Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

If I remember well, I already posted a solution for this problem. I don't understand way it isn't here anymore. Could someone, please, explain what is going on? Ciao sprmnt21

sprmnt21 wrote: If I remember well, I already posted a solution for this problem. I don't understand way it isn't here anymore. Could someone, please, explain what is going on? Ciao sprmnt21 Sorry, your solution was lost to an attack on the server. Please be so kind rewrite it. [Edit by Darij: Indeed, (at least) 3 messages from this thread were lost: IMO ShortList 1999, geometry problem 6 4 orl 86 Sat Nov 27, 2004 3:33 pm sprmnt21 |> 4 replies means 5 messages including the original post. The first two messages are here (these are the two posts by Orl), the other 3 were lost.]

{U,V} = o1^o2; E=AN^o2; F=BN^o2;p=DF^CE ; X = centre(o1); Y = centre(o2). AC*AM = AV*AU = AE*AN ==> MNEC cyclic; <ACE = <ANM = <ABM = <CDM ==> CE tangent to o1; <AEC = <AMN = <ABN = <EFN ==> EC tangent to o2; G= o2^XY, H = foot(Y) on CX; YH//CE ==> XH = r1 – r2 = XG ==> CXG == YXH ==> <CGX = 90° XY symmetry axes of CD and EF ==> <DGX = 90°.

sprmnt21 you have proved that $CD$ is parallel to $AB$. Can you post the proof?

According to this figure we only need to show that EF||AB, right? Then you can draw a tangent line throught the point M. And let X be a point on that line. Thus we have <EFM=<EMX(i) and <EMX=<ABM(ii), so from i and ii we have <EMX=<ABM, similarly <FEM=<BAM. Note: All typos used in this observation are according the figure. davron

It’s a nice problem , which with $INVERSION$, will be more nice. (i changed names of the circles from $\omega$ to $O$) Let me invert all of the shape(not just a part of it), round point $N$. $O,O_2$, turn into two parallel lines . the inversive form of $O_1$, is a circle which goes through the respect of $N$, to $O_1$, and is also tangent to $O$. the inversive form of $AB$, is acircle which goes through the intersection of $O_1,O_2$, and also through$N$. and it is known that its radius is equal to the radius of $O_1$. $AN,BM$ are tangent to the two equal circles , which goes through $N,M,A,B$. $C,D$ , are the intersection of circles $AM,BM$ with $O_1$. Now we should prove that: The circum circle of triangle $C,D,N$, is tangent to $O_2$. To prove it , we should say that, $\angle NDT=90$, or the other words $T,D,B$ should be collinear. We know that $\angle NMB=90$, then we draw $TB$, and intersect it with $O_1$ at $R$. And then we should prove , $RBNM$ is a cyclic quadrilateral. and then $\angle NRB=\angle NMB=90$. And it means that: $R=D$. Then we should prove that: $TR.TB=TM.TN$ Suppose centers of circles $O_1,NAB$ are $V,U$ respectively, and consider $d$, as the distance of these centers. And the two circles , intersect at $S,Q$. And also suppose that the radius of these two circles is $r$. We have these facts: $TU=\frac{d}{2}$ $TS^2=r^2-\frac{D^2}{4}$ $TM=r-\frac{d}{2}$ $TN=r+\frac{d}{2}$ $TM.TN=r^2-\frac{D^2}{4}=TS^2$ then we should prove:$TS^2=TR.TB$ if the recent equality holds, triangles $TSR,TSB$ will be similar to each other.and therefore we should have: $\angle SBT=\angle RST$. The places of $SQ$ , in circles $O_1,NAB$ , are equal, and $\angle SBT$=arc$\frac{SH}{2}$ And $\angle RST$=arc$\frac{QR}{2}$. We can prove the equality of the arcs $SH,QR$, instead of the equality of the arcs $QH,SR$, Now look at the shape, $T$ is the midpoint of $QS$ and $UV$. We draw two parallel radius from $U,V$. and let them intersects their circles at $W,X$ respectively. (NOTE:both of them are in to apposite side) then we have: .......................................................................................................................................................................................... $UW$ is parallel to $VX$, and then $\angle TVX=\angle TUW$ $TU=TV$ $WU=VX=r$ .......................................................................................................................................................................................... And conclude is : .......................................................................................................................................................................................... Triangles $WUT,TVX$ are equal. $\angle UTW=\angle VTX$ $WT=TX$ ………………………………………………………………………………………………………………………….. then $WX$ goes through$T$. Now, to prove the equality of the arcs $QH,RS$, we do like this: Consider a homotechy with center $T$, and the radius $(-1)$. $Q$, turns to $S$. and each point $W$, from circle $NAB$, turns to a point $X$, from circle $O_1$. Then point $R$, from $O_1$, turns to point $H$ , from $NAB$,and it means that: Arcs$QH,RS$ are to homothetic arc, and they are equal to each other: arc$QH$=arc$RS$. And every thing is ok. It was hard for me to find an inversive proof for it.

@Ashegh: I have no comments to you, you are really good in Inversion. I hope that you will teach me using it, too. davron

uew wellcom dear Darvron... hmmmm,,,, i know nothing about geometry, then how can i teach some thing if i dont know it?

ashegh wrote: i know nothing about geometry, yes, I think,too.

