It can be done nicely using complex numbers. Let $O$ be the origin of the complex plane, and $|a|=|b|=|c|=1$. Observe that $M$ is the refflection of $B$ wrt $HC$. So: $m=a+c-\dfrac{ab}{c}$. By the same way: $n=a+b-\dfrac{ac}{b}$. Now, we're supposed to find $O'(o')$ (the circumcenter of $\triangle{MHN}$). Using the well-known lemma: If $X(x), Y(y)$ and $Z(z)$ are three distinct points in plane, then the circumcenter $W$ of $\triangle {XYZ}$ satisfy: \[ w\left (\dfrac{\overline{x}-\overline{y}}{x-y}-\dfrac{\overline{x}-\overline{z}}{x-z}\right )=\dfrac{|x|^2-|y|^2}{x-y}-\dfrac{|x|^2-|z|^2}{x-z} \] After expanding both sides we found that: \[ o'=\dfrac{(b^2+c^2)(a+b+c)}{b^2+bc+c^2} \] $O'\in OH$ if and only if: $\dfrac{h}{o'}=\dfrac{\overline{h}}{\overline{o'}}$. But this is obvious, because $\dfrac{b^2+c^2}{b^2+bc+c^2}\in\mathbb{R}$.

This problem can also be solved by inversion. Choose O as the inversion point.

litongyang wrote: This problem can also be solved by inversion. Choose O as the inversion point. How? I absolutely don't see how this inversion simplifies the problem.

Is there another solution ? (without complex number)

litongyang wrote: This problem can also be solved by inversion. Choose O as the inversion point. Can you post your solution use inversion ?

It's enough to prove that $ \angle{MHO}+\angle{MNH} = 90^{\circ} $. But I dunno how to prove!

let circle WMH intersect AB and AC at points P and Q easy to see that angle QPH = PQH = QWH = HCA , let line parallel to QP intersects HP and HQ at points P' and Q' easy to see that BHAP' and AHCQ' are cyclic and HP' = HQ' so HP*PP' = AP*PB = power of point P wrt circumcercle of ABC and HQ*QQ' = CQ*QA = HP*PP' = PB*PA = power of point Q wrt circumcercle of ABC so power of point Q = power of point P wrt circumcercle of ABC so OQ = OP this idea you can see in IMO 2009 geometry problem

Stifler wrote: If $X(x), Y(y)$ and $Z(z)$ are three distinct points in plane, then the circumcenter $W$ of $\triangle {XYZ}$ satisfy: \[ w\left (\dfrac{\overline{x}-\overline{y}}{x-y}-\dfrac{\overline{x}-\overline{z}}{x-z}\right )=\dfrac{|x|^2-|y|^2}{x-y}-\dfrac{|x|^2-|z|^2}{x-z} \] I'd like to point out that there's a slightly simpler way to carry out this computation: first note that the circumcenter of the triangle with vertices $0,x,y$ is \[\frac{xy(\overline{x}-\overline{y})}{\overline{x}y-x\overline{y}},\]so by translating we have \[o'=h+\frac{(m-h)(n-h)(\overline{m}-\overline{n})}{(\overline{m}-\overline{h})(n-h)-(m-h)(\overline{n}-\overline{h})}.\]Now we just use \begin{align*} m-h=-\frac{b}{c}(a+c) &\implies \overline{m}-\overline{h}=-\frac{a+c}{ab},\\ n-h=-\frac{c}{b}(a+b) &\implies \overline{n}-\overline{h}=-\frac{a+b}{ac},\\ &\implies \overline{m}-\overline{n}=\frac{(b-c)(a+b+c)}{abc}. \end{align*}

It is true that < BMH = < ACH = < MBH = < OBC = 90 - <A. Thus ∆HMB ~ ∆OBC, now of the similarity OH/BC = BH/BM →OH/BN = BH/BM →OH/BH = BN/WB and < MBC = < C - <A = < OBH this ∆MBN ~ ∆HBO → < AMN = < OHQ = α, where BQ is height. then < NHQ = A → < OHN = α-A also < HMB = 90 - A → < HMN = 90 + A-α → < HMN + < OHN = 90 this complete solution for If draw the MHN circuncirculo to prolong OH until that cut to the circuncirculo in S →El arc SN is equal to 2(α-A) and the NH arc is equal to 180 - 2(α-A) →el arc SH is 180 this implies that OH is diameter and contains the circumcenter of MHN.

