[asy][asy] size(8cm); pair A=dir(150); pair B=dir(30); pair C=dir(90); pair P=dir(-50); pair M=midpoint(A--B); pair Q=1/3*P+2/3*C; pair N=extension(M, Q, P, P+dir(90)*(A-B)); draw(unitcircle); draw(circumcircle(A, B, N), dotted); draw(A--B--C--cycle); draw(M--N); draw(P--N); draw(C--P); draw(C--M); string[] names = {"$A$", "$B$", "$C$", "$P$", "$M$", "$Q$", "$N$"}; pair[] pts = {A, B, C, P, M, Q, N}; pair[] labels = {A, B, C, P, dir(270), dir(30), N}; for(int i=0; i<names.length; ++i){ dot(names[i], pts[i], dir(labels[i])); } [/asy][/asy] By similar triangles, $PN=2CM$. Hence $N$ lies on $(ABC)$ translated by the vector $2\vec{MC}$. (Note that we may eliminate essentially all of the assumptions of the problem. All that is important is that $CM\parallel PN$ and $\frac{CQ}{QP}$ is fixed.)

Let $\overline{CD}$ be a diameter of $\odot(ABC)$. Let $R=\overline{CP} \cap \overline{AB}$. Note $\overline{PN} \perp \overline{RM}$ and $\overline{MQ} \perp \overline{RP}$ since $\triangle CQM \sim \triangle CPD$. Thus, $N$ is the orthocenter of $\triangle RPM$. Now $NPDM$ is a parallelogram hence $P$ moves on a circle implies $N$ also moves on a circle.

Since $\vec N = 3\vec Q - 2\vec M$, it suffices to show $Q$ moves on a fixed circle, which follows by homothety...?

We know $\triangle CQM\sim \triangle PQN,$ then $PN=CA=CB.$ Apply Sine Law for triangles $CBP,CPN,$ we can get easily $PN=NC.$ Then $P$ vary on $(ABC)$ then $N$ lies on the circle with center $C,$ and radii $CA.$

v_Enhance wrote: Since $\vec N = 2\vec Q - \vec M$, it suffices to show $Q$ moves on a fixed circle, which follows by homothety...? Maybe you mean $\vec N = 3\vec Q - 2\vec M$?

Yes, fixed. Of course, the constants don't matter very much

anantmudgal09 wrote: Let $\overline{CD}$ be a diameter of $\odot(ABC)$. Let $R=\overline{CP} \cap \overline{AB}$. Note $\overline{PN} \perp \overline{RM}$ and $\overline{MQ} \perp \overline{RP}$ since $\triangle CQM \sim \triangle CPD$. Thus, $N$ is the orthocenter of $\triangle RPM$. Now $NPDM$ is a parallelogram hence $P$ moves on a circle implies $N$ also moves on a circle. Maybe NPDO is parallelogram,not NPDM?

