whooo let's go bary Denote by $X$ the second intersection of $(BPMQ)$ with $\overline{AC}$. Let $T = (u,v,w)$ thus $a^2vw+b^2wu+c^2uv=0$. Then $\overline{PT} \parallel \overline{BC} \implies P = (u:w+v:0)$. Analogously, $Q = (0:u+v,w)$. So, $$AX = \frac{AB \cdot AP}{AM} = \frac{2c^2}{b} (v+w) \quad\text{and}\quad CX = \frac{CB \cdot CQ}{CM} = \frac{2a^2}{b} (v+u).$$ Adding these implies that $\frac{1}{2} b^2 = (v+w)c^2 + (v+u)a^2$. However, by barycentric distance formula, $$BT^2 = -a^2(v-1)w-b^2wv-c^2u(v-1) = a^2w+c^2u + \underbrace{-a^2vw-b^2wu-c^2uv}_{=0}.$$ Thus, adding gives $\frac{1}{2} b^2 + BT^2 = a^2+c^2$, so $BT^2 = a^2+c^2 - \frac{1}{2} b^2 = 2BM^2$, thus $BT/BM =\sqrt2$ is the only possible value.

Let $N$ be the midpoint of $PQ$ and let $E$ $\equiv$ $BM$ $\cap$ $\odot (ABC),$ $S$ $\equiv$ $BN$ $\cap$ $\odot (BPQ).$ Since $$\left\{\begin{array}{cc} \measuredangle MPQ = \measuredangle MBC = \measuredangle EAC, \measuredangle SPQ = \measuredangle SBC = \measuredangle TAC \\\\ \measuredangle MQP = \measuredangle MBA = \measuredangle ECA, \measuredangle SQP = \measuredangle SBA = \measuredangle TCA \end{array}\right\| \Longrightarrow ACETM \stackrel{+}{\sim} PQMSN ,$$ so $\measuredangle BMN$ $=$ $\measuredangle (EM, MN)$ $=$ $\measuredangle (TM,SN)$ $=$ $\measuredangle MTB$ $\Longrightarrow$ $BM$ is tangent to $\odot (MNT)$ at $M,$ hence we conclude that $$\frac{1}{2} \cdot {BT}^2 = BN \cdot BT = {BM}^2 \Longrightarrow \frac{BT}{BM} = \sqrt{2}.$$

Let $D=AC\cap PQ$. Let $N$ be midpoint of $PQ$ and let $S$ be a point such that $\triangle ACS\stackrel{+}{\sim}\triangle PQT$. Let $R$ be Miquel point of $CAPQ\implies R\in \odot ABC$, $R\in \odot CDQ$. $R$ is center of spiral similarity sending $CQ\rightarrow AP$, so it's also center of spiral similarity sending $MN\rightarrow AP\implies R\in \odot DMN$. Let $T'=MS\cap BT$. Obviously $\triangle NPT\sim \triangle MAS\implies \angle DNT'=\angle DMT'\implies T'\in \odot DMN$. $\angle BT'R=\angle NT'R= \angle NDR=\angle QDR=\angle QCR=\angle BCR\implies T'\in \odot ABC\implies T\equiv T'$ $\angle MDR=\angle CDR=\angle BQR=\angle BMR \implies BM$ touches $\odot TMN$ $\implies BM^2=BN\cdot BT=\frac{BT^2}{2}\implies \frac{BT}{BM}=\sqrt{2} \blacksquare$

Answer. $\sqrt{2}$. Solution. Let us prove a well known spiral similarity lemma. Lemma. Let $AB$ and $CD$ be two segments, and let lines $AC$ and $BD$ meet at $X$. Let circumcircle of $ABX$ and $CDX$ meet again at $O$. Then $O$ is the center of the spiral similarity that carries $AB$ to $CD$. Proof of Lemma. Since $ABOX$ and $CDXO$ are cyclic. we have $\angle OBD= \angle OAC$ and $\angle OCA=\angle ODB$. It follows that $\triangle AOC$ and $\triangle BOD$ are similar, thus the result. Back to the main proof. Suppose that $T$ lies on $(ABC)$. Let $PQ\cap BT=N$, $(ABC)\cap (APQ)=X\neq A$, $PQ\cap AC=Y$. Clearly since $BPTQ$ is a parallelogram, we have $BT=2BN$. We will prove that points $X, Y$ lies on $(TMN)$. By the lemma, $X$ is the center of spiral similarity that maps segment $PQ$ to $AC$. Note that this spiral similarity also maps the midpoint of $PQ$ to midpoint of $AC$, which is $N$ to $M$. So it also maps segment $PN$ to $AM$, which implies $\triangle XMA\sim\triangle XNP$. Similarly, $X$ is also the center of spiral similarity that maps $AP$ to $BQ$. Notice that this spiral similarity also maps $A$ to $M$ and $P$ to $N$ because $M$ and $N$ are midpoints of $AP$ and $BQ$ respectively, so we also conclude that $\triangle XPA\sim\triangle XNM$. (Also known as the Averaging Principle.) So using directed angles, $\angle (XM, MN)=\angle (XA, AP)=\angle (XT, TB)=\angle (XT, TN)$, so $X$ lies on $(TMN)$. Likewise, $\angle (YN, NX)=\angle (PN, NX)=\angle (AM, MX)=\angle (YM,MX)$, so $Y$ lies on $(TMN)$. Now, note that $\angle (BM,MX)=\angle (BA,AX)=\angle (PA,AX)=\angle (NM,MX)$, so $BM$ is tangent to circle $(TMN)$. This means $2BM^2=2BN\cdot BT=BT^2$, which gives $\displaystyle \frac{BT}{BM}=\sqrt{2}$, as desired. Q.E.D

Let $X$ be the Miquel point of complete quadrilateral $\{PC, CA, AQ, QP\}$ with $R$ as the midpoint of $PQ$. Since spiral similarities are linear functions in the plane, it follows that $X$ is also the spiral center sending $AQ$ to $MR$, so $G\equiv PQ\cap AC$ gives $X\in\odot(GRM)$. Then $$\angle BMX=\angle BPX=\pi-\angle XPC=\pi-\angle XRM$$ so $BM$ is tangent to $\odot(XRM)$. Since $T\in \odot(XRM)$, $BM^2=BR\cdot BT=\frac{BT^2}{2}$, so the answer is $\sqrt{2}$.

