Only problem I solved Just note that if $Q_A$ is the miquel point of $BCEF$ , then the spiral similarity $f$ centered at $Q_A$ that maps $F \mapsto E $ ,$B \mapsto C $ also maps $P$ to $Q$ . So $HK$ passes through $Q_A$ as $AH$ and $AK$ are diameters of $(AQ_AEF)$$(AQ_APQ)$ .So since $Q_A , H ,M$ are collinear , we're done This idea shows up in ISL 2005 G5 too

I had the courage to barycentric bash this one And the bash is one of the most elegant ones I have ever seen (if it is OK to call bashes elegant) Will post my solution in a while

We begin with a few claims. [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 = -0.8180960403987637, xmax = 19.75989821315811, ymin = -8.074152801420684, ymax = 5.515340806285347; /* image dimensions */ pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen qqqqcc = rgb(0,0,0.8); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen ffqqff = rgb(1,0,1); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); /* draw figures */ draw((6.088458641577638,1.3168499898692518)--(2.42,-6.22), linewidth(1) + dtsfsf); draw((2.42,-6.22)--(14.08,-5.9), linewidth(1.2) + dtsfsf); draw((14.08,-5.9)--(6.088458641577638,1.3168499898692518), linewidth(1) + dtsfsf); draw((4.779223485042442,-1.3729742497806712)--(14.08,-5.9), linewidth(1) + qqqqcc); draw((2.42,-6.22)--(7.816767030526164,-0.24391805472439435), linewidth(1) + qqqqcc); draw((6.088458641577638,1.3168499898692518)--(6.292385235510393,-6.113725276555461), linewidth(1) + green); draw((5.407333073423285,1.9319479827556738)--(6.964915610863294,-6.095268182206153), linewidth(1) + green); draw(circle((5.3387990590064085,1.1713877709826779), 0.7636417660554436), linewidth(1) + rvwvcq); draw(circle((6.134731324041502,-0.3692108774077651), 1.6866957073833229), linewidth(1) + rvwvcq); draw((4.786380996257173,0.6441468101188115)--(4.779223485042442,-1.3729742497806712), linewidth(1) + qqqqcc); draw((4.786380996257173,0.6441468101188115)--(7.816767030526164,-0.24391805472439435), linewidth(1) + qqqqcc); draw((4.786380996257173,0.6441468101188115)--(5.686804794082172,0.49165197288267126), linewidth(1) + ffqqff); draw((2.9960596440555958,4.109475176050405)--(6.088458641577638,1.3168499898692518), linewidth(1) + dtsfsf); draw((2.9960596440555958,4.109475176050405)--(8.25,-6.06), linewidth(1.2) + dtsfsf); draw((4.589139476435177,1.0259255520961055)--(5.686804794082172,0.49165197288267126), linewidth(1) + dbwrru); draw((5.407333073423285,1.9319479827556738)--(4.589139476435177,1.0259255520961055), linewidth(1)); draw((5.686804794082172,0.49165197288267126)--(7.305875991485536,0.21744776002407873), linewidth(1) + ffqqff); /* dots and labels */ dot((6.088458641577638,1.3168499898692518),dotstyle); label("$A$", (6.143679937590282,1.453191291297507), NE * labelscalefactor); dot((2.42,-6.22),dotstyle); label("$B$", (2.4690381065979836,-6.083165045719407), NE * labelscalefactor); dot((14.08,-5.9),dotstyle); label("$C$", (14.134355628257172,-5.762469031378262), NE * labelscalefactor); dot((6.181004006505365,-2.055271744684782),linewidth(4pt) + dotstyle); label("$H$", (6.23721627510645,-1.9542038610771617), NE * labelscalefactor); dot((7.816767030526164,-0.24391805472439435),linewidth(4pt) + dotstyle); label("$E$", (7.867421014673943,-0.1369264464773385), NE * labelscalefactor); dot((6.292385235510393,-6.113725276555461),linewidth(4pt) + dotstyle); label("$F$", (6.344114946553499,-6.002991042134121), NE * labelscalefactor); dot((4.779223485042442,-1.3729742497806712),linewidth(4pt) + dotstyle); label("$G$", (4.834171212363936,-1.272724830602228), NE * labelscalefactor); dot((5.407333073423285,1.9319479827556738),linewidth(4pt) + dotstyle); label("$Q$", (5.462200907115347,2.0411339842562732), NE * labelscalefactor); dot((5.686804794082172,0.49165197288267126),linewidth(4pt) + dotstyle); label("$P$", (5.742809919663849,0.5980019197211194), NE * labelscalefactor); dot((6.964915610863294,-6.095268182206153),linewidth(4pt) + dotstyle); label("$J$", (7.012231643097554,-5.98962870820324), NE * labelscalefactor); dot((4.589139476435177,1.0259255520961055),linewidth(4pt) + dotstyle); label("$K$", (4.6470985373316,1.1324952769563617), NE * labelscalefactor); dot((4.786380996257173,0.6441468101188115),linewidth(4pt) + dotstyle); label("$L$", (4.834171212363936,0.744987592960811), NE * labelscalefactor); dot((8.25,-6.06),linewidth(4pt) + dotstyle); label("$N$", (8.30837803439302,-5.949541706410597), NE * labelscalefactor); dot((2.9960596440555958,4.109475176050405),linewidth(4pt) + dotstyle); label("$M$", (3.0436184656258702,4.219194414989885), NE * labelscalefactor); dot((5.288228233514515,-0.3272237566756608),linewidth(4pt) + dotstyle); label("$O$", (5.341939901737417,-0.21710045006262482), NE * labelscalefactor); dot((7.305875991485536,0.21744776002407873),linewidth(4pt) + dotstyle); label("$I$", (7.359652325300462,0.33075524110349835), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Claim 1: $AQKP$ is a harmonic quaderilateral. Proof: Firstly note that $\angle AQK +\angle APK=90+90=180 \implies AKQP$ is cyclic. Now by the defination of harmonic quaderilaterals we need to prove that $AQ \cdot KP =QK \cdot AP$. Now note that $\angle AGH+\angle AEH=90+90=180 \implies AGHE $ cyclic $\implies \angle BHC=\angle GHE=180-\angle A$. Now let $J$ be the foot of the angle bisector of $\angle BHC$ onto $BC$ . Clearly $\angle BHJ= \frac{\angle BHC}{2}= 90-\frac{\angle A}{2}\implies \angle QHE=90-\frac{\angle A}{2}$ and now since $\angle HEQ=90\implies \angle HQE=\frac{\angle A}{2}$ and now since $\angle QAP=180-\angle A \implies AQ=AP \implies \angle KQP=\angle KPQ=90-\frac{\angle A}{2} \implies KQ=KP$. Now we can easily see that $AQ \cdot KP=AP \cdot QK$ . We are done with the claim. $\square$. Now back to the main problem. Note that since $\angle KQE+\angle QEH=90+90=180\implies KQ \parallel EH$. So if $\odot(AGEH)\cap \odot(AQKP)=L \neq A $ by reims theorem we have that $K-L-H$ collinear.Now by our claim since $(A,K;Q,P)=-1$ (just writing in harmonic notation) by taking perspective from $L$ and projecting this harmonic bundle onto line $AQ$ we have $(A,K;Q,P) \overset{L}{=} (A,LK \cap AQ;Q,LP \cap AQ)=-1$. Let $LK \cap AQ=M$ and $LP \cap AQ=I$. So we can rephrase the last result we obtained as $(A,M;Q,I)=-1$. Now by taking perspective from $P$ and projecting this bundle onto line $KH$ we have $(A,M;Q,I)\overset{P}{=} (PA \cap KH,M;H,L)=-1$. Now let $PA \cap KH=O$ .Now rephrasing last condition we have $(O,M;H,L)=-1$. Now taking projeting harmonic bundle onto $\odot(AGEH)$ and taking the perspectivity from $A$ we have $(G,E;H,L)=-1 \implies GHEL$ is a harmonic quaderilateral. Now by definition we have that the tangent at $G,E$ and $LH \equiv KH$ are concurrent. Now we state a lemma. Lemma: Let $N$ be the midpoint of $BC$ then $NG,NE$ are tangents. Proof: Its well known that $NFGE$ is cyclic (all lie on 9-point circle). So $\angle NGE=\angle NFE=\angle A$ since $AEFB$ is cyclic (as $\angle AEB=\angle AFB=90$) $\implies NG$ tangent to $AGEH$ similiarly we can prove $NE$ tangent to $AGEH$. Done with the claim. $\square$ Now back to the main problem . By claim 2 we know that the tangent at $G,E$ intersect at the midpoint of $BC$. But as we claimed earlier we know that $KH$ and the tangents at $G,E$ are concurrent $\implies KH$ passes through midpoint of $BC$. We are done with the problem. $\blacksquare$

