TwoTimes3TimesSeven wrote: Let $\triangle ABC$ be an acute-angled triangle with $AB<AC$. Let $M$ and $N$ be the midpoints of $AB$ and $AC$, respectively; let $AD$ be an altitude in this triangle. A point $K$ is chosen on the segment $MN$ so that $BK=CK$. The ray $KD$ meets the circumcircle $\Omega$ of $ABC$ at $Q$. Prove that $C, N, K, Q$ are concyclic. Redefine $Q \ne K$ as the common point of circles $\odot(CNK), \odot(BMK)$. Let $X$ and $Y$ be the midpoints of $BD$ and $CD$. The projections of $XK$ and $CN$ along $BC$ have the same length (namely $\tfrac{1}{2}CD$) but opposite direction so $CNKX$ is an isosceles trapezoid. Thus, $X$ lies on $\odot(CNK)$ and also $Y$ lies on $\odot(BMK)$. Now $DX \cdot DC=DB \cdot DY$ proves that $D$ lies on the radical axis $\overline{KQ}$ of these two circles. Finally, $\angle BQC=\angle BQK+\angle CQK=\angle AMN+\angle ANM=180^{\circ}-\angle BAC$ hence $Q$ lies on $\odot(ABC)$ as desired.

TwoTimes3TimesSeven wrote: Let $\triangle ABC$ be an acute-angled triangle with $AB<AC$. Let $M$ and $N$ be the midpoints of $AB$ and $AC$, respectively; let $AD$ be an altitude in this triangle. A point $K$ is chosen on the segment $MN$ so that $BK=CK$. The ray $KD$ meets the circumcircle $\Omega$ of $ABC$ at $Q$. Prove that $C, N, K, Q$ are concyclic. From $K \in MN$ and $BK=CK$ $KO \perp BC$.From $BC||MN$, $\angle ACB=\angle KNA=\gamma$.Hence, we need to prove $\angle KQC=\gamma$. We take any $G$ such that $AG||BC$.We need to show $G,K,D-$collinear.Now, $MN\perp AD$ and $MN \cap AD=T$,where $T \equiv \frac{AD}{2}$.We know $AGCB$ isosceles trapezoid.So, $AG=BC-2BD$,other hand $TK=\frac{BC}{2}-BD$ $\implies AG=2TK$. Hence, we have $AG||TK$ and $\frac{AD}{TD}=\frac{AG}{TK} \implies G,K,D-$collinear. Hence, $\angle ACB=\angle GBC=\angle GQC=\gamma$. Done

Let $AK$ meets $BC$ and $(ABC)$ at $E$ and $P$ respectively. Since $A$ and $D$ are symmetric with respect to $MN$, so $\angle KED = \angle AKM = \angle DKM = \angle KDE$, which means $KD=KE$. Since $K$ lies on perpendicular bisector of $BC$, this implies $D,E$ are symmetric with respect to perpendicular bisector of $BC$, and therefore the same also applies for $Q$ and $P$. Hence, $Q$ and $P$ is isogonal with respect to $\angle BAC$. Therefore $\angle QCN =\angle QCA= \angle BEA = \angle QKM \implies C,N,Q,K$ are concyclic.

TwoTimes3TimesSeven wrote: Let $\triangle ABC$ be an acute-angled triangle with $AB<AC$. Let $M$ and $N$ be the midpoints of $AB$ and $AC$, respectively; let $AD$ be an altitude in this triangle. A point $K$ is chosen on the segment $MN$ so that $BK=CK$. The ray $KD$ meets the circumcircle $\Omega$ of $ABC$ at $Q$. Prove that $C, N, K, Q$ are concyclic. Let $G$ be the centroid of $\triangle{ABC}$ and $\triangle{MNP}$, where $P$ is the midpoint of $BC$. Also, let $A’ \in \odot(ABC)$ such that $AA’ \parallel BC$. From a very known Lemma( Problem 3.24, EGMO) in $\triangle{MNP}$ and $\triangle{ABC}$ we obtain that points $D, G, K, A’$ are on the same line. Now, $\angle{A’KN}=\angle{A’DC}=\angle{AA’D}=\angle{NCQ}$.