Proof: Apply an inversion $ I(H, HA\cdot HB) $ where $HA \cdot HB $ is the directed product the two lengths. This inversion sends the vertices $ ABC $ onto the corresponding vertices of the orthic triangle $ DEF $ ( $ D $ , $ E $ and $ F $ are the foot of perpendiculars from $ A $ , $ B $ and $ C$ to the opposite sides respectively ). The lines of triangle $ ABC $ which are $ BC$ , $ CA $ and $ AB $ go to the circles $ HEAF$ , $ HDFB $ and $ HDEC$ respectively. The circles $ HAB $ and $ HAC $ go to the lines $ DE $ and $ DF $ respectively. Denoting the image of $X $ as $ X' $ , we get that the image $M' $ , $ N' $ of points $ M $ , $N $ is the intersection point of the circle $ HDEC $ and $ HDFB $ with $ DF $ and $ DB $ . The statement is then equivalent to proving that $ TH $ , where $ T $ is the nine point centre of triangle $ ABC $, is orthogonal to $ M'N'$; this must happen since the centre of the circle $ HMN $ must lie on the foot of perpendicular of $ H $ to the line $ M'N' $ which is by the statement of the problem must be shown to be the Euler line of triangle $ ABC $ . But $ H $ is the incentre of triangle $ DEF $ and $ T $ is it's circumcentre. Now $ M' $ and $ N' $ are such that $ E $ lie outside the segment $ M' D $ and $ F$ outside segment $ N'D $ and such that $ EM' = FN' = EF $ . This is a well know result that the circle passing through two vertices of the triangle and the incenter is symmetric in the angle bisector of the remaining vertex. To show that $ HT $ is orthogonal to $ M'N' $ , one just computes the expression $ HM'^{2} + TM'^{2} - HN'^{2} - TN'^{2} $ . We have $ TM'^{2} = M'D \cdot M'E + \frac{TD^{2}}{4} = (M'E - DE) M'E + \frac{TD^{2}} { 4 } = M'E^{2} - M'E \cdot DE + \frac{TD^{2} }{4} = EF^{2} - |DE \cdot EF| + \frac {TD^{2}}{4} $ ( the products are directed). Similarly $ TN'^{2} = EF^{2} - | DF \cdot EF| + \frac{TD^{2}}{4} $ . Next denoting $ H' $ and $ H'' $ the feet of perpendicular of $H $ onto $ DE $ and $ DF $ respectively and note that $ HH' $ = $ HH'' $ , We get from the right angled triangle $ HH' M' $ that $ HM' ^{2} = HH'^{2} + H'M'^{2} = HH'^{2} + ( EM' - EH' ) ^{2} = HH'^{2} + ( |EF| - \frac{| EF|+| ED| - |FD|}{2} )^{2} $ . Similarly $ HN'^{2} = HH''^{2} + ( |EF| - \frac { |EF | + | FD | - | ED | }{2} ) ^ {2} $ . All it remains to plug this into the expression which indeed proves that they are orthogonal . The proof is complete .

I'd like to contribute a solution Let $O_{1}$ and $O_{2}$ be the curcumcenters of $AHC$ and $AHB$, respectively. Let $P$ and $Q$ be the midpoints of $MH$ and $NH$, respectively. Since $M$ and $N$ are points on a circle with center $O_{1}$, we have that $O_{1}P$ is the perpendicular bisector of $MH$. Similarly, $O_{2}Q$ is the perpendicular bisector of $NH$. Hence, $O_{1}P$ and $O_{2}Q$ meet at the circumcenter of $MNH$. Thus, it suffices to show that $O_{1}P$, $O_{2}Q$ and $OH$ are concurrent. First observe that \begin{eqnarray*} \angle AO_{2}B &=&\angle AO_{2}H+\angle BO_{2}H=2\left( \angle ABH+\angle BAH\right) \\&=&2\left( 90^{\circ }-\angle CAB+90^{\circ }-\angle ABC\right) =2\angle BCA=\angle AOB. \end{eqnarray*} And since $AO_{2}=BO_{2}$ and $AO=BO$, we have that $APBO$ is a rhombus. Similarly, $AQCO$ is also a rhombus. In particular, $AO_{1}=AO=A_{2}O$, so that $O$ is the circumcenter of $OO_{1}O_{2}$, and we have also $\angle OAO_{2}=2\angle OAB$ dan $\angle OAO_{1}=2\angle OAC$. Thus \begin{eqnarray*} \angle OO_{2}O_{1} &=&\frac{1}{2}\angle OAO_{1}=\angle OAC=90^{\circ}-\angle ABC \\ &=&\angle BAH=\angle MAH=\frac{1}{2}\angle MO_{1}H=\angle PO_{1}H. \end{eqnarray*} Similarly, we have $\angle OO_{1}O_{2}=\angle QOH$. Now since $O_{1}$ is the circumcenter of $AMHC$, $O_{2}$ is the circumcenter of $ANHB$, and $O$ is the circumcenter of $OO_{1}O_{2}$ we have \[HO_{2}=AO_{2}=AO_{1}=HO_{1},\] and hence $O_{1}O_{2}H$ is an isosceles triangle with $\angle O_{1}O_{2}H=\angle O_{2}O_{1}H$. Since $ABC$ is acute and $AB>BC$, $AC>BC$, then $O$ lies insides the triangle $O_{1}O_{2}H$. The conclustion then follows from the following simple lemma. Lemma. Let $ABC$ be an isosceles triangle with $\angle ABC=\angle ACB$ and let $P$ be a point inside $ABC$. Let $BE$ and $CF$ be cevians of $ABC$ such that \[\angle ABE=\angle PCB\text{ \ and \ }\angle ACF=\angle PBC.\] Then the lines $BE$, $CF$ and $AP$ are concurrent. Proof. This is a consequence of Ceva's theorem (the trigonometric version).