Let $N'$ be the symmetric of $O$ with respect to the midpoint $R$ of $CP$. Then $COPN'$ is rombus, so it $N'P$ is perpendicular to $AB$. So it suffices to prove that the points $N'$, $Q$, $M$ are collinear which follows immediately by Menelaus theorem on $CRO$ with traversal the line $N'QM$. (Here we use that $CM=MO$, $CQ=2QR$ and $N'O=N'R$. [asy][asy] size(11.224253480894681cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; real xmin = 2.458849573335637, xmax = 13.683103054230319, ymin = -1.1685089558279511, ymax = 9.487980730313296; draw(circle((8.245672866014798,2.805837607446039), 4.), linewidth(0.8)); draw((8.182851857090862,6.805344267127676)--(11.67793673144217,4.85999552690636), linewidth(0.8)); draw((8.182851857090862,6.805344267127676)--(4.750587991663486,4.75118634766736), linewidth(0.8)); draw((4.750587991663486,4.75118634766736)--(11.67793673144217,4.85999552690636), linewidth(0.8)); draw((8.245672866014798,2.805837607446039)--(8.182851857090862,6.805344267127676), linewidth(0.8)); draw((4.362806728588357,1.8449262906081807)--(8.182851857090862,6.805344267127676), linewidth(0.8)); draw(circle((8.182851857090858,6.805344267127673), 4.), linewidth(0.8) + linetype("2 2")); draw((8.182851857090862,6.805344267127676)--(4.299985719664417,5.844432950289805), linewidth(0.8)); draw((8.245672866014798,2.805837607446039)--(4.299985719664417,5.844432950289805), linewidth(0.8)); draw((4.362806728588357,1.8449262906081807)--(4.299985719664417,5.844432950289805), linewidth(0.8)); draw((4.299985719664417,5.844432950289805)--(8.21426236155283,4.80559093728686), linewidth(0.8) + linetype("2 2")); dot((8.245672866014798,2.805837607446039),linewidth(3.pt) + dotstyle); label("$O$", (8.354858211159693,2.4346074339534134), NE * labelscalefactor); dot((11.67793673144217,4.85999552690636),linewidth(3.pt) + dotstyle); label("$B$", (11.761440979680257,4.989544510343835), NE * labelscalefactor); dot((8.182851857090862,6.805344267127676),linewidth(3.pt) + dotstyle); label("$C$", (8.26750993504378,6.933043653922875), NE * labelscalefactor); dot((4.750587991663486,4.75118634766736),linewidth(3.pt) + dotstyle); label("$A$", (4.467859924001611,4.771173820054056), NE * labelscalefactor); dot((8.21426236155283,4.80559093728686),linewidth(3.pt) + dotstyle); label("$M$", (7.896279761551154,4.465454853648364), NE * labelscalefactor); dot((4.362806728588357,1.8449262906081807),linewidth(3.pt) + dotstyle); label("$P$", (4.074792681480007,1.5611246727942947), NE * labelscalefactor); dot((6.909503480923359,5.151871608287843),linewidth(3.pt) + dotstyle); label("$Q$", (7.1756564835948815,5.207915200633615), NE * labelscalefactor); dot((4.299985719664417,5.844432950289805),linewidth(3.pt) + dotstyle); label("$N'$", (3.7,5.600982443155218), NE * labelscalefactor); dot((6.272829292839609,4.325135278867929),linewidth(3.pt) + dotstyle); label("$R$", (6.258499584377806,3.897691058894937), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]

BarishNamazov wrote: Let $ABC$ be a triangle with $CA=CB$ and $\angle{ACB}=120$, and let $M$ be the midpoint of $AB$. Let $P$ be a variable point of the circumcircle of $ABC$, and let $Q$ be the point on the segment $CP$ such that $QP = 2QC$. It is given that the line through $P$ and perpendicular to $AB$ intersects the line $MQ$ at a unique point $N$. Prove that there exists a fixed circle such that $N$ lies on this circle for all possible positions of $P$. Using a homothety with center $C$ and coefficient $1/3$, we can see that the image of $P$ is $Q$. So $Q$ is on a fixed circle. Now take a homothety with center $M$ and coefficient $3$. As the image of $Q$ is $N$, we are done.

Take $O$ be center of $(ABC)$, then $AM=MO=\frac{1}{2}PN$, so $CNPO$ is parallelogram, so $CN=OP=OA$, which is fixed.

First note that $\angle ACB = 120$ implies that if $P$ is on the bigger arc $AB$ then $\angle APB=60$, from here, we may use the lemma that was found in another EGMO problem, that the reflection of the circumcenter lies on the circle , let $O$ be the circumcenter of $(ABC)$ then it is easy to see that $M$ is the midpoint of $CO$, then, it's easy to see that $\triangle PNQ$ is similar to $\triangle CMQ$, then for the condition, we have that $PN=2CM=CO$. It's well known that if we have a triangle $APB$, such that $\angle APB=60$ and ortocenter $N$, then $PN=OP$ (Where $O$ is circumcenter), as $PN$ is perpendicular to $AB$, it's easy to see that $N$ is ortocenter of $APB$, Now, let's consider the reflection of $(ABC)$ wrt $AB$, then it's clear that all possible $N$ are on the smaller arc $AB$ of the new circle. Then, consider that the new circle and see that we may re-do the previous steps to prove that if we choose a point $N$ (as it is $P$) on the bigger arc $AB$, all possible points $P$ (as it is $N$) lie on the smaller arc $AB$ of the original circle, then it's clear that if we originally choose $P$ on such arc, it gives us that $N$ is on the circle with center $C$ and radius $CA$