I have a fairly simple proof using coordinate geometry- let the center of $ \omega_{2}$ and have radius r. Let $ \omega{1}$ have radius R. So the diameter of $ \omega$ is r+2R, and its radius is R+r/2. Since the circles are symetric about the x-axis, I will just work with the top half of each for now. With some simplifying, the equations for the top parts of $ \omega_{2}$, $ \omega{1}$, and $ \omega$ are $ y^{2}=r^{2}-x^{2}$, $ y^{2}=R^{2}-(x-R)^{2}=2xR-x^{2}$, and $ y^{2}=(R+r/2)^{2}-(x-R+r/2)^{2}=(x+r)(2R-x)$, respectively. Now $ \omega_{2}$ and $ \omega_{1}$ intersect at $ r^{2}-x^{2}=2xR-x{2}$, or $ x=r^{2}/2R$. Since this is the same for the bottom also, the common chord is a vertical line, and it intersects $ \omega$ at $ x=r^{2}/2R$, so $ y^{2}=(r^{2}/2R+r)(2R-r^{2}/2R)$, or $ (1+r/2R)^{2}(r(2R-r)$. The sdquare of the line from $ (2R,0)$ to this point is $ y^{2}=r/(2R-r)(2R-x)^{2}$, which meets $ \omega_{1}$ at the point when $ x(2R-x)=r/(2R-r)(2R-x)^{2}$, or $ 2Rx-rx=2Rr-rx$. So the x-coordinate of point C is r. The same can be proven using the same methods for point D. Thus, CD is a vertical line where $ x=r$, and thus it is tangent to $ \omega_{2}$ at the point $ (r,0)$.

Let X,Y be the centres of $\Omega_{1}$, $\Omega_{2}$ respectively. Lemma: the tangent line to $\Omega_{1}$ through C is also tangent to $\Omega_{2}$. Let the tangent line to $\Omega_{1}$, $\Omega_{2}$ touch $\Omega_{1}$, $\Omega_{2}$ at C', E respectively. I will show that C' = C. Let the tangent line to $\Omega_{1}$, $\Omega_{2}$ intersect $\Omega$ at J and K. Let the midpoint of arc JK be A'. The homothety centered at M that takes $\Omega_{1}$ to $\Omega$ also takes C' to A'. Similarly, the homothety centered at N takes E to A'. Therefore, lines MC' and NE intersect on the circumcircle of $\Omega$, at A'. Since the homothety centered at N takes E to A', the measures of arcs EN and A'N are equal. Therefore, ∠KEN = ∠A'MN, meaning MC'EN is cyclic. Therefore, A' lies on the radical axis of $\Omega_{1}$, $\Omega_{2}$. However, since A' lies on the $\Omega$, this means A' = A. This means that the lines MC' and NE intersect at A. However, MA intersects $\Omega_{1}$ in only one place other than M, namely C, so C' = C. The homothety from M taking $\Omega_{1}$ to $\Omega$ also takes C to A and D to B. Therefore, CD is parallel to the radical axis, and since the radical axis is perpendicular to XY, CD is perpendicular to XY. Let CD intersect XY at L. Let $\Omega_{2}$ intersect XY at L'. Let the foot of the perpendicular from Y to XC be P. Since we have ▲XCL~▲XYP, and XY = XC, we have ▲XCL≅▲XYP. Therefore, we have XL = XP = XC - YE = XY - L'Y = XL'. Therefore, L = L', and therefore CD is tangent to $\Omega_{2}$.

By homothety $CD || AB$ and, if $MN$ is the common chord of $\Omega_1,\Omega_2$, then $A-M-N-B$ collinear. Easily $M\Omega_1N\Omega_2$ is a kite, so $\Omega_1\Omega_2\bot AB, \therefore\Omega_2$ is the midpoint of the arc $CD$. Let the tangent at $C$ to $\Omega_1$ intersect the circle $\Omega$ at $E,F$ and $AN$ at $P$. We see that $\angle ANM=\angle ACP$, so $MNPC$ is cycli, ınferring $AC\cdot AM=AP\cdot AN$; but $A$ has equal powers to the cırcles $\Omega_1,\Omega_2$, hence $P$ lies onto $\Omega_2$, i.e. $CP$ is the common tangent to the 2 circles. Further $\angle PC\Omega_2=\angle BC\Omega_2$ and, $BC$ is symmetrical of $CP$ through $C\Omega_2$, so it is tangent to the cırcle $\Omega_2$ as claımed. Best regards, sunken rock

My solution... $\mathrm{Let:\,}G_{1}\cap G_{2}=\left\{ P,Q\right\},\;NA\cap G_{2}=\left\{N,E\right\},\;NB\cap G_{2}=\left\{N,F\right\}\\ I\,\mathrm{is\,the\,midpoint\,of\,}CD\\ H\,\mathrm{is\,the\,foot\,of\,altitude\,from\,}G_{2}\,\mathrm{to\,}CG_{1}$ $\mathrm{First\,we\,can\,see\,that\,the\,tangents\,to\,}G\,\mathrm{at\,}M,N\,\mathrm{are\,concurrent\,with\,}AB\,(\mathrm{because\,they\,are\,3\,radical\,axises})\\ \mathrm{so\,}AB\,\mathrm{is\,symmidian\,in\,}\Delta MAN,\,\mathrm{now\,we\,have\,}AE.AN=AP.AQ=AC.AM\Leftrightarrow NECM\,\mathrm{is\,cyclic,}\\ \mathrm{and\,by\,hemothiety\,we\,have\,}CD\parallel AB,\,EF\parallel AB\Rightarrow \angle ECA=\angle ANM=\angle AMB=\angle CDM\Leftrightarrow \\ EC\,\mathrm{is\,tangent\,to\,}G_{1},\, \mathrm{similarly\,we\,can\,get\,}CE\,\mathrm{is\,tangent\,to\,}G_{2}\Rightarrow G_{2}HCE\,\mathrm{is\,rectangle}\Rightarrow G_{2}H=CE.\\ \mathrm{and\,since\,}G_{1}G_{2}=G_{1}C\,\mathrm{hence\,}CI=G_{2}H=CE\,(\mathrm{because\,we\,have\,}CI\perp G_{1}G_{2}),\,\mathrm{so\,we\,get:}\\ G_{2}E^{2}=CG_{2}^{2}-CE^{2}=CG_{2}^{2}-CI_{2}=G_{2}I^{2}\Leftrightarrow I\in G_{2},\,\mathrm{but\,since\,}CE=CI\,\mathrm{and\,}CE\,\mathrm{is\,tangent\,to\,}G_{2}\\ \mathrm{hence\,}CI\,\mathrm{is\,also\,tangent\,to\,}G_{2},\,\mathrm{and\,we\,are\,done}$