Bary was the first thing that came to my mind when I saw this one in the TST. Since v_Enhance already posted the bash I won't do it again but just remark that this is an excellent bary tutorial problem.

Here is my approach. Firstly observe that since $B,P,M,Q$ are concyclic, the midpoint $K$ of $PQ$ varies on a line as $P,Q$ vary on $BA,BC$ respectively. Indeed, this follows since triangle $MPQ$ has a fixed shape and spiral similarity moves each linear combination uniformly. Let $(BAM)$ meet $BC$ again at $X$ and $(BCM)$ meet $BA$ again at $Y$. Let $U,V$ be the midpoints of $AX,CY$. Then, the locus of $K$ is the line $UV$. Let $L,N$ be the midpoints of $BA,BC$. Consider the circle $\gamma=(BLN)$ and $\omega=(B,\frac{BM}{\sqrt{2}})$. We claim that the line $UV$ is the radical axis of $\omega$ and $\gamma$. This shall establish the result; $\frac{BT}{BM}=\frac{1}{\sqrt{2}}$. Indeed, we define the function $f$ from the Euclidean plane to the set of real numbers as follows: $f(Z)=p(Z,\gamma)-p(Z,\omega)$. It is clear that $f$ is a linear function in $Z$. We want to establish that $f(U)=f(V)=0$. Proving $f(U)=0$ suffices, since $f(V)=0$ shall follow analogously. Notice that $f(U)=\frac{f(A)+f(X)}{2}$. We only need to ascertain ourselves that $f(A)=-f(X)$. It is evident that $f(A)=-\frac{c^2}{2}+\frac{2a^2+2c^2-b^2}{8}$ and $f(X)=XB.XN-XB^2+\frac{2a^2+2c^2-b^2}{8}$. It is equivalent to showing that $XB.XN-XB.XB=-\frac{2a^2-b^2}{4}$. Notice that $XB-XN=-(BX+XN)=-BN=-\frac{a}{2}$ and $XB=a-\frac{b^2}{2a}$ and hence the claim holds. Comment. This problem came in India TST 2016 and Romania TST 2016 as per my knowledge.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(9.cm); 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 = 2., xmax = 11., ymin = 0., ymax = 8.; /* image dimensions */ pen ffxfqq = rgb(1.,0.4980392156862745,0.); /* draw figures */ draw(circle((6.331211478628279,3.9090896183427657), 3.6128252550638846), blue); draw(circle((5.406292594169632,4.696263381502455), 2.503407971114143), ffxfqq); draw((4.532022448266124,7.042047510918392)--(6.747139816608675,0.320286204767108)); draw((6.747139816608675,0.320286204767108)--(3.4209710969677025,3.1712879644593737)); draw((4.532022448266124,7.042047510918392)--(3.1442802912498613,2.207332899377859)); draw((3.1442802912498613,2.207332899377859)--(9.70248113005252,2.6102257836729033)); draw((4.532022448266124,7.042047510918392)--(9.70248113005252,2.6102257836729033)); draw((3.4209710969677025,3.1712879644593737)--(7.858191167907094,4.191045751226126)); draw(circle((5.431616963447202,5.47556856463058), 1.8064126275319419), linetype("4 4")); draw((7.858191167907094,4.191045751226126)--(6.747139816608675,0.320286204767108)); /* dots and labels */ dot((4.532022448266124,7.042047510918392),linewidth(3.pt) + dotstyle); label("$A$", (4.660735465368374,7.223225436821573), NE * labelscalefactor); dot((3.1442802912498613,2.207332899377859),linewidth(3.pt) + dotstyle); label("$B$", (3.266868968405002,2.389177798416677), NE * labelscalefactor); dot((9.70248113005252,2.6102257836729033),linewidth(3.pt) + dotstyle); label("$C$", (9.821007177530646,2.774715340129951), NE * labelscalefactor); dot((6.423380710651191,2.4087793415253813),linewidth(3.pt) + dotstyle); label("$M$", (6.529109705978852,2.596774936262286), NE * labelscalefactor); dot((6.747139816608675,0.320286204767108),linewidth(3.pt) + dotstyle); label("$T$", (6.855333779736237,0.4911468238282515), NE * labelscalefactor); dot((3.4209710969677025,3.1712879644593737),linewidth(3.pt) + dotstyle); label("$P$", (3.533779574206499,3.3381932857108896), NE * labelscalefactor); dot((7.858191167907094,4.191045751226126),linewidth(3.pt) + dotstyle); label("$Q$", (7.982289670898113,4.376178974938934), NE * labelscalefactor); dot((3.838151369757993,4.624690205148125),linewidth(3.pt) + dotstyle); label("$X$", (3.948973849897716,4.791373250630153), NE * labelscalefactor); dot((7.117251789159322,4.826136647295648),linewidth(3.pt) + dotstyle); label("$Y$", (7.24087132144951,4.998970388475762), NE * labelscalefactor); dot((5.639581132437399,3.68116685784275),linewidth(3.pt) + dotstyle); label("$Z$", (5.758034622552305,3.8720144973138844), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] angle chasing yields $\Delta BTC \sim \Delta AQT$ so that $$\dfrac{BT}{AQ}= \dfrac{TC}{QT}= \dfrac{BC}{AT}$$ Now take $X,Y,Z$ to be the midpts of $AB,AC,PQ$ resp. Similar angle chasing gives $\Delta PMQ \sim \Delta AYM$ so that $$\dfrac{AC}{PM}=\dfrac{AB}{MQ}=\dfrac{2AM}{PQ}$$ multiplying these two gives $$\dfrac{2AM.BC}{AT.PQ}=\dfrac{BT.AC}{AQ.PM}=\dfrac{TC.AB}{QT.MQ}=\dfrac{BT.AC+TC.AB}{AQ.PM+QT.MQ}=\dfrac{AT.BC}{AM.PQ}$$ so that $2AM^2=AT^2$ as desired

