Let $O$ be the common center of the concentric circles $c_1, c_2$. The tangency point B is the midpoint of the chord AC, because AC is perpendicular to the radius OB of the circle $c_2$, and O is also the center of the circle $c_1$. The power of the point A to the circle $c_2$ is $AE \cdot AF = AB^2$. But since B is the midpoint of AC and D the midpont of AB, $AD \cdot AC = \frac{AB}{2} \cdot 2 AB = AB^2$ as well. Hence, the quadrilateral CDEF is cyclic. The intersection M of the perpendicular bisectors of its diagonals CE, DF is its circumcenter. If this circumcenter is to be on one of its side CD, it must be the midpoint of this side, $DM = MC = \frac{DC}{2}$. Since $DC = \frac 3 2 AB$, $DM = MC = \frac 3 4 AB$ and $AM = AD + DM = \frac{AB}{2} + \frac 3 4 AB = \frac 5 4 AB$ and $\frac{AM}{MC} = \frac 5 3$.

Quote: But since M is the midpoint of AC But since B is the midpoint of AC.

Problem: Two circles are concentric. A chord $AC$ of the outer circle touches the inner circle at $Q$. $P$ is the midpoint of $AQ$. A line through $A$ intersects the inner circle at $R$ and $S$. The perpendicular bisectors of $PR$ and $CS$ meet at $T$ on the line $AC$. What is the ratio $AT/TC$? By Power of a Point we have $(AR)(AS)=AQ^2 = \frac{1}{2}2(AQ)^2 = (AP)(AC)$, so $\triangle ARP\sim\triangle ACS$. Hence we have $\angle ACS=\angle ARP\implies PRSC$ is cyclic. Hence $T$ must be the center of its circumcircle and must also lie on the perpendicular bisector of $CP$. Hence, $T$ is the midpoint of $CP$. This yields $CT=\frac{3}{8}(CA)\implies\frac{AT}{TC}=\frac{5}{3}$.

You guys are wrong its actually 3/5...

Why, maplestory? Is there something wrong with the other solutions?

yes, they assumed that M is not on AB but rather reflected across line OB (O is the center of the concentric circles) so when they found AM/MC they actually found MC/AM, now they have to take the reciprocal to get 3/5.

I don't follow

can any one please give a bash type solution as well I tried both complex and bary bash but the calculations are getting huge(alas!)

Between I agree with maplestory wrote: You guys are wrong its actually 3/5... The answer is coming 3/5 See p8 in the attachment

Yufei probably just made a typo. Note that from $\tfrac{MC}{AC}=\tfrac38$ we actually get \[\dfrac{AM}{MC}=\dfrac{AC-MC}{MC}=\dfrac{AC}{MC}-1=\dfrac83-1=\dfrac53,\]not $\tfrac35$ as he suggests.

Wait it says that M is on AB, so I took it to be M on segment AB. However, according to the solution, M is on segment BC. Did they mess up the problem or did they mean line AB?

Phie11 wrote: Wait it says that M is on AB, so I took it to be M on segment AB. However, according to the solution, M is on segment BC. Did they mess up the problem or did they mean line AB? I believe they are referring to line $AB$, not segment $AB$ because that is how they defined point $C$ as well.

Some wishful thinking tells us that M should be the midpoint of $\overline{DC}$. Since $M$ is the intersection of the perpendicular bisectors of $\overline{DE}$ and $\overline{CF}$, it suffices to show that quadrilateral $DEFC$ is cyclic. However notice that $\overline{AC}$ is the radical axis of $C_1$ and $C_2$, hence by PoP, we have that $\overline{AE}\cdot\overline{AF}=\overline{AD}\cdot\overline{AC}$, implying that $DEFC$ is cyclic. Through direct computation, we see that $\frac{\overline{AM}}{\overline{MC}}=\frac{5}{3}$. $\square$

Umm... $\overline{AC}$ is not the radical axis of $C_1$ and $C_2$; the power of point $A$ with respect to $C_1$ is $0$ and with respect to $C_2$ is $\overline{AB}^2$. Not the same it seems. Correct me if I'm wrong; if it actually is the radical axis then the solution would be quite simple and elegant.