Since $A$ lies on the radical axis of $\Omega_1$ and $\Omega_2$, set $\varrho = \text{pow}(A, \Omega_1) = \text{pow}(A, \Omega_2).$ Then the inversion $\Psi$ with center $A$ and power $\varrho$ fixes $\Omega_1$ and $\Omega_2.$ Hence, $\Psi$ interchanges $\{M, C\}.$ Then since $\Omega$ is tangent to $\Omega_1$ at $M, \Psi(\Omega)$ is the line tangent to $\Omega_1$ at $C.$ However, $\Psi(\Omega)$ is also tangent to $\Omega_2$ because $\Omega$ and $\Omega_2$ are tangent. Therefore, $\Psi(\Omega)$ is a common tangent to $\Omega_1$ and $\Omega_2$ passing through $C.$ Similarly, we can show that there exists a common tangent to $\Omega_1$ and $\Omega_2$ passing through $D.$ Let these two tangents meet at $E$, and denote by $S$ and $T$ the points of tangency of $EC$ and $ED$ with $\Omega_2.$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(15cm); 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 = 125.90308049860157, xmax = 243.67692567282657, ymin = 88.52559844242623, ymax = 148.12940530451223; /* image dimensions */ Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax, Ticks(laxis, Step = 10., Size = 2, NoZero),EndArrow(6), above = true); yaxis(ymin, ymax, Ticks(laxis, Step = 10., Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw(circle((162.81796556704418,117.30338148064997), 22.272952473938865)); draw(circle((159.77238711367855,125.42492402295828), 13.59914123653413), linetype("2 2") + blue); draw(circle((173.27868183110525,123.8385301950535), 9.938660703570978), linetype("2 2") + blue); draw((171.274578812264,137.90848046917992)--(166.26990936151225,95.29955236055868)); draw((154.99740984006004,138.15819675260755)--(171.274578812264,137.90848046917992)); draw((154.99740984006004,138.15819675260755)--(166.26990936151225,95.29955236055868)); draw((164.9357204453974,138.0057281020608)--(209.94995472824428,119.53127330465229), red); draw((161.88003185707186,111.99010076372485)--(209.94995472824428,119.53127330465229), red); /* dots and labels */ dot((162.81796556704418,117.30338148064997),linewidth(3.pt) + dotstyle); label("$O$", (163.18106279722585,117.91785371634148), NW * labelscalefactor); dot((154.99740984006004,138.15819675260755),linewidth(3.pt) + dotstyle); label("$M$", (153.3495070261601,138.80990972985617), NW * labelscalefactor); dot((159.77238711367855,125.42492402295828),linewidth(3.pt) + dotstyle); label("$O_1$", (160.00628957948587,126.41805297674208), N * labelscalefactor); dot((173.27868183110525,123.8385301950535),linewidth(3.pt) + dotstyle); label("$O_2$", (173.72950284326512,124.47222423038531), E * labelscalefactor); dot((181.7076627244061,129.10438792965618),linewidth(3.pt) + dotstyle); label("$N$", (182.12729006438377,129.4904141552001), ESE * 1.5); dot((171.274578812264,137.90848046917992),linewidth(3.pt) + dotstyle); label("$A$", (171.6812620576264,138.50267361201037), NE * labelscalefactor); dot((166.26990936151225,95.29955236055868),linewidth(3.pt) + dotstyle); label("$B$", (166.2534239756839,93.44137632795909), S * labelscalefactor); dot((164.9357204453974,138.0057281020608),linewidth(3.pt) + dotstyle); label("$C$", (164.71724338645487,139.01473380842003), N * 2); dot((161.88003185707186,111.99010076372485),linewidth(3.pt) + dotstyle); label("$D$", (161.23523405086908,109.92971465235058), SW * labelscalefactor); dot((209.94995472824428,119.53127330465229),linewidth(3.pt) + dotstyle); label("$E$", (210.8026610633255,118.73715003059696), E * labelscalefactor); dot((177.05220084902166,133.03295953037014),linewidth(3.pt) + dotstyle); label("$S$", (175.7777436289038,133.8941318443233), ENE * 4); dot((174.8190141045,114.01995835041882),linewidth(3.pt) + dotstyle); label("$T$", (174.65121119680254,111.7731313594254), S * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $\Gamma \equiv \odot(ESO_2T)$ be the circle of diameter $\overline{EO_2}.$ Clearly the center $X$ of $\Gamma$ lies on the perpendicular bisector of $\overline{ST}.$ Hence, $O_1, O_2, X$ are collinear, implying that $\Omega_1$ and $\Gamma$ are tangent. Now consider the inversion $\Upsilon$ with center $E$ and power $EC \cdot ES = ED \cdot ET.$ Clearly $\Upsilon$ interchanges $\{C, S\}$ and $\{D, T\}$ and therefore $\Upsilon$ interchanges $\{\Omega_1, \Omega_2\}.$ Moreover, $\Upsilon(\Gamma) \equiv CD.$ Then since $\Omega_1$ and $\Gamma$ are tangent, it follows under inversion that $\Omega_2$ and $CD$ are tangent, as desired.