Rename $A$ and $B$. Let tangent at $A$ on $\omega$ cuts cicumcircle $\mathcal{K}$ of $ABC$ at $A'$ and let $AM$ cut $\mathcal{K}$ at $M'$. Let $R$ halves $PQ$ and let $\mathcal{K}$ meets $\omega$ second time at $S$. Then $S$ is center of spiral similarity $\mathcal{P}$, which maps $\omega \longmapsto \mathcal{K}$ and $$\begin{eqnarray*} % \omega &\longmapsto &\mathcal{K} \\ P &\longmapsto & B\\ Q &\longmapsto & C \\ A &\longmapsto & A' \\ R &\longmapsto & M \\ M &\longmapsto & M' \end{eqnarray*}$$ Say $AR$ meets $A'M$ at $T'$. Since $\mathcal{P}$ send $AR$ to $A'M$ and $SA$ to $SA'$ we have $\angle AT'A' = \angle (AR,AM')= \angle ASA'$, thus $T'\in \mathcal{K}$, and so $T'=T$. Finally since $\mathcal{P}$ preserves angles we have $$\angle RMA \stackrel{\mathcal{P}}{=} \angle MM'A' = \angle AM'A' = \angle ATA' =\angle RTM$$ and this means $AT$ is tangent on circumcircle $\triangle MTR$, so $AM^2 = AT \cdot AR = {AT^2 \over 2} \Longrightarrow {AT \over AM} =\sqrt{2}$.

This is also India TST 2016 Day 3 Problem 2.

Can we use Cartesian coordinate on this problem?

kapilpavase wrote: [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(9.cm); 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 = 2., xmax = 11., ymin = 0., ymax = 8.; /* image dimensions */ pen ffxfqq = rgb(1.,0.4980392156862745,0.); /* draw figures */ draw(circle((6.331211478628279,3.9090896183427657), 3.6128252550638846), blue); draw(circle((5.406292594169632,4.696263381502455), 2.503407971114143), ffxfqq); draw((4.532022448266124,7.042047510918392)--(6.747139816608675,0.320286204767108)); draw((6.747139816608675,0.320286204767108)--(3.4209710969677025,3.1712879644593737)); draw((4.532022448266124,7.042047510918392)--(3.1442802912498613,2.207332899377859)); draw((3.1442802912498613,2.207332899377859)--(9.70248113005252,2.6102257836729033)); draw((4.532022448266124,7.042047510918392)--(9.70248113005252,2.6102257836729033)); draw((3.4209710969677025,3.1712879644593737)--(7.858191167907094,4.191045751226126)); draw(circle((5.431616963447202,5.47556856463058), 1.8064126275319419), linetype("4 4")); draw((7.858191167907094,4.191045751226126)--(6.747139816608675,0.320286204767108)); /* dots and labels */ dot((4.532022448266124,7.042047510918392),linewidth(3.pt) + dotstyle); label("$A$", (4.660735465368374,7.223225436821573), NE * labelscalefactor); dot((3.1442802912498613,2.207332899377859),linewidth(3.pt) + dotstyle); label("$B$", (3.266868968405002,2.389177798416677), NE * labelscalefactor); dot((9.70248113005252,2.6102257836729033),linewidth(3.pt) + dotstyle); label("$C$", (9.821007177530646,2.774715340129951), NE * labelscalefactor); dot((6.423380710651191,2.4087793415253813),linewidth(3.pt) + dotstyle); label("$M$", (6.529109705978852,2.596774936262286), NE * labelscalefactor); dot((6.747139816608675,0.320286204767108),linewidth(3.pt) + dotstyle); label("$T$", (6.855333779736237,0.4911468238282515), NE * labelscalefactor); dot((3.4209710969677025,3.1712879644593737),linewidth(3.pt) + dotstyle); label("$P$", (3.533779574206499,3.3381932857108896), NE * labelscalefactor); dot((7.858191167907094,4.191045751226126),linewidth(3.pt) + dotstyle); label("$Q$", (7.982289670898113,4.376178974938934), NE * labelscalefactor); dot((3.838151369757993,4.624690205148125),linewidth(3.pt) + dotstyle); label("$X$", (3.948973849897716,4.791373250630153), NE * labelscalefactor); dot((7.117251789159322,4.826136647295648),linewidth(3.pt) + dotstyle); label("$Y$", (7.24087132144951,4.998970388475762), NE * labelscalefactor); dot((5.639581132437399,3.68116685784275),linewidth(3.pt) + dotstyle); label("$Z$", (5.758034622552305,3.8720144973138844), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] angle chasing yields $\Delta BTC \sim \Delta AQT$ so that $$\dfrac{BT}{AQ}= \dfrac{TC}{QT}= \dfrac{BC}{AT}$$ Now take $X,Y,Z$ to be the midpts of $AB,AC,PQ$ resp. Similar angle chasing gives $\Delta PMQ \sim \Delta AYM$ so that $$\dfrac{AC}{PM}=\dfrac{AB}{MQ}=\dfrac{2AM}{PQ}$$ multiplying these two gives $$\dfrac{2AM.BC}{AT.PQ}=\dfrac{BT.AC}{AQ.PM}=\dfrac{TC.AB}{QT.MQ}=\dfrac{BT.AC+TC.AB}{AQ.PM+QT.MQ}=\dfrac{AT.BC}{AM.PQ}$$ so that $2AM^2=AT^2$ as desired Really GREAT Solution. İn the last step you used Ptolemy's theorem.

