My solution : Let $ P, Q $ be the projection of $ X $ on $ AC, AB $, respectively . Let $ M $ be the midpoint of $ BC $ and $ T $ be a point such that $ TB \parallel AC, TC \parallel AB $ . Since $ \{ BT, BX \}, \{ CT, CX \} $ are isogonal conjugate WRT $ \angle CBA, \angle ACB $, respectively , so $ X, T $ are the isogonal conjugate of $ \triangle ABC $ $ \Longrightarrow P, Q, M, H' $ are concyclic ( pedal circle of $ \{ X, T \} $ ) . Since $ B, M, X, Q $ are concyclic and $ C, M, X, P $ are concyclic , so $ \angle QH'P=\angle QMP=\angle QBX+\angle PCX=\angle ACB+\angle CBA $ . ... $ (\star) $ Since $ H', X, Q, Y $ are concyclic and $ H', X, P, Z $ are concyclic , so $ \angle ZXY=\angle ZXH'+\angle YXH'=\angle APH'+\angle AQH'=\angle QH'P-\angle BAC $ , hence combine $ (\star) $ we get $ \angle ZXY=180^{\circ}-2\angle BAC=\angle CXB $ $ \Longrightarrow \angle ZXC=\angle YXB $ . Q.E.D

beautifull prove thanks

Another solution: Let $M$ be the midpoint of $BC$. $U, V$ are the projection of $X$ on $AB, AC$, respectively. $T$ is the projection of $A$ on $OX$. Then $AMH'T$ is the parallelogram. We have $\angle AUM=\angle BXM=90^{\circ}-\angle BAC \Longrightarrow UM\perp AC$. Similarly, $VM\perp AB$ $\Longrightarrow M$ is the orthocenter of $\triangle AUV$. On the other hand, $A, T, U, V, X$ are concyclic in the circle $\odot (AX)$, $AMH'T$ is the parallelogram. $\Longrightarrow H'$ is the orthocenter of $\triangle TUV$. $\Longrightarrow \angle UH'V=180^{\circ}-\angle BAC$ $\Longrightarrow \angle YXZ=180^{\circ}-\angle XYZ-\angle XZY=180^{\circ}-\angle H'UX-\angle H'VX=\angle UH'V+\angle UXV-180^{\circ}=180^{\circ}-2.\angle BAC=\angle BXC$ $\Longrightarrow \angle ZXC=\angle YXB$. Q.E.D

Too easy problem for problem 6 My solution: Lemma1: let $R,T$ projections of $X$ on $AB,AC$ and assume that $M$ is midpoint of $BC$ then $M$ is orthocenter of $\triangle ART$ for prove see Russia 2013 grade 9. Lemma2: in cyclic quadrilateral $ABCD$ points $M,N$ are midpoints of $AB$ and $CD$ and $AC\cap CD=P$ also $H_1,H_2$ are projections of $P$ on $AD,BC$ then $\triangle H_1MN=\triangle H_2MN$.(it's well known) Lemma3: in triangle $ABC$ ,$AD,BE,CF$ are altitudes and $M,N,S,T$ are midpoints of $BC,CE,BF,EF$ and $AL$ is altitude in $\triangle AEF$ and let $R$ be the reflection of $L$ WRT midpoint of $EF$ then quadrilateral $NLRS$ is cyclic. Prove: note that $R$ is foot of $H$(orthocenter of $\triangle ABC$) on $EF$.then using lemma 2 for the cyclic quadrilateral $BFEC$ we get that $\angle SRN=\angle SMN$ note that $\angle ADM=\angle MSA=90\longrightarrow AMDS$ is cyclic. So $\angle SMN=\angle 180-A$ so $\angle SRN=180-A$ also using lemma 1 we get that $T$ is orthocenter of $ASN$ so $\angle STN=180-A$ and the lemma proved. BACK TO THE MAIN PROBLEM: Let $M$ midpoint of $BC$ then from the lemma3,1 $RMH'T$ is cyclic and $M$ is orthocenter of $ARZ$ so $\angle RMT=\angle RH'T=180-A$ but because $XH'ZT,XH'YR$ are cyclic we get that $180-A=\angle RH'T=\angle XZT+\angle XYR=\angle A+\angle TXY\longrightarrow \angle ZXY=180-2A=\angle CXB\longrightarrow \angle CXZ=\angle YXB$ DONE

Attachments:

andria, can you provide a link for Lemma 2?