A nice solution: Let $O$ be the center of $\Omega_{2}$.Join $NA$ and $NB$. Let $NA\cap \Omega_{2}$ = $G$ and $NB\cap \Omega_{2}$ = $H$. Join $OG$ and $OH$. Also, join $CG$ and $DH$. Let the tangent from $C$ to $\Omega_{2}$ meet it at $X$. Join $DX$. Also, join$GX$, $HX$, $OD$ and $OC$. We shall show that $C$, $X$ and $D$ are collinear. Now, $\angle OGC$ = $\angle OXC$ = $90$°. So, $OGCX$ is cyclic. We have, $CG$ = $DH$, $OG$ = $OH$ and $\angle OGC$ = $\angle OHD$ = $90$°. By congruence, $\angle ODH$ = $\angle OCG$...($1$) Again, $OX$ = $OG$, or, $\angle OXG$ = $\angle OGX$, or, $\angle OCG$ = $\angle OCX$...($2$) ($1$) and ($2$) gives, $\angle ODH$ = $\angle OCX$ But, $\angle ODH$ = $\angle OCF$ (alternate segment theorem) Thus, $C$, $X$ and $D$ are collinear and hence the result follows.

Let $Z=(a,b), Z_1=(0,0), Z_2=(1,0)$ be the centers of circles $\Omega, \Omega_1, \Omega_2$. $Z, Z_1, M$ are collinear, and $Z, Z_2, N$ are collinear. $\angle BMA = \angle DMC$, so $\angle BZA = \angle DZ_1C$. The triangles $MZ_1C$ and $MZA$ are similar, so $\angle MCZ_1 = \angle MAZ$. So, $\angle MCD = \angle MAB$. Since $AB$ is perpendicular to the $x$-axis, that means that $CD$ is perpendicular to the $x$-axis. Let $P$ be the point where $CD$ intersects the $x$-axis. So the problem is to show that $|Z_2P| = |Z_2N|$. $|ZZ_1| + |Z_1Z_2| = \sqrt{a^2+b^2} + 1$, and $ZZ_2 = \sqrt {(a-1)^2 + b^2}$. So $|Z_2N| = \sqrt{a^2+b^2} + 1 - \sqrt {(a-1)^2 + b^2}$. Let $A^\prime$ and $B^\prime$ be the points of intersection of $\Omega_1$ and $\Omega_2$. By the law of cosines, $|Z_1Q| = \cos{(\angle A^\prime Z_1 Z_2)} = 1 - \dfrac {|Z_2N|^2} 2 $. So the distance of $Z$ from the common chord of circles $\Omega_1$ and $\Omega_2$ is $a - 1 + \dfrac {|Z_2N|^2} 2 $. $|Z_1P| = \dfrac 1 {|ZM|}\left (a - 1 + \dfrac {|Z_2N|^2} 2 \right )$, because triangles $BZA$ and $DZ_1C$ are similar. So $|Z_2P| = 1 + \dfrac 1 {|ZM|}\left (a - 1 + \dfrac {|Z_2N|^2} 2 \right ) =$ $\displaystyle 1+ \dfrac {2a-2 +|ZM|^2 - 2|ZM||ZZ_2| + |ZZ_2|^2} {2|ZM|} =$ $\displaystyle 1 - |ZZ_2| + \dfrac {|ZM|} 2 + \dfrac {2a-2+(a-1)^2+b^2} {2(\sqrt{a^2+b^2}+1)} =$ $1 - |ZZ_2| + \dfrac {|ZM|} 2 + \dfrac {a^2+b^2-1} {2(\sqrt{a^2+b^2}+1)} =$ $1 + \sqrt{a^2+b^2} - |ZZ_2| = |Z_2N|$.

Tumon2001 wrote: A nice solution: Let $O$ be the center of $\Omega_{2}$.Join $NA$ and $NB$. Let $NA\cap \Omega_{2}$ = $G$ and $NB\cap \Omega_{2}$ = $H$. Join $OG$ and $OH$. Also, join $CG$ and $DH$. Let the tangent from $C$ to $\Omega_{2}$ meet it at $X$. Join $DX$. Also, join$GX$, $HX$, $OD$ and $OC$... What do you mean by "joining" two line segments? Quote: But, $\angle ODH$ = $\angle OCF$ (alternate segment theorem) Thus, $C$, $X$ and $D$ are collinear and hence the result follows. And where is the point "F"?

My solution: Let $\Omega_1 \cap \Omega_2 = U,V$ and $NA \cap \Omega_2 = P,NB \cap \Omega_2 = Q$. Also, Let $MN \cap \Omega_1 = R,MN \cap \Omega_2 = S$. Lastly, Let $O_1$ and $O_2$ be the centers of $\Omega_1$ and $\Omega_2$ respectively, and let $O_1O_2 \cap CD = T$. We denote the tangent at a point $K$ to a circle $\alpha$ by $f_{\alpha}(K)$. Let $f_{\Omega}(M) \cap f_{\Omega}(N) = X, f_{\Omega_1}(C) \cap f_{\Omega_1}(D) = Y, f_{\Omega_2}(P) \cap f_{\Omega_2}(Q) = Z$. By Radical Axes Theorem on $\Omega, \Omega_1, \Omega_2$, we get that $MX,NX,UV$ concur $\Rightarrow X$ lies on $UV$ (i.e. $AB$) $\Rightarrow AMBN$ is harmonic. $\Rightarrow -1 = (A,B;M,N) \overset{\text{M}}{=} (C,D;M,R)$ and $-1 = (A,B;M,N) \overset{\text{N}}{=} (P,Q;S,N)$ $\Rightarrow CMDR$ and $PNQR$ are harmonic $\Rightarrow Y$ lies on $MR$ and $Z$ lies on $NS$ $\Rightarrow Y,Z,M,N$ are collinear. Now, By Reim's Theorem, $PQ \parallel AB \parallel CD$ $\Rightarrow PUVQ$ and $CUVD$ are isosceles trapeziums. And, As $O_1O_2$ is the perpendicular bisector of $UV$, so $O_1O_2$ is the perpendicular bisector of $CD$ and $PQ$ also. But, $YC = YD$ and $ZP = ZQ$ $\Rightarrow Y,Z$ lie on $O_1O_2$. As $Y,Z$ lie on $MN$ also, thus $Y = Z$, i.e. $CP$ and $DQ$ are common tangents of $\Omega_1$ and $\Omega_2$. This means that $O_2P \perp CZ$ and $O_2Q \perp DZ$. Also, $\angle ZCO_2 = \angle CDO_2 = \angle DCO_2 \Rightarrow O_2$ lies on the angle bisector of $\angle ZCD$. As $O_2T \perp CD$ and $O_2P \perp CZ$, we get that $O_2P = O_2T$ $\Rightarrow T$ lies on $\Omega_2$ $\Rightarrow CD$ is tangent to $\Omega_2$ at $T$. $\blacksquare$