Nice and Easy and one of my favourites from this year! Sharygin 202 CR P12 wrote: Let $H$ be the orthocenter of a nonisosceles triangle $ABC$. The bisector of angle $BHC$ meets $AB$ and $AC$ at points $P$ and $Q$ respectively. The perpendiculars to $AB$ and $AC$ from $P$ and $Q$ meet at $K$. Prove that $KH$ bisects the segment $BC$.

Introduce the Orthic triangle $H_AH_BH_C$ of $\triangle ABC$. Firstly we will exploit the condition (Angle Bisector of $\angle BHC$). Let $HY$ be the Bisector of $\angle BHC$ where $Y\in BC$. Now drop a Perpendicular from $A$ to $KH$. Let $AX\perp KH$ where $X\in KH$. Now notice that $\angle AXK=\angle AQK=90^\circ\implies X\in\odot(AH)$ and similarly $\angle AXH=\angle AH_BH\implies X\in\odot(AH)$. So, $\odot(AK)\cap\odot(AH)=\{A,X\}$. Let $AX\cap QH=J$. Claim 1:-. $AP=AQ$

Claim 2:-. $(Q,P;J,H)$ is Harmonic.

Claim 3:- $AX,H_BH_C,BC$ are concurrent.

Claim 4:- $X\in\odot(ABC)$.

Now let $XH\cap \odot(ABC)=A'$. So as $\angle AXA'=90^\circ$, $A'$ is the $A-\text{ antipode}$ of $\odot(ABC)$. So, now notice that $CH\|BA'$ and $BH\|CA'\implies BHCA'$ is a parallelogram. Now in a parallelogram diagonals bisect each other so $HA'$ bisects $BC$ , hence $KH$ bisects $BC$. $\blacksquare$

I did this one partially. I was so close

amar_04 wrote: Solution for 12:-

Introduce the Orthic triangle $H_AH_BH_C$ of $\triangle ABC$. Firstly we will exploit the condition (Angle Bisector of $\angle BHC$). Let $HY$ be the Bisector of $\angle BHC$ where $Y\in BC$. Now drop a Perpendicular from $A$ to $KH$. Let $AX\perp KH$ where $X\in KH$. Now notice that $\angle AXK=\angle AQK=90^\circ\implies X\in\odot(AH)$ and similarly $\angle AXH=\angle AH_BH\implies X\in\odot(AH)$. So, $\odot(AK)\cap\odot(AH)=\{A,X\}$. Let $AX\cap QH=J$. Claim 1:-. $AP=AQ$

Claim 2:-. $(Q,P;J,H)$ is Harmonic.

Claim 3:- $AX,H_BH_C,BC$ are concurrent.

Claim 4:- $X\in\odot(ABC)$.

Now let $XH\cap \odot(ABC)=A'$. So as $\angle AXA'=90^\circ$, $A'$ is the $A-\text{ antipode}$ of $\odot(ABC)$. So, now notice that $CH\|BA'$ and $BH\|CA'\implies BHCA'$ is a parallelogram. Now in a parallelogram diagonals bisect each other so $HA'$ bisects $BC$ , hence $KH$ bisects $BC$. $\blacksquare$ I have written the same solution!

I guess this was the only nice problem for 10th graders other than 23 and 24.Rest all were very doable.