Let $L$ be the second intersection of $DG$ with $\Omega.$

Then from $Claim1$ and $Claim 2$ we can find easily $\angle KQC=\angle CAL=\angle ACB=\angle ANK.$ As desired.

One homothety is all it takes! TwoTimes3TimesSeven wrote: Let $\triangle ABC$ be an acute-angled triangle with $AB<AC$. Let $M$ and $N$ be the midpoints of $AB$ and $AC$, respectively; let $AD$ be an altitude in this triangle. A point $K$ is chosen on the segment $MN$ so that $BK=CK$. The ray $KD$ meets the circumcircle $\Omega$ of $ABC$ at $Q$. Prove that $C, N, K, Q$ are concyclic. Let $L$ be the midpoint of $BC$ and $G$ be the centroid of $\triangle ABC$. Let $Z$ be a point on $\odot(ABC)$ such that $AZ \parallel BC$. Consider the homothety $\mathcal{H}(G,-2)$. Clearly, the medial triangle is mapped to $\triangle ABC$ and the nine-point circle to $\odot(ABC)$ and thus $\mathcal{H}: K \mapsto D$ and $\mathcal{H}: D \mapsto Z$, and so $Q, D, G, K, Z$ are all collinear. So, $\angle CQK=\angle CAZ=\angle C$ and $\angle KNC=\pi-\angle C$ and so $KNCQ$ is cyclic. $\square$ [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(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -10.663655155185191, xmax = 13.93247296550686, ymin = -9.256284736570779, ymax = 8.32837881055424; /* image dimensions */pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen ttqqqq = rgb(0.2,0,0); draw((-4.72,4.91)--(-8.12,-3.61)--(6.32,-4.37)--cycle, linewidth(0.5) + rvwvcq); draw(arc((-4.72,4.91),0.8960338113184718,319.9502722342918,356.98721249581666)--(-4.72,4.91)--cycle, linewidth(0.5) + ttqqqq); draw(arc((-7.230895102893001,-5.640254925172773),1.0080380377332807,365.3552320046501,402.392172266175)--(-7.230895102893001,-5.640254925172773)--cycle, linewidth(0.5)); draw(arc((6.32,-4.37),0.8960338113184718,139.95027223429176,176.98721249581666)--(6.32,-4.37)--cycle, linewidth(0.5)); draw(arc((0.8,0.27),0.8960338113184718,139.95027223429176,176.98721249581666)--(0.8,0.27)--cycle, linewidth(0.5)); draw((-5.151598959880638,-3.2903802377321254)--(-5.626135849220336,-3.2654046119774045)--(-5.651111474975057,-3.739941501317102)--(-5.176574585635359,-3.764917127071823)--cycle, linewidth(0.5)); draw((-0.6966883329370405,-0.12707832580378603)--(-0.22215144359734285,-0.15205395155850715)--(-0.1971758178426219,0.3224829377811905)--(-0.6717127071823195,0.34745856353591154)--cycle, linewidth(0.5)); /* draw figures */draw((-4.72,4.91)--(-8.12,-3.61), linewidth(0.5) + rvwvcq); draw((-8.12,-3.61)--(6.32,-4.37), linewidth(0.5) + rvwvcq); draw((6.32,-4.37)--(-4.72,4.91), linewidth(0.5) + rvwvcq); draw(circle((-0.7743659244917712,-1.602952565343659), 