Anyone succesfully bash this? I noticed that inverting about $A$ with radius $AP$ made something interesting, but I am not too sure.

How much would I get for pointing out the similar triangles and claiming that $M'$, $M$, and the midpoints of $BQ$ and $BP$ are concyclic?

DrMath wrote: Anyone succesfully bash this? Yes. Toss on the complex unit circle with $a = -1$, $b=1$, $z = -\tfrac12$. Let $s$ and $t$ be on the unit circle. We claim $Z$ is the center. It follows from standard formulas that \[ x = \frac 12 \left( s + t - 1 + s/t \right) \] thus \[ 4\operatorname{Re} x + 2 = s + t + \frac 1s + \frac 1t + \frac st + \frac ts \] which depends only on $P$ and $Q$, and not on $X$. Thus \[ 4\left\lvert z-\frac{s+t}{2} \right\rvert^2 = \left\lvert s+t+1 \right\rvert^2 = 3 + (4\operatorname{Re}x+2) \] does not depend on $X$, done.

brian22 wrote: I KNEW IT WAS AT -1/2! I'm just bad at geo so I couldn't prove it Did you write that down? I think that might be the kind of thing that would be worth a point. Actually IMO realizing the center is $-\tfrac 12$ is the entire difficulty of the problem -- it's totally unexpected that the center doesn't even depend on $P$ and $Q$! (This is why drawing good diagrams is important!) I cheated and used GeoGebra so I saw this in about five minutes, but once you realize this it's really quite direct if you've done enough complex numbers.

Sketch: $AX\cdot AS=AP^2$ is fixed. So if $N$ is the midpoint of $AS$ then $AX\cdot AN$ is fixed, i.e. the power of $A$ wrt the nine point circle $\gamma$ of $AST$ is fixed. Let $R$ be the radius of $\omega$. The radius of $\gamma$ is $\tfrac{R}{2}$. If $R$ is the nine point center of $AST$, then the power of $A$ wrt $\gamma$ is $AR^2-\tfrac{R^2}{4}$. Since $R$ is fixed, so is $AR$. If $G$ is the centroid of $AST$, then $G$ lies on segment $RO$ with $GR:GO=\tfrac{1}{2}$. Also $\tfrac{AG}{GM}=2$. $R$ lies on the circle $\Omega$ centered at $A$ with radius $AR$ ($AR$ is fixed). Two homotheties, centered at $O$ and $A$, send the locus of $P$ to the locus of $M$. Since the locus of $P$ is a subset of $\Omega$, the locus of $M$ is a subset of some circle (circles are sent to circles under homotheties). So done.

@v_Enhance I wrote that it had to pass through $P$ and $Q$... . I was less sure of the $-\frac{1}{2}$ thing, so I didn't write it down.

v_Enhance wrote: brian22 wrote: I KNEW IT WAS AT -1/2! I'm just bad at geo so I couldn't prove it Did you write that down? I think that might be the kind of thing that would be worth a point. Actually IMO realizing the center is $-\tfrac 12$ is the entire difficulty of the problem -- it's totally unexpected that the center doesn't even depend on $P$ and $Q$! (This is why drawing good diagrams is important!) I cheated and used GeoGebra so I saw this in about five minutes, but once you realize this it's really quite direct if you've done enough complex numbers. What is the intuition after you realize the center is at -1/2? I realize that, but couldn't continue anywhere with my bash. What made you set s,t and solve for x, as opposed to setting x and solve for m?

I tried coordinate bashing and got a big expression for $MZ$ where $Z=-1/2$ but didn't explicity show that $MZ$ was constant. How many points would that get?

If you do special points on the segment then you get that it's the circumcircle of PQC where C is the midpoint of BP or something like that, so then you can get the center is 1/2

Yeah, take $X\rightarrow P, Q, N$ where $N$ is the midpoint of $PQ$.

mota60ceng wrote: What is the intuition after you realize the center is at -1/2? I realize that, but couldn't continue anywhere with my bash. What made you set s,t and solve for x, as opposed to setting x and solve for m? The point is that the center doesn't even depend on $P$ and $Q$! Clearly, then $P$ and $Q$ cannot be very important In other words, with this insight the problem becomes ``show that $ZM$ depends only on the real part of $x$''. For this $s$ and $t$ make much nicer free variables.

Is it true that A, M, the midpoint of BP, and the midpoint of BQ are concyclic?

math2468 wrote: Is it true that A, M, the midpoint of BP, and the midpoint of BQ are concyclic? No, unfortunately.