v_Enhance wrote: Let $T = (u,v,w)$ thus $a^2vw+b^2wu+c^2uv=0$. Then $\overline{PT} \parallel \overline{BC} \implies P = (u:w+v:0)$. . Teacher Evan,How can we get $P=(u:w+v;0)$ and why $T(u,v,w)?$ Can you tell me,please.

Please,help me.

Pure Bary: No synthetic observation whatsoever! So $B = (0, 1, 0)$ and $M = (\frac {1} {2}, 0, \frac {1} {2})$. Take the arbitrary circle through $B$ and $M$, which has equation $-a^2yz-b^2zx-c^2xy + (ux+vy+wz)(x+y+z) = 0$. We will now find relations of these constants. Plugging in $B$ yields $v = 0$, and then plugging in $M$ implies $\frac {-b^2} {4} + \frac {u + w} {2} = 0$, so $u + w = \frac {b^2} {2}$. To find $P$, note this equation intersects $AB$, or the line $z = 0$, at $B$ and $P$. $z = 0$ implies in our circle equation that $-c^2xy + (ux)(x + y) = 0$. Let $P$ be normalized coordinates $(x, y, 0)$, so that $x + y = 1$, and thus we get $c^2xy = ux$. $x$ is not $0$ lest $P$ be congruent to $B$, so dividing by $x$ we have $y = \frac {u} {c^2}$. Thus $P = (1 - \frac {u} {c^2}, \frac {u} {c^2}, 0)$, and by similar arguments, $Q = (0, \frac {w} {a^2}, 1 - \frac {w} {a^2})$. $BPTQ$ is a parallelogram, so $B + T = P + Q$. Thus solving for $T$, $T = (1 - \frac {u} {c^2}, \frac {u} {c^2} + \frac {w} {a^2} - 1, 1 - \frac {w} {a^2})$. $T$ lies on the circumcircle of $\triangle ABC$, so it satisfies $a^2yz + b^2zx + c^2xy = 0$. Thus, $a^2 \cdot (\frac {u} {c^2} + \frac {w} {a^2} - 1) \cdot (1 - \frac {w} {a^2}) + b^2 \cdot (1 - \frac {w} {a^2}) \cdot (1 - \frac {u} {c^2}) + c^2 \cdot (1 - \frac {u} {c^2}) \cdot ({\frac {u} {c^2} + \frac {w} {a^2} - 1}) = 0$. Keep this equation in mind , and call it $f(T)$. Now look at ${TB}^2$. The displacement vector $TB$ is $(1 - \frac {u} {c^2}, \frac {u} {c^2} + \frac {w} {a^2} - 2, 1 - \frac {w} {a^2})$. The distance formula is $-a^2yz-b^2zx-c^2zy$,, and plugging this in, ${BT}^2 = -a^2 \cdot (\frac {u} {c^2} + \frac {w} {a^2} - 2) \cdot (1 - \frac {w} {a^2}) -b^2 \cdot (1 - \frac {w} {a^2}) \cdot (1 - \frac {u} {c^2}) -c^2 \cdot (1 - \frac {u} {c^2}) \cdot (\frac {u} {c^2} + \frac {w} {a^2} - 2)$. This equals: $-a^2 \cdot (\frac {u} {c^2} + \frac {w} {a^2} - 1 -1) \cdot (1 - \frac {w} {a^2}) -b^2 \cdot (1 - \frac {w} {a^2}) \cdot (1 - \frac {u} {c^2}) -c^2 \cdot (1 - \frac {u} {c^2}) \cdot (\frac {u} {c^2} + \frac {w} {a^2} - 1 - 1)$ $= -[a^2 \cdot (\frac {u} {c^2} + \frac {w} {a^2} - 1) \cdot (1 - \frac {w} {a^2}) + b^2 \cdot (1 - \frac {w} {a^2}) \cdot (1 - \frac {u} {c^2}) + c^2 \cdot (1 - \frac {w} {a^2}) \cdot ({\frac {u} {c^2} + \frac {u} {c^2} - 1})] + (-a^2)(1 - \frac {w} {a^2}) + (-c^2)(1 - \frac {u} {c^2})$. But the part in brackets is just $f(T) = 0$! Thus ${BT}^2 = (-a^2)(1 - \frac {w} {a^2}) + (-c^2)(1 - \frac {u} {c^2}) = -a^2-c^2+u+w$, but remember $u+w = \frac {b^2} {2}$, so ${BT}^2 = -a^2-c^2+\frac {b^2} {2}$. We now find ${BM}^2$ using the distance formula and displacement vector $MB = (\frac {1} {2}, -1, \frac {1} {2})$. Thus plugging in, ${BM}^2 = -a^2 \cdot \frac{1} {2} + b^2 \cdot \frac{1} {4} - c^2 \cdot \frac {1} {2}$. Magically, ${BT}^2 = 2 \cdot {BM}^2$, and we conclude $\frac {BT} {BM} = \sqrt 2$. Done!

