Omid Hatami wrote: Let $O$, $H$ be the circumcenter and the orthogonal center of triangle $\triangle ABC$, respectively. Let $M$ and $N$ be the midpoints of $BH$ and $CH$. Define $B'$ on the circumcenter of $\triangle ABC$, such that $B$ and $B'$ are diametrically opposed. If $HONM$ is a cyclic quadrilateral, prove that $\overline{B'N}=\frac12\overline{AC}$. Proof. Let $K$ be the intersection of lines $B'H$ and $BC$. Let $L$ be the midpoint of $\overline{AH}$ and let $X$, $Y$ and $Z$ be the midpoints of sides $\overline{BC}$, $\overline{CA}$ and $\overline{AB}$, respectively. Let $D$ be the foot of perpendicular through $A$ of $BC$ and let $k$ be the circle through $H$, $M$ and $N$. From the condition, $O \in k$. Since $M$ and $N$ are the midpoints of $\overline{BH}$ and $\overline{CH}$, $MN \perp BC$. Since $Y$ and $Z$ are the midpoints of $\overline{AC}$, $\overline{AB}$, $BC\perp YZ$, thus, $MN \perp YZ$. Since $M$ and $Z$ are the midpoints of $\overline{BH}$ and $\overline{AB}$, $MZ \perp AH$. Since $N$ and $Y$ are the midpoints of $\overline{HC}$, $\overline{AC}$, $AH \perp NY$, thus, $MZ \perp NY$ and quadrilateral $MNYZ$ is a rectangle. Moreover, $MH \parallel YO$, $HN \parallel OZ$ and $NM \parallel ZY$ and hence triangles $\triangle HMN$ and $\triangle OYZ$ are similar. Thus, $OH \parallel MN$. Since $2\overline{OX} =\overline{AH}$, we get $AH/OX=AH/HD=2$. Since $\angle BCB'=90^{\circ}$, $B'C\parallel AH$. Since $\angle B'AB=90^{\circ}$, $AB' \parallel HC$. Therefore, $AHCB'$ is a parallelogram and hence $\overline{AH}=\overline{B'C}$. Therefore, since $\overline{LD}=\overline{AH}$, $DBB'L$ is a rectangle and $\overline{B'L}/\overline{LD}=2$, which means that $\overline{DK}=\overline{CD}$, since $\overline{DH}=\overline{HL}$. We have \[ \frac{\overline{CY}}{\overline{YA}}\cdot \frac{\overline{AH}}{\overline{HB}} \cdot \frac{\overline{DK}}{\overline{KC}}=-1,\] and hence by Menelaos' theorem applied in triangle $\triangle ADC$, $K$, $H$ and $Y$ are collinear. Since $K$, $H$ and $B'$ are collinear, $H$, $Y$ and $B'$ are collinear too. Since $YN \parallel B'C$, $\overline{NB'}=\overline{CY}=\frac{1}{2}\overline{CA}$, which is what we had to prove. $\square$

I don't have so much time so I will post only the hints for my solution: Let $B_1$ and $C_1$ be the midpoints of $AC$ and $AB$ respectively. 1. Prove that if $HONM$ is a cyclic then $HC_1AB_1$ is cyclic. 2. Prove that in any triangle, $B_1N$ is parallel to $B'C$ 3.Prove that $B_1H=NH$. Then you will solve the problem.

This problem is really nice. It's long time I have never seen a problem with so many reflections: Let $A_{1},B_{1}$ be midpoints of $BC,AC$ respectively First Reflection Reflects $H$ through the point $O$, named $H'$. By Homothethy(center $H$), we have $HH'BC$ is concyclic (now we don't use $M,N$ any more). We see that $OA_{1}$ pass through the center this circle. Then, it's trivial to prove that $OA_{1}\perp HH'$ Second Reflection---- Now we can see; $wrt$ the line $OA_{1}$, $H',B$ are symmetry to $H,C$ respectively That gives us $H'B=HC$ Third Reflection Don't forget that $B,H$ are reflected (with center $O$) to $B',H'$. Hence, $HB'=H'B=HC$ Fourth Reflection Remarkably, $OB'=OC , HB'=HC$,then $B'$ is the reflection of $C\ \ wrt$ to the line $OH$ And the fourth Reflection gives us priceless information , $B'N=CB_{1}=\frac{1}{2}AC$

easy to prove that angle BAC = 90

skytin wrote: easy to prove that angle BAC = 90 Dear skytin, I think you're wrong. Due to my proof, all the condition we need is $AH=2HD$ where $D$ is the foot of altitude(from $A$) of triangle $ABC$, and we can construct some different triangles with acute $A$.