Not A... Anyways, it was actually pretty easy by trying trivial possible X to figure out that the center was the midpoint of AO. Then 4 hours later, no progress (darn I thought geo was my strength). Finally, in the last half hour I proceed that if you extend TX to T', call M' the midpoint of ST' (they were trying to trick you into only working with the given T), then the midpoint of AO is equidistant from M and M'. I thought continuity with proof by contradiction might work but that seemed like fake-solve, but time was up anyways. But you find that P, Q, M, M', midpoint of BP, BQ lie on the circle. I guess since I ended with midpoint of AO equidistant from M, M', I would get 2 or 3? Any guesses? Pythagoras I pray to you please help me with geo, like you and Euclid did last year.

WOW I had the right center and yet since I bashed based on P, S it was not feasible. Probably still get some point for the right circle hopefully.

I can't believe I'm saying this, but... I complex bashed this problem. Well ok, not completely. By homothety and "all you have to do is construct a parallelogram" I reduced it to the following complex numbers problem: Quote: Let $a=1+0i$ and let $s$ and $t$ move freely along the unit circle. Let $o'$ be the reflection of the origin across $\overline{st}$ and let $x$ be the foot of the altitude from $t$ to $\overline{sa}$. Then the distance from $o'$ to $a$ is uniquely determined by $\Re(x)$. Then after a good amount of bashing, I managed to derive something that was similar to v_Enhance (in that there were invariants and what not). It took at least three hours to recognize that i could bash and then execute though

So I just drew the diagram and wrote the correct locus without proof (derp) any points? Edit: to clarify I drew diagram, showed locus was independent from P and Q, and then said it is well known that it's -1/2

Synthetic: Let $Y$ be the second intersection of $AT$ and $PQ$. Let $PQ$ meet $ST$ at $Z$. Let $A$ project onto $ST$ at $R$. Note that $STXY$ is cyclic by a well known-lemma, which you can prove with inversion or other things. So this has diameter $ST$ and center $M$. By nine-point circle $MRXY$ is cyclic, and we know $STXY, STPQ$ cyclic. So by POP \[ ZM\cdot ZR = ZX\cdot ZY= ZS\cdot ST = ZP \cdot ZQ \] and $MRPQ$ is cyclic. The center is clearly the midpoint of $AO$ since it lies on the perpendicular bisectors of $MR$ and $PQ$. So we're done. (except the parallel case, but this can be handled with continuity.)

pi37 wrote: Let $A$ project onto $ST$ at $R$. Motivation?

Here is what I think is a motivated synthetic solution to this beast of a problem. In any locus problem, we want to figure out what the locus is. In this case, we let $X$ be $P$ to make $M$ the midpoint of $BP$, let $X$ be the midpoint of $PQ$ to make $M$ the midpoint of $BQ$, and let $X$ approach $Q$ to get $M=Q$. Thus, $M$ should lie on the circle passing through $Q$ and the midpoints of $BP$ and $BQ$, which also passes through $P$ because isosceles trapezoids are cyclic. Our problem now is that $B$ doesn't appear at all in the original problem statement and is something we want to remove. However, $B$ is used here to describe the midpoints of $BP$ and $BQ$, so we can't remove it...yet. To remove $B$, we look at the center of the circle that is the locus of $M$. By perpendicular bisectors, this center must be the midpoint $O_1$ of $AO$, where $O$ is the center of $\omega$. Thus, it suffices to prove that $MPQ$ has circumcenter $O_1$, and we can remove $B$ and the midpoints of $BP$ and $BQ$ from our diagram. Think about how tempting it is to use shooting lemma $A$; we have all the necessary ingredients, such as $AP=AQ$ and $X=AS \cap PQ$. So, we let $Y=AT \cap PQ$, which means $SXYT$ is cyclic and $\angle SYT=90^\circ$. Symmetry is nice, so $Y$ is probably a good point to add in our diagram. The definitions of $X$, $Y$, and $M$ make it clear what we want to do: this is an orthocenter configuration with reference triangle $AST$. With a good diagram, we can see that the other vertex of the orthic triangle, $Z$, satisfies $O_1M=O_1Z$. Thus, proving that $O_1$ is the circumcenter of $MPQ$ is equivalent to proving that $O_1$ is the circumcenter of $ZPQ$. In this orthocenter configuration, $P$ and $Q$ are the weirdest points; I don't know of any problems that uses $EF \cap (ABC)$ in a legit way. Thus, we want to remove $P$ and $Q$. There is a nice way to do so: $\overline{PQ}$ is the radical axis of $(APQ)$ and $(ZPQ)$. Thus, we can reduce the problem to the following: Reduced problem wrote: Let $AST$ be a triangle with orthic triangle $ZYX$, circumcircle $\omega$, and circumcenter $O$. Let $O_1$ be the midpoint of $AO$. Prove that $X$ and $Y$ lie on the radical axis of $\omega$ and the circle $\gamma$ centered at $O_1$ passing through $Z$. We want to show that $R=ZX \cap (AYZS)$ lies on $\gamma$, after which we finish by power of a point at $X$ (which is forced due to the radical axis condition). Since proving that a point lies on $\gamma$ ultimately boils down to proving a length condition ($O_1Z=O_1R$), we use symmetry. Let $K$ be the midpoint of $AS$, and notice that $K$ is the circumcenter of $AYZS$. Then, showing that $O_1K \perp XZ$ suffices. But we also have $O_1K \parallel SO$, and $XZ$ is perpendicular to $SO$ by angle chase, so we are done.