Lemma: In $\triangle XYZ$, there are points $U$ and $V$ on $XY$ and $XZ$, respectively, such that $UVZY$ is cyclic. Then if $O$ is the circumcenter of $\triangle XUV$, then $XO \perp YZ$. Proof: Let $H$ be the orthocenter of $\triangle XYZ$. Let the reflection about the angle bisector of $\angle X$ send $U$ and $V$ to $U'$ and $V'$, respectively. Because $\triangle XUV \sim \triangle XZY$, $U'V' || YZ$. Since $O$ and $H$ are isogonal, line $XH$ is mapped to line $XO$. Since $XH \perp UV$, $XO \perp U'V'$, so $XO \perp YZ$. Let $E$ be the foot of the altitude from $B$ to $AC$, and let $F$ be the foot of the altitude from $C$ to $AB$, and let $O'$ be the circumcenter of $\triangle MNH$. $\angle CNH = \angle ABH = \angle HCE$, so $N$ is the reflection of $C$ across $E$; similarly, $M$ is the reflection of $B$ across $F$. We will first solve this problem in the case such that $NH || AB$ or $MH || AC$. Note that \[ NH || AB \iff \angle BHN = \angle HNA \iff \angle ENH = \angle HEN \iff \angle HCE = 45^{\circ}. \] Since $\angle HCE = 45^{\circ} \iff \angle FBH = 45^{\circ}$, $NH || AB$ if and only if $MH || AC$, so we may assume that $NH || AB$ and $MH || AC$. Because $ABHN$ is an isosceles trapezoid, the perpendicular bisector of $NH$ is the perpendicular bisector of $AB$, so it passes through $O$; similarly, the perpendicular bisector of $MH$ passes through $O$, so $O = O'$, so $O$, $O'$, and $H$ all lie on a line. We may suppose now that neither $NH || AB$ nor $MH || AC$. Let $P = HN \cap AB$ and $Q = HM \cap AC$. $PN \cdot PH = PA \cdot PB$ and $QM \cdot QH = QA \cdot QC$, since $ANHB$ and $AMHC$ are cyclic, so line $PQ$ is the radical axis of the circumcircles of $\triangle HMN$ and $\triangle ABC$. Hence, $OO' \perp PQ$. But $\angle QNP = \angle CNH = \angle HCN = \angle MBH = \angle HMB = \angle QMP$, so $PQMN$ is cyclic. The lemma thus reveals that $HO' \perp PQ$, so $H$, $O$, and $O'$ are indeed collinear.

Stifler wrote: It can be done nicely using complex numbers. Let $O$ be the origin of the complex plane, and $|a|=|b|=|c|=1$. Observe that $M$ is the refflection of $B$ wrt $HC$. So: $m=a+c-\dfrac{ab}{c}$. By the same way: $n=a+b-\dfrac{ac}{b}$. can you explain me please why $ m=a+c-\dfrac{ab}{c} $ if $ M $ is the refflection of $B$ wrt $HC$? I'm new in complex numbers.