7.6148919477710395), linewidth(0.5) + wrwrwr); draw((-6.42,0.65)--(0.8,0.27), linewidth(0.5) + wrwrwr); draw((0.8,0.27)--(-0.9,-3.99), linewidth(0.5) + wrwrwr); draw((-0.9,-3.99)--(-6.42,0.65), linewidth(0.5) + wrwrwr); draw((-4.72,4.91)--(-5.176574585635359,-3.764917127071823), linewidth(0.5) + wrwrwr); draw((-7.230895102893001,-5.640254925172773)--(3.8331491712707173,4.459834254143646), linewidth(0.5) + wrwrwr); draw((-7.230895102893001,-5.640254925172773)--(6.32,-4.37), linewidth(0.5) + wrwrwr); draw((3.8331491712707173,4.459834254143646)--(-4.72,4.91), linewidth(0.5) + wrwrwr); draw((-0.6717127071823195,0.34745856353591154)--(-0.9,-3.99), linewidth(0.5) + wrwrwr); /* dots and labels */dot((-4.72,4.91),dotstyle); label("$A$", (-5.018642143878818,5.371467233203281), NE * labelscalefactor); dot((-8.12,-3.61),dotstyle); label("$B$", (-8.490773162737897,-3.583297083753258), SW * labelscalefactor); dot((6.32,-4.37),dotstyle); label("$C$", (6.2265821881680035,-5.022524978090997), NE * labelscalefactor); dot((-6.42,0.65),linewidth(4pt) + dotstyle); label("$M$", (-7.079519909911303,0.756893104913149), NE * labelscalefactor); dot((0.8,0.27),linewidth(4pt) + dotstyle); label("$N$", (0.8951810108230958,0.4432812709516837), NE * labelscalefactor); dot((-0.9,-3.99),linewidth(4pt) + dotstyle); label("$L$", (-0.9192874570968097,-4.619309762997684), NE * labelscalefactor); dot((-2.1733333333333325,-1.0233333333333332),linewidth(4pt) + dotstyle); label("$G$", (-2.196135638225632,-1.662398185646726), NE * labelscalefactor); dot((-5.176574585635359,-3.764917127071823),linewidth(4pt) + dotstyle); label("$D$", (-5.242650596708436,-4.44010300073399), NE * labelscalefactor); dot((3.8331491712707173,4.459834254143646),linewidth(4pt) + dotstyle); label("$Z$", (4.053700195720709,4.833846946412198), NE * labelscalefactor); dot((-7.230895102893001,-5.640254925172773),linewidth(4pt) + dotstyle); label("$Q$", (-7.66194188726831,-5.532170623370934), SW * labelscalefactor); dot((-0.6717127071823195,0.34745856353591154),linewidth(4pt) + dotstyle); label("$K$", (-1.0536925287945804,0.6000871879324163), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Pretty easy Let $A'$ be the other intersection of $\overleftrightarrow{DK}$ and $\odot (ABC),$ let $D'$ be the reflection of $D$ over the perpendicular bisector of $\overline{BC},$ and let $O$ be the circumcenter of $\triangle ABC$. Lemma: $A,K,D'$ are collinear. Proof: Note that $K$ is the projection of $O$ onto $\overline{MN}.$ Thus $\triangle OKN \sim \triangle OMA$ and $\triangle OKM \sim \triangle ONA$ which yield $\dfrac{KN}{ON}=\frac{MA}{OA},\frac{KM}{OM}=\frac{NA}{OA}.$ Dividing, $$\frac{KN}{KM}=\frac{ON}{OM}\cdot\frac{MA}{NA}=\frac{R\cos \angle B}{R\cos\angle C}\cdot\frac{c}{b}=\dfrac{BD}{CD}=\frac{CD'}{BD'},$$and so $K,D'$ are corresponding points on the homothety that sends $\overline{MN}\to \overline{BC}.$ It follows that $ADD'A'$ is a rectangle, and so $\overline{AA'}\parallel \overline{BC}.$ Thus $$\angle KQC=\angle A'QC=\angle C=\angle ANM=\angle ANK,$$which implies the conclusion.