wsas proposed by pranjal and anant....thats why @below i cant see it

How is my solution Solution : Lemma 1: In any $\triangle XYZ$ , consider a point $M$ on $YZ$ such that $\frac{XY}{XZ}=\frac{sin \angle ZXM}{sin \angle YXM}$ , then $M$ is the midpoint of $YZ$ . Proof: Applying sine rule in $\triangle XYM$ and $\triangle XZM$ , we get ; $\frac {YM}{sin\angle YXM}=\frac {XY}{sin\angle XMY}=\frac{XY}{sin \angle XMZ}$ ---- (1) $\frac {ZM}{sin\angle ZXM}=\frac{XZ}{sin \angle XMZ}$ ---- (2) Dividing (1) and (2) and from the given conditions in the lemma ; we get $YM=ZM$ . $\implies M $ is the midpoint of $YZ$. Now coming back to the original problem ; let $KH\cap BC=M$ and let $HD$ be the angle bisector of $\angle BHC$ , with $D$ on $BC$. Let $E$ and $F$ be the foot of altitude from $B$ and $C$ to $AC$ and $AB$ respectively . Let $\angle BAC=A ,\angle ABC=B , \angle ACB=C $ As $H$ is the orthocentre , $\angle BHC=180^\circ-A$ ,so $\angle BHD=90^\circ-\frac{A}{2}$. $\implies \angle BHP=90^\circ+\frac{A}{2}$ . In $\triangle BHP$ , $\angle PBH=90^\circ-A$ , so $\angle BPH=\frac{A}{2}=\angle QPA=\angle QKA$ . $\angle QAP=180^\circ-A \implies \angle PQA=\frac{A}{2}$. So, $\angle KQP=\angle KPQ=90^\circ-\frac{A}{2}$ . So , $AP=AQ ; KP=KQ$. As $BH\perp AC$ and $KQ\perp AC , BH\parallel KQ$ . Similarly, $CH\parallel KP$. Applying Sine Rule in $\triangle HQK , \triangle KPH$ , we get $\frac{HQ}{sin \angle HKQ}=\frac{HQ}{sin \angle KHE}=\frac{HQ}{sin \angle BHM}=\frac{KQ}{sin \angle KHQ}=\frac{KP}{sin \angle KHP}=\frac{PH}{sin \angle PKH}=\frac{PH}{sin \angle KHC}=\frac{PH}{sin \angle CHM}$ . $\implies \frac{HQ}{HP}=\frac{sin \angle BHM}{sin \angle CHM}$ ---- (3) Now , in $\triangle HBP$ and $\triangle HCQ$ , $\angle HBP=\angle HCQ=90^\circ-A , \angle BHP=\angle CHQ=90^\circ+\frac{A}{2}$. $\implies \triangle HBP \sim \triangle HCQ \implies \frac{HQ}{HP}=\frac{HC}{HB}=\frac{sin \angle BHM}{sin \angle CHM}$ --- (from (3)) . Thus , by Lemma 1 , we get that $M$ is the midpoint of $BC \implies KH$ bisects $BC$. $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(22.692cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -5.72, xmax = 32.1, ymin = -8.64, ymax = 8.88; /* image dimensions */ pen ududff = rgb(0.30196078431372547,0.30196078431372547,1.); pen zzttqq = rgb(0.6,0.2,0.); pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); draw((8.64,4.58)--(5.18,-5.94)--(20.88,-5.86)--cycle, linewidth(2.) + zzttqq); draw((3.3598423201644065,3.099201148276634)--(6.314614616879968,6.563416944425912)--(8.64,4.58)--(7.685091981184514,1.6766380468384605)--cycle, linewidth(2.) + zzttqq); draw((7.685091981184514,1.6766380468384605)--(8.672739181695123,-1.8450644076677885)--(5.18,-5.94)--cycle, linewidth(2.) + green); draw((6.314614616879968,6.563416944425912)--(8.672739181695123,-1.8450644076677885)--(20.88,-5.86)--cycle, linewidth(2.) + green); /* draw figures */ draw((8.64,4.58)--(5.18,-5.94), linewidth(2.) + zzttqq); draw((5.18,-5.94)--(20.88,-5.86), linewidth(2.) + zzttqq); draw((20.88,-5.86)--(8.64,4.58), linewidth(2.) + zzttqq); draw((xmin, -0.9306157610068706*xmin + 6.225923365919522)--(xmax, -0.9306157610068706*xmax + 6.225923365919522), linewidth(2.) + dotted + red); /* line */ draw((5.18,-5.94)--(8.672739181695123,-1.8450644076677885), linewidth(2.)); draw((8.672739181695123,-1.8450644076677885)--(20.88,-5.86), linewidth(2.)); draw((xmin, -3.565749442397591*xmin + 29.079750493521335)--(xmax, -3.565749442397591*xmax + 29.079750493521335), linewidth(1.6) + dotted); /* line */ draw((xmin, -0.8529411764705884*xmin + 11.949411764705886)--(xmax, -0.8529411764705884*xmax + 11.949411764705886), linewidth(2.)); /* line */ draw((3.3598423201644065,3.099201148276634)--(6.314614616879968,6.563416944425912), linewidth(2.) + zzttqq); draw((6.314614616879968,6.563416944425912)--(8.64,4.58), linewidth(2.) + zzttqq); draw((8.64,4.58)--(7.685091981184514,1.6766380468384605), linewidth(2.) + zzttqq); draw((7.685091981184514,1.6766380468384605)--(3.3598423201644065,3.099201148276634), linewidth(2.) + zzttqq); draw((3.3598423201644065,3.099201148276634)--(8.64,4.58), linewidth(2.)); draw(shift((5.999921160082205,3.8396005741383172)) * scale(2.7419350084927276, 2.7419350084927276)*unitcircle, linewidth(2.)); draw((8.672739181695123,-1.8450644076677885)--(8.64,4.58), linewidth(2.)); draw((7.685091981184514,1.6766380468384605)--(8.672739181695123,-1.8450644076677885), linewidth(2.) + green); draw((8.672739181695123,-1.8450644076677885)--(5.18,-5.94), linewidth(2.) + green); draw((5.18,-5.94)--(7.685091981184514,1.6766380468384605), linewidth(2.) + green); draw((6.314614616879968,6.563416944425912)--(8.672739181695123,-1.8450644076677885), linewidth(2.) + green); draw((8.672739181695123,-1.8450644076677885)--(20.88,-5.86), linewidth(2.) + green); draw((20.88,-5.86)--(6.314614616879968,6.563416944425912), linewidth(2.) + green); /* dots and labels */ dot((8.64,4.58),linewidth(3.pt) + ududff); label("$A$", (8.72,4.7), NE * labelscalefactor,ududff); dot((5.18,-5.94),linewidth(3.pt) + ududff); label("$B$", (5.26,-5.82), NE * labelscalefactor,ududff); dot((20.88,-5.86),linewidth(3.pt) + ududff); label("$C$", (20.96,-5.74), NE * labelscalefactor,ududff); dot((8.672739181695123,-1.8450644076677885),linewidth(3.pt) + uuuuuu); label("$H$", (8.38,-2.46), NE * labelscalefactor,uuuuuu); dot((13.03,-5.9),linewidth(3.pt) + uuuuuu); label("$M$", (13.1,-5.78), NE * labelscalefactor,uuuuuu); dot((11.831266900350528,1.8580370555833747),linewidth(4.pt) + uuuuuu); label("$E$", (11.92,2.02), NE * labelscalefactor,uuuuuu); dot((6.7362862314704595,-1.2081701863961771),linewidth(3.pt) + uuuuuu); label("$F$", (6.82,-1.08), NE * labelscalefactor,uuuuuu); dot((7.685091981184514,1.6766380468384605),linewidth(3.pt) + uuuuuu); label("$P$", (7.76,1.8), NE * labelscalefactor,uuuuuu); dot((6.314614616879968,6.563416944425912),linewidth(3.pt) + uuuuuu); label("$Q$", (6.4,6.68), NE * labelscalefactor,uuuuuu); dot((3.3598423201644065,3.099201148276634),linewidth(4.pt) + uuuuuu); label("$K$", (2.94,2.68), NE * labelscalefactor,uuuuuu); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* re-scale y/x */ currentpicture = yscale(1.6666666666666667) * currentpicture; /* end of picture */ [/asy][/asy]

Let $M$ be the midpoint of $BC$. Let the line through $M$ parallel to $HP$ intersects $BH$ and $CH$ at $P_1$ and $Q_1$ respectively. Let $N$ be the midpoint of arc $\widehat{BHC}$ in $\odot(BHC)$. Claim: $H,P_1,Q_1,N$ are concyclic. Proof: It suffices to show that $BP_1 = CQ_1$ by spiral similarity. Reflect $Q_1$ across $M$ to get $Q_1'$. Notice that $\triangle BP_1Q_1'$ is $B$-isosceles hence $BP_1 = BQ_1' = CQ_1$ as desired. To finish, note that $HP_1=HQ_1$ so $\triangle HP_1Q_1\cup N\cup M$ and $\triangle KQP\cup A\cup H$ are homothetic. Hence $KH\parallel HM$ or $K\in HM$.

Nice and easy problem.