I just found a purely synthetic solution to this problem, here it is: We first prove a lemma: $ \textit{ Lemma:} $ Let $ ABC $ be a triangle with circumcenter $ O$ and orthocenter $ H $ . Suppose that the circumcircle of $ BOC $ intersects $ AB $ and $ AC $ again in $ Y $ and $ X $ respectively. Then the lines $ CY $ , $ BX $ and $ OH $ are concurrent. $ \textit{Proof:} $ Let $ CY $ and $ BX $ intersect the circumcircle $ \gamma $ of $ ABC $ again in $ E' $ and $ F' $. Here, working with oriented angles modulo 180, we have, $ \angle XBA = 180 - \angle BAX - \angle AXB = - \angle BAC - CXB = $ $ -\angle BAC - \angle COB = -\angle BAC + 2\angle BAC = \angle BAC = \angle BAX . $ So, $ \triangle XAB $ is isosceles with $ AX = XB $ and similarly $ \triangle ACY $ is isosceles with $ AY = Y C $. This implies that XF'C is isosceles or $ CF' || AB $ and similarly $ BE' || AC $. Let $ CF' $ and $ BE' $ intersect at $A'$ and let $ A'B'C' $ be the anti - medial triangle of $ ABC$. Consider the circumcircle $ \gamma' $ of $ \triangle A'B'C' $. This circle $ \gamma' $ is the image of $ \gamma $ under the homothety $ \mathcal { H } ( G , {-2} ) $ where $ G $ is the centroid of $ \triangle ABC $. So, the circle $ \gamma $ is the nine point circle of $ \triangle A'B'C'$ and so $ E' $ and $ F' $ are the foot of perpendiculars from $ B' $ and $ C' $ to lines $ C'A' $ and $ A' B' $ respectively. Consider the two triads of collinear points, $ B'CF' $ and $ C' B E' $. By Pappus' Theorem, $ G = B'C \cap C'B $ and $ H' = B'E' \cap C'F' $ and $ K = BE' \cap CF' $ must be collinear or in other words, the Euler line $ H'G' $ of $ \triangle A'B'C' $ which is the same as the Euler line $ HGO $ of $ \triangle ABC $ passes through $ BX \cap CY $ and the lemma follows. $ \square $ We know prove that the result of the problem is equivalent to that of the lemma. Let the images of the midpoints $ L $, $ M $ and $ N $ of $ BC $ , $ CA $ and $ AB $ respectively under the homothety $ \mathcal { H} ( O , 2 ) $ be $ O _ { A } $ , $ O _ { B } $ and $ O _ { C } $ respectively. It is well known that these are respectively the circumcenters of triangles $ HBC $ , $ HAC $ and $ HAB $ and the three circumradii are equal so that $ AH $ , $ BH $ and $ CH $ are respectively the perpendicular bisectors $ O_{B} O _ { C} $ , $ O _ { A } O _ { C } $ and $ O _ { B } O_{ A } $ making $ H $ the circumcenter of the $ \triangle O_{A} O _ {B} O _ { C } $ and clearly $ O $ is it's orthocenter. Consider the spiral symmetry having centre $ H $ which sends $ B $ to $ N $ and $ M $ to $ C $. This symmetry must send $ O_{C} $ to $ O_{B} $ and the perpendicular bisectors $ b $ and $ m $ of $ HB $ and $ HM $to the perpendicular bisectors , $ n $ and $ c $ of $ HN $ and $ HC $ respectively. We have $ b \cap m = O _ {A} $ , $ b \cap n = X $ , $ b \cap c = O _ { a } $ , $ m \cap n = O _ { H M N } $ , $ m \cap c = Y $ and $ n \cap c = O _ { B } $ . Now, as $ O_{B} $ goes to $ O _ { C} $ and the line $ b $ goes to the line $ n $, $ H \in $ circumcircle of $ \triangle X O_ { A} O _ { B} $ and similarly $ H \in $ circumcircle of $ \triangle Y O _ { B } O _ { C } $ , in other words, the five points $ H $, $ O _ { B} $ , $ O _ { C } $ , $ X $ and $ Y $ are concyclic. Applying the lemma to the $ \triangle O_{A} O_{B} O _ {C }$ , we have the result of the problem ....

it's trivial that MH=BH,NH=CH. let MH,AC;NH,AB intersect at S,T;$HU\bot OC,HV\bot OB$intersecting AB,AC at U,V,respectively.by Gosk-Dubo theorem,$OH\bot UV$. since $\frac{AU}{AV}=\frac{\frac{AU}{AH}}{\frac{AV}{AH}}=\frac{\frac{cosA}{sinB-A}}{\frac{cosA}{sinC-A}}=\frac{sinC-A}{sinB-A},\frac{AT}{AS}=\frac{\frac{AT}{AH}}{\frac{AS}{AH}}=\frac{sin2A+B}{sin2C+B}$,so $\frac{AU}{AV}=\frac{AT}{AS}$,ST is parallel to UV. hence $OH\bot ST$.On the other hand,by equal power theorem it's easy to porve that ST is the equal-power axis of circle Q and O(circum of MNH),hence $OQ\bot ST$ yielding O,Q,H collinear.

about Gosk-Dubo theorem,see here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=50&t=449575

Let $O'$ be the circumcentre of $\triangle HMN$. Let $B', C'$ be the foots of the perpendicular to $\triangle ABC$ from $B, C$ respectively. We have $\angle OCH = \angle OCB - \angle HCB = (90 - \angle A) - (90 - \angle B) = \angle B - \angle A$. Note that $\angle BMC = 180 - \angle CMA = 180 - \angle CHA = \angle B$, we have $\triangle BCM$ is isoceles. Therefore, $\angle MCN = \angle BMC - \angle A = \angle B - \angle A = \angle OCH$. Since $H$ lies on $BB'$ in isoceles $\triangle CBN$, we have $\triangle CHN$ is isoceles. Since $\angle HCN = \angle HNC = \angle B'BA = 90 - \angle A = \angle OCB$, and $\triangle OCB$ is isoceles too, we have that $\triangle CHN \sim \triangle OCB$. Therefore, we have $\frac{CH}{CN}=\frac{OC}{BC}=\frac{OC}{CM}$. Combining this with $\angle MCN = \angle OCH$, we get that $\triangle OHC \sim \triangle CMN$, and therefore $\angle OHC = \angle MNC$. Now, $\angle OHM = (180 - \angle OHC) - \angle MHC' = (180 - \angle MNC) - \angle A =$ $90 - (\angle MNC - (90 - \angle A)) = 90 - (\angle MNC - \angle HNC) = 90 - \angle HNM = \angle O'HM$, so $O', O, H$ are collinear.