Invert at $N$. We get the following: Let $N$ be a point and let $\Omega_2, \Omega$ be lines so that $N$ is on the opposite side of $\Omega$ as $\Omega_2$. Let $O$ be the reflection of $N$ over $\Omega_2$, and let $\Omega_1$ be a circle through $O$ tangent to $\Omega$ at $M$ and intersecting $\Omega_2$ at $X,Y$. Let $\Omega$ meet $(NXY)$ at $A,B$, and let $(MNA), (MNB)$ meet $\Omega_1$ again at $C,D$, Show that $(NCD)$ is tangent to $\Omega_2$. Note that now $(NXY)$, $(OXY)$ are now reflections over $\Omega_2$. Let $\Omega'$ be the reflection of $\Omega$ over $\Omega_2$. This is still too hard, so we invert again, this time at $X$: Let $X$ be a point on a fixed line $\Omega_2$. Let $\Omega, \Omega'$ be circles tangent to $\Omega_2$ at $X$ which are reflections of each other over $\Omega_2$. Let $N, O$ be reflections of each other over $\Omega_2$ where $N$ lies inside $\Omega$. Let the $N$-tangent to $\Omega'$ and the $O$-tangent to $\Omega$ meet at $Y$ (then $Y \in \Omega_2$ by symmetry) and let the $O$-tangent be tangent to $\Omega$ at $M$. Let the $N$-tangent to $\Omega'$ intersect $\Omega$ at $A,B$. Let $(MNA), (MNB)$ meet $YM$ again at $C,D$. Show that $(NCD)$ is tangent to $\Omega_2$. To do this, reflect $A,B,X,M$ over the perpendicular to $\Omega_2$ at $Y$ to get $A', B', X', M'$. Then, let $h$ be the homothety centered at $Y$ taking $M'\to N$. It follows that $h\colon A'\to C$ since \[ \frac{YA'}{YC} = \frac{YA}{YC} = \frac{YM}{YN} = \frac{YM'}{YN} \]and similarly $h\colon B'\to D$. Say $h\colon X'\to Z$, then $(A'B'X'M')$ is cyclic and tangent to $\Omega_2$ at $X'$, which implies that $(NCDZ)$ is cyclic and tangent to $\Omega_2$ at $Z$ as desired. $\blacksquare$

Let $O,O_1,O_2$ be the center of $\Omega,\Omega_1,\Omega_2$. Then by Reim's theorem, $AB \parallel CD$ implying that $O_1O_2 \perp CD$. Let $K,L$ be the midpoint of $AB$ and $CD$. Thus $O_2L \perp CD$. Consider the homothety of center $M$ sending $\odot O_1$ to $\odot O$. Thus $O_1 \rightarrow O$, $C \rightarrow A$, $D \rightarrow B$, $L \rightarrow K$. Let $Q = MO_2 \cap \odot O$, then $O_2 \rightarrow Q$ and $QK \perp AB$, meaning that $Q,K,O$ are collinear. Thus we have \[ \frac{O_2L}{O_2M} = \frac{QK}{QM}. \]Consider the tangent line at point $M$ and $N$, by radical axis theorem, they intersect at $T$ which lies on $AB$. Hence $MOKNT$ are concyclic. Thus $\angle MNO = \angle MKO$. Now we prove that $MQ$ bisects the $\angle KMN$. Indeed \[ \angle QMN = \angle OMN - \angle OMQ = \angle MNO - \angle MQO = \angle MKO - \angle MQO = \angle KMQ\]Therefore, by law of sines \[ \frac{QK}{QM} =\frac{\sin \angle KMQ}{\sin \angle MKO} = \frac{\sin \angle QMN}{\sin MNO} = \frac{O_2N}{O_2M}\]Therefore $O_2L = O_2N$ implies that $ L \in \odot O_2$ and $\odot O_2$ tangent to $CD$.

This problem is significantly easier if you think of what works well with tangent circles.

Let $AN$ and $BN$ intersect $\Omega_{2}$ with center $O$ at $E$ and $F$, respectively. Note that there exists a homothety with center $M$ that takes $\Omega_{1}$ to $\Omega$, thus $\overline{CD}\parallel \overline{AB}$. Similarly, there exists a homothety with center $N$ that takes $\Omega_{2}$ to $\Omega$, thus $\overline{EF}\parallel \overline{AB}$. Since $AB$, $CM$, and $EN$ concur, by the converse of the radical axis theorem, $CEMN$ is cyclic. Thus $180^{\circ}-\angle ECM = \angle ENM = \angle ABM = \angle CDM$, thus $EC$ is tangent to $\Omega_{1}$. By symmetry, we may claim and prove in the same way that $DFMN$ is cyclic, $FD$ is tangent to $\Omega_{1}$, and $EC$ and $FD$ are tangent to $\Omega_{2}$. Thus $CDEF$ is an isosceles trapezoid, with $O$ lying on the perpendicular bisectors of its bases. Thus $\angle ECO = \angle CMO = \angle CDO = \angle DOC$. Similarly, $\angle FDO = \angle DMO = \angle DCO =\angle CDO$. Thus $\overline{CD}$ is tangent to $\Omega_{2}$, as desired. $\blacksquare$