Lemma:Let the $\angle A$-bisector of $ABC$ meet $BC$ at $X$.Let $D$ be the foot of the perpendicular from $A$ to $BC$. $M$ is the midpoint of $BC$. Then, $$\frac {XA^2}{XB \cdot XC}=\frac{XD}{XM}$$ Proof:

We add the point $N$ which is the the intersection of $AX$ and the circumcircle of $ABC$.It is well known that $NM \perp BC$. Hence, we can conclude $\triangle AXD \sim \triangle NXM$. This yields: $$\frac{XD}{XM}=\frac{XA}{XN}=\frac{XA^2}{XA \cdot XN}=\frac{XA^2}{XB \cdot XC}$$

Claim 1:$AXPQ$ is cyclic. Proof:After an inversion centred at $H$ with radius $\sqrt{HA \cdot HD}$followed by a $-1$ homothety, the problem reads as follows: Quote: In triangle $HBC$ the $H$-angle bisector intersects the circles with diameters $HB$ and $HC$ at $Q',P'$ respectively. The angle bisector $BC$ at $T$.$M$ and $D$ are the midpoint and the foot of the altitude of $BC$ respectively. Prove that $MDP'Q'$ is cyclic.

We obtain the following relations using the Power of a point: $$TD \cdot TB=TH \cdot TQ'$$$$TD \cdot TC=TH \cdot TP'$$Multiplying, $$TD^2=\frac{TH^2}{TB \cdot TC} \cdot TP' \cdot TQ'$$Invoking the lemma we proved at the beginning, $$TD^2=\frac{TD}{TM}\cdot TP' \cdot TQ' \implies TD \cdot TM=TP' \cdot TQ'$$Since $MDP'Q'$ is cyclic, $AXPQ$ is cyclic. Main proof: Observe that $K$ also lies on the circumcircle of $AXPQ$.In fact, $K$ is the $A$-antipode in $(AXKPQ)$. Therefore, $$\measuredangle AXK=90^0=\measuredangle AXH$$So, $K,X,H,$ are collinear but it is well known that $M$ lies on $XH$. Therefore, $KH$ bisects $BC$.

My first bary bash Problem 12. Let $H$ be the orthocenter of a nonisosceles triangle $ABC$. The bisector of angle $BHC$ meets $AB$ and $AC$ at points $P$ and $Q$ respectively. The perpendiculars to $AB$ and $AC$ from $P$ and $Q$ meet at $K$. Prove that $KH$ bisects the segment $BC$. [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 = -11.18265488812662, xmax = 7.232793336551774, ymin = -5.091717656662791, ymax = 6.57353839646101; /* image dimensions */ pen qqccqq = rgb(0,0.8,0); draw((-3.59,3.35)--(-6.25,-4.63)--(4.07,-4.55)--cycle, linewidth(0.8) + red); /* draw figures */ draw((-3.59,3.35)--(-6.25,-4.63), linewidth(0.8) + red); draw((-6.25,-4.63)--(4.07,-4.55), linewidth(0.8) + red); draw((4.07,-4.55)--(-3.59,3.35), linewidth(0.8) + red); draw((-6.25,-4.63)--(-0.890879510032572,0.5663117662215824), linewidth(0.8) + linetype("2 2")); draw((-3.522070373032119,-1.9849441844843072)--(-3.510139841924269,-2.0940333222974243), linewidth(0.8) + linetype("2 2")); draw((-3.522070373032119,-1.9849441844843072)--(-3.630739668108302,-1.969654911480994), linewidth(0.8) + linetype("2 2")); draw((-5.194,-1.462)--(4.07,-4.55), linewidth(0.8) + linetype("2 2")); draw((-7.753032520796283,2.4010725639715442)--(-1.09,-4.59), linewidth(1.2) + dotted + blue); draw((-3.59,3.35)--(-5.133082217153884,4.941429440667845), linewidth(0.8) + red); draw((-5.133082217153884,4.941429440667845)--(-7.753032520796283,2.4010725639715442), linewidth(0.8) + linetype("4 4") + blue); draw((-4.290981482888165,1.2470555513355052)--(-7.753032520796283,2.4010725639715442), linewidth(0.8) + linetype("4 4") + blue); draw((-5.133082217153884,4.941429440667845)--(-3.25,-3.37), linewidth(0.8) + qqccqq); /* dots and labels */ dot((-3.59,3.35),dotstyle); label("$A$", (-3.534148169026744,3.480769739912235), NE * labelscalefactor); dot((-6.25,-4.63),dotstyle); label("$B$", (-6.203425681732741,-4.501396668275888), NE * labelscalefactor); dot((4.07,-4.55),dotstyle); label("$C$", (4.127191615038066,-4.424398278486292), NE * labelscalefactor); dot((-3.548445595854923,-2.0105181347150256),linewidth(4pt) + dotstyle); label("$H$", (-3.5854804288864743,-1.6396231810959014), NE * labelscalefactor); dot((-5.133082217153884,4.941429440667845),linewidth(4pt) + dotstyle); label("$Q$", (-5.0869490297835975,5.046403665634021), NE * labelscalefactor); dot((-4.290981482888165,1.2470555513355052),linewidth(4pt) + dotstyle); label("$P$", (-4.239966742098041,1.3504809557334114), NE * labelscalefactor); dot((-7.753032520796283,2.4010725639715442),linewidth(4pt) + dotstyle); label("$K$", (-8.012887841788249,2.312960828103362), NE * labelscalefactor); dot((-1.09,-4.59),linewidth(4pt) + dotstyle); label("$M$", (-1.044533565829804,-4.488563603310955), NE * labelscalefactor); dot((-3.25,-3.37),dotstyle); label("$I$", (-3.200488479938494,-3.2437563017124855), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Solution. Notations. Throughout the proof, we assume, $\angle BHC=A, \angle HBC=B, \angle HCB=C$ and we denote lengths of the segments by $BC=a, HB= c, HC= b$. Also, $I$ denotes the incenter of $\triangle BHC$ and $M$ denotes mid point of $BC$. Also, if a point $X$ has barycentric cordinates $(x:y:z)$, we will denote it by $X:(x:y:z)$. Lemma. We have $\sin(B)\cos(B)(\tan(A)+\tan(B))=\sin(C)\cos(C)(\tan(A)+\tan(C))$. Proof of lemma. We know that$$\frac{\cos(2B)-\cos(2C)}{\sin(2B)-\sin(2C)}=\frac{-2\sin(B+C)\sin(B-C)}{2\sin(B-C)\cos(B+C)}=-\tan(B+C)=\tan(A)$$Thus, we must have, $\cos(2B)-\tan(A)\sin(2B)=\cos(2C)-\tan(A)\sin(2C)$. But we also have $1-2\sin^(\theta)=\cos(2\theta)$ and $\sin(2\theta)=2\sin(\theta)\cos(\theta)$, hence, we get,$$\tan(A)\sin(2B)+2\sin^2(B)=\tan(A)\sin(2C)+2\sin^2(C)\implies \sin(B)\cos(B)(\tan(A)+\tan(B))=\sin(C)\cos(C)(\tan(A)+\tan(C))$$completing the proof of lemma. Proof. We first note that $KQ||HB, KP||CH$ as both $KQ, HB$ are perpendicular to $AC$ and both $KP,HC$ are perpendiular to $AB$. Also, $Q,P,H,I$ are collinear due to obvious reasons and $A$ is the orthocenter of $\triangle BHC$ because $BH\perp AC, CH\perp AB$. Now, we will be using barycentric coordinates to solve this problem. We set $\triangle BHC$ as the reference triangle with $$H:(1,0,0)\quad B:(0,1,0)\quad C: (0,0,1)$$Now, as $A$ and $I$ are the orthocenter and incenter of the reference triangle, we have, $A:(\tan(A):\tan(B),\tan(C))$ and $I:(a:b:c)$. Now, note that $BA$ is the cevain of the reference triangle and $P$ lies on this cevian. Thus, $P:(\tan(A):t:\tan(C))$ for some $t$. Also, $P$ lies on the cevian $HI$ and so $P:(t':b:c)$ for some $t'$. So, the barycentric coordinates $(\tan(A):t:\tan(C))$ and $(t':b:c)$ are the same and so, $\frac{t}{\tan(C)}=\frac{b}{c}\implies P:(\tan(A):\frac{b\tan(C)}{c}:\tan(C))=(c\tan(A):b\tan(C):c\tan(C))$. Similarly, as $Q$ is the intersection of cevians $HI,AC$, we get $Q:(b\tan(A):b\tan(B):c\tan(B))$. Now, note that the equation of the line $BH$ wil be $z=0$ and let equation of line $KQ$ be $a'x+b'y-z=0$. As $KQ||BH\implies$ they meet at $P_{\infty}$. However, the intersection of lines $KQ$ and $BH$ will be thier cross product, which in this case will be $P_{\infty}:(-b':a':0)\implies -b'+a'=0\implies a'=b'=m$(say). Thus, equation of line $KQ$ is $m(x+y)=z$ for some real constant $m$. Similarly as $KP\cap HC=P'_{\infty}$, we get equation of line $KP$ as $n(x+z)=y$ for some real constant $n$. Now, the intersection of lines $KQ$ and $KP$ whihc will be their cross prduct, is barycentric coordinate of $K$. Thus, $K:(mn-1:n-mn:m-mn)$. Now, obviously, $M:(0:1:1)$. So, to prove $K,H,M$ collinear, it is enough to show that $$0=\begin{vmatrix} 1 & 0 & 0\\ 0 & 1 & 1\\ mn-1 & n-mn & m-mn\\ \end{vmatrix}=m-n\implies m=n$$Thus, it is enough to prove $m=n$. Now, we first find the values of $m$ and $n$. For that, we note that the equations if lines $KQ$ and $KP$ are $m(x+y)=z$ and $n(x+z)=y$ respectively. Also, these lines passes through $Q$ and $P$ respectively and we know that $P:(c\tan(A):b\tan(C):c\tan(C))$ and $Q:b\tan(A):b\tan(B):c\tan(B))$. Hence, we must have $$m=\frac{z}{x+y}=\frac{c\tan(B)}{b\tan(A)+b\tan(B)}\quad n=\frac{y}{x+z}=\frac{b\tan(C)}{c\tan(A)+c\tan(C)}$$So, it is enough to show prove that, $$\frac{c\tan(B)}{b\tan(A)+b\tan(B)}=\frac{b\tan(C)}{c\tan(A)+c\tan(C)}$$Also, by sine rule, we know that $\frac{c}{b}=\frac{\sin(C)}{\sin(B)}$. Thus, we get, \begin{align*} &\frac{c}{b}\cdot\frac{\tan(B)}{\tan(A)+\tan(B)}=\frac{b}{c}\cdot\frac{\tan(C)}{\tan(A)+\tan(C)}\\ &\iff \frac{\sin(C)}{\sin(B)}\cdot\frac{\sin(B)}{\cos(B)}(\tan(A)+\tan(C))=\frac{\sin(B)}{\sin(C)}\cdot\frac{\sin(C)}{\cos(C)}(\tan(A)+\tan(B))\\ &\iff \frac{\sin(C)}{\cos(B)}(\tan(A)+\tan(C))=\frac{\sin(B)}{\cos(C)}(\tan(A)+\tan(B))\\ &\iff \sin(B)\cos(B)(\tan(A)+\tan(B))=\sin(C)\cos(C)(\tan(A)+\tan(C)) \end{align*}and clearly the last equation is true due to our lemma which completes the proof. Feedback: Using bary bash by taking reference triangle $ABC$ would be too hectic task, however the observation that $A$ is orthocenter of $BHC$ makes the bash simple and less calculative.