Although this question was asked above could someone reiterate why this is true? Specifically how the reflection about a perpendicular works? Stifler wrote: It can be done nicely using complex numbers. Let $O$ be the origin of the complex plane, and $|a|=|b|=|c|=1$. Observe that $M$ is the refflection of $B$ wrt $HC$. So: $m=a+c-\dfrac{ab}{c}$. By the same way: $n=a+b-\dfrac{ac}{b}$.

JUST FINISHED MY 5 PAGE LONG BASH,FEELS GOOD .I used all the 3 hints,I didn't have to use the last one if I hadnt sillied about the fact that 2 minuses yielded to a plus and instead wrote minus instead.I think problem was pretty good candidate for a nice bash if you can prove $M=$reflection of B over AC.This can be showed by $\measuredangle AMH=\measuredangle ACH=90-\measuredangle CAB,\measuredangle CMH=\measuredangle CAH=90-\measuredangle BCA$ or $\measuredangle CMA=\measuredangle ABC$ which implies $CM=CB$,which inturn implies our result.And same goes for $N$.Still,I cant deny that it was the $a+b+c$ map which gave it all. .

Dear Mathlinkers, here p. 10... Sincerely Jean-Louis

Here's an approach that I believe is new.

I found this problem months ago while searching for nice weird geo problems. I finally got to trying this today, and I'm not disappointed . Let $O_1$ be the circumcenter of $MNH$, and let the circumcircle of $MNH$ intersect $\overline{AB}$ and $\overline{AC}$ again at $X$ and $Y$, respectively. Let $\omega$ be the circumcircle of $ABC$. We claim that $H$, $O_1$, and $O$ lie on the perpendicular bisector of $\overline{XY}$. $O_1X=O_1Y$ is obvious. We have \[\measuredangle HXY=\measuredangle HNY=\measuredangle HBA=\measuredangle ACH=\measuredangle XMH=\measuredangle XYH,\]so $HX=HY$. To prove that $OX=OY$, we claim that $X$ and $Y$ have the same power with respect to $\omega$, which means $XA \cdot XB=YA \cdot YB$. Since $\tfrac{XA}{XN}=\tfrac{YA}{YM}$ by similar triangles, it suffices to prove that $XB \cdot XN=YC \cdot YM$. By angle chasing, $H$ is the center of spiral similarity taking $BXN$ to $MYC$, and since $\measuredangle HBA=\measuredangle ACH=\measuredangle BMH$, $HB=HM$, so $BXN \cong MYC$. Thus, $XB=YM$ and $XN=YC$, so we are done.

To compute $m, n$, notice that the center of $(AHC)$ is the reflection of $O$ over $\overline{AC}$, which is just $a+c$. Then, the foot is given by $\frac 12(o_B+a+b-ab\overline{o_B})$, so$$n = 2a+c - \frac{ab}c - a = a+c - \frac{ab}c.$$Now, shift triangle $MNH$ by $\overrightarrow{HO}$. Notice that the desired condition is equivalent to the shifted circumcenter lying on $\overline{OH}$. This shifted circumcenter is$$\frac{mn(\overline n - \overline m)}{m \overline n - \overline mn} = -\frac{bc(a+b+c)}{b^2+bc+c^2}$$after much simplification. Then,$$\frac oh = -\frac{bc}{b^2+bc+c^2}$$has conjugate equal to itself, so we are done.