Delray wrote: However notice that $\overline{AC}$ is the radical axis of $C_1$ and $C_2$, hence by PoP, we have that $\overline{AE}\cdot\overline{AF}=\overline{AD}\cdot\overline{AC}$ amackenzie1 wrote: $\overline{AC}$ is not the radical axis of $C_1$ and $C_2$ Correct me if I'm wrong; if it actually is the radical axis then the solution would be quite simple and elegant. Yes, you are correct in your objection. The equality \[ AE\cdot AF = AD\cdot AC \]is true but does not follow from the 1st poster's logic. Instead, the equality is achieved as follows: \[ AE\cdot AF = AB^2 = \frac12AB\cdot AC = AD\cdot AC, \]where the first equality follows from Radical Axis on $A$ w.r.t. $\mathcal{C}_2$.

By Power of a Point, $AE \cdot AF=AB^2=AD \cdot AC$. Therefore, $CDEF$ is cyclic, and $M$ is the center of this circle. Thus, $M$ is midpoint of $CD$. So, $MC=\frac{1}2CD=\frac{3}8AB$ and $AM=\frac{5}8AB$. Therefore, $\frac{AM}{MC}=\frac{5}3$. $\square$

By power of a point we have \[ AE \cdot AF = AB^2 = \left( \frac{1}{2} AB \right) \cdot \left( 2AB \right) = AD \cdot AC \]and hence $CDEF$ is cyclic. Then $M$ is the circumcenter of quadrilateral $CDEF$. [asy][asy] pair A = dir(170); pair C = dir(40); pair B = midpoint(A--C); pair D = midpoint(A--B); pair M = midpoint(C--D); pair O = origin; draw(unitcircle, heavycyan); draw(CP(O, B), heavycyan); pair E = IP(CP(M, D), CP(O, B)); pair F = OP(CP(M, D), CP(O, B)); draw(CP(M, D), red); draw(A--C, heavygreen); draw(A--F, heavygreen); draw(D--E, red); draw(C--F, red); pair U = midpoint(D--E); pair V = midpoint(C--F); draw(U--M--V, red+dotted); dot("$A$", A, dir(A)); dot("$C$", C, dir(C)); dot("$B$", B, dir(B)); dot("$D$", D, dir(D)); dot("$M$", M, dir(M)); dot("$O$", O, dir(45)); dot("$E$", E, dir(220)); dot("$F$", F, dir(F)); /* TSQ Source: A = dir 170 C = dir 40 B = midpoint A--C D = midpoint A--B M = midpoint C--D O = origin R45 unitcircle 0.1 lightcyan / heavycyan CP O B heavycyan E = IP CP M D CP O B R220 F = OP CP M D CP O B CP M D 0.1 lightred / red A--C heavygreen A--F heavygreen D--E red C--F red U := midpoint D--E V := midpoint C--F U--M--V red dotted */ [/asy][/asy] Thus $M$ is the midpoint of $\overline{CD}$ (and we are given already that $B$ is the midpoint of $\overline{AC}$, $D$ is the midpoint of $\overline{AB}$). Thus a quick computation along $\overline{AC}$ gives $AM/MC = 5/3$.