See that $D$ is the anticomplement of $K$ wrt triangle $ABC$. It is well known and easy to prove that $DG$, the line parallel to $BC$ through $A$ meet on the circumcircle (for a proof of this fact, consider $D'$ to be the isotomic point in $BC$ and complete the rectangle $ADD'E$. Use ratios to show that $E$ lies on $DG$). Now we see that if $Q'$ is the second intersection of $DG$ with $\odot (ABC)$, $\angle CQK = \angle Q'AC = \angle ANK$. So, $CNKQ$ is cyclic. $\blacksquare$.

Construct $X$ on $\omega_{ABC}$ such that $AX || BC$. Claim: $D,K,X$ are collinear (also $DK = KX$, but that's not needed). Proof: Toss $\omega_{ABC}$ into the unit circle in complex plane, with WLOG assuming $A = a$ and $B = b$ and $C = \bar{b}$. Now $M = \frac{a+b}{2}$ and $N = \frac{a+\bar{b}}{2}$. Let $Z = \frac{a+\bar{a}+b+\bar{b}}{4}$. First as $Im(Z) = 0$, we see $BZ = CZ$. Also note that $Z \in MN \Leftrightarrow Re(Z) = Re(M)$ (since $MN$ is parallel to the imaginary axis), but $Re(M) = \frac{a+b+\bar{a}+\bar{b}}{4} = Re(Z)$, so $Z \in MN$. Hence $Z = K$. Now note that by perpendicular formula $D = \frac{a+b+\bar{b}-\bar{a}}{2}$ and $X = \bar{a}$. As $\frac{D+X}{2} = \frac{a+b+\bar{b}+\bar{a}}{4} = K$ so $K \in DX$ and we're done. Now returning to the main problem, $$ \angle 180 - KNC = \angle ANK = \angle CAX = \angle XQC = \angle KQC $$, so $CNKQ$ is cyclic as desired.

there are better solutions above but in this one we don't use any angles chasing etc. take $\odot ABC$ the unit circle and then let $Q'$ be the send intersection of $(KG \cap \odot CNK) \implies \triangle KGB \sim \triangle NGQ'$ indirectly which implies $\frac{g-n}{g-q}=\frac{\overline{g}-\overline{k}}{\overline{g}-\overline{b}} \implies q=\frac{g\overline{b}+n\overline{g}-n\overline{b}-g\overline{k}}{\overline{g}-\overline{k}}$ and now we have to show $q\overline{q}=1$ which just algebra and by the way $k=\frac{(a+b)(a+c)}{4a}$

My solution: Let $P$ be the midpoint of $BC$. Then $KP \perp BC \Rightarrow KP \perp MN$. From here, we get that $AKOQ$ is cyclic, where $O$ is the circumcenter of $\triangle ABC$. As $O$ lies on $PK$, we have $\angle QKN = 90^{\circ}+\angle QKO=90^{\circ}+\angle QAO=90^{\circ}+(90^{\circ}-\angle QCA)=180^{\circ}-\angle QCN$, giving that $CNKQ$ is cyclic.

let A' be the reflection of D across K. then note that AA" is parallel to BC and A' lies on the circumcircle of ABC. we are thus done by reim.

Let $P$ be the midpoint of $BC$ and $R$ be the midpoint of $AD$. We have $MN \parallel BC$, $R$ is on $MN$, and $PK$ is the perpendicular bisector of $BC$, so $PKRD$ is a rectangle. Let $QD$ meet $\Omega$ at $L$.Then $\angle LKN = \angle DKR = \angle AKR$. Since $PK$ is the perpendicular bisector of $BC$ and $MN \parallel BC$, $L$ is the reflection of $A$ over the line $PK$, so arc $\widehat{AB} =$ arc $\widehat{LC}$. Then $\angle KQC = \angle LQC = \frac{1}{2} \widehat{LC} = \frac{1}{2} \widehat{AB} = \angle ACB = \angle ANK = 180 - \angle CNK$. Thus, $CNKQ$ is cyclic