Simple angle chasing shows that $M$ and $N$ are the reflections of $B$ and $C$ across $\overline{CH}$ and $\overline{BH}$ respectively. Setting $(ABC)$ as the unit circle: \[h=a+b+c,\quad m = a+c-\frac{ab}{c},\quad n=a+b-\frac{ac}{b}\]Now if $O'$ is the circumcenter of $\triangle MNH$: \begin{align*} o' &= \frac{\begin{vmatrix} m & m\overline{m} & 1\\ n & n\overline{n} & 1\\ h & h\overline{h} & 1\\ \end{vmatrix}}{\begin{vmatrix} m & \overline{m} & 1\\ n & \overline{n} & 1\\ h & \overline{h} & 1\\ \end{vmatrix}}\\ &= \frac{\begin{vmatrix} a+c-ab/c & (a+c-ab/c)(1/a+1/c-c/ab) & 1\\ a+b-ac/b & (a+b-ac/b)(1/a+1/b-b/ac) & 1\\ a+b+c & (a+b+c)(1/a+1/b+1/c) & 1\\ \end{vmatrix}}{\begin{vmatrix} a+c-ab/c & 1/a+1/c-c/ab & 1\\ a+b-ac/b & 1/a+1/b-b/ac & 1\\ a+b+c & 1/a+1/b+1/c & 1\\ \end{vmatrix}}\\ &= \frac{\begin{vmatrix} ac+c^2-ab & (ac+c^2-ab)(bc+ab-c^2) & c\\ ab+b^2-ac & (ab+b^2-ac)(bc+ac-b^2) & b\\ a+b+c & (a+b+c)(ab+bc+ca) & 1\\ \end{vmatrix}}{\begin{vmatrix} ac+c^2-ab & c(bc+ab-c^2) & c\\ ab+b^2-ac & b(bc+ac-b^2) & b\\ a+b+c & ab+bc+ca & 1\\ \end{vmatrix}}\\ &= \frac{\begin{vmatrix} -ab-bc & (ac+c^2-ab)(bc+ab-c^2)-c(a+b+c)(ab+bc+ca) & 0\\ -ac-bc & (ab+b^2-ac)(bc+ac-b^2)-b(a+b+c)(ab+bc+ca) & 0\\ a+b+c & (a+b+c)(ab+bc+ca) & 1\\ \end{vmatrix}}{\begin{vmatrix} -ab-bc & c(-c^2-ac) & 0\\ -ac-bc & b(-b^2-ab) & 0\\ a+b+c & ab+bc+ca & 1\\ \end{vmatrix}}\\ \end{align*}\begin{align*} &= \frac{\begin{vmatrix} -b(a+c) & (ac+c^2-ab)(bc+ab-c^2)-c(a+b+c)(ab+bc+ca)\\ -c(a+b) & (ab+b^2-ac)(bc+ac-b^2)-b(a+b+c)(ab+bc+ca)\\ \end{vmatrix}}{\begin{vmatrix} -b(a+c) & -c^2(a+c)\\ -c(a+b) & -b^2(a+b)\\ \end{vmatrix}}\\ &= \frac{\begin{vmatrix} -b(a+c) & -(a + c)^2 (b^2 + c^2)\\ -c(a+b) & -(a + b)^2 (b^2 + c^2)\\ \end{vmatrix}}{\begin{vmatrix} -b(a+c) & -c^2(a+c)\\ -c(a+b) & -b^2(a+b)\\ \end{vmatrix}}\\ &= \frac{(b^2+c^2)\begin{vmatrix} b & a + c\\ c & a + b\\ \end{vmatrix}}{\begin{vmatrix} b & c^2\\ c & b^2\\ \end{vmatrix}}\\ &= \frac{(b^2+c^2)(b-c)(a+b+c)}{b^3-c^3}\\ &= (a+b+c)\cdot \frac{b^2+c^2}{b^2+bc+c^2}\\ \end{align*}Clearly $o'$ lies on $\overline{OH}$ as a point $x$ lies on the Euler line iff $x=r(a+b+c)$ for some real $r$ and $o'$ is of that type as $\frac{b^2+c^2}{b^2+bc+c^2}$ is real.

Solved with Math4Life2020

Note that $\measuredangle AMC=\measuredangle AHD=\measuredangle CBA$ so $\triangle CMB$ is isosceles, and $HM=HB$. Symmetrically, $NH=HC$.

$OH\perp M'N'$. Now, we can finish using (a relatively clean) complex bash. Set $(ABC)$ as the unit circle, so $H=a+b+c$ and $F=\frac{a+b+c-\frac{ab}{c}}{2}$ and $E=\frac{a+b+c-\frac{ac}{b}}{2}$. We can find $$N'=\frac{2c^2-b^2+bc+ac-ab}{2c}$$and $$M'=\frac{2b^2-c^2+bc+ab-ac}{2b}.$$Thus, $M'-N'=\frac{b^2-c^2}{2bc}(a+b+c)$, and we find that $\frac{M'-N'}{H}=\frac{b^2-c^2}{2bc}$ is pure imaginary, as it's conjugate is equal to $\frac{c^2-b^2}{2bc}$, as desired.

huh Quote: In $\triangle ABC$ with altitudes $BE$ and $CF$, let $X$ and $Y$ be variable points on $AB$ and $AC$ with $\frac{\overline{XB}}{\overline{YC}} = \frac{\overline{FB}}{\overline{EC}}$, and let $P$ be the point satisfying $\measuredangle AXP = \measuredangle PYA = \measuredangle CAB$. Then the locus of $P$ is the Euler line of $\triangle ABC$. The locus is clearly a line, so it suffices to check two cases. Case 1: Set $X = F$ and $Y = E$. Then $P$ is collinear with $H$ and $N_9$ by Pascal's on $LFH_CKEH_B$ where $K$, $L$, $H_B$, $H_C$ are the midpoints of $AB$, $AC$, $HB$, and $HC$ respectively. Case 2: Let $X$ and $Y$ be on $AB$ and $AC$ such that $\measuredangle AXH = \measuredangle HYA = \measuredangle CAB$. Then $P=H$, and $X$ and $Y$ satisfy the length condition as degenerate quadrilaterals $HFXB$ and $HEYC$ are similar. When $X = M$ and $Y = N$, $P$ is the antipode of $H$ in $\triangle HMN$, which solves the original problem.