Standard Solution Let $G$ and $N$ be the centroid and nine point center of $\triangle AST$ respectively. Then the nine point circle intersects $AS$ at $X$ and $Y$ respectively, where $Y$ is the midpoint of $AS$. Note that since $\measuredangle PSX = \measuredangle APX$, $AP$ is perpendicular to $(PSX)$. Let $R_N$ be the radius of the ninepoint circle. Then, \[ AN^2 = AX \cdot AY + R_N^2 = AP^2 + R_N^2 \]so $AN$ is constant and $N$ moves along a circle centered at $A$. Taking a homothety of $\frac{2}{3}$ at $O$ means that $G$ does so as well. Then a homothety of $\frac{3}{2}$ at $A$ gives the result.

Let $Y = AT\cap PQ$. Then $STYX$ is cyclic by shooting lemma and has center $M$ since $\angle SXT = 90^\circ$. But then if $\omega$ is centered at $O$ with radius $R$ we have $$AP^2 = \text{pow}_{(ST)}(A) = AM^2 - MS^2 = AM^2 - (R^2 - OM^2)$$ so $AM^2 + OM^2$ is fixed, and hence $M$ lies on a circle centered at the midpoint of $AO$, as desired.

$\underline{Claim:}$ $AX\cdot AS$ is independent of $X$. $\underline{Proof:}$ Denote $Y=AB\cap PQ$ and $R$ as the radius of $\omega$. By Power of a Point, we can say that: \begin{align*} AX\cdot AS & =AX^2+AX\cdot XS\\ &= AX^2+R^2-OX^2\\ &=AX^2+R^2-YO^2-YX^2\\ &=AY^2-YO^2+R^2. \end{align*} Therefore, as $Y$ is fixed, $AX\cdot AS$ is independent of $X$. $\underline{Claim:}$ If $N_9$ is the nine-point center of $\triangle{AST}$, the locus of $N_9$ is a circle with a center at $A$. $\underline{Proof:}$ Denote $S'$ as the midpoint of $AS$. By Power of a Point, and the fact that the radius of the nine-point circle is half of $R$, we can say that: \[AX\cdot AS'=AN_9^2-\frac{R^2}{4},\]\[\frac{AX\cdot AS}{2}=AN_9^2-\frac{R^2}{4},\]\[AN_9^2=\frac{AY^2-YO^2}{2}+\frac{3R^2}{4}.\]Thus, $AN_9^2$ does not depend on the position of $X$ so it is fixed, as desired. Now, consider a homothety centered at $H$, the orthocenter of $\triangle{AST}$, with a scale factor of $\frac{4}{3}$. This maps $N_9$ to $G$ and $A$ to a point $A'$. Now, consider a homothety centered at $A$ with a scale factor of $\frac{3}{2}$. This maps $G$ to $M$ and $A$ to some point $A''$. Thus, $M$ lies on a circle with center $A''$ as a homothety maps a circle to a circle.