Proposed by Anant Mudgal(anantmudgal09 on AoPS) and Pranjal Srivastava(p_square on AoPS)

mathsworm wrote: I had the courage to barycentric bash this one And the bash is one of the most elegant ones I have ever seen (if it is OK to call bashes elegant) Will post my solution in a while I have used the same idea as @ 2 above, but my solution is shorter

Attached is my soln.

The simplest(one line) solution is by dropping perpendiculars from K to BH and HC and after noting two rectangles use similar triangles and locus of points on a median on K in triangle HBC.

I'll post my bary bash I guess [asy][asy] import olympiad; size(7cm); defaultpen(fontsize(10pt)); pair A = dir(115), B = dir(215), C = dir(325), H = orthocenter(A,B,C), M = (B+C)/2, P = extension(A,B,H,incenter(H,B,C)), Q = extension(A,C,H,incenter(H,B,C)), K = extension(Q,Q+dir(H--B),P,P+dir(C--H)), R = extension(H,incenter(H,B,C),B,C); dot("$A$", A, dir(60)); dot("$B$", B, dir(300)); dot("$C$", C, dir(330)); dot("$H$", H, dir(45)); dot("$P$", P, dir(0)); dot("$Q$", Q, dir(90)); dot("$K$", K, dir(150)); dot("$M$", M, dir(270)); dot("$R$", R, dir(270)); draw(B--H--C^^R--Q--K--P^^A--B--C--A--Q); draw(K--M, dashed); [/asy][/asy] Let $M$ be the midpoint of $\overline{BC}$ and $R$ be the point where the bisector of angle $BHC$ meets $\overline{BC}$. Claim: $K$ lies on the external bisector of $A$. Proof. Since $\overline{KP} \perp \overline{AB} \perp \overline{CH}$ and $\overline{KQ} \perp \overline{CA} \perp \overline{BH}$, we have $$\angle KPQ = \dfrac{1}{2} \angle BHC = \angle KQP,$$so $KQ = KP$, implying the result. $\square$ Now we use barycentric coordinates with respect to $ABC$. Set $A = (1,0,0)$, $B = (0,1,0)$, $C = (0,0,1)$, with $a = BC$, $S_A = \tfrac{1}{2} (-a^2+b^2+c^2)$, etc. By Angle Bisector Theorem and Law of Sines, we have $$\dfrac{CR}{BR} = \dfrac{CH}{BH} = \dfrac{\sin \angle HBC}{\sin \angle HCB} = \dfrac{\cos C}{\cos B},$$so $$R = (0:CR:BR) = (0:\cos C: \cos B) = \left(0:\dfrac{S_C}{ab}:\dfrac{S_B}{ca}\right) = (0:cS_C:bS_B).$$Recall that $H = (S_{BC}:S_{CA}:S_{AB})$. Let $P = (p,1-p,0)$. $P$ lies on $\overline{HR}$ implies \begin{align*} 0 &= \begin{vmatrix} 0 & cS_C & bS_B \\ S_{BC} & S_{CA} & S_{AB} \\ p & 1-p & 0 \end{vmatrix} \\ &= S_{BC}[(1-p)bS_B] + p[(cS_C)(S_{AB})-(bS_B)(S_{CA})] \\ &= S_{BC}[(1-p)bS_B + pS_A(c-b)], \end{align*}which simplifies to $$p(b(S_A+S_B)-cS_A) = bS_B \implies p = \dfrac{bS_B}{c(bc-S_A)}.$$Thus $$P = \left(\dfrac{bS_B}{c(bc-S_A)}, 1-\dfrac{bS_B}{c(bc-S_A)}, 0\right) = \left(\dfrac{bS_B}{c(bc-S_A)}, \dfrac{S_A(b-c)}{c(bc-S_A)}, 0\right).$$By the result of our claim, we must have $K = (k:b:-c)$ for some $k$. It follows that $$\overrightarrow{PK} = \left(\dfrac{bS_B}{c(bc-S_A)} - \dfrac{k}{k+b-c}, \dfrac{S_A(b-c)}{c(bc-S_A)} - \dfrac{b}{k+b-c}, \dfrac{c}{k+b-c}\right).$$Recall that by EFFT, a displacement vector $(x_1,y_1,z_1)$ is perpendicular to $\overrightarrow{AB} = (1,-1,0)$ if and only if \begin{align*}0 &= a^2[(z_1)(-1)] + b^2[(z_1)(1)] + c^2[(1)(y_1)+(-1)(x_1)] \\ &= a^2(x_1+y_1) - b^2(x_1+y_1) + c^2(-x_1+y_1) \\ &= x_1(a^2-b^2-c^2)+y_1(a^2-b^2+c^2) \\ \implies x_1S_A &= y_1S_B.\end{align*}We know that $\overrightarrow{PK} \perp \overrightarrow{AB}$, so \begin{align*} S_A\left(\dfrac{bS_B}{c(bc-S_A)} - \dfrac{k}{k+b-c}\right) &= S_B\left(\dfrac{S_A(b-c)}{c(bc-S_A)} - \dfrac{b}{k+b-c}\right) \\ \implies \dfrac{S_{AB}}{c(bc-S_A)}(b-(b-c)) &= \dfrac{kS_A-bS_B}{k+b-c} \\ \implies \dfrac{S_{AB}}{bc-S_A} &= \dfrac{kS_A-bS_B}{k+b-c} \\ \implies kS_{AB} + S_{AB}(b-c) &= kS_A(bc-S_A) + bS_B(S_A-bc) \\ \implies kS_A(S_B + S_A - bc) &= S_B(bS_A-b^2c-bS_A+cS_A) \\ \implies kcS_A(c-b) &= cS_B(-S_C) \\ \implies k &= \dfrac{S_{BC}}{S_A(b-c)}.\end{align*}Substituting yields $$K = \left(\dfrac{S_{BC}}{S_A(b-c)}:b:-c\right) = (S_{BC}:bS_A(b-c):cS_A(c-b)).$$Note that $M = (0:1:1)$. To show that $K$, $H$, $M$ are collinear, we calculate \begin{align*} \begin{vmatrix} S_{BC} & bS_A(b-c) & cS_A(c-b) \\ S_{BC} & S_{CA} & S_{AB} \\ 0 & 1 & 1 \end{vmatrix} &= S_{BC} \begin{vmatrix} 1 & bS_A(b-c) & cS_A(c-b) \\ 1 & S_{CA} & S_{AB} \\ 0 & 1 & 1 \end{vmatrix} \\ &= S_{BC}[(S_{CA}-S_{AB})+(cS_A(c-b)-bS_A(b-c)] \\ &= S_{ABC}(S_C-S_B+c^2-bc-b^2+bc) \\ &= S_{ABC} (b^2-c^2+c^2-b^2) \\ &= 0,\end{align*}as desired. $\blacksquare$

My solution (ofc Humpty point) All of the following steps are either use of a property of humpty point or simson line. Let $Z$ be humpty of $HBC$ Since $ZABC$ cyclic, $Z$ reflected across $AB, AC, H$ and $Z$ reflected across $BC$ collinear. This means $Z$ reflected across $AP, AQ$ is $H-$ symmedian, and it follows that $ZAPQ$ is cyclic. We're done as $\angle KZA = \dfrac{\pi}{2} = \angle MZA$

Nice and easy! I'll just give a quick sketch: Note that it suffices to show that $KH$ passes through the $A$-Queue point $Q_A$. Since $\measuredangle AQ_AH=90^{\circ}$, so it suffices to show that $Q_A \in \odot (AK)$. But $\odot (AK)$ is nothing but $\odot (APQ)$. I didn't get a synthetic solution on first glance , but this can easily be Menelaus bashed after $\sqrt{AH \cdot AD}$ inversion (use the fact that the angle bisectors of $\angle BHC$ and $\angle BAC$ are parallel to find the inverse of $P,Q$).

Here's a short synthetic solution: Let $AX$ be $\perp$ to $KH$ and $X$ lies on $KH$;First of all observe that the angle bisector of $BHC$ is parallel to angle bisector of $BAC$ So we have $AP$=$AQ$ and $KP=KQ$.So $XA$ is internal angle bisector and $XK$ is external angle bisector of $\angle XPQ$ Let $XA\cap PQ=Y$;Then $(P,Q;Y,H)=-1$.Projecting from $A$ on $(ABC)$ we get $X$ is the $A$ Queue point and we are done.

math_pi_rate wrote: Nice and easy! I'll just give a quick sketch: Note that it suffices to show that $KH$ passes through the $A$-Queue point $Q_A$. Since $\measuredangle AQ_AH=90^{\circ}$, so it suffices to show that $Q_A \in \odot (AK)$. But $\odot (AK)$ is nothing but $\odot (APQ)$. I didn't get a synthetic solution on first glance , but this can easily be Menelaus bashed after $\sqrt{AH \cdot AD}$ inversion (use the fact that the angle bisectors of $\angle BHC$ and $\angle BAC$ are parallel to find the inverse of $P,Q$). Menelaus bash afer inversion is quite similar to using cross ratios before inverting