Headsolving without coherency :D Write $AC\cdot AD=AB^2=AE\cdot AF$ so that $CDEF$ is cyclic. It follows that $M$ is the center with $MC=MD$, whence it's clear that $\tfrac{AM}{MC}=\tfrac{5}{3}$. $\square$

[asy][asy] size(250,250); draw(circle((0,0),5)); draw((-4,3)--(4,3)); draw(circle((0,0),3)); label("$A$", (-4,3), NW); filldraw(circle((-4,3),0.05),black); label("$B$", (0,3), N); filldraw(circle((0,3),0.05),black); label("$C$", (4,3), NE); filldraw(circle((4,3),0.05),black); label("$D$", (-2,3), NW); filldraw(circle((-2,3),0.05),black); label("$M$", (1,3), N); filldraw(circle((1,3),0.05),black); label("$E$", (-1.918677,2.306225775), W); filldraw(circle((-1.918677,2.306225775),0.05),black); label("$F$", (2.918677,0.6937742252), SE); filldraw(circle((2.918677,0.6937742252),0.05),black); draw(circle((1,3),3)); draw((-4,3)--(2.918677,0.6937742252)); [/asy][/asy] I claim that the ratio $\frac{AM}{MC}$ is always $\frac{5}{3}$. 1. $B$ is the midpoint of $AC$. Note that since $AC$ is tangent to $\mathcal C_2$, we have that $\angle OBA=90$. However, since $AC$ is also a chord of $\mathcal C_1$, which has center $O$, this also means that $B$ is the midpoint of $AC$. 2. (Power of a Point Converse) $CDEF$ is cyclic. Note that the power of $A$ with respect to $\mathcal C_2$ is \[AB^2=BC^2=AD(AC),\]since by \textbf{(1)}, $B$ is the midpoint of $AC$, and by problem conditions, $D$ is the midpoint of $AB$. However, note by Power of a Point on $A$ with respect to $\mathcal C_2$, we have \[AB^2=AE(AF) \iff AE(AF)=AD(AC).\]Finally, using the converse of Power of a Point, we have that $CDEF$ must be cyclic. 2. (Similar Triangles) $CDEF$ is cyclic. Similarly, as in the first proof of \textbf{(2)}, we find that \[AE(AF)=AC(AD),\]which actually gives us that \[\frac{AD}{AF}=\frac{AE}{AC},\]meaning that by SAS similarity, $\triangle ADE \sim \triangle AFC$. Through angle chasing, we then have that \[\angle AFC + \angle EDC = \angle AFC + (180-\angle ADE) = \angle AFC + (180-\angle AFC) = 180,\]meaning that $CDEF$ is cyclic. 3. (Finishing) $M$ is the midpoint of $CD$. Note that since $CDEF$ is cyclic, and the perpendicular bisectors of $DE$ and $CF$ intersect at $M$, $M$ must be the circumcenter of $(CDEF)$. However, since $M$ is on $CD$, this means that $M$ must be the midpoint of $CD$. Therefore, since $D$ is the midpoint of $AB$, and $B$ is the midpoint of $AC$, we find that $\frac{AM}{MC}=\frac{5}{3}$, finishing the problem.

Let the common centre be $O$ and the radius of $C_1$ be $R_1$ and of $C_2$ be $R_2$. Now from PoP we have: $AB^2 = AO^2-R_2^2 = R_1^2-R_2^2 = CO^2-R_2^2 = CB^2$. So: $AB=BC$. Again from a PoP: $AB.AC=2AB.\frac{AC}{2}=AB^2=AE.AF$. $ \Leftrightarrow DEFC$ is cyclic. Now we have that $M$ is the midpoint of $DC=3AD$, so $AM=\frac{5}{2}AD$ and because of $AC=4AD$, $MC=\frac{3}{2}AD$. So: $\frac{AM}{MC}=\frac{5}{3}$

We first claim that $CFED$ is cyclic. This is because $AB^2 = AE \cdot AF \Rightarrow AD \cdot AC = AE \cdot AF \Rightarrow \triangle ACF \sim \triangle AED$ This proves our claim. From the given perpendicularity constraints we can see that $M$ is the circumcenter of $CFED$. This gives our answer of $\boxed{\frac{5}{3}}$ . $\blacksquare$

Notice that \[AE \cdot AF = AB^2=AB \cdot BC = AD \cdot AC\] implying that $CDEF$ is cyclic. Hence, it is clear that $M$ must be the midpoint of $CD$ at which point we easily find the ratio to be $\boxed{\frac{5}{3}}$. [Refer to #48 for a diagram, I was too lazy to make one]

From Power of a point, $AD \cdot AC = AB^2 = AE \cdot AF$, so quadrilateral $CDEF$ is cyclic. Point $M$ must be the circumcenter of $CDEF$, so $M$ is the midpoint of $CD$, and the ratio $AM/MC = \frac{5}{3}$.