Let $N$ and $O$ be the midpoints of $AB$ and $AN$ respectively, and let $\Omega$ be the circle centered at $O$ with radius $AO$. We shall prove that $M$ moves along a circle centered at $O$ by showing that $\text{Pow}_\Omega(M)=MO^2-AO^2$, and in turn $MO$, is constant. Define $f:\mathbb{R}^2\rightarrow\mathbb{R}$ by $f(P)=\text{Pow}_\Omega(P)-\text{Pow}_\omega(P)$, so that $f$ is linear by Linearity of Power of a Point. Hence, $f(M)=\tfrac{f(S)+f(T)}{2}$, which rewrites as \[\text{Pow}_\Omega(M)-\text{Pow}_\omega(M)=\frac{\text{Pow}_\Omega(S)+\text{Pow}_\Omega(T)}{2},\]because $S,T\in\omega\implies\text{Pow}_\omega(S)=\text{Pow}_\omega(T)=0$. Now $\text{Pow}_\omega(M)=SM\cdot TM=\tfrac{ST^2}{4}$. Additionally, taking a homothety of scale factor $\tfrac{1}{2}$ at $A$, we find that $AS$ and $AT$ intersect $\Omega$ for a second time at their midpoints; so $\text{Pow}_\Omega(S)=\tfrac{AS}{2}\cdot AS=\tfrac{AS^2}{4}$ and $\text{Pow}_\Omega(T)=\tfrac{AT^2}{4}$. Therefore, it remains to show that \begin{align*} \text{Pow}_\Omega(M)&=\frac{\text{Pow}_\Omega(S)+\text{Pow}_\Omega(T)}{2}+\text{Pow}_\omega(M)\\ &=\frac{AS^2+AT^2-ST^2}{4} \end{align*}is constant. The rest is direct by Pythagoras: $AT^2-ST^2=AX^2-SX^2$ since $AS\perp TX$, and $AS^2=(AX+SX)^2=AX^2+SX^2+2\cdot AX\cdot SX$. Summing gives $2(AX^2+AX\cdot SX)=2(AX^2+AN^2-XN^2)$ by Power of a Point. But $AX^2-XN^2=AQ^2-QN^2$ by Pythagoras, and adding $AN^2$ leaves $AQ^2$, constant.

Our synthetic solution follows with a series of claims. Claim 1: $\triangle APX \sim \triangle ASP$. We have $\angle PAX = \angle SAP$, and \begin{align*} \angle APX &= \angle APQ = \angle ABQ = \angle ABP = \angle ASP \text{ } \Box \\ &\implies \boxed{AP^2 = AX \cdot AS.} \end{align*} Claim 2: The length $AN$ is fixed. Denote $Y$ as the midpoint of $AS$, implying $(XYM)$ is the nine-point circle of $\triangle AST$. Then let $H$ and $N$ be the orthocenter and nine-point center of $\triangle AST$, respectively. Using Power of a Point, we find \begin{align*} AN^2 &= pow(A, (XYM)) + R^2_{(XYM)} \\ &= AX \cdot AY + \frac{1}{4} R^2_{(APBQ)} \\ &= \frac{1}{2} AP^2 + \frac{1}{4} R^2_{(APBQ)}, \end{align*}which is a fixed quantity. $\Box$ Claim 3: $AN = KM$, where $K$ is the midpoint of $AO$. This follows from a stronger statement - $AKMN$ is a parallelogram. We have $AK = NM$ as $R_{(APBQ)} = 2R_{(XYM)}$, and $AK \parallel NM$ due to a homothety at $H$ with ratio $2$. $\Box$ Hence $KM$ is a fixed quantity, meaning $M$ moves along a circle centered at $K$. $\blacksquare$ [asy][asy] size(256); defaultpen(linewidth(0.4)+fontsize(8)); pair O, K, A, P, B, Q, S, X; O = 0; K = -.5; A = -1; P = dir(135); B = 1; Q = dir(225); S = dir(40); X = extension(A, S, P, Q); pair [] T=intersectionpoints(circle(O, 1), 3X-2*(rotate(90, X)*S)--rotate(90, X)*S); pair M, Y, H, N; M = .5S + .5T[1]; Y = .5S + .5A; H = orthocenter(A, S, T[1]); N = .5H; fill(A--P--S--cycle, palegreen); fill(A--K--M--N--cycle, lightyellow); draw(circle(O, 1)); draw(A--P--B--Q--A--P--Q); draw(S--A--B); draw(A--T[1]--S); draw(T[1]--X); draw(circumcircle(X, Y, M), blue); draw(A--H--B, dashed); draw(2N-M--M); draw(P--S); draw(A--N^^M--K); label("$O$", O, dir(90)); label("$A$", A, W); label("$P$", P, NW); label("$B$", B, E); label("$Q$", Q, SW); label("$S$", S, NE); label("$X$", X, NW); label("$T$", T[1], dir(270)); label("$Y$", Y, dir(90)); label("$H$", H, dir(225)); label("$N$", N, SE); label("$M$", M, dir(0)); label("$K$", K, dir(90)); dot(O); dot(A); dot(P); dot(B); dot(Q); dot(S); dot(X); dot(T[1]); dot(Y); dot(H); dot(N); dot(M); dot(2N-M); dot(K); [/asy][/asy]