Err... All we need to observe is that \(N\) is the orthocenter of \(\Delta APB.\)

Take $C’$ on the circumcircle of $ABC$ such that $CAPC’$ is isosceles trapezoid. Then you easily prove that points $N-A-C’$ are collinear using the fact that $PC’=CA=NP$. Then you find that $\angle PNC’=\angle PC’N=\angle PCA$. Then a few angle chasings lead to the fact that $CN=CA$. Point $N$ is on a circle centered at $C$ with radius $CA$. P.S: I’l try to elaborate when I switch to PC, if needed.

I tossed it on a coordinate plane and it was way nicer than I thought.

By homothecy centered at $A$ with ratio $\frac 13$, $Q$ lies on a fixed circle; by homothecy centered at $M$ with ratio $3$, $N$ also lies on a fixed circle, done!

Toss the figure on the coordinate plane, and easily get $PN=R$, a fixed circumradius of $ABC$.

Let $O$ be the circumcenter of triangle $CAB$. Note that $NP \parallel CM \implies$ $\frac{CQ}{QP}=\frac{CM}{NP}=\frac{1}{2}$, hence $CO=NP$ and $CO \parallel NP \implies NCOP$ is a parallelogram, then $CN=OP=CO$. So $N$ lies on the circle with center $C$ and radius $CA$.

Let $C'$ be reflection of $C$ over $M$. $\triangle CMQ \sim \triangle PNQ$, so it follows that $CC'=PN$. But $CC' \parallel PN$, therefore $CC'PN$ is a parallelogram. Now, $C'$ lies on perpendicular bisector of $AB$ and $\angle AC'B=120^{\circ}$, so $C'$ is the center of $(ABC)$. Since $C'$ is the center, $CC'=CP$, thus $CC'PQ$ is a rhombus, and therefore $CN=CC'$. Now, $CAC'$ is equilateral, so $CA=CC'$, and thus $CA=CB=CN$. By this; $N$ lies on a circle with center $C$ and radius $CA$. $\square$

Trivial for P1. With a lot of extraneous conditions. Anyway, note by similar triangle that $PN = 2CM$ so $N$ lies on the translation image of $\odot(ABC)$ with vectors $2\overrightarrow{MC}$ so we are done.

Really simple problem. We see that $PN$ is fixed as $P$ varies to $(ABC)$, and from here it is obvious that $N$ also varies to the circle.

Easy with complex. Let $\omega^3=1$ be a root of unity and $C=-1,A=\omega,B=\omega^2,P=p$ and so $\odot ABC$ is the unit circle then: $$M=\frac{\omega+\omega^2}{2}=-\frac{1}{2}$$$$Q=\frac{2}{3} \left (-1 \right) +\frac{1}{3} \left (p \right)=\frac{p-2}{3}$$From $PN \bot AB$ we see as $AB$ is perpenciular to the real axis: $$n-p \in \mathbb{R} \Leftrightarrow n-p=\overline{n}-\frac{1}{p}$$From $MNQ$ colinear we see: $$\frac{z-m}{q-m} \in \mathbb{R} \Leftrightarrow n \left (\frac{2}{p}-1 \right)+\frac{1}{p}=\overline{n} \left (2p-1 \right)+p$$Combining these we get: $$n=\frac{(p-1)(p^2-1)}{p^2-1}=p-1$$Where we use $p^2-1=0 \Leftrightarrow p=\pm 1$ which would mean $N$ isn't unique as the lines concur so $p^2-1 \neq 0$ As $P$ is varying on $\odot ABC$ it follows $N$ also varies on a circle (with centre $C$ passing through the centre of $\odot ABC$)