Solution using just the usual angle chase, power of point, homothety. Let $P$ and $Q$ be the centers of $\Omega_1$ and $\Omega_2$. Let line $MQ$ meet $\Omega_1$ again at $W$, the homothetic image of $Q$ under $\Omega_1 \to \Omega$. Meanwhile, let $T$ be the intersection of segment $PQ$ with $\Omega_2$, and let $L$ be its homothetic image on $\Omega$. Since $\overline{PTQ} \perp \overline{AB}$, it follows $\overline{LW}$ is a diameter of $\Omega$. Let $O$ be its center. [asy][asy] size(10cm); pen xfqqff = rgb(0.49803,0.,1.); pen qqffff = rgb(0.,1.,1.); pen cqcqcq = rgb(0.75294,0.75294,0.75294); pair O = (0.,0.), M = (-0.54799,0.83648), P = (-0.16957,0.25884), Q = (-0.16861,-0.43170), A = (-0.97449,-0.22439), B = (0.97512,-0.22166), C = (-0.84251,0.10388), D = (0.50379,0.10577), T = (-0.16936,0.10483), L = (-0.00139,0.99999); draw(circle(O, 1.), linewidth(0.6)); draw(circle(P, 0.69055), linewidth(0.6) + xfqqff); draw(circle(Q, 0.53653), linewidth(0.6) + xfqqff); draw(O--M, linewidth(0.6) + red); draw(M--A, linewidth(0.6) + qqffff); draw(M--B, linewidth(0.6) + qqffff); draw(O--(-0.36380,-0.93147), linewidth(0.6) + red); draw(P--Q, linewidth(0.6)); draw(M--(0.00139,-0.99999), linewidth(0.6) + green); draw((0.00139,-0.99999)--L, linewidth(0.6)); draw(L--(-0.36380,-0.93147), linewidth(0.6) + green); draw(M--(-0.36380,-0.93147), linewidth(0.6)); draw(A--B, linewidth(0.6)); dot("$O$", O, dir((1.868, -2.897))); dot("$N$", (-0.36380,-0.93147), dir((-5.700, -12.058))); dot("$M$", M, dir((-8.656, 10.478))); dot("$P$", P, dir((-2.789, 6.521))); dot("$Q$", Q, dir((3.600, -6.558))); dot("$A$", A, dir((-15.478, -5.915))); dot("$B$", B, dir((4.982, -4.507))); dot("$C$", C, dir((-8.023, -3.199))); dot("$D$", D, dir((1.442, 0.454))); dot("$T$", T, dir((-7.133, 1.989))); dot("$W$", (0.00139,-0.99999), dir((-2.113, -15.052))); dot("$L$", L, dir((1.046, 1.812))); dot("$E$", midpoint(A--B), dir(-45)); [/asy][/asy] Claim: $MNTQ$ is cyclic. Proof. By Reim: $\measuredangle TQM = \measuredangle LWM = \measuredangle LNM = \measuredangle TNM$. $\blacksquare$ Let $E$ be the midpoint of $\overline{AB}$. Claim: $OEMN$ is cyclic. Proof. By radical axis, the lines $MM$, $NN$, $AEB$ meet at a point $R$. Then $OEMN$ is on the circle with diameter $\overline{OR}$. $\blacksquare$ Claim: $MTE$ are collinear. Proof. $\measuredangle NMT = \measuredangle TQN = \measuredangle LON = \measuredangle NOE = \measuredangle NME$. $\blacksquare$ Now consider the homothety mapping $\triangle WAB$ to $\triangle QCD$. It should map $E$ to a point on line $ME$ which is also on the line through $Q$ perpendicular to $\overline{AB}$; that is, to point $T$. Hence $TCD$ are collinear, and it's immediate that $T$ is the desired tangency point.

Denote by $X,Y$ meet-points of $\Omega_1,$ $\Omega_2.$ Due to homothety $AB$ $\parallel$ $CD,$ and center of $\Omega_2$ is the midpoint of arc $XY$ of $\Omega_1.$ We invert wrt $M$ with power $|MX|\cdot |MY|,$ and reflect over the bisector of angle $XMY.$ Well-known that $\Omega_2$ is fixed. Note that $\Omega_1,AB$ are swapped, and so the same $A,D$ and $B,C.$ Hence $\Omega,CD$ are swapped and conclusion is trivial.

Six hundred sixty-five. Let $O,O_1,$ and $O_2$ be the centers of $\Omega,\Omega_1,$ and $\Omega_2.$ Let $\overline{BN}$ and $\overline{CN}$ intersect $\Omega_2$ at $F$ and $E$ so that $C$ and $E$ are on the same side of $\overline{O_1O_2}.$ Claim: $\overline{CE}$ is a common external tangent of $\Omega_1$ and $\Omega_2.$ Proof. By homothety, $\overline{CD}\parallel\overline{AB}\parallel\overline{EF},$ and ny Radical Axis, $BCMN$ is cyclic. Hence, $$\measuredangle(\overline{AE},\overline{CE})=\measuredangle NEC=\measuredangle NMC=\measuredangle NBA=\measuredangle NFE.$$Similarly, $\measuredangle(\overline{CE},\overline{AC})=\measuredangle CDM.$ $\blacksquare$ Claim: $O_2$ is the incenter of $\triangle(\overline{CD},\overline{CE},\overline{DF}).$ Proof. Notice $$\angle DCO_2=90-\angle CO_2O_1=\tfrac{1}{2}\angle O_2O_1C=\angle O_2CE$$and similarly $\overline{O_2D}$ bisects $\angle FDC.$ $\blacksquare$ Since $\Omega_2$ is tangent to $\overline{CE},$ it is also tangent to $\overline{CD}.$ $\square$