Here's a short solution I found: Let $\Omega$ be the circle with center $A$ passing through $P,Q$. Note that as $AP^2 = AX\cdot AS$ we have that $(XST)$ is orthogonal to $\Omega$. Hence, w.r.t. $\Omega$, the power of the center of $(XST)$ i.e. $M$ is $MS^2$. So, as the power of $M$ w.r.t. $\omega$ is $MS\cdot MT = - MS^2$, we have that the ratio of powers of $M$ w.r.t. $\omega,\Omega$ is constant i.e. $-1$. Hence, $M$ moves along a fixed circle coaxial with $\omega,\Omega$ and we are done! Remark: For any $2$ circles, $\mathcal{C}_1,\mathcal{C}_2$ the locus of all points s.t. the powers w.r.t. the $2$ circles have ratio $-1$ is actually a circle with its center at the midpoint of the centers of $\mathcal{C}_1,\mathcal{C}_2$ and coaxial with $\mathcal{C}_1,\mathcal{C}_2$ (if it exists). So, in particular for the given problem, the center of the required locus is actually at the midpoint of $AO$!

Here's a solution with Inversion at $P$. (elaborated with motivation) Invert at $P$. Then the problem becomes: (this is the literal translation into the inversion so it is an easy exercise to check that this is the reduction) Inverted Problem: Let $P,Q,A$ be points so that $QP = QA$. $X$ is a variable point on ray $\overrightarrow{PQ}$ not lying on $\overline{PQ}$. Let the circumcircle of $\displaystyle \triangle PAX$ be $\Gamma$. Suppose $\Gamma$ intersects $AQ$ at $S$. Let $K$ be the point of intersection of the tangent to $\Gamma$ at $P,X$, and let $\gamma$ be the circle with center $K$ and radius $KP$. Let $T$ be the point of intersection of $AQ$ and $\gamma$ so that $T,S$ are on the same side of $PQ$. Finally let $M$ be the point on circumcircle of $PST$ so that $PSMT$ is a harmonic quadrilateral. Show that as $X$ varies, $M$ moves along a line. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8.69579111791154cm); 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 = -12.588921244761162, xmax = 34.10686987315038, ymin = -10.97652652920867, ymax = 8.59096386647165; /* image dimensions */ pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); pen ttzzqq = rgb(0.2,0.6,0.); pen zzttqq = rgb(0.6,0.2,0.); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); draw(arc((-4.73285770201925,-1.5851882676025821),1.2742938752613824,-94.88382104067733,-17.15501704176266)--(-4.73285770201925,-1.5851882676025821)--cycle, linewidth(0.8) + ttzzqq); draw(arc((3.5993457452603925,-1.55082866575813),1.5574702919861338,187.95251875876014,200.22371475984545)--(3.5993457452603925,-1.55082866575813)--cycle, linewidth(0.8) + zzttqq); /* draw figures */ draw(circle((-0.5587056160162932,-3.5202213397406914), 4.600858466346371), linewidth(0.8) + wvvxds); draw(circle((-3.86334837257875,-3.2378535050286303), 1.8674444733326263), linewidth(0.8) + yqqqyq); draw((-4.73285770201925,-1.5851882676025821)--(-5.000632730256479,-4.719046656093717), linewidth(0.8) + wrwrwr); draw((-4.73285770201925,-1.5851882676025821)--(-2.2135750277227,-2.3628696357012853), linewidth(0.8) + wrwrwr); draw((-2.2135750277227,-2.3628696357012853)--(3.5993457452603925,-1.55082866575813), linewidth(0.8) + wrwrwr); draw((3.5993457452603925,-1.55082866575813)--(-5.000632730256479,-4.719046656093717), linewidth(0.8) + wrwrwr); draw((1.3624776541290877,0.6603232982059701)--(3.5993457452603925,-1.55082866575813), linewidth(0.8) + wrwrwr); draw((-0.6034183727430487,7.322622166497663)--(-4.73285770201925,-1.5851882676025821), linewidth(0.8) + wrwrwr); draw((-0.6034183727430487,7.322622166497663)--(3.5993457452603925,-1.55082866575813), linewidth(0.8) + wrwrwr); draw((1.3624776541290877,0.6603232982059701)--(-5.000632730256479,-4.719046656093717), linewidth(0.8) + wrwrwr); draw((-4.73285770201925,-1.5851882676025821)--(3.5993457452603925,-1.55082866575813), linewidth(0.8) + wrwrwr); draw((-2.2135750277227,-2.3628696357012853)--(1.8091785098197786,-9.94773124493867), linewidth(0.8) + wrwrwr); draw((1.8091785098197786,-9.94773124493867)--(-5.000632730256479,-4.719046656093717), linewidth(0.8) + wrwrwr); draw(shift((-0.6034183727430492,7.3226221664976645))*xscale(9.818419216148714)*yscale(9.818419216148714)*arc((0,0),1,212.00934216225082,330.46867914558794), linewidth(0.8) + wrwrwr); draw((-1.9080693022464674,0.14256281609147714)--(1.8091785098197786,-9.94773124493867), linewidth(0.8) + linetype("4 