Here's my solution [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(1) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.88, xmax = 8.12, ymin = -3.8, ymax = 9.36; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-2.38,5.72)--(-6.38,1.36), linewidth(2) + wrwrwr); draw((-6.38,1.36)--(-0.32,1.46), linewidth(2) + wrwrwr); draw((-0.32,1.46)--(-2.38,5.72), linewidth(2) + wrwrwr); draw((-6.38,1.36)--(-1.429304369573761,3.753998356497195), linewidth(2) + wrwrwr); draw((-3.5606590192404375,4.433081669027923)--(-0.32,1.46), linewidth(2) + wrwrwr); draw((-6.38,1.36)--(-4.359711624189135,-0.4934755741384077), linewidth(2) + wrwrwr); draw((-4.359711624189135,-0.4934755741384077)--(-0.32,1.46), linewidth(2) + wrwrwr); draw((-2.38,5.72)--(-1.9113879556341498,6.230787128358778), linewidth(2) + wrwrwr); draw((-2.3402883758108652,3.313475574138408)--(-1.9113879556341498,6.230787128358778), linewidth(2) + wrwrwr); draw((xmin, 1.8851675355000737*xmin + 7.725311243925247)--(xmax, 1.8851675355000737*xmax + 7.725311243925247), linewidth(2) + linetype("4 4") + wrwrwr); /* line */ draw((-2.0782303455219133,5.095952073749199)--(-1.15895686910294,5.5404833792475765), linewidth(2) + wrwrwr); draw((-1.9113879556341498,6.230787128358778)--(-1.15895686910294,5.5404833792475765), linewidth(2) + wrwrwr); /* dots and labels */ dot((-2.38,5.72),dotstyle); label("$A$", (-2.3,5.92), NE * labelscalefactor); dot((-6.38,1.36),dotstyle); label("$B$", (-6.3,1.56), NE * labelscalefactor); dot((-0.32,1.46),dotstyle); label("$C$", (-0.24,1.66), NE * labelscalefactor); dot((-2.3402883758108652,3.313475574138408),linewidth(4pt) + dotstyle); label("$H$", (-2.26,3.48), NE * labelscalefactor); dot((-4.359711624189135,-0.4934755741384077),linewidth(4pt) + dotstyle); label("$A_1$", (-4.28,-0.34), NE * labelscalefactor); dot((-3.5606590192404375,4.433081669027923),linewidth(4pt) + dotstyle); label("$F$", (-3.48,4.6), NE * labelscalefactor); dot((-1.429304369573761,3.753998356497195),linewidth(4pt) + dotstyle); label("$E$", (-1.34,3.92), NE * labelscalefactor); dot((-1.9113879556341498,6.230787128358778),linewidth(4pt) + dotstyle); label("$Q$", (-1.84,6.4), NE * labelscalefactor); dot((-2.0782303455219133,5.095952073749199),linewidth(4pt) + dotstyle); label("$P$", (-2,5.26), NE * labelscalefactor); dot((-1.15895686910294,5.5404833792475765),linewidth(4pt) + dotstyle); label("$K$", (-1.08,5.7), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $A_1$ be the antipode of $A$ in $\odot(ABC)$. Then, we want to show $A_1$, $H$,$K$ collinear. Note that $B, F, Q$ / $C, E, P$are the foots of perpendicular lines which passes $A_1, H, K$. So, It's enough to show ${BF \over BQ} = {CE \over EP}$. Now, let's look at $\triangle BHQ$ and $\triangle CHP$. $\angle QBH$ = $\angle PCH$, $\angle BHQ$ = $\angle CHP$ because the line $HPQ$ is the angle bisector of $\angle BHC$. So we get $\triangle BHQ \sim \triangle CHP$ . The point $E$ and $F$ are corresponding points, and we are done.

Let the two angular bisectors of BHC intersect perp bisector at M,N and say X,Y,Z are the midpoints of line BC, minor arc BC and major arc BC Then it suffices to prove MX/ NX = YX/XZ which is just pop.

Nothing to see here……. Move along

Aryan-23 wrote: Only problem I solved Just note that if $Q_A$ is the miquel point of $BCEF$ , then the spiral similarity $f$ centered at $Q_A$ that maps $F \mapsto E $ ,$B \mapsto C $ also maps $P$ to $Q$ . So $HK$ passes through $Q_A$ as $AH$ and $AK$ are diameters of $(AQ_AEF)$$(AQ_APQ)$ .So since $Q_A , H ,M$ are collinear , we're done This idea shows up in ISL 2005 G5 too Why $f$ maps $P$ to $Q$?

We will use Cartesian coordinates. We translate, rotate, and scale $ABC$ until the $y$-axis is the bisector of angle $BHC$, $H = (0,0), C = (1,a)$ for some real $a \neq 0$. If the slope of $CH$ is $\frac{a-0}{1-0} = a$, then the slope of $BH$ must be $-a$. So the coordinates of $B$ must be $(-b,ab)$ for some real $b \neq 1$ (because $AB \neq AC$). $P$ and $Q$ are both on the $y$-axis such that $PB \perp HC$ and $QC \perp HB$. $P = (0,p)$ and $Q =(0,q)$ while we will solve for $p$ and $q$. Since the slope of $HC$ is $a$, the slope of $PB$ must be $-\frac{1}{a}$. Then $\frac{p - ab}{b} = -\frac{1}{a} \Leftrightarrow p = ab - \frac{b}{a}$ Since the slope of $HB$ is $-a$, the slope of $QC$ must be $\frac{1}{a}$. Then $\frac{q - a}{-1} = \frac{1}{a} \Leftrightarrow q = a - \frac{1}{a}$ Since $PK \perp AB$, $PK$ must be parallel to $HC$ and must have the same slope, which is $a$. Similarly, $QK$ must be parallel to $HB$ and must have the same slope, which is $-a$. So the equation of $PK$ is $y = ax + p$ which is $y = ax + ab - \frac{b}{a}$, and the equation of $QK$ is $y = -ax + q$ which is $y = -ax + a - \frac{1}{a}$. Solving the system we get $x = \frac{(a - \frac{1}{a})(-b + 1)}{2a}$ and $y = \frac{(a - \frac{1}{a})(b + 1)}{2}$. Thus, $K = \left(\frac{(a - \frac{1}{a})(-b + 1)}{2a}, \frac{(a - \frac{1}{a})(b + 1)}{2}\right)$. The slope of $KH$ is $\frac{(a - \frac{1}{a})(b + 1)}{2} \div \frac{(a - \frac{1}{a})(-b + 1)}{2a} = \frac{ab + a}{-b + 1}$. Letting $M$ as the midpoint of $BC$, $M = \left(\frac{-b + 1}{2},\frac{ab + a}{2}\right)$. The slope of $MH$ is $\frac{ab + a}{2} \div \frac{-b + 1}{2} = \frac{ab + a}{-b + 1}$. Since $KH$ and $MH$ have the same slope, $K, M, H$ are collinear, and we are done.

Sol:- Let (AH) intersect (ABC) at Qa .Clearly Qa-H-M are collinear where M is the midpoint of BC. If AH intersect (ABC) at Ha. Projecting through A gives (Qa,Ha;B,C) =-1 . Hence the angle bisectors of BQaC & BHC meet at BC. Let this point be S. We get that BQaPS is cyclic. Hence Qa is the center of spiral sim. mapping SC to PA. From here we get that angle SQaC=angle PQaA. So, APQaQR is cyclic. Hence R-Qa-H-M are collinear.