Achillys wrote: andria, can you provide a link for Lemma 2? See http://www.artofproblemsolving.com/community/c6h296040 I also solved the problem 3 in second round exam using this lemma.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */import graph; size(300); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.3, xmax = 9.5, ymin = -6.58, ymax = 6.3; /* image dimensions */pen zzttqq = rgb(0.6,0.2,0); pen wwzzqq = rgb(0.4,0.6,0); pen qqwuqq = rgb(0,0.39215686274509803,0); draw((2.92,4.04)--(1.04,-0.08)--(6.24,-0.04)--cycle, zzttqq); draw(arc((3.680402725208175,-5.312354277062829),0.6,60.844980720518215,64.10457654225013)--(3.680402725208175,-5.312354277062829)--cycle, qqwuqq); draw(arc((3.680402725208175,-5.312354277062829),0.6,473.51728378153996,476.7768796032719)--(3.680402725208175,-5.312354277062829)--cycle, qqwuqq); draw((3.9073430096064508,-0.0028055971098801327)--(3.8554441513586597,-0.4238833910061812)--(4.27652194525496,-0.47578224925397233)--(4.328420803502752,-0.054704455357671315)--cycle, qqwuqq); draw((-0.09509926716624112,-3.589552397047317)--(0.08102693760250201,-3.203573692979646)--(-0.3049517664651692,-3.0274474882109033)--(-0.48107797123391227,-3.4134261922785742)--cycle, qqwuqq); /* draw figures */draw((1.04,-0.08)--(6.24,-0.04), zzttqq); draw(circle((3.6300996777336425,1.2270418946263961), 2.901202311956972)); draw((4.328420803502752,-0.05470445535767114)--(3.680402725208175,-5.312354277062829)); draw((2.92,4.04)--(6.46639011001663,-0.3182143520686293)); draw((6.46639011001663,-0.3182143520686293)--(1.2260256631944162,0.327673261894146)); draw((1.04,-0.08)--(3.680402725208175,-5.312354277062829)); draw((3.680402725208175,-5.312354277062829)--(6.24,-0.04)); draw((1.2260256631944162,0.327673261894146)--(3.680402725208175,-5.312354277062829)); draw((3.680402725208175,-5.312354277062829)--(6.46639011001663,-0.3182143520686293)); draw((2.92,4.04)--(-0.48107797123391227,-3.4134261922785742)); draw((-0.48107797123391227,-3.4134261922785742)--(3.680402725208175,-5.312354277062829)); draw((-0.48107797123391227,-3.4134261922785742)--(4.328420803502752,-0.05470445535767114), wwzzqq); draw((2.92,4.04)--(3.680402725208175,-5.312354277062829), wwzzqq); /* dots and labels */dot((2.92,4.04),dotstyle); label("$A$", (2.68,4.4), NE * labelscalefactor); dot((1.04,-0.08),dotstyle); label("$B$", (0.48,-0.28), NE * labelscalefactor); dot((6.24,-0.04),dotstyle); label("$C$", (6.44,-0.04), NE * labelscalefactor); dot((2.951579196497248,-0.06529554464232885),linewidth(3pt) + dotstyle); label("$H$", (2.84,-0.62), NE * labelscalefactor); dot((3.680402725208175,-5.312354277062829),linewidth(3pt) + dotstyle); label("$X$", (3.84,-5.72), NE * labelscalefactor); dot((4.328420803502752,-0.05470445535767114),linewidth(3pt) + dotstyle); label("$H'$", (4.32,0.18), NE * labelscalefactor); dot((1.2260256631944162,0.327673261894146),linewidth(3pt) + dotstyle); label("$Y$", (0.98,0.48), NE * labelscalefactor); dot((6.46639011001663,-0.3182143520686293),linewidth(3pt) + dotstyle); label("$Z$", (6.62,-0.76), NE * labelscalefactor); dot((-0.48107797123391227,-3.4134261922785742),linewidth(3pt) + dotstyle); label("$P$", (-0.8,-4.06), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Let the projection of $X$ on $AB$ be $P$. Now we observe that $\frac{CA}{CX} = \frac{b}{CX} = \frac{b \cos{C}}{CX \cos{C}} = \frac{BH'}{BX \cos{C}} = \frac{BH'}{BP}$ Also obviously, $\angle H'BP = \angle ACX$ Thus we have $\triangle H'BP \sim \triangle ACX$ Since $Y,H',X,P$ are concyclic, Thus angle $\angle CXA = \angle BPH' = \angle H'XY$ That is, $XC,XY$ and $XH',XA$ are isogonal lines with respect to $\angle CXY$ By our lemma here ($AY \cap CH' = B$, $YH' \cap AC = Z$), $XB$ and $XZ$ are also a pair of isogonal lines with respect to $\angle CXY$ Thus $\angle BXY = \angle ZXC$ hence proved.

ATimo wrote: $AH$ is the altitude of triangle $ABC$ and $H^\prime$ is the reflection of $H$ trough the midpoint of $BC$. If the tangent lines to the circumcircle of $ABC$ at $B$ and $C$, intersect each other at $X$ and the perpendicular line to $XH^\prime$ at $H^\prime$, intersects $AB$ and $AC$ at $Y$ and $Z$ respectively, prove that $\angle ZXC=\angle YXB$. Let $A’$ be the point such that $AHMA’$ form a rectangle. Let $P, Q$ be the projection of $X$ on $AB, AC$. Claim : $H’$ is the orthocentre of $\triangle A’PQ$ Proof : Note that $A’H’MA$ forms a parallelogram. So, $A’H’ \perp PQ$. Note that $M $ is the orthocentre of $APQ$. So, by the converse of a well-known lemma, we have $H’$ to be the orthocentre of $\triangle A’PQ$. $\square$ So, we have $\measuredangle{PH’Q} = 180 - \measuredangle{BAC}$ $$\measuredangle{ZXY} = \measuredangle{H’QA} + \measuredangle{H’PA} = \measuredangle{PH’Q} - \measuredangle{BAC} = 180 - 2 \measuredangle{BAC} = \measuredangle{BXC}$$Thus, we have $\measuredangle{ZXC} = \measuredangle{YXB}$. $\blacksquare$