Ack this took 40 minutes [asy][asy] size(250); defaultpen(linewidth(0.7)+fontsize(11)); pair A = dir(125), B = dir(210), C = dir(330), G = centroid(A,B,C), D = foot(A,B,C), O = circumcenter(A,B,C), M = (A+B)/2, N = (A+C)/2, K = foot(O,M,N), P=(A+D)/2; pair X = extension(A,K,B,C), Ap = A+X-D, Y = 2*D-G, Q = intersectionpoint(G--Y,unitcircle); draw(A--B--C--A--D,rgb(0.1,0.3,0.7)); draw(unitcircle,rgb(0.1,0.7,0.3)); draw(M--N,rgb(0.7,0.3,0.7)); draw(B--G--N,purple); draw(Q--Ap,red); draw(K--O,orange); draw(circumcircle(A,M,N),gray); dot("$A$",A,dir(O--A),linewidth(3.5)); dot("$B$",B,dir(O--B),linewidth(3.5)); dot("$C$",C,dir(O--C),linewidth(3.5)); dot("$D$",D,S,linewidth(3.5)); dot("$M$",M,W,linewidth(3.5)); dot("$N$",N,E,linewidth(3.5)); dot("$O$",O,S,linewidth(3.5)); dot("$K$",K,NW,linewidth(3.5)); dot("$Q$",Q,dir(O--Q),linewidth(3.5)); dot("$A'$",Ap,dir(O--Ap),linewidth(3.5)); dot("$G$",G,S,linewidth(3.5)); dot("$P$",P,NE,linewidth(3.5)); dot("$P'$",intersectionpoint(P--D,circumcircle(A,M,N)),SW,linewidth(3.5)); [/asy][/asy] Let $A'$ be the other intersection point of $KD$ with $\Omega$, let $G$ be the intersection point of $BG$ and $KD$, and let $P$ and $P'$ be the intersection points of $AD$ with $MN$ and the circumcircle of $\triangle AMN$, respectively. Observe that $AP'$ and $AO$ are isogonal lines with respect to $\angle MAN$; since $P$ and $K$ are the projections of $P'$ and $O$, respectively, onto $MN$, $MP = KN$. Then \[ \frac{BG}{GN} = \frac{BD}{KN} = \frac{2MP}{KN} = 2, \]so $G$ is the point on the median $\overline{BN}$ with $BG = 2GN$. This means that $G$ is the centroid of $\triangle ABC$. It follows that $A'$, being the intersection point of ray $\overrightarrow{DG}$ with $\Omega$, is the unique point such that $ABCA'$ is an isosceles trapezoid with $AA'\parallel BC$. (To prove this, let $A_0$ be a point such that $ABCA_0$ is such an isosceles trapezoid, and show that $A_0$, $G$, and $D$ are collinear.) To finish, observe that \[ \angle KQC\equiv\angle A'QC = \angle A'BC = \angle ACB = \angle ANK, \]whence $CNKQ$ is cyclic. $\blacksquare$

We use complex numbers. We know that $K$ is on the perpendicular bisector of $BC$, so $k=bc\bar{k}$. We know that $K\in{MN}$ so computing the determinant of $[KMN]=0$ and combining with the previous equation yields $k=\frac{(a+b)(a+c)}{4a}$. Let $E$ be the point such that $AE//BC$ and $E\in{(ABC)}$. We have that $\frac{e-a}{c-b}=\overline{\frac{e-a}{c-b}}$. Solving this we have that $e=\frac{bc}{a}$ and $\frac{a}{bc}=\bar{e}$. Now we solve the determinant of $[EKD]$ and discover it equals $0$. With angle chasing, $\angle{ANM}=\angle{NAE}=\angle{CAE}=\angle{CQE}=\angle{CQK}\Rightarrow K,N,C,Q$ are concyclic.

I spent more than $2$ hours in total for this problem

Diagram for reference

Denote reflection of $D$ wrt $K$ by $P$. Note that homothety with coefficient $-2$ wrt centroid of $\triangle ABC$ maps $K$ onto $D$, and so $D$ onto $P$. Hence $AP\parallel BC,P\in \odot (ABC)$, implying $\measuredangle CQK=\measuredangle ACB=\measuredangle CNK$ and we are done.

forgor :skull:, please someone check this and tell if its a fakesolve or not :skull:, ill be tensed until someone checks this plix :skull:

By the way, gravity of ABC lies on the line (QDKA'). As said in All-Russian 2015 Grade 9 P7