the center of 9 points circle (O') is midpoint o segment OH so O' lies on perpendicular bisector of OH and MN so O' is center of circle around of MNH so HO is diameter so angle HNO and HMO = 90 and HM = MB so H lies on circle so angle CAB = 90

skytin wrote: the center of 9 points circle (O') is midpoint o segment OH so O' lies on perpendicular bisector of OH and MN so O' is center of circle around of MNH so MN is diameter so angle HNO and HMO = 90 and HM = MB so H lies on circle so angle CAB = 90 This is not making any sense. Why must $O'$ be the circumcentre of $MNH$ and even if it is, why must $MN$ be the diameter?!?!

skytin wrote: the center of 9 points circle (O') is midpoint o segment OH so O' lies on perpendicular bisector of OH and MN so O' is center of circle around of MNH so HO is diameter so angle HNO and HMO = 90 and HM = MB so H lies on circle so angle CAB = 90 That's wrong. It is true that $O'$ is on both the perpendicular bisectors of $OH$ and $MN$, but those perpendicular bisectors can be the same line.(which means $O'$ needn't be the circumcenter$) Dear skytin, you forgot to consider this case.

The following is enough right. We don't need the cyclic quadrilateral: Let $H$ be the orthogonal center of triangle $ABC$. $N$ is the midpoint of $CH$. $BB'$ is the diagonal of circumcircle. Prove that $B'N=\frac12AC$.

Pikachu!!! wrote: And the fourth Reflection gives us priceless information , $B'N=CB_{1}=\frac{1}{2}AC$ Sorry for the revive, but I'm not sure how you get that B'N = CB_1. Could someone explain please? Thanks. litongyang wrote: The following is enough right. We don't need the cyclic quadrilateral: Let $H$ be the orthogonal center of triangle $ABC$. $N$ is the midpoint of $CH$. $BB'$ is the diagonal of circumcircle. Prove that $B'N=\frac12AC$. Actually, it won't work for any triangle.

Let $o$ be the circumcircle of triangle $HMN$. We know that in any triangle cirumcircle of triangle $HBC$ has the same radius what circumcircle of $ABC$, thus $o$ radius is two time smaller than radius of circumcirle of $ABC$. But $o$ passes through $O$ so, $o$ is tangent to circumcircle of $ABC$. Let E be the tangency point. Consider nine point circle of $ABC$. We know that $H$ is the center of homothety which maps nine point circle to circumcircle of $ABC$ and nine point circle has two times smaller radius, thus nine point circle has the same radius what $o$. Now we will show that $AH$ contains $E$. Consider homothety which maps $o$ to circumcircle of $ABC$ (center of this homothety is $E$) and homothety which maps circumcircle of $ABC$ to nine point circle (center of this homothety is $H$). Nowe we know that center of composition these homotheties lies on $EH$. This composition maps $o$ to nine point circle, so this center must lies on line containing center of these circles, but they have the same radius, so $EH$ is parallel to line containing center of these circles, so $EH$ is perpendicular to $MN$, but $MN$ is parallel to $BC$, so $AH$ contains $E$. Let $F$ be the second intersection point of circumcircle of $ABC$ and $EN$. We know that $HN$ is parallel to $FA$, but $HN$ is perpendicular to $AB$, so $F=B'$. And it's almost everything. We know that $B'N=\frac{1}{2}BE$ and by some angle chasing we know that $\sphericalangle EBB'= \sphericalangle CBA$ and we're done.

it can be quite well done by complex numbers. let A(a) , B(b),C(c) be the numbers.and the circumcentre be the origin of the argand plane.then H=a+b+c ,B' = -b , we can easily find M and N .and use ptolemy's theorem for cyclic quards and after a little calculation, u r done.

Let $AD$ be the A-altitude, $F$ the midpoint of $AC$. It is well known that $B', F, H$ are collinear and $B'CHA$ is a parallelogram. Notice that $MO\parallel HB'$. Therefore, as $MN\parallel BC$ and $HONM$ cyclic, $\angle FHO = \angle HOM = \angle HNM = \angle HCB = 90^{\circ} - B = \angle OAF$, $HOFA$ is cyclic, then $OH\perp AL$ and $HO\parallel MN\parallel BC$. Now $\angle OHN = \angle OMN = \angle FHO$, $HO$ is the angle bisector of $\angle FHN$, also notice that $FN\parallel AL\parallel B'C\perp HO$. Thus both $\Delta FHN$ and $\Delta B'HC$ are isosceles and $FNCB'$ is a isosceles trapezium, it follows that $B'N = FC = 1/2 AC$ Note: Another property: The circumcircle of $HONM$ is tangent to the circumcircle of $ABC$ at point $L$, where $L\equiv AH\cap B'N$.

