and $DP \perp BE$ $\implies MO \parallel DP$ $OP \perp AC$ and $MD \perp AC$ $\implies OP \parallel MD$ $\implies MDOP$ is paralelogram. Then: $M,N,P$ are collinear.$\blacksquare$

Extend $ AO $ so that it meets $\omega $ at $F$. As $\angle B A E=\angle C A F$ so, $BE=CF$ $\Rightarrow$ $BCEF$ is an isoceles trapezoid with $ BC \parallel EF $. Now $\angle A E F=90^{\circ} \Rightarrow \angle A D C=90^{\circ}$. Let $\angle B A E=\angle C A O=x$ and $\angle M E D=\angle M D E=\theta$, then by some angle chase, we get $\angle B M D=2 \theta$ $\Rightarrow$ $\angle D M O=90-2 \theta$. Similarly, we get $\angle D P O=\angle D P C-\angle O P C=180-2 \theta-90=90-2 \theta$. Also $\angle M D P=360-(\angle A D P+\angle A D B+\angle B D M)=90+2 \theta$ and $\angle M O P=360-(\angle M O E+\angle C O E+\angle P O C)=90+2 \theta$. As the opposite angles are equal, $MDOP$ is a parallelogram and $N$ is the midpoint of the diagonal $OD$ and so we get $M,N,P$ are collinear.

trigisfun wrote: Hello, Can someone explain why 1/2BH=OP? see Property 10.3.2.

Note that by isogonality $AD\perp BC$. Claim: $MDPO$ is a parallelogram. Extend $DM$ to meet $AC$ at $Q$. Note that since $DM$ is a median in $\triangle DBE$, $DQ$ is a symmedian in $\triangle ADC$, and is therefore also an altitude (since it is a right triangle). Therefore, $DM\perp AC$, so $DM\parallel OP$. Let $H$ be the orthocenter. Furthermore, $BH=2OP$, so $BE=2OP$ since $E$ is the reflection of $H$ over $D$. Then $MD=ME=\frac{1}{2}BE$, so $MD=OP$ which shows the claim. Since $MDPO$ is a parallelogram, $MP$ passes through the midpoint of $OD$, so we are done.

Clearly, we have $AD \perp BC$. Now, apply complex numbers with $(ABC)$ as the unit circle. We then have $d=\frac{a+b+c-\frac{bc}{a}}{2}, e=\frac{bc}{a}$. The three midpoints are just \begin{align*} \frac{a+b+c-\frac{bc}{a}}{4} \\ \frac{ab-bc}{2a} \\ \frac{a+c}{2} \\ \end{align*} These are collinear by the Collinearity Criterion. $\blacksquare$

Wow so nice Consider the rectangle passing through $A, B, C$ and $E$ such that the sides are parallel to $BC$ and $AE$. Note that a homothety of factor $2$ centered at $D$ maps $M$ and $P$ to opposite vertices of the rectangle and $N$ to $O$. It suffices to prove that $O$ lies on the diagonal from the opposite vertices, which is true because $O$ lies on the perpendicular bisector of $BC$ and $AE$.

from $\angle BAH$=$\angle CAO$ lemma $AD$ is perpendicular to $BC$ from a little angle chasing you will observe that if $MDPO$ is parallelogram then we are done we can see that $OM$ and $DP$ are perpendicular to $BE$ similarly $MD$ and $OP$ are perpendicular to $AC$ Thus $OM//DP$ and $MD//OP$ since that $MDPO$ is parallelogram.

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It is well known that $D$ is the foot of the perpendicular from $A$ to $BC$. Then $(ABC)$ unit circle and braindead complex bash easily kills.