Different solution from post #8. IMO ShortList 2015 G4 wrote: Let $ABC$ be an acute triangle and let $M$ be the midpoint of $AC$. A circle $\omega$ passing through $B$ and $M$ meets the sides $AB$ and $BC$ at points $P$ and $Q$ respectively. Let $T$ be the point such that $BPTQ$ is a parallelogram. Suppose that $T$ lies on the circumcircle of $ABC$. Determine all possible values of $\frac{BT}{BM}$. Henceforth $\Gamma$ denotes the circumcircle of triangle $ABC$. Let $K$ be the midpoint of $\overline{PQ}$, $L=\overline{TM} \cap \Gamma$, $N=\overline{BT} \cap \omega$ and $R=\overline{BM} \cap \Gamma$ with $B \not \in \{L, R, N\}$. Let $X \ne B$ be the intersection of $\omega$ and $\Gamma$; then it is the spiral center for the similarity $\mathcal{S}: \omega \mapsto \Gamma$. Notice that $P \mapsto A, Q \mapsto C$ so $K \mapsto M$. Also, $M \mapsto R$ since $\triangle PMQ \sim \triangle ARC$ and $N \mapsto T$ since $\triangle PNQ \sim \triangle ATC$. As $B, K, N$ are collinear; $B \mapsto L$. Therefore, $$\measuredangle BMK=\measuredangle LNM=\measuredangle LNB=\measuredangle MTB,$$ so $\overline{BM}$ is tangent to the circumcircle of triangle $MKT$. Finally, we have $$BM^2=BK \cdot BT=\frac{1}{2}BT^2 \Longrightarrow \frac{BT}{BM}=\sqrt{2}.$$

$A$-index everything. Let $M'$ be the midpoint of $\overline{PQ},$ $U$ be the second intersection of line $AM$ with $\odot (ABC),$ and $A'$ be the second intersection of the tangent to $\omega$ at $A$ with $\odot (ABC).$ Lemma: Let $\mathcal{S}$ be the spiral similarity sending $PQ\to BC.$ Then $\mathcal{S}$ sends $R\in \omega$ to $AR\cap \odot (ABC).$ Proof: Just angle chasing. Thus $A\to A',M'\to M,M\to U$ under $\mathcal{S}.$ Then the preimage of $T$ under $\mathcal{S}$ is $AT\cap \omega,$ so $AM'\mathcal{S}^{-1}(T)$ maps to $A'MT$ and so $A'MT$ is a line. Then PoP at $M$ gives $\triangle A'MU\sim \triangle AMT.$ But $\triangle AM'M\sim \triangle A'MU$ because of $\mathcal{S},$ so we conclude $\triangle AM'M\sim \triangle AMT\implies AM^2=AM'\cdot AT\implies \tfrac{AT}{AM}=\sqrt{2}.$

As $P$ varies linearly on $BA$, we have that $Q$ varies linearly on $BC$ due to spiral similarity. This means that $T=P+Q-B$ varies along a line. In particular, this line passes through $X$ and $Y$, where $X$ is the intersection of the circle through $B$ and $M$ tangent to $BC$, and $Y$ is defined similarly (this is by taking the limiting cases of $P=B$ and $Q=B$). Thus, the $T$ given in the problem is found by intersecting line $XY$ with $(BAC)$. We now apply a $\sqrt{BA\cdot BC}$ inversion followed by a reflection about the angle bisector of $\angle ABC$ to get the following new problem. Restated Problem wrote: Let $BAC$ be a triangle, and let $K$ be the harmonic conjugate of $B$ in $AC$. Let $X$ and $Y$ be on $BA$ and $BC$ such that $BYKX$ is a parallelogram, and let $T$ be an intersection of $(BAC)$ and $AC$. Find all possible values of $BK/BT$. [asy][asy] unitsize(2.5inches); pair B=dir(100); pair A=dir(195); pair C=dir(-15); pair H=foot(B,A,C); pair K=2*foot(0,2*A*C/(A+C),B)-B; pair X=extension(K,B+K-A,B,C); pair Y=extension(K,B+K-C,B,A); pair U=foot(H,B,C); pair V=foot(H,A,B); pair HH=2*circumcenter(B,X,Y)-B; pair T1=intersectionpoints(circumcircle(B,X,Y),A--C)[0]; pair T2=intersectionpoints(circumcircle(B,X,Y),A--C)[1]; pair M=(A+C)/2; draw(circumcircle(A,B,C)); draw(circumcircle(B,X,Y)); draw(B--H); draw(A--B--C--cycle); draw(K--X); draw(K--Y); draw(U--H--V); draw(H--HH); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$H$",H,dir(H)); dot("$K$",K,dir(K)); dot("$U$",U,dir(U)); dot("$V$",V,dir(V)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$H'$",HH,dir(HH)); dot("$T_1$",T1,dir(240)); dot("$T_2$",T2,dir(-60)); dot("$M$",M,dir(M)); [/asy][/asy] We'll solve this restated problem. Let $H$ be the foot from $B$ to $AC$, and let $H'$ denote the second intersection of $BH$ with $(BXY)$. Let $U$ and $V$ be the feet from $H$ to $BA$ and $BC$, and let $\{T_1,T_2\}=(BXY)\cap AC$. Note that the diameter line of $\triangle BUV$ is $BH$, which is the altitude line of $\triangle BAC$. Thus, $UV$ and $AC$ are anti-parallel. Furthermore, by the law of sines on $\triangle BYK$, we have $$YK=\frac{BK}{\sin B}\sin\angle ABK=\frac{BK\cdot AK}{AC},$$ and similarly $XK=\frac{BK\cdot CK}{AC}$. Thus, $$\frac{BY\cdot BA}{BX\cdot BC}=\frac{XK\cdot BA}{YK\cdot BC}=\frac{BA\cdot CK}{AK\cdot BC}=1,$$ as $(AC;KB)=-1$, so $YXCA$ is cyclic, so $YX$ is also anti-parallel to $AC$. This implies that $BH'$ is a diameter of $(BXY)$ and that $UV\parallel XY$. Let $\alpha=BX/BU=BY/BV=BH'/BH$ be the scale factor in the similarity between $\triangle BUV$ and $\triangle BXY$. Note that $BU=BC-BC\cos^2 C=BC\sin^2 C$. Thus, $$\alpha=\frac{YK}{BC\sin^2 C}=\frac{BK\cdot AK}{AC\cdot BC\cdot AB^2}(2R)^2,$$ where $R$ is the circumradius of $(BAC)$. We also have the similarity $\triangle BT_1H\sim\triangle BH'T_1$ with scale factor $\sqrt{\alpha}$, so we have $BT_1^2=BH^2\cdot\alpha$. Thus, $$BT_1^2=\frac{BK\cdot AK\cdot BH^2}{AC\cdot BC\cdot AB^2}(2R)^2,$$ By computing the area of $\triangle BAC$ in two ways, we have $$AB\cdot BC\cdot CA=(2R)\cdot AC\cdot BH,$$ so it's not hard to see that $$BT_1^2=\frac{BK\cdot AK\cdot BC}{AC}.$$ Thus, $$(BK/BT_1)^2=\frac{BK\cdot AC}{AK\cdot BC}.$$ Note that $AK$ and $AM$ are isogonal, so $\triangle BAK\sim\triangle BCM$, so $\frac{BK\cdot CM}{AK\cdot BC}=1$, so $(BK/BT_1)^2=2$, so the answer is $\boxed{\sqrt{2}}$.