Taking $R=1$ we've $\angle{HON}=90+C$ so $4OH^2.ON^2Sin^2C=(c.OH.ON)^2=OH^2+ON^2-\frac {CH^2}{4}$.Now $CH=2CosC$, and $4ON^2=2(OH^2+1)-CH^2$ , so using both two equations we've $OH$ in terms of $a,b,c$ but depending on cyclic condition. Now for any triangle $4OH^2=2(BH^2+HB'^2)-4$ and which is same as $8Cos^2B-4+2HB'^2=8Cos^2B-4+8Cos^2C+2a^2+8Cos^2C.a(2Cos^2A-1)$. Now by both two equation we get $b^2+4Cos^2C=10Cos^2C++16Cos^2C(2Cos^2-1)+2a^2$ and then so we get $b^2=4B'N^2$ so done.

skytin wrote: the center of 9 points circle (O') is midpoint o segment OH so O' lies on perpendicular bisector of OH and MN so O' is center of circle around of MNH so HO is diameter so angle HNO and HMO = 90 and HM = MB so H lies on circle so angle CAB = 90 skytin is wrong, but no so far to the answer. There are basically 2 cases: 1. When $HO$ and $MN$ (or $BC$) are not parallel, and so $\triangle{ABC}$ has $\widehat{BAC}=90$ and $H=A, C=B'$ and the conclusion is obvious. 2. When $HO$ and $MN$ are parallel, but then let $U$ be the reflection of $A$ in $H$, and $U$ obviously lies on the circumcircle of $\triangle{ABC}$, $AHB'C$ is obviously a parallelogram, so $B'C=AH=HU$, so $B'HUC$ is a parallelogram, so $N$ is the midpoint of $B'U$, and $AB'CU$ is cyclic and has $B'C || AU$, so $B'U=AC$ and we are done.

I thought this day was pretty easy - here's a simple solution (pretty much the same as vsimat's): First, it's easy to see that $OHMN$ must be an isosceles trapezoid, as $MN || BC$ and looking at the projections $O',H',M',N'$ of $O,H,M,N$ onto $BC$, $M'H' = O'N'$. Thus, $\angle ONM = \angle HMN = 90 - \angle C$. Let $P$ be the midpoint of $AC$, so $PN \perp BC$. Then, $\angle O'ON = \angle C = 180 - \angle O'OP$, so $\triangle PON$ is isosceles. Furthermore, $\triangle B'OC$ is isosceles, and since $PN || B'C$, $PNCB'$ is an isosceles trapezoid and $B'N = PC = \frac{1}{2}AC$, as desired.

colinhy wrote: I thought this day was pretty easy - here's a simple solution (pretty much the same as vsimat's): First, it's easy to see that $OHMN$ must be an isosceles trapezoid, as $MN || BC$ and looking at the projections $O',H',M',N'$ of $O,H,M,N$ onto $BC$, $M'H' = O'N'$. Thus, $\angle ONM = \angle HMN = 90 - \angle C$. Let $P$ be the midpoint of $AC$, so $PN \perp BC$. Then, $\angle O'ON = \angle C = 180 - \angle O'OP$, so $\triangle PON$ is isosceles. Furthermore, $\triangle B'OC$ is isosceles, and since $PN || B'C$, $PNCB'$ is an isosceles trapezoid and $B'N = PC = \frac{1}{2}AC$, as desired. How can $OHMN$ be an isosceles trapezoid?

@above how do u get from those magnitudes to those equations? (on lines 2,3,4)