Claim: $DEFC$ is cyclic Proof: By PoP $$AE \cdot AF=AB^2=2AB \cdot \tfrac{AB}{2}=AD \cdot AC$$hence $DEFC$ is cyclic. $\square$ Since $M$ is the intersection of the perpendicular bisectors of $DE$, and $CF$, $M$ is the center of $(DEFC)$. So: $$MC=\frac{AC-AD}{2}=\frac{AC-\tfrac{1}{2}AB}{2}=\frac{AC-\tfrac{3}{4}AC}{2}=\frac{3}{8}AC.$$$$\frac{AM}{MC}=\frac{AC-MC}{MC}=\frac{AC-\tfrac{3}{8}AC}{\tfrac{3}{8}AC}=\frac{\tfrac{5}{8}AC}{\tfrac{3}{8}AC}=\frac{5}{3}.$$

Note $B$ is the midpoint of $\overline{AC}$ so \[AF\cdot AE=AB^2=(\tfrac{1}{2}AB)(2AB)=AD\cdot AC\]and $CDFE$ is cyclic. Hence, $M$ is the center of this circle and thus the midpoint of $\overline{CD}$. WLOG if $AD=2$, then $DB=2$ so $BC=4$ so $CM=3$ so $AM/CM=5/3$. $\square$

By power of a point $AE*AF=AB^2=AD*AC$, so $DEFC$ is cyclic. Because $M$ is the intersection of the perpendicular bisectors of $DE$ and $FC$, $M$ is the center of the circumcircle, so $MD=MC$ Let $AD=a, BM=x$ $MC=MD=a+x$, so $MC+MB=a+2x=2a$, so $x=\frac{a}{2}$. Then $\frac{AM}{MC}=\frac{2.5a}{1.5a}=\boxed{\frac{5}{3}}$

By Power of a Point with respect to $\mathcal{C}_2,$ we get $$AB^2 = AE \cdot AF.$$ Then, since $D$ is the midpoint of $\overline{AB},$ it follows that $$AD \cdot AC = AB^2 \implies AD \cdot AC = AE \cdot AF.$$ Hence, by the converse of Power of Point, quadrilateral $CDEF$ is cyclic. Since the perpendicular bisectors of $\overline{DE}$ and $\overline{CF}$ intersect at $M,$ this tells us that the circumcenter of $CDEF$ is $M,$ so $M$ is on the midpoint of $\overline{CD}.$ Therefore, if we let $AD=x,$ we then get \begin{align*} CD=3x &\implies DM=MC=1.5x \\ &\implies AM=2.5x \\ &\implies \frac{AM}{MC} = \frac{2.5x}{1.5x} = \boxed{\frac{5}{3}}. \end{align*}$\square$