Linearity of PoP makes this problem (which is also Moldova TST 2023 #1) quite straightforward: Let $S$ be the midpoint of $AD$ and $\omega_1,\omega_2$ be the circles $(MKB),(NKC)$ respectively. Define $f(X)={\rm pow} (X,\omega_1)-{\rm pow} (X,\omega_2)$. It suffices to prove that $f(D)=0$. Claim: $MK=\dfrac{DC}{2}$ and $NK=\dfrac{BD}{2}$. Proof: Let $D'$ be the reflection of $D$ across $M$. Then, $ADBD'$ is a rectangle and so triangle $D'BC$ is right, hence if $K'$ is the midpoint of $D'C$ then $MK' \parallel BC$ and $K'B=K'C$, implying that $K \equiv K'$. Therefore, $MK=\dfrac{DC}{2}$ and similarly $NK=\dfrac{BD}{2}$, as desired $\blacksquare$ Now, note that $f(A)={\rm pow} (A,\omega_1)-{\rm pow} (A,\omega_2)=AM \cdot AB-AN\cdot AC=\dfrac{c^2-b^2}{2}$ and $f(S)={\rm pow} (S,\omega_1)-{\rm pow} (S,\omega_2)=-SM \cdot SK-SK \cdot SN=-SK \cdot MN=-(MK-MS) \cdot MN=-\dfrac{(DC-BD)a}{4}=\dfrac{c^2-b^2}{4},$ and since $S$ is the midpoint of $AD$, linearity of PoP implies that $f(D)=2f(S)-f(A)=0,$ and we are done.

Orestis_Lignos wrote: Linearity of PoP makes this problem (which is also Moldova TST 2023 #1) quite straightforward: Let $S$ be the midpoint of $AD$ and $\omega_1,\omega_2$ be the circles $(MKB),(NKC)$ respectively. Define $f(X)={\rm pow} (X,\omega_1)-{\rm pow} (X,\omega_2)$. It suffices to prove that $f(D)=0$. Claim: $MK=\dfrac{DC}{2}$ and $NK=\dfrac{BD}{2}$. Proof: Let $D'$ be the reflection of $D$ across $M$. Then, $ADBD'$ is a rectangle and so triangle $D'BC$ is right, hence if $K'$ is the midpoint of $D'C$ then $MK' \parallel BC$ and $K'B=K'C$, implying that $K \equiv K'$. Therefore, $MK=\dfrac{DC}{2}$ and similarly $NK=\dfrac{BD}{2}$, as desired $\blacksquare$ Now, note that $f(A)={\rm pow} (A,\omega_1)-{\rm pow} (A,\omega_2)=AM \cdot AB-AN\cdot AC=\dfrac{c^2-b^2}{2}$ and $f(S)={\rm pow} (S,\omega_1)-{\rm pow} (S,\omega_2)=-SM \cdot SK-SK \cdot SN=-SK \cdot MN=-(MK-MS) \cdot MN=-\dfrac{(DC-BD)a}{4}=\dfrac{c^2-b^2}{4},$ and since $S$ is the midpoint of $AD$, linearity of PoP implies that $f(D)=2f(S)-f(A)=0,$ and we are done. Is it okay to take tst questions from former olympiads especially this well-known?

@above I'm not from Moldova though

is this right? feels too easy Let $E$ be the point on $(ABC)$ such that $ABCE$ is an isosceles trapezoid with $\overline{BC} \parallel \overline{AE}$. Let $Q'=\overline{ED} \cap (ABC) \neq E$, and let $K'=\overline{ED} \cap \overline{MN}$. By Reim's, $CNK'Q'$ is cyclic. Further, it is clear that $\overline{ED}$ bisects the midline of trapezoid $ABCE$ (draw a rectangle with three vertices being $A,E,D$), hence $BK'=CK'$, so $K'=K$ and hence $Q'=Q$, so $CNKQ$ is cyclic. $\blacksquare$

Let $L$ be the midpoint of $\overline{BC}$, $G$ be the centroid of $\triangle ABC$, and $P \neq Q$ be the second intersection of line $DG$ with $(ABC)$. Since $K$ is the foot of the $L$-altitude in $\triangle LMN$, by a negative homothety at $G$ we find that $D$, $G$, and $K$ are collinear. However, $G$ is also the center of the negative homothety which takes the nine-point circle of $\triangle ABC$ to $(ABC)$, so $\overline{AP} \parallel \overline{BC} \parallel \overline{MN}$. Thus, $$\measuredangle KNC = \measuredangle BCA = \measuredangle PAC = \measuredangle PQC = \measuredangle KQC,$$so we are done.

Easy problem... Clearly $AD=2KP$, this immediately yields $AP \cap DK = G$ the centroid. This implies $Q$ is the ISL 2011 G4 point, and hence if $DK \cap (ABC) = A'$ be the other intersection then $AA' || BC || MN$ By Reim's we're done