We will use complex numbers. Since $\angle AMC=\angle AHC=180^{\circ}-\angle ABC$, we get that $M$ is the reflection of $B$ over $CH$. Similarly, $N$ is the reflection of $C$ over $BH$. Therefore, we have $m=a+c-\frac{ab}c$ and $n=a+b-\frac{ac}b$. It suffices to show $\angle OHN+\angle NMH=90^{\circ}$, or $\frac{(m-h)(n-h)}{(o-h)(m-n)}$ is imaginary. We have $m-h=-\frac{b(a+c)}c$, $n-h=-\frac{c(a+b)}b$, $o-h=-a-b-c$, and $m-n=-\frac{(b-c)(bc+ab+ac)}{bc}$, which implies $\frac{(m-h)(n-h)}{(o-h)(m-n)}=\frac{bc(a+b)(a+c)}{(a+b+c)(bc+ab+ac)(b-c)}$ and the conjugate of this is $-\frac{bc(a+b)(a+c)}{(a+b+c)(bc+ab+ac)(b-c)}$, implying the circumcenter of $MNH$ lies on $OH$. Remark: imagine actually computing the circumcenter

Set $(ABC)$ as the complex unit circle. We make the following claim: Claim: $M$ and $N$ are the reflections of $B$ and $C$ over lines $CH$ and $BH$ respectively. Hence one obtains that $m = a + c - ab/c$ and $n = a + b - ac/b$. Proof: Observe that, \[\angle B = 180^{\circ} - \angle AHC = 180^{\circ} - \angle AMC = \angle BMC, \]so that $BMC$ is isosceles. A similar proof for $N$ is immediately provides $BNC$ as isosceles also. $\square$ Translating the entire figure so that the orthocenter $h$ of $\triangle ABC$ is at the origin, we obtain that the circumcenter of $M'N'H'$ is (where $m' = m -h; n' = n - h; h' = h - h = 0$), \begin{align*} x -h &= \frac{m'n'(\bar m' - \bar n')}{\bar m' n' - m' \bar n'} \\ &= \frac{\left(-\frac{ab}{c} \right) \left(-c - \frac{ac}{b}\right) \left(\frac{1}{c} - \frac{c}{ab} - \frac{1}{b} + \frac{b}{ac} \right)}{\left(-\frac{c}{ab} - \frac{1}{b}\right) \left(-\frac{ac}{b} - c\right) - \left(-\frac{ab}{c} - b \right) \left(-\frac{b}{ac} - \frac{1}{c}\right)} \\ &= \frac{1}{bc} \cdot \frac{(b - c)(a + b + c)}{\frac{c^3 - b^3}{b^2c^2}} \\ &= \frac{bc(b - c)(a + b + c)}{(c - b)(c^2 + cb + b^2)} \\ &= -\frac{bc(a + b + c)}{c^2 + bc + b^2}. \end{align*}Hence $x = h \left(1 - \frac{bc}{c^2 + bc + b^2} \right)$, and so it suffices to now show $X, H, O$ are collinear, which is equivalent to showing that $x/h$ is real. Indeed, \[\frac{x}{h} = \frac{bc}{c^2 + bc + b^2} = \overline{\left(\frac{bc}{c^2 + bc + b^2} \right)} = \frac{\frac{1}{b} \cdot \frac{1}{c}}{\frac{1}{c^2} + \frac{1}{bc} + \frac{1}{b^2}}, \]which holds for all $b, c$. $\blacksquare$

I think this is worth a double post. We want to show that the circumcenter of $a+b+c$, $a+c-\frac{ab}{c}$, and $a+b-\frac{ac}{b}$ lies on $OH$. Applying a quick translation and reflection, we want to show that the circumcenter of $0$, $b+\frac{ab}{c}$, and $c+\frac{ac}{b}$ lies on $OH$. Note that $\frac{ac}{b}$ is the point $P$ such that $APBC$ is an isosceles trapezoid, and similarly $\frac{ab}{c}$ is the point $Q$ such that $AQCB$ is an isosceles trapezoid. The circumcenter of $O, B+Q, C+P$ is just $X = BQ\cap CP$. Now, we have $\angle XCB = \angle PQB = \angle C - \angle A$ so $\angle ACX = \angle A$, and similarly $\angle ABX = \angle A$. Hence the isogonal conjugate of $X$ is the orthocenter of $BOC$, which lies on the Jerabek Hyperbola, so $X$ lies on $OH$ as desired.