4") + dbwrru); draw((1.8091785098197786,-9.94773124493867)--(-4.73285770201925,-1.5851882676025821), linewidth(0.8) + wrwrwr); /* dots and labels */ dot((-4.73285770201925,-1.5851882676025821),linewidth(3.pt) + dotstyle); label("$P$", (-5.396240259952471,-1.1219872271873232), NE * labelscalefactor); dot((-1.27681893696828,-1.5709365613549493),linewidth(3.pt) + dotstyle); label("$Q$", (-1.3751351424609977,-1.2635754355496989), NE * labelscalefactor); dot((1.3624776541290877,0.6603232982059701),linewidth(3.pt) + dotstyle); label("$A$", (1.4849466664589936,0.8319300482134614), NE * labelscalefactor); dot((3.5993457452603925,-1.55082866575813),linewidth(3.pt) + dotstyle); label("$X$", (3.948581491964333,-1.291893077222174), NE * labelscalefactor); dot((-5.000632730256479,-4.719046656093717),linewidth(3.pt) + dotstyle); label("$S$", (-5.8776401683845485,-4.973186494643942), NE * labelscalefactor); dot((-0.6034183727430487,7.322622166497663),linewidth(3.pt) + dotstyle); label("$K$", (-1.3751351424609977,6.891905366123141), NE * labelscalefactor); dot((-2.2135750277227,-2.3628696357012853),linewidth(3.pt) + dotstyle); label("$T$", (-2.479523167687529,-3.2174927109504834), NE * labelscalefactor); dot((1.8091785098197786,-9.94773124493867),linewidth(3.pt) + dotstyle); label("$U$", (2.022981858236022,-10.013726712344518), NE * labelscalefactor); dot((-2.468559581939059,-4.4795914365127425),linewidth(3.pt) + dotstyle); label("$M$", (-2.677746659394855,-5.482904044748494), NE * labelscalefactor); dot((-3.3637792092231593,-3.335251606389178),linewidth(3.pt) + dotstyle); label("$V$", (-3.498958267896635,-2.905998652553257), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Okay, this is quite intimidating at first, for example the definition of $M$ is quite scary. But then we realize that $M$ is supposed to move along a line through $Q$ (since that earlier locus should have been a circle passing through $P$ and $Q$), so all we need to show is that $QM$ is a fixed line independent of $M$. Now, let the tangents to $(PST)$ at $T,S$ intersect at $U$, then $P,M,Z$ are collinear. Let $PZ$ intersect $ST$ at $V$. Then $(P,M; V,U) = -1$, hence $(QP, QM; QV, QU) = -1$. Now $QP, QV$ are fixed lines, so it is enough to show that $QU$ is a fixed line. Inspecting the diagram it appears as if $QU$ seems to be the angle bisector of $PQA$, so let's try to make that claim. (note that $QA = QP$, therefore $PAXS$ is an isosceles trapezium) Reduced problem: Let $PAXS$ be an isosceles trapezium, with circumcircle $\Gamma$, and diagonals intersecting at $Q$. Suppose tangents to $\Gamma$ at $P,X$ intersect at $K$, and let $T$ be the point on segment $QS$ so that $KT = KP = KX$. If the tangents to $(PST)$ at $S,T$ intersect at $U$ then $U$ lies on the angle bisector of $PQA$. Proof: Note that the angle bisector of $PQA$ is the perpendicular bisector of $SX$. Thus, we are reduced to proving that $U$ is the circumcenter of $\displaystyle \triangle STX$. This is just angle chasing: we already know that $US = UT$, therefore it is enough to show that $\angle TUS = 2 \angle TXS$. But $\angle TUS = 180^\circ - 2 \angle TPS$, thus it is enough to show that $\angle TPS + \angle TXS = 90^\circ$. Now, $\angle PKX = 180^\circ - 2 \angle PSX$ thus $\angle PTX = 90^\circ + \angle PSX$. Therefore $\angle TPX+ \angle TXP = 90^\circ - \angle PSX$, and finally we have $$\angle TPS + \angle TXS = 180^\circ - (\angle TPX+ \angle TXP + \angle PSX) = 90^\circ.$$And we are through. $\square$