We use bary wrt $ABC$. $A=(1,0,0),B=(0,1,0),C=(0,0,1)$, and $M=(1/2,0,1/2)$. Let $T=(k,j,m)$, so that $a^2jm + b^2km + c^2kj=0$. We will now get $P$ and $Q$ in terms of $T$, and then enforce that $BPMQ$ is cyclic to get an equation. Let $Q=(0,r,1-r)$. Since $QT\parallel AB$, then $[BQA]=[BTA]$. So $$\begin{vmatrix} 0&1&0\\0&r&1-r\\1&0&0\end{vmatrix} = \begin{vmatrix}0&1&0\\k&j&m\\1&0&0\end{vmatrix} \implies r=1-m \implies Q=(0,1-m,m).$$ Similarly, $P=(k,1-k,0)$. Now, $M\in (BPQ)$. It is not hard to get that the equation of $(BPQ)$ is $$-a^2yz-b^2xz-c^2xy + (c^2(1-k)x+a^2(1-m)z)(x+y+z) = 0.$$ Since $M=(1:0:1)$ is on this circle, $$-b^2 + 2(c^2(1-k) + a^2(1-m))=0 \implies c^2k + a^2m = \frac{2a^2+2c^2-b^2}{2}.$$ The displacement vector for $BT$ is $\overleftrightarrow{BT} = (k,j-1,m)$, so $$\begin{align*} BT^2 &= -a^2(j-1)m - b^2km - c^2(j-1)k \\ &= (-a^2jm-b^2km-c^2jk) + (a^2m+c^2k) \\ &= \frac{2a^2+2c^2-b^2}{2}. \end{align*}$$ It is well-known that $$BM^2 = \frac{2a^2+2c^2-b^2}{4} \implies \frac{BT}{BM} = \boxed{\sqrt2}.$$

wtf why why? Why? WHY? WHY?!!!!!!!!!!

DottedCaculator wrote:

How do you learn to make diagrams like this also this got liked by tapir orz orz orz

oh, great, here we go again. i am now thoroughly convinced that in this white wasteland of geometry, bary is just problem suicide - your last resort, unless it's not. We use barycentric coordinates. Preliminaries: Let $ABC$ be the reference triangle, and $P = (p, 1-p, 0), Q = (0, 1-q, q)$. The circle through $B, M$ is characterized by $$\begin{align*}-a^2yz-b^2zx-c^2xy+(x+y+z)(ux+wz) &= 0,\\ u+w &= \frac{b^2}{2}.\end{align*}$$ So after plugging in $P$ and $Q$ and heavily simplifying, $$\begin{align*}c^2p &= c^2-u \\ a^2q &= a^2-w.\end{align*}$$ Actual Problem: By distance formula, we can now proceed: $$\begin{align*}\frac{BT^2}{BM^2} &= \frac{a^2p-wp+a^2q-wq+c^2p-up+c^2q-uq-b^2pq}{\frac 12\left(a^2+c^2-\frac{b^2}{2}\right)} \\ &= \frac{(a^2-w+c^2-u)(p+q)+a^2q-a^2pq-a^2q^2+c^2p-c^2p^2-c^2pq}{\frac 12\left(a^2+c^2-\frac{b^2}{2}\right)} \\ &= \frac{\left(a^2+c^2-\frac{b^2}{2}\right)(p+q)+(a^2q+c^2p)(1-p-q)}{\frac 12\left(a^2+c^2-\frac{b^2}{2}\right)} \\ &= \frac{\left(a^2+c^2-\frac{b^2}{2}\right)(p+q)+\left(a^2+c^2-\frac{b^2}{2}\right)(1-p-q)}{\frac 12\left(a^2+c^2-\frac{b^2}{2}\right)} \\ &= \frac{\left(a^2+c^2-\frac{b^2}{2}\right)}{\frac 12 \left(a^2+c^2-\frac{b^2}{2}\right)} = 2. \end{align*}$$ Since length is positive, the only possible value is $\sqrt 2$, done. $\square$