Geo sol is not that obvious(maybe finding root axis?), so why not bash it

Let $AD, BE, CF$ be altitudes of $\triangle ABC;$ $Y \equiv DF \cap CA, Z \equiv DE \cap AB$. Consider the inversion $$\mathcal{I}^{k = \overline{AH} \cdot \overline{AD}}_A: H \longleftrightarrow D, B \longleftrightarrow F, C \longleftrightarrow E$$So $$\mathcal{I}^{k = \overline{AH} \cdot \overline{AD}}_A: (AHC) \longleftrightarrow DE, (AHB) \longleftrightarrow DF, M \longleftrightarrow Z, N \longleftrightarrow Y$$This means $M, N, Y, Z$ lie on a circle. From this, if we let $S$ be Anti - Steiner point of $HO$ WRT $\triangle ABC,$ $S_a, S_b \in (O)$ such as $AS_a \perp HO, BS_b \perp HO$ then $$(MN, AB) \equiv (YZ, CA) \equiv (AS_a, CA) \equiv (AB, AS) \pmod \pi$$or $\ell_a \parallel AS$. Suppose that $J$ is center of $(MNH)$. We have $$(HO, HJ) \equiv (HO, HM) + (HM, HJ) \equiv (HO, AB) + (AB, HM) + \dfrac{\pi}{2} + (NM, NH) \equiv (HO, AB) + (CA, CH) + \dfrac{\pi}{2} + (AS, NH)$$$$\equiv (AC, AS) + (AS, NH) + (CA, CH) \equiv (AC, NH) + (CA, CH) \equiv 0 \pmod \pi$$or $H, O, J$ are collinear

same as the above ones

hi orz Very different complex bash: Let $H$ be the origin. Denote each point by their lowercase. Clearly, we have $\angle HCA=\angle ABH=\angle CNH$. Hence $\triangle CNH$ is isosceles, and analogously $\triangle BHM$ is isosceles. Now, letting the foot of $H$ onto $AB$, $AB$ be $E$, $F$ respectively, we have \[e=\frac{ \overline{a}c-a\overline{c} }{2(\overline{a}-\overline{c})}\]and \[f=\frac{ \overline{a}b-a\overline{b} }{2(\overline{a}-\overline{b})}\]By parallelogram properties, $n=2e-c$ and $m=2f-b$. Evaluating, \[m=\frac{ \overline{a}b-a\overline{b} }{\overline{a}-\overline{b}}-b=\frac{\overline{a}b-\overline{a}b+\overline{b}b-a\overline{b}}{\overline{a}-\overline{b}}=\frac{\overline{b}(b-a)}{\overline{a}-\overline{b}} \]However, by perpendicularity, \[\frac{c}{b-a}=\frac{\overline{c}}{\overline{a}-\overline{b}}\quad \text{or}\quad \frac{c}{\overline{c}}=\frac{b-a}{\overline{a}-\overline{b}}\]Hence \[m=\frac{c\overline{b}}{\overline{c}}\]Similarly \[n=\frac{b\overline{c}}{\overline{b}}\]Now, using the circumcentre formula with the origin, we have the centre of $(MNH)$ being \[\frac{bc(\frac{\overline{c}b}{c}-\frac{\overline{b}c}{b})}{\frac{\overline{c}^2b^2}{\overline{b}c}-\frac{\overline{b}^2c^2}{b\overline{c}}}=\frac{bc\overline{b}\overline{c}(b^2\overline{c}-c^2\overline{b})}{\overline{c}^3b^3-\overline{b}^3c^3}\]Now, the centroid of $\triangle ABC$ is still $\frac{a+b+c}{3}$. Now, because $A$ is the orthocentre of $\triangle HBC$, we actually have by shifting: \[b+c-3o=b-o+c-o-o=a-o\]where $o$ is the circumcentre of $\triangle HBC$. Hence $b+c-2o=a$. Now, by circumcentre formula again, \[o=\frac{bc(\overline{b}-\overline{c})}{\overline{b}c-\overline{c}b}\]Hence, we actually have \[a+b+c=2b+2c-2o\]Because in the end we want to show that the centre of $(MNH)$ lies on $OH$, which passes through $H$ and the centroid by Euler Line, there should be a real scalar that sends the centroid to the centre of $(MNH)$. Thus, we ignore real constant scalars. Consequentially, evaluating \[b+c-o=\frac{b\overline{b}c+\overline{b}c^2-b^2\overline{c}-\overline{c}cb-b\overline{b}c+\overline{c}cb}{\overline{b}c-\overline{c}b}=\frac{\overline{b}c^2-b^2\overline{c}}{\overline{b}c-\overline{c}b}\]Consequentially it suffices to show the quotient of our two complex numbers is a real number, or self conjugate. Thus we check whether \[\frac{bc\overline{b}\overline{c}(b\overline{c}-\overline{b}c)}{\overline{c}^3b^3-\overline{b}^3c^2}\]is self conjugate, but this is trivial.