Somehow this works, okay. Draw in the points $E = \overline{AN} \cap \Omega_2$ and $F = \overline{BN} \cap \Omega_2$. Claim. $\overline{CE}$ is a common external tangent to $\Omega_1$ and $\Omega_2$. First, notice that for homothety reasons, $\overline{CD} \parallel \overline{AB} \parallel \overline{EF}$ and that $CENM$ is cyclic for radical axis reasons. Now angle chase: $$\measuredangle NEC = \measuredangle NMC = \measuredangle NBA = \measuredangle NFE.$$The other tangency follows because $$\measuredangle ECD = \measuredangle CEF = \measuredangle ENF = \measuredangle AMB = \measuredangle CMD. \ \blacksquare$$ Thus, it suffices to show that $\overline{CO_2}$ bisects the angle $\angle NCD$. Notice that $\overline{CO_2}$ bisects $\angle PCQ$, so it suffices to show $\angle ECP = \angle QCD$. But this is obviously true as $\angle ECP = \angle CDP = \angle QCD$ by the previous claim.

v_Enhance wrote: Now consider the homothety mapping $\triangle WAB$ to $\triangle QCD$. It should map $E$ to a point on line $ME$ which is also on the line through $Q$ perpendicular to $\overline{AB}$; that is, to point $T$. Hence $TCD$ are collinear, and it's immediate that $T$ is the desired tangency point. What if the line $ME$ coincides the perpendicular bisector of CD? Is it an easy casework?

what is this solution Let $O_1$ and $O_2$ be the centers of $\Omega_1$ and $\Omega_2$ respectively. Observe that $\overline{AB}$ and the tangents to $\Omega$ at $M$ and $N$ concur by radical center, hence $AMBN$ is harmonic, so $\overline{MN}$ passes through the intersection of the tangents to $\Omega$ at $A$ and $B$. By taking a homothety at $M$ sending $\Omega$ to $\Omega_1$, it follows that $\overline{MN}$ also concurs with the tangents to $\Omega_1$ at $C$ and $D$ at some point $X$, which also lies on $\overline{O_1O_2}$. On the other hand, by Monge, the exsimilicenter of $\Omega_1$ and $\Omega_2$ lies on $\overline{MN}$, and since it also lies on $\overline{O_1O_2}$, it must be $X$, hence the tangents to $\Omega_1$ at $C$ and $D$ are also tangent to $\Omega_2$. Now, it is clear that $O_1CDX$ is cyclic, and that $O_1$ is the midpoint of arc $CD$, hence $\overline{XO_1}$ is the $\angle CXD $-bisector. Therefore, by incenter-excenter, $\Omega_1 \cap \overline{XO_1}=O_2$ is the incenter of $\triangle CXD $, which implies the desired result. $\blacksquare$ Remark: To top off such a strange solution, I first noticed the concurrency of the tangents at $M$ and $N$ with $\overline{AB}$ because of... AIME I 2019/15. what.

!?!?!?!?!?! Let $XY$ be the common chord, let $O,O_1,O_2$ be centers, let $C',D'$ be intersections of external tangents with $\Omega_1$ and let $T_C,T_D$ be intersections with $\Omega_2$, also by angle chasing note that $C'D'$ is tangent to $\Omega_2$ at $T$, by Monge also let $MN$, $C'T_C$, and $D'T_D$ intersect at $K$. Now we need to realize that $KM\cdot KN=KC'\cdot KT_C=KD'\cdot KT_D$. To see this draw $KM\cap \Omega_1=M'$ and define $N'$ on $\Omega_2$ similarly. By homothety note \[\frac{KM}{KM'}=\frac{KN'}{KN}\]which gives what we need. This basically implies by radical axis that $MC',XY,T_CN$ and $MD',XY,T_DN$ concur at $A',B'$ respectively. Now it just suffices to show $MA'NB'$ is cyclic which is trivial by angle chase. The one that works for me is \[\angle MA'B'=\angle MC'D'=\angle T_DD'B'=\angle MNB'\]and now this given circle must be tangent to $\Omega_1$ and $\Omega_2$ by homothety and is therefore $\Omega$. wow that was fun

Dukejukem wrote: Let $\Gamma \equiv \odot(ESO_2T)$ be the circle of diameter $\overline{EO_2}.$ Clearly the center $X$ of $\Gamma$ lies on the perpendicular bisector of $\overline{ST}.$ Hence, $O_1, O_2, X$ are collinear, implying that $\Omega_1$ and $\Gamma$ are tangent. Now consider the inversion $\Upsilon$ with center $E$ and power $EC \cdot ES = ED \cdot ET.$ Clearly $\Upsilon$ interchanges $\{C, S\}$ and $\{D, T\}$ and therefore $\Upsilon$ interchanges $\{\Omega_1, \Omega_2\}.$ Moreover, $\Upsilon(\Gamma) \equiv CD.$ Then since $\Omega_1$ and $\Gamma$ are tangent, it follows under inversion that $\Omega_2$ and $CD$ are tangent, as desired. yikes this is really smart what the heck