Siddharth03 wrote: Here's a short solution I found: Let $\Omega$ be the circle with center $A$ passing through $P,Q$. Note that as $AP^2 = AX\cdot AS$ we have that $(XST)$ is orthogonal to $\Omega$. Hence, w.r.t. $\Omega$, the power of the center of $(XST)$ i.e. $M$ is $MS^2$. So, as the power of $M$ w.r.t. $\omega$ is $MS\cdot MT = - MS^2$, we have that the ratio of powers of $M$ w.r.t. $\omega,\Omega$ is constant i.e. $-1$. Hence, $M$ moves along a fixed circle coaxial with $\omega,\Omega$ and we are done! Remark: For any $2$ circles, $\mathcal{C}_1,\mathcal{C}_2$ the locus of all points s.t. the powers w.r.t. the $2$ circles have ratio $-1$ is actually a circle with its center at the midpoint of the centers of $\mathcal{C}_1,\mathcal{C}_2$ and coaxial with $\mathcal{C}_1,\mathcal{C}_2$ (if it exists). So, in particular for the given problem, the center of the required locus is actually at the midpoint of $AO$! forgotten coaxiality lemma: ratio of locus of points such that power of one circle is k* power of another is another coaxal circle

In my honest opinion. guessing the locus was trivial but proving it required some ingenuity (since I forgot that complex numbers existed). We claim that as $X$ varies along $PQ$, the point $M$ varies along the circle passing through $P$ , $Q$ , and the midpoints of segments $PB$ and $QB$. To see why this is true, we let $T_1$ and $T_2$ be the intersections of the line perpendicular to $AS$ at $X$ with $\omega$ and rewrite the problem in terms of the reference triangle $\triangle ST_1T_2$. Then, it suffices to show the following. Rephrased Problem wrote: Let $\triangle ABC$ be an acute scalene triangle with $M_B$ and $M_C$ being the midpoints of $AC$ and $AB$. Let $D$ be the foot of the perpendicular from $A$ to $BC$ and $H_A$ the intersection of $\overline{AD}$ with $(ABC)$. Let $A'$ be the intersection of the line parallel to side $BC$ through $A$ and $(ABC)$. Let $P$ and $Q$ be the intersections of the line through $D$ perpendicular to segment $A'H_A$, with $(ABC)$. Let $M_P$ and $M_Q$ be the midpoints of sides $A'Q$ and $A'P$ respectively. Show that the points, $P$ , $Q$ , $M_B$ , $M_C$ , $M_P$ and $M_Q$ lie on the same circle. Let $X$ denote the intersection of the tangent to $(ABC)$ at $A$ and $\overline{PQ}$. Then, note that \[\measuredangle DAX = \measuredangle H_AAX = \measuredangle H_AA'A = \measuredangle QDH_A = \measuredangle XDA\]so $XD=XA$. Thus, $X$ lies on the perpendicular bisector of segment $AD$. Further, points $M_B$ and $M_C$ also lie on this perpendicular bisector so points $X$ , $M_B$ and $M_C$ are collinear. We also know that circles $(ABC)$ and $(AM_BM_C)$ are tangent at $A$ (due to homothety reasons). Thus, \[XP \cdot XQ = XA^2 = XM_B \cdot XM_C\]so quadrilateral $PQM_BM_C$ is indeed cyclic. Further, let $N$ be the midpoint of $A'D$. Then, $N$ clearly lies on $M_PM_Q$ since this is the $A-$midline of $\triangle A'PQ$ and $D$ is a point lying on side $PQ$. Further, let $D'$ be the reflection of $D$ across the perpendicular bisector of side $BC$. $N$ is clearly the center of rectangle $AA'D'D$ so it is also the midpoint of $AD'$. Thus, $N$ also lies on $M_BM_C$ as it is the $A-$midline of $\triangle ABC$ and $D'$ lies on side $BC$. Thus, lines $\overline{M_BM_C}$ , $\overline{M_PM_Q}$ and $\overline{A'D}$ concur , at $N$. Then, \[4NM_P \cdot NM_Q = DP \cdot DQ = DB \cdot DC = D'B \cdot D'C = 4NM_C \cdot NM_B\]so quadrilateral $M_CM_PM_BM_Q$ is cyclic. Now, since lines $\overline{M_BM_C}$ and $\overline{PQ}$ are not parallel ($BC$ and $PQ$ intersect at $D$), it follows that points $P$ , $Q$ , $M_B$ , $M_C$ , $M_P$ and $M_Q$ lie on the same circle as desired.