[asy][asy] draw(circle((2,1),1)); draw(circle((2,1), sqrt(5))); pair O=(2,1), A=(0,0), B=(2,0), C=(4,0), D=(1,0), M=(5/2,0); draw(A--C); dot(D); dot(B); dot(M); dot(O); dot(A); dot(C); label("$A$", A, SW); label("$B$", B, S); label("$C$", C, SE); label("$D$", D, S); label("$M$", M, S); label("$O$", O, S); pair E=(2-2sqrt(5)/5, 1-sqrt(5)/5), F=(2+2sqrt(5)/5, 1+sqrt(5)/5); dot(E); dot(F); label("$E$", E, NW); label("$F$", F, NE); draw(A--F); pair G=D/2+E/2, H=F/2+C/2; draw(G--M--H); draw(D--E); draw(F--C); [/asy][/asy] We claim the answer is $\boxed{\dfrac{5}{3}}.$ Let $O$ be the center of both $C_2$ and $C_1.$ Since $AB$ is tangent to $C_2,$ $\angle OBA=90^{\circ},$ so $AC$ is bisected by $OB$ and $AB=BC.$ By power of a point, $AE\cdot AF=AB^2=\dfrac{AB}{2}\cdot 2AB=AD\cdot AC,$ so $DEFC$ is cyclic. Note that the perpendicular bisectors of chords $DE$ and $FC$ in the circumcircle of $DEC$ intersect at the circumcenter of $DEC,$ so $M$ is the circumcenter of $DEC$ and $DM=CM.$ As such, $\dfrac{AM}{MC}=\dfrac{AD+DM}{MC}=\dfrac{\left(AD+\dfrac{DC}{2}\right)}{\left(\dfrac{DC}{2}\right)}=\dfrac{\left(AD+\dfrac{3AD}{2}\right)}{\left(\dfrac{3AD}{2}\right)}=\dfrac{5}{3},$ as desired. $\Box$

We claim the requested ratio $\frac{AM}{MC} = \boxed{\frac 53}$. Let the length $AB = 2x$, then $AD = DB = x$ and $BC = 2x$. Additionally, $DC = 3x$. By Power of a Point, $AB^2 = AF \cdot FE = (2x)^2$. On the other hand, we compute \[ AD \cdot AC = x \cdot 4x = 4x^2.\]Since \[ AF * FE = (2x)^2 = 4x^2 \]from before, we have $AD \cdot AC=AF \cdot AE$ and quadrilateral $CDEF$ is cyclic by Converse of Power of a Point. Let $\omega$ be the circumcircle of quadrilateral $CDEF$. Then, all chords of $\omega$ must intersect at the center of $\omega$. We know that the perpendicular bisectors of $DE$ and $CF$ intersect at a point $M$ on $CD$, therefore this point $M$ must be the center of $\omega$. Finally, since $CD$ is a chord of $\omega$ and its perpendicular bisector intersects itself at its midpoint, $M$ must be the midpoint of $DC$. We now finish with \[\frac{AM}{MC} = \frac{x+\frac{3x}{2}}{\frac{3x}{2}} = \frac{\frac{5x}{2}}{\frac{3x}{2}} = \boxed{\frac 53} \].

$AD \cdot AC=AB \cdot AB = AE \cdot AF$ so $CDEF$ cyclic with $CD$ as a diameter. Thus $C$ is the midpoint of $DC$ and $MC=1.5AD,$ $AM=2.5AD$ so the ratio is $\dfrac{5}{3}.$

We claim that $CDEF$ is cyclic, namely by proving that $AE \cdot AF = AD \cdot AC$. Notice that $AD \cdot AC = \text{Pow}(B, C_1)$ using the fact that $1 \cdot 4 = 2 \cdot 2$. Also, $AE \cdot AF = \text{Pow}(F, C_1) = \text{Pow}(B, C_1)$ say by extending $AE \cap C_1 = X$ and noting that $AE = FX$. So it is cyclic and hence $M$ must be the midpoint of $DC$ from which we can extract $\boxed{\frac53}$

Note that $AE \cdot AF = AB^2 = AD \cdot (4AD) = AD \cdot AC$ by Power of Point. Therefore by Power of Point $FCDE$ is cyclic, so $M$ is its center. Hence $CM=MD$ and thus $\frac{AM}{MC} = \frac{AD+DM}{MC} = \frac{\frac14 AC+\frac12 DC}{\frac12 DC} = \frac{\frac14 AC + \frac12 \frac34 AC}{\frac12 \frac34 AC} = \frac53.$