G #1/100 Let the midpoint of $BT$ be $N$, Let $PQ$ intersect $AC$ at $D$ and the Miquel point of $APQC$ be $R$ Claim: $DRNM$ cyclic Proof: Due to the definition of a Miquel point, $APRD$ cyclic, $QCDR$ cyclic So, $R$ is the center of the spiral similarity sending $QC$ to $PA$, which means that it is the center of the spiral similarity sending $NM$ to $PA$ Hence, our claim must be true. Claim: $MB$ tangent to $MNRT$ Proof: $$\angle BMR=\angle BPR$$ $$=180-\angle APR$$ $$=\angle MDR$$ Hence, our claim must be true. Using the tangency, we have $MB^2=(NB)(BT)=\frac{1}{2} BT^2$ So, $$\frac{BT}{MB}=\sqrt{2}$$

Solved with a hint from Evan Chen's solution. Let $X$ be the second intersect of $\omega$. Set $T=(p,q,r)$. The parallelogram condition yields $P=(p:q+r:0)$ and $Q=(0:p+q:r)$. Then, we have $$b^2=qc^2+rc^2+a^2p+a^2q.$$ By distance formula, we get $BT^2=a^2r+c^2p$, which gives us $BT^2=2BM^2$, so $\sqrt2$ is the only value.

We will use barycentric coordinates. Let $$T=(r,s,t)$$ for which $r+s+t=1.$ Then, from the parallelogram, $$P=(r,s+t,0),Q=(0,r+s,t).$$ Consider the equation of $(BQMP)$. By putting in $B$, $v=0$. By putting in $P$, we have $u=c^2(1-r)$. By plugging in $Q$, we have $w=a^2(1-t).$ Therefore, the equation of the circle is $$(x+y+z)(c^2(1-r)x+a^2(1-t)z)=a^2yz+b^2xz+c^2xy.$$ Plugging in $(1/2,0,1/2)$ for $M$ reveals that $$b^2=2(c^2(1-r)+a^2(1-t)) (*).$$ We will return to the starred equation later. For now, note that $$\overrightarrow{BT}=(r,s-1,t),$$ so $$BT^2=-(a^2(s-1)t+b^2rt+c^2r(s-1)).$$ Let's expand this: $$-(\mathbf{a^2st+b^2rt+c^2rs}-a^2t-c^2r).$$ Since $(r,s,t)$ lies on the circumcircle, the bolded things sum to 0, so this is just $$BT^2=a^2t+c^2r.$$ Let's now go back to the starred equation: $$b^2=2(c^2(1-r)+a^2(1-t)).$$ Expanding, this becomes $$b^2=2(a^2+c^2-a^2t-c^2r)$$ $$b^2=2(a^2+c^2-BT^2)$$ $$BT^2=a^2+c^2-\frac{1}{2}b^2.$$ It is well known that $$BM^2=\frac{1}{2}a^2+\frac{1}{2}c^2-\frac{1}{4}b^2,$$ so our answer is $\sqrt{2}$ and we are done.

Similar bary bash: $A=(1,0,0),B=(0,1,0),C=(0,0,1),M=(\frac{1}{2},0,\frac{1}{2}),T=(m,n,k)$. Then equation of $PT$ and $QT$ is: $$PT:(n+k)x-m(y+z)=0 \implies P=(m,1-m,0) \qquad QT:(m+n)z-k(x+y)=0 \implies Q=(0,1-k,k)$$ Let $(BPMQ):-a^2yz-b^2zx-c^2xy+(x+y+z)(ux+wz)=0.$ By plugging coordinates of $M,P$and $Q$ we get: $$u+w=\frac{b^2}{2} \qquad c^2m=c^2-u \qquad a^2k=a^2-w$$ Also since $T \in (ABC)$ we have $a^2nk+b^2mk+c^2mn=0$. Thus by distance formula: $$BT^2=-a^2(n-1)k-b^2mk-c^2m(n-1)=-a^2nk-b^2mk-c^2mn+a^2k+c^2m=a^2k+c^2m=a^2+c^2-(u+w)=a^2+c^2-\frac{b^2}{2}$$ On the other hand $BM^2=\frac{1}{2}(a^2+c^2-\frac{b^2}{2}).$ Hence $\frac{BT}{BM}=\sqrt 2 \qquad \blacksquare$

Let $X$ be $BM$ intersect $(ABC)$. Let $Z$ be center of $BPTQ$ and $Y$ be $BZ$ intersect $(BPQ)$. Let $(ABC)$ and $(BPQ)$ intersect again at $E$ then $E$ is center of spiral similarity taking $ATXCM$ to $PYMQZ$. Clearly, $\angle ZMB$ is the angle of the spin about $E$, since $ZM$ is taken to $MX$. Since $ZY$ is taken to $MT$, $\angle ZTM$ is equal to that. Thus, $BM$ is tangent to $ZMT$ and so the answer is just $\sqrt{2}$.