this aint 15m Taking reference triangle $ABC$ and a bit of relabelling, we have the following problem. Let $ABC$ be a triangle and $D,E,F$ be the feet from $A,B,C$ to the opposite sides. Note by a simple angle chase $EF$ is the same line as $PQ$ in the problem. Suppose now this intersects $(ABC)$ at $P,Q$. Let $A'$ be the $A$-antipode, and midpoints of $PA',QA'$ be $U,V$. Let $M$ be the midpoint of $BC$. We want to prove $PQVUM$ is cyclic, as in the problem taking $X$ as $P,AB\cap PQ, Q$ give two points from the locus. Clearly, $PQMD$ is cyclic, as if $EF\cap BC=Y$, $YD\times YM=YE\times YF=YB\times YC=YP\times YQ$ by power of a point on nine point circle, semicircle on $BC$, circumcircle. Thus we actually need $PQDUV$ cyclic. We do this by forgotten coaxiality lemma. Consider $\omega=(A,P)$ and $(ABC)$. Checking the power of $D$, we see that $AD^2-AH\times AD=AD(DH)$, $H$ is orthocentre. Now check if this is $DB\times DC$ which is negative. The ratio is $-1$ by similar triangles $\triangle BDH\sim\triangle ADC$. However, the ratio for $U,V$ is just $-1\times \frac{UP^2}{UP\times UA'}=-1$ etc, so we're done as then $D$ lies on a circle coaxial to $(ABC),\omega$, hence passing through $P,Q$.

TestX01 wrote: this aint 15m yes it is (in particular, forgotten coaxiality lemma is a >= 15m technique) We complex bash. Let \(a = 1\), \(b = -1\) and \(r = 1 / 2\). We wish to show that if \(\operatorname{Re}(x)\) is constant, then \(|r - m|\) is constant. The complex foot formula yields that \[\frac{1}{2} (s + t + 1 - s \overline{t}) = x;\]therefore \begin{align*} \operatorname{Re}(x) &= \operatorname{Re}\left(\frac{1}{2} (s + t + 1 - s \overline{t})\right) \\ &= \frac{1}{2} \operatorname{Re}(s + t + 1 - s \overline{t}) \\ &= \frac{1}{4} (s + t + 1 - s \overline{t} + \overline{s} + \overline{t} + 1 - \overline{s} t) \\ &= \frac{1}{4} (2 + s + t + \overline{s} + \overline{t} - s \overline{t} - \overline{s} t) \\ &= \frac{1}{2} + \frac{c}{4} \end{align*}where \(c = s + t + \overline{s} + \overline{t} - s\overline{t} - \overline{s}t\). However, \begin{align*} |r - m|^2 &= (r - m)(\overline{r - m}) \\ &= \left(\frac{s + t}{2} - \frac{1}{2}\right) \left(\overline{\frac{s + t}{2} - \frac{1}{2}}\right) \\ &= \frac{1}{4} (s + t - 1) (\overline{s} + \overline{t} - 1) \\ &= \frac{1}{4} (3 - s - t - \overline{s} - \overline{t} + s \overline{t} + \overline{s} t) \\ &= \frac{1}{4} (3 - c) \end{align*}which is constant, as desired.

Let $O$ be the center of $\omega$. We will show that the fixed circle is the circle centered at $\overline{AO}$ passing through $P$ and $Q$. Let $T_1$ and $T_2$ be the points on $\omega$ for which $\angle TX_1 S = \angle T_2 XS = 90^{\circ}$, and let $M_1$ and $M_2$ be the midpoints of $\overline{ST_1}$ and $\overline{ST_2}$. Claim: We have that $PM_2 M_1 Q$ is cyclic. Proof: Invert at $S$ and refer to points' images by their original names (except for $S$) (just for this claim). We will show that $PM_2 M_1 Q$ is an isosceles trapezoid (after inversion) – since $\overline{PQ} \parallel \overline{M_1 M_2}$, it suffices to show that $XP = XQ$ and $XM_1 = XM_2$. The former is true since $X$ is now the arc midpoint of arc $PQ$ in $(SPQ)$, while the latter is true since $XM_1 = XS = XM_2$. The claim is proven. Claim: The perpendicular bisector of $\overline{M_1 M_2}$ always bisects $\overline{AO}$. Proof: Let $A'$ be the midpoint of $\overline{AS}$. Since $\angle OA'A = 90^{\circ}$, the perpendicular bisector of $\overline{A'O}$ bisects $\overline{AO}$. Since $O$ is the antipode of $S$ in $(SM_1 M_2)$ and $A'$ lies on $(SM_1 M_2)$ such that $\overline{SA_1} \perp \overline{M_1 M_2}$, it follows that $A'OM_1 M_2$ is an isosceles trapezoid. Therefore, the perpendicular bisector of $\overline{A'O}$ coincides with the perpendicular bisector of $\overline{M_1 M_2}$. These two claims are enough to show the problem.