Direct application of Moving Points. ( This solution is wrong in many ways, as I did not understand Moving Points very well, but I didnt find a button to remove it, so just ignore.) Let's animate A around the circle $\omega$ (contains $B, C$ and is tangent to $XB, XC$). First, let's define some points, let $K = YX \cap BC, K'$ is the reflection of K wrt. the midpoint $M$ of BC, and $O$ the center of $\omega$ The first transformation is $A \rightarrow H \rightarrow H' \rightarrow H'X \rightarrow H'Y \rightarrow Y \rightarrow K \rightarrow K' \rightarrow XZ_1 \rightarrow Z_1$, with $XZ_1$ being the reflection of $XK'$ wrt. $XC$, $Z_1 \in AC$. The second is $A \rightarrow H \rightarrow H' \rightarrow H'X \rightarrow HZ \rightarrow Z$, so it suffices to prove that these transformations coincide for 3 points, ie. $Z = Z_1$ for three choices of $A$. 1) Let $A$ be the second intersection of $XO$ and $\omega$, so $H = M = H' \rightarrow Y=B, Z=C \rightarrow K=B \rightarrow K' = C \rightarrow Z_1 = C = Z$ 2) Let $A = B$, this gives us that $H' = C$, implying that the line perpendicular to $H'X$ is, in fact, line $OC \rightarrow Z = C$. As line $AB$ is $XB \rightarrow K'=C \rightarrow Z_1=C=Z$ 3) The case $A=C$ is analogous, so we are done

Let $A'$ is the reflection of $A$ over the perpendicular bisector of $BC$. BY DDIT, $\square BYZC$, $(XB,XZ) , (XY,XC) , (XA,XH')$ are reciprocal pairs. So, It suffices to show that $\angle A'XC =\angle AXB = \angle ZXH'$. Which is true since $Z$ the isogonal conjugate of $A'$ wrt. $\triangle XH'C$. ($\angle XCZ = \angle ABC = \angle A'CB$ and $\angle XH'Z = \angle AH'C = 90^{\circ}$)

Felt somewhat easy for a P6

Let $E$ and $F$ be the feet of perpendicular from $X$ to $AB$ and $AC$ CLaim 1:$\triangle EBH' \sim \triangle XCA$

Proof

Now, $$\measuredangle BXH'=\measuredangle BEH' = \measuredangle AXC$$, now by Isogonal Lines lemma we're done.

andria wrote: Too easy problem for problem 6 My solution: Lemma1: let $R,T$ projections of $X$ on $AB,AC$ and assume that $M$ is midpoint of $BC$ then $M$ is orthocenter of $\triangle ART$ for prove see Russia 2013 grade 9. Lemma2: in cyclic quadrilateral $ABCD$ points $M,N$ are midpoints of $AB$ and $CD$ and $AC\cap CD=P$ also $H_1,H_2$ are projections of $P$ on $AD,BC$ then $\triangle H_1MN=\triangle H_2MN$.(it's well known) Lemma3: in triangle $ABC$ ,$AD,BE,CF$ are altitudes and $M,N,S,T$ are midpoints of $BC,CE,BF,EF$ and $AL$ is altitude in $\triangle AEF$ and let $R$ be the reflection of $L$ WRT midpoint of $EF$ then quadrilateral $NLRS$ is cyclic. Prove: note that $R$ is foot of $H$(orthocenter of $\triangle ABC$) on $EF$.then using lemma 2 for the cyclic quadrilateral $BFEC$ we get that $\angle SRN=\angle SMN$ note that $\angle ADM=\angle MSA=90\longrightarrow AMDS$ is cyclic. So $\angle SMN=\angle 180-A$ so $\angle SRN=180-A$ also using lemma 1 we get that $T$ is orthocenter of $ASN$ so $\angle STN=180-A$ and the lemma proved. BACK TO THE MAIN PROBLEM: Let $M$ midpoint of $BC$ then from the lemma3,1 $RMH'T$ is cyclic and $M$ is orthocenter of $ARZ$ so $\angle RMT=\angle RH'T=180-A$ but because $XH'ZT,XH'YR$ are cyclic we get that $180-A=\angle RH'T=\angle XZT+\angle XYR=\angle A+\angle TXY\longrightarrow \angle ZXY=180-2A=\angle CXB\longrightarrow \angle CXZ=\angle YXB$ DONE Very interesting that $RMH'T$ is cyclic is just All Russian MO 2015, grade 10, problem 7 https://artofproblemsolving.com/community/c6h1126020p5204821