Let $E,F$ be foot from $B,C$ to $AC,AB$, respectively. Let $R=(ABC)\cap(AHEF)$. I contend that $R$ lies on $(APQ)$. Observe that $\triangle PHB\sim\triangle QHC$. Indeed, \begin{align*} \frac{\mathbb{P}(P,(ABC))}{\mathbb{P}(P,(AHEF))}=\frac{PB}{PF}=\frac{QE}{QC}=\frac{\mathbb{P}(Q,(ABC))}{\mathbb{P}(Q,(AHEF))}, \end{align*}thus by the coaxiality lemma, $(APQ)$ passes through $R$. It is well-known that $R,H,M$ are collinear and by Reim's thereom, $K,R,H$ are collinear, we are done. $\blacksquare$

BALAMATDA wrote: Aryan-23 wrote: Only problem I solved Just note that if $Q_A$ is the miquel point of $BCEF$ , then the spiral similarity $f$ centered at $Q_A$ that maps $F \mapsto E $ ,$B \mapsto C $ also maps $P$ to $Q$ . So $HK$ passes through $Q_A$ as $AH$ and $AK$ are diameters of $(AQ_AEF)$$(AQ_APQ)$ .So since $Q_A , H ,M$ are collinear , we're done This idea shows up in ISL 2005 G5 too Why $f$ maps $P$ to $Q$? $BHP$ is similar to $CHQ$, so $BF:FP=CE:EQ$. Since $f$ maps $B$ to $C$, $F$ to $E$, it also maps $P$ to $Q$.

Here's a proof with inversion: [asy][asy] size(200); pair A= dir(110),B=dir(-150),C=dir(-30),H=A+B+C,E=foot(H,A,C),F=foot(H,A,B),M=1/2*(B+C),XA=extension(E,F,B,C),QA=extension(M,H,A,XA),T=extension(QA,dir(-90),B,C),P=extension(H,T,A,B),Q=extension(H,T,A,C),K=2*circumcenter(A,P,Q)-A; draw(circumcircle(A,B,C)^^circumcircle(A,H,QA)); dot("$A$",A,dir(70)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$H$",H,dir(-90)); dot("$M$",M,dir(M)); dot("$E$",E,dir(0)); dot("$F$",F,dir(200)); dot("$Q_A$",QA,dir(180)); dot("$P$",P,dir(0)); dot("$Q$",Q,dir(Q)); dot("$K$",K,dir(K)); draw(A--B--C--A--Q,red); draw(B--E^^C--F,blue); draw(H--Q,blue); draw(M--K,green); draw(P--K--Q,brown); [/asy][/asy] Let $Q_A$ be the $A$-Queue point of $\triangle ABC$ (i.e. $Q_A = \odot(ABC) \odot(AH) \ne A$) and $\overline{BE},\overline{CF}$ be altitudes. We know $\overline{HQ_A}$ bisects segments $BC$, so it suffices to show $Q_A \in \odot(APQ)$. As $(H,Q_A ; E,F) = -1$, thus our problem is reduced to following: Reduced Problem wrote: Four points $H,Q_A,E,F$ lie on a circle $\Gamma$ with $(H,Q_A ; E,F) = -1$. $A$ is antipode of $H$ wrt $\Gamma$. Angle bisector of $\angle EFH$ intersects $\overline{AF},\overline{AE}$ at points $P,Q$, respectively. Prove that points $A,P,Q,Q_A$ are concyclic. If we invert at $H$ with squared radius $HE \cdot HF$ followed by reflection in angle bisector of $\angle EHF$, our problem becomes: Inverted Problem wrote: Given $\triangle HEF$, let $\ell$ be angle bisector of $\angle EHF$ ; $A,P,Q$ be projections of $H,F,E$ onto $EF,\ell,\ell$, respectively ; $Q_A$ midpoint of segment $EF$. Prove points $A,P,Q,Q_A$ are concyclic. [asy][asy] size(200); pair H = dir(135),E=dir(-160),F=dir(-20),QA = 1/2*(E+F),A=foot(H,E,F),D=extension(H,dir(-90),E,F),P=foot(F,H,D),Q=foot(E,H,D),T=foot(QA,P,Q); dot("$H$",H,dir(H)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$A$",A,dir(A)); dot("$D$",D,dir(50)); dot("$P$",P,dir(P)); dot("$Q$",Q,dir(40)); dot("$Q_A$",QA,dir(50)); dot("$T$",T,dir(T)); draw(H--P,linewidth(0.8)); draw(H--E--F--H,red); draw(E--Q^^P--F^^QA--T,brown); draw(circumcircle(A,P,Q),dotted); draw(H--A,blue); [/asy][/asy] Let $D = \ell \cap \overline{EF}$. Projecting $\ell$ onto $EF$ through point at infinity in the direction perpendicular to $\ell$ gives $(H,D ; P,Q) = -1$. Let $T$ be projection of $Q_A$ onto $\ell$. Note $T$ is the midpoint of segment $PQ$. From $(A,D ; P,Q) = -1$ we get $DP \cdot DQ = DH \cdot DT$. Now as points $A,H,T,Q_A$ lie on $\odot(HQ_A)$, so $DH \cdot DT = DA \cdot DQ_A$, as desired. $\blacksquare$

Very clean bash! Let $KQ$ meet $AB$ and $BC$ at $D$ and $U$ respectively. Similarly, let $KP$ meet $AC$ and $BC$ at $E$ and $V$ respectively. Finally, let $KH$ meet $BC$ at $L$. Because $KU\parallel HB$ and $KV\parallel HC$, it can easily be seen that the desired reduces to showing that $BU=CV$. By the law of sines in $\triangle BUD$ it follows that $$\frac{BU}{BD}=\frac{\cos{A}}{\cos{C}}$$and by the law of sines in $\triangle CVE$ we have $$\frac{CV}{CE}=\frac{\cos{A}}{\cos{B}},$$so the desired reduces to showing that $\frac{BD}{CE}=\frac{\cos{C}}{\cos{B}}$. Now denote $AP=AQ=x$. Then $AD=AE=x/\cos{A}$, so the desired becomes $$\frac{c+x/\cos{A}}{b-x/\cos{A}}=\frac{\cos{C}}{\cos{B}}$$or equivalently $$b\cos{C}-c\cos{B}=\frac{x}{\cos{A}}\left(\cos{B}+\cos{C}\right)\Leftrightarrow 2R\cos{A}\sin{\left(B-C\right)}=2x\cos{\frac{B+C}{2}}\cos{\frac{B-C}{2}}$$$$\Leftrightarrow 2R\cos{A}\sin{\frac{B-C}{2}}=x\sin{\frac{A}{2}}.$$ However Menalaus gives $$\frac{BL}{CL}\cdot\frac{CQ}{AQ}\cdot\frac{CP}{AP}=1\Leftrightarrow\frac{BH}{CH}\cdot\frac{b+x}{x}\cdot\frac{x}{c-x}=1$$or $$\frac{b+x}{c-x}=\frac{\cos{C}}{\cos{B}}\Leftrightarrow x\left(\cos{B}+\cos{C}\right)=c\cos{C}-b\cos{B}$$$$\Leftrightarrow 2x\cos{\frac{B+C}{2}}\cos{\frac{B-C}{2}}=R\left(\sin{2C}-\sin{2B}\right)=2R\cos{A}\sin{\left(B-C\right)}$$$$\Leftrightarrow x\sin{\frac{A}{2}}=2R\cos{A}\sin{\frac{B-C}{2}},$$which is what we wanted.