Let $O_1$ and $O_2$ be centers of $\Omega_1$ and $\Omega_2$. Let the two intersections of $\Omega_1$ and $\Omega_2$ be $P$ and $Q$, with $P$ closer to $A$ than $Q$. Let $NA$ and $NB$ intersect $\Omega_2$ at $E$ and $F$. Let $CE$ and $DF$ intersect at $G$. Claim 1: $AB\parallel CD\parallel EF$. Consider the homothety centered at $M$ that takes $\Omega_1$ to $\Omega$. $C$ is taken to $A$ and $D$ to $B$, so $CD\parallel AB$. $EF\parallel CD$ follows similarly. Claim 2: $MCEN$ is cyclic. By Power of a Point, $AC\cdot AM=AP\cdot AQ=AE\cdot AN$, so the claim is true. Claim 3: $EC$ is tangent to both $\Omega_1$ and $\Omega_2$. We have \[\angle ACE=\angle ANM=\angle ABM=\angle CDM\]so $EC$ is tangent to $\Omega_1$, and tangency to $\Omega_2$ follows similarly. Now, note that $CP=DQ$ because $CD\parallel PQ$, and $O_2Q=O_2P$, so $O_2C=O_2D$. Thus, $\angle ECO_2=\angle O_2CD$ by inscribed angles theorem. In particular, $CO_2$ bisects $\angle DCG$ Similarly, $DO_2$ bisects $\angle CDG$. This implies that $O_2$ is the incenter of $\triangle CDG$. $O_2E$ is the length of the perpendicular from the incenter to one of the sides, so $\Omega_2$ is the incircle of $\triangle CDG$. We are done.

Let $r,r_1,r_2$, be the radii of the obvious circle. Define $O,O_1,O_2$ as the centers of the obvious circles. Let the tangents at $M,N$ and line $AB$ concur (by radax) at point $X$. It is trivial to see that $AB$ is parallel to $CD$ by homothety centered at $M$. Let $C'D'$ be the unique chord of $\Omega_1$ tangent to $\Omega_2$ parallel to $AB$. We want to show $CD = C'D"$, equivalently that the unique homothety centered at $M$ taking $\Omega_1$ to $\Omega$ takes $C'D'$ to $AB$. Let $A'B'$ be the result of this homothety, it suffices to prove $A'B' =AB$. Let $M_{AB}$ be the midpoint of $AB$. By La Hire, since $X$ lies on the polar of the inverse of $M_AB$ we know that the inverse of $M_{AB}$ lies on $MN$. Let $Y$ be the intersection of $MO_2$ and $\Omega$, then clearly $M_{AB}$ lies on $OY$. In addition, letting $M_{A'B'}$ be the midpoint of $A'B'$, we see since $A'B'$ parallel to $AB$ this $M_{A'B'}$ also lies on $OY$. Thus it suffices to prove $OM_{AB} = OM_{A'B'}$. Since $O_1O_2 = r_1 - r_2$, homothety gives $OM_{A'B'} = \frac{r(r_1 - r_2)}{r_1}$. Let $Z$ be the inverse of $M_AB$, then clearly we have $OZ \cdot OM_{AB} = r^2$, then let $O_1O_2$ meet $MN$ at $Q$, then we have $O_1Q \cdot \frac{r}{r_1} = OZ$ by homothety, so $OM_{AB} = r^2 \frac{1}{OZ} = rr_1 \frac{1}{O_1Q}$ , so it suffices to prove that $O_1Q = \frac{r_1^2}{r_1 - r_2}$, or equivalently $O_2Q = \frac{r_1r_2}{r_1 - r_2}$. To do this, we invoke cartesian coordinates. Let $O = (0, a)$, $M = (-1 , 0), N = (1, 0 )$. Now let $O_2 = (1 - k, ka), O_1 = (l - 1, la)$, now we find the equation of line $O_1O_2$, slope is clearly $\frac{(l - k)a}{l + k - 2}$. We clearly get the final equation as $\frac{(l -k)a}{l + k - 2}x - \frac{(l - k)(l - 1)a}{l + k - 2} + la = y$. Setting $y = 0$ to intersect with the line $MN$ gives $x = (l - 1) - \frac{l(l + k + 2)}{l - k}$. Thus $O_1Q^2 = (la)^2 + \frac{l^2(l + k -2)^2}{(l - k)^2}$. Now $(\frac{r_1^2}{r_1 - r_2})^2 = (a^2 + 1)(\frac{l^2}{l - k})^2$. We want to prove these are equal, or that $l^4(a^2 + 1) = (la)^2(l - k)^2 + (l)^2(l + k -2)^2$, dividing out $l^2$ yields that we want to prove $l^2(a^2 + 1) = a^2(l - k)^2 + (l + k - 2)^2$. Now we also have the condition $O_1O_2 = O_1M$, so we can write (squaring both sides) $(l + k -2)^2 + a^2(l - k) = l^2(a^2 + 1)$, so the thing we wanted is obviously true.

blud what the sigma is this Let $\Omega_1 \cap \Omega_2=X,Y$, and their centres be $O_1,O_2$ respectively. Let $\Omega$ have centre $O$ Note that the tangents at $M,N$ to $\Omega$ concur at a point on $XY$, call it $P$. It is well-known thus that $AMBN$ is harmonic. This implies that the tangents at $A,B$ to $\Omega$ concur on $MN$, say at $S_1$. Now, by Monge on the three circles (insimilicentre, insimilicentre, exsimilicentre variant), the external tangents to $\Omega_1,\Omega_2$ concur on $MN$, say at $S_2$. Yet clearly $S_1$ lies on the perpendicular bisector of $AB$, and $S_2$ on the perpendicular bisector of $XY$. Then $O_1S_2\parallel OS_1$. Thus the homothety at $M$ sending $S_2$ to $S_1$ takes $\Omega_1$ to $\Omega$. Let $MD\cap \Omega_1=D'$. By homothety, the tangent at $D'$ passes through $S_2$, thus $D'=D$. Similarly $C'=C$. For the finish, we note that we can cut out $\Omega$ completely, as $C,D$ are just the tangency points on $\Omega_1$ of the common external tangent. We claim that $O_2$ is the incentre of $CDS_2$. This is trivial as $O_2$ by symmetry is the midpoint of arc $CD$, note $\angle O_2DS_2=\angle O_2CD=\angle CDO_2$ and analogous results give us our claim. This implies as $\Omega_2$ is tangent to $CS_2,DS_2$, it is the incircle, which finishes.