Great. The answer is $\sqrt{2}$ only. Let $E$ and $F$ lie on $BA$ and $BC$ such that $M$ is the $B$-Dumpty point of $BEF$, and let $K$ be the reflection of $B$ over $M$, hence on $(BEF)$. We now claim that $T$ lies on $EF$. Let $P'$ and $Q'$ denote the images of $P$ and $Q$ under a scale 2 homothety at $B$, so it suffices to prive $EF$ bisects $P'Q'$. But applying phantom points we observe that the intersection of $P'Q'$ and $EF$ is on $(KEP')$ and $(KFQ')$. The result is true after a very brief law of sines computation. Take an inversion at $B$ with radius $BM\sqrt{2}$. $BEF$ becomes through a line antiparallel to $EF$ through $M$, whence $(E, A)$ amd $(F, C)$ swap. Hence line $EF$ maps to $(ABC)$. But this implies if $T$ lies on said circumcircle, $BT = BM\sqrt{2}$, as desired.

Actually trivial by bary. The answer is $\sqrt{2}$ only. We set $A=(1,0,0)$ and etc. Then, $M=(1:0:1)$. Now, let $T=(r,s,t)$ for real numbers $r,s,t$ such that $r+s+t=1$. Now, since $\overline{PT} \parallel \overline{BC}$, $P=(r,t_1,0)$ for some real number $t_1$ which then implies $t_1=s+t$ so $P=(r,s+t,0)$. Similarly, $Q=(0,r+s,t)$. Now, by Stewart's theorem, we can compute $BM^2 = \frac{2a^2-b^2+2c^2}{4}$. We consider the displacement vector $\overrightarrow{BT} = (r,s-1,t)=(r,-(r+t),t)$. Then, by the barycentric distance formula, we have that $$BT^2 = a^2(r+t)t-b^2rt+c^2(r+t)t = a^2t+c^2r$$ where the second equality follows from the fact that $a^2st+b^2rt+c^2rs=0$ since $T$ lies on the circumcircle. Now, since $BPMQ$ is cyclic, we consider the equation of this circle $$-a^2yz-b^2xz-c^2xy+(ux+vy+wz)(x+y+z)=0$$ where $v=0$ immediately since $B$ lies on the circle. Further, since $P$ lies on the circle we have $$\begin{align*} -c^2r(s+t) + ur &=0\\ u &= c^2(s+t) \end{align*}$$ similarly, we also have that since $Q$ lies on the circle, $w=a^2(r+s)$. Thus, the equation of the circle is simply, $$-a^2yz-b^2xz-c^2xy+(c^2(s+t)x+a^2(r+s)z)(x+y+z)=0$$ which since $M$ lies on this circle implies, $$\begin{align*} -b^2 + 2(c^2(s+t)+a^2(r+s)) &=0\\ -b^2 + 2(c^2 (1-r)+ a^2 (1-t)) &= 0\\ c^2 + a^2 - (a^2t+c^2r) &= \frac{b^2}{2}\\ a^2t + c^2 r &= \frac{2a^2 - b^2 + 2c^2}{2} \end{align*}$$ But this is precisely the length of $BT^2$! So, we have that $$BT^2 = a^2t + c^2 r = \frac{2a^2 - b^2 + 2c^2}{2} = 2\left(\frac{2a^2-b^2+2c^2}{4}\right) = 2BM^2$$ Thus, $\frac{BT}{BM} = \sqrt{2}$ which finishes the problem.

I changed it to being $A$-centered for convenience, $P$ lies on $AB$ and $Q$ lies on $AC$. Let $(A)$ be the circle with centre $A$ and radius $0$. Denote by $Pow_{\omega}(P)$ as the power of a point $P$ wrt circle $\omega$. Define $f(X)=Pow_{(A)}(X)-Pow_{(ABC)}(X)$. By linearity of pop, we have that $f(A)+f(T)=f(P)+f(Q)$. So $$AT^2=PA^2-(-PA \cdot PB)+QA^2-(-QA \cdot QB)=PA \cdot AB + QA \cdot AC.$$ Now let $R \neq M$ be the second intersection of $\omega$ with $BC$, then $$PB \cdot AB + QC \cdot AC = BM \cdot BR + CM \cdot CR = \frac{BC^2}{2}.$$ It follows that $AT^2+\frac{BC^2}{2}=AB^2+AC^2 \implies AT^2=AB^2+AC^2-\frac{BC^2}{2}$. By Stewart's theorem, $\frac{BC}{2}(AB^2 +AC^2)=BC \cdot (AM^2+ \frac{BC^2}{4}) \implies 2AM^2=AB^2+AC^2-\frac{BC^2}{2}=AT^2$. Hence $\frac{AT}{AM}=\sqrt{2}$, so we are done. $\square$

WHAT We employ barycentric coordinates wrt $ABC$. The equation of the circle is $$-a^2yz-b^2xz-c^2xy+(ux+wz)(x+y+z)=0$$ where $u+w=\frac{b^2}{2}$. Let $P=(p,1-p,0)$ and $Q=(0,1-q,q)$. Then $$-c^2p(1-p)+up=0\Longrightarrow c^2(1-p)=u$$ $$-a^2q(1-q)+wq=0\Longrightarrow a^2(1-p)=w$$ Therefore, $$c^2(1-p)+a^2(1-q)=\frac{b^2}{2}\Longrightarrow c^2p+a^2q=c^2+a^2-\frac{b^2}{2}.$$ Also, note that $T=(p,1-p-q,q)$, so $$a^2(1-p-q)q+b^2pq+c^2(1-p-q)p=0$$ $$-a^2(-p-q)q-b^2pq-c^2(-p-q)p=a^2q+c^2p=c^2+a^2-\frac{b^2}{2}.$$ That looks suspiciously like the distance formula! Indeed, we get that $$BT=\sqrt{c^2+a^2-\frac{b^2}{2}}.$$ Also, it is well known that $$BM=\frac{1}{\sqrt2}\cdot\sqrt{c^2+a^2-\frac{b^2}{2}}.$$ Therefore, $\frac{BT}{BM}=\boxed{\sqrt2}$.