bever209 wrote: @above how do u get from those magnitudes to those equations? (on lines 2,3,4) A number is purely real if and only if it's equal to its complex conjugate, that's the equation for concyclicity. And when it comes to the segment length equality I just square both sides. Since absolute value is a square root from a positive real it's always positive that's why squaring is equivalent. Then I open the brackets using the identity $|z|^2=z \overline{z}$.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.7418091502722035, xmax = 11.103538582407108, ymin = -5.3171356334118975, ymax = 5.532204094332087; /* image dimensions */ pen ccqqqq = rgb(0.8,0,0); pen qqccqq = rgb(0,0.8,0); pen ttffcc = rgb(0.2,1,0.8); pen yqqqyq = rgb(0.5019607843137255,0,0.5019607843137255); /* draw figures */ draw(circle((0,0), 5), linewidth(2) + ccqqqq); draw((0,5)--(-4.707074511417593,-1.6862531079179148), linewidth(2) + qqccqq); draw((-4.707074511417593,-1.6862531079179148)--(3.8718940415703527,-3.1636113115950253), linewidth(2) + qqccqq); draw((0,5)--(3.8718940415703527,-3.1636113115950253), linewidth(2) + qqccqq); draw((3.8718940415703527,-3.1636113115950253)--(-2.5596249879204396,1.3641322249999375), linewidth(2) + blue); draw((-4.707074511417593,-1.6862531079179148)--(1.7244445180731987,1.3641322249999375), linewidth(2) + blue); draw((-4.707074511417593,-1.6862531079179148)--(4.707074511417593,1.6862531079179148), linewidth(2.4) + ttffcc); draw(circle((-0.8351804698472399,-2.34986441951294), 2.5), linewidth(2.4) + yqqqyq); draw((0,5)--(-1.6995719587844729,-4.695675015842676), linewidth(2) + red); draw((4.707074511417593,1.6862531079179148)--(3.8718940415703527,-3.1636113115950253), linewidth(2) + red); draw((1.9359470207851763,0.9181943442024871)--(1.5183567858615563,-1.5067378655539827), linewidth(2) + red); draw((0,0)--(1.9359470207851763,0.9181943442024871), linewidth(2) + blue); /* dots and labels */ dot((0,0),linewidth(4pt) + dotstyle); label("$O$", (-0.07744035098842501,0.14701508827575704), NE * labelscalefactor); dot((0,5),dotstyle); label("$A$", (0.15944479590554814,5.105810829922935), NE * labelscalefactor); dot((-4.707074511417593,-1.6862531079179148),dotstyle); label("$B$", (-4.957274377004272,-1.9691588906437394), NE * labelscalefactor); dot((3.8718940415703527,-3.1636113115950253),dotstyle); label("$C$", (4.012776518714178,-3.3588850857550505), NE * labelscalefactor); dot((-0.83518046984724,0.15013558048705972),linewidth(4pt) + dotstyle); label("$H$", (-1.0723579679431123,0.4470696076747901), NE * labelscalefactor); dot((-1.2538263156950662,-2.2809265257971543),linewidth(4pt) + dotstyle); label("$D$", (-1.1829043698269663,-2.158667008158918), NE * labelscalefactor); dot((1.7244445180731987,1.3641322249999375),linewidth(4pt) + dotstyle); label("$E$", (1.7860561379108306,1.4893642540082734), NE * labelscalefactor); dot((-2.5596249879204396,1.3641322249999375),linewidth(4pt) + dotstyle); label("$F$", (-2.4936688493069514,1.4893642540082734), NE * labelscalefactor); dot((-2.7711274906324164,-0.7680587637154276),linewidth(4pt) + dotstyle); label("$M$", (-2.904269770589838,-0.5794326955324283), NE * labelscalefactor); dot((1.5183567858615563,-1.5067378655539827),linewidth(4pt) + dotstyle); label("$N$", (1.6913020791532412,-1.4795962537295275), NE * labelscalefactor); dot((4.707074511417593,1.6862531079179148),linewidth(4pt) + dotstyle); label("$B'$", (4.660262586891038,1.9157575184174256), NE * labelscalefactor); dot((-1.6995719587844729,-4.695675015842676),dotstyle); label("$G$", (-1.9567291830139455,-4.96970408463407), NE * labelscalefactor); dot((1.9359470207851763,0.9181943442024871),linewidth(4pt) + dotstyle); label("$K$", (1.9913565985522739,1.0471786464728563), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] By Angle Chasing, we can say that $B'CHA$ is a parallelogram. Notice that $B',F, H$ are collinear. So, we can get $MN || BC$ and $HONM$ is cyclic quadrilateral. By doing some more angle chasing, we can get $\angle KHO=\angle HOM$, then $\angle HOM=\angle HNM$. Now, since $MN|| BC$, we get $\angle HNM=\angle HCB=90^{\circ}-\angle B=\angle OAK$. Hence, we can say that $HOKA$ is a cyclic quadrilateral. The last two results are the properties in Pedal Triangle. Hence, we get $OH\perp AG$ and then $HO||MN \implies HO||MN||BC$. Lastly, we can say that $\Delta KHN$ and $\Delta B'HC$ are isosceles triangles and hence $B'KNC$ is an isosceles trapezium. $\therefore B'N=FC, FC=\frac{1}{2} \implies B'N=\frac{1}{2}. \qquad \blacksquare$