Let $O$ be the circumcenter of $\triangle ABC$. As, $\angle ACB = 120^{\circ}$, we can deduce that $C$ is the reflection of $O$ wrt $AB$, so $MC=MO$. Again, $\triangle CQM \sim \triangle PQN \Rightarrow 2CM = PN \Rightarrow 2OM=PN$ and $N$ is the point on the perpendicular from $P$ to $AB$. So, we can say that $N$ is the orthocenter of $\triangle APB$. Now it's trivial that if $P$ moves along the circumciecle of $\triangle APB$, the orthocenter will move along a circle. $Q.E.D$

This is so, so easy to do a Cartesian bash on. Let $(ABC)$ be the unit circle, with $AB$ as the $x$-axis. It follows that $C = (0, 1), A = (-\frac{\sqrt{3}}{2}, \frac{1}{2}), B = (\frac{\sqrt{3}}{2}, \frac{1}{2}), M = (0, \frac{1}{2})$. We let $P = (\sin\theta, \cos\theta)$. Then we get $Q = (\frac{1}{3}\sin\theta, \frac{1}{3}\cos\theta + \frac{2}{3})$. Then the point $N$ is where $x = \sin\theta$ and $N$ lies on $MQ$. We can deduce that the line $MQ$ is where $y=\frac{\cos\theta+\frac{2}{3}}{\sin\theta}x+\frac{1}{2}$. This means that $N = (\sin\theta, \cos\theta+\frac{7}{6})$ and so it obviously lies on a circle.

Let $O$ be the circumcenter of $\triangle CAB.$ Now $CQ:QP=1:2, PN \parallel CM \implies PN=2CM=CO.$ Hence $COPN$ is a parallelogram and so $CN=OP$ $=CA=CB$ and so $N$ lies on the circle $\mathcal{C}(C, CA).$ $\square$

So basically the easiest olympiad geometry problem,ever.....

I will prove that fixed circle is circle with center $C$ and radius $CA$ (say this circle $\omega$). Take any point on $\odot (ABC)$. Let $L$ be foot of perpendicular from $P$ to $AB$, and $N=PL\cap \omega$. Now we need to prove that $PQ=2\cdot QC$ or $PN=2\cdot CM$ (Since $PN||CM$). Or we need to prove $P$ is orthocenter of $\triangle NAB$. We will prove this with $\text{Phantom Point}$. Let $P'$ be orthocenter of $\triangle NAB$. Since $\angle ACB=120$ we have $\angle ANB=60$ $\implies$ $\angle BP'A=120=\angle BCA$ $\implies$ $ABCP'$ is cyclic and $P'\in NL$ $\implies$ $P'\equiv P$. So we are done!

reflect P across M to a point P'. Now, since NQ: QC=2:1 and NQ: QM=2:1 hence, Q is the Centroid of Triangle (P'NP) This implies P' ,C ,N are collinear and NC= P'C. Now , it is obvious that P' must lie on the circle which has C as Centre and A, B lies on it.(By symmetry) hence, as C is centre of the circle we also get P'N is the diameter of this circle.

Quite easy.

Wait, when did I ever use the fact that ACB=120? Denote by $\omega$ the circumcircle of $\triangle ABC$, and let $\frac{1}{3}\omega$ be the image of $\omega$ under a homothety with ratio $1/3$ centered at $C$. Clearly, the locus of $Q$ is $\frac{1}{3}\omega$ (more specifically, $Q$ is the second intersection of $CP$ with $\frac{1}{3}\omega$.) Note that $CM\parallel PN$ since they are both perpendicular to $AB$. Thus, $\triangle QNP\sim\triangle QMC$. Since $$\frac{CQ}{PQ}=\frac{1}{2},$$we have $$\frac{QM}{QN}=\frac{1}{2},$$so $NM=3QM$. Thus, the locus of $N$ is the circle that is the image of $\frac{1}{3}\omega$ under a homothety centered at $M$ with ratio 3, so we are done.

Let $CM\cap\omega=K$, $CK$ is diameter. Let $X$ be the center of $\odot(ABC)$. Since $CK=2BC$ from $60-30-90$,we see that $BC=AC=CX=r$,and at the same time from similarity we get $NP=2CM=BC$. $CX||PN$,$CX=PN$, and $XP=r=NP$ so $CXNP$ is a rhombus.So we conclude that $\odot(AXNB)$ is a circle with center $C$, radius $r$.