Sharky devil, projective, brianchon, radical (will add sol later probably) @below https://artofproblemsolving.com/community/c946900h1911664_properties_of_the_sharkydevil_point https://artofproblemsolving.com/community/c6h2165146p16078079 https://artofproblemsolving.com/community/c776104

Did the problem say that the circles are externally tangent?

@above they can be tangent internally and externally.

geometry6 wrote: @above they can be tangent internally and externally. ThAT's a lOt Of dAMaGe. Here are pictures of both cases.

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@above there are two more cases too.

Kar-98k wrote: @above there are two more cases too. I don't sure. But $\square ABCD$ has to be a convex quadrilateral.

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@above Ah! I didn't see that "convex" word in the problem. So we basiccaly need to check two cases right? For the First Diagram which you posted in #6 the tangency point is basically the Miquel Point $ABCD$ and $ABCD$ must be bicentric. (An easy angle chase will suffice)

As there is no solution, Here is a hint from my solution during the contest Use Casey Theorem [+straight forward length bash]

Solution from Twitch Solves ISL: We are going to prove both conditions are equivalent to $\overline{KM} \perp \overline{LN}$. We introduce the following notation. $\Omega$ denotes the circumcircle of $KLMN$. Rays $EN$ and $EL$ meet $\Omega$ again at $N'$ and $L'$. Points $X'$, $Y'$, $F'$, $W$ are the midpoints of $\overline{NN'}$, $\overline{LL'}$, $\overline{NL}$, $\overline{N'L'}$. Note that $X'$ and $Y'$ lie on the circle with diameter $\overline{IE}$. Let $G = \overline{LN} \cap \overline{KM}$ and $V = \overline{X'Y'} \cap \overline{MK}$. The relevance of $F'$, $X'$, $Y'$ is explained as follows: Claim: The points $F'$, $X'$, $Y'$ are the inverses of $F$, $X$, $Y$ with respect to $\Omega$. Proof. For $F$ it's clear. For $X$, the inverses of $\overline{MK}$ and $\overline{NN}$ are the circles with diameter $\overline{IE}$ and $\overline{IN}$. $\blacksquare$ Hence, we are interested in when $(F'X'Y')$ is tangent to $\overline{MK}$, the latter being the inverse of the circle with diameter $\overline{EI}$. [asy][asy] import graph; size(12cm); pen qqwuqq = rgb(0.,0.39215,0.); pen xfqqff = rgb(0.49803,0.,1.); pen dcrutc = rgb(0.86274,0.07843,0.23529); pen qqffff = rgb(0.,1.,1.); pen ffxfqq = rgb(1.,0.49803,0.); pen yqqqqq = rgb(0.50196,0.,0.); pen cqcqcq = rgb(0.75294,0.75294,0.75294); draw(circle((0.,0.), 1.), linewidth(0.6) + qqwuqq); draw((-0.32917,0.94426)--(-0.31711,-0.94838), linewidth(0.6) + red); draw(circle((0.,-0.83333), 0.83333), linewidth(0.6) + xfqqff); draw(circle((-0.31933,-0.30201), 0.29798), linewidth(0.6) + dcrutc); draw(circle((-0.15855,-0.47419), 0.5), linewidth(0.6) + qqffff); draw(circle((-0.16458,0.47213), 0.5), linewidth(0.6) + qqffff); draw((0.,-1.66666)--(-0.91446,0.40466), linewidth(0.6) + ffxfqq); draw((0.,-1.66666)--(-0.32917,0.94426), linewidth(0.6) + ffxfqq); draw((-0.08450,-0.99642)--(-0.91446,0.40466), linewidth(0.6) + ffxfqq); draw((-1.16175,-0.6)--(-0.20684,-0.02607), linewidth(0.6) + red); draw((-0.31933,-0.6)--(-0.61578,-0.27186), linewidth(0.6) + yqqqqq); draw((-0.31933,-0.6)--(-0.20684,-0.02607), linewidth(0.6) + yqqqqq); draw((-1.16175,-0.6)--(0.8,-0.6), linewidth(0.6) + red); draw((0.,-1.66666)--(-0.8,-0.6), linewidth(0.6) + qqwuqq); draw((0.,-1.66666)--(0.8,-0.6), linewidth(0.6) + qqwuqq); dot("$I$", (0.,0.), dir((1.160, 2.848))); dot("$K$", (0.8,-0.6), dir((4.948, -3.148))); dot("$M$", (-0.8,-0.6), dir((-11.235, -8.026))); dot("$E$", (0.,-1.66666), dir((-3.143, -12.694))); dot("$N$", (-0.31711,-0.94838), dir((-9.882, -9.342))); dot("$N'$", (-0.91446,0.40466), dir((-12.414, 5.711))); dot("$L$", (-0.32917,0.94426), dir((1.080, 2.827))); dot("$X'$", (-0.61578,-0.27186), dir((-13.014, 0.767))); dot("$Y'$", (-0.20684,-0.02607), dir((1.185, 2.300))); dot("$F'$", (-0.32314,-0.00206), dir((1.051, 2.194))); dot("$G$", (-0.31933,-0.6), dir((-9.086, -8.599))); dot("$L'$", (-0.08450,-0.99642), dir((2.725, -9.703))); dot("$V$", (-1.16175,-0.6), dir((1.124, 2.304))); dot("$W$", (-0.49948,-0.29588), dir((1.181, 2.308))); [/asy][/asy] The first critical lemma is the following: Claim: [A generalization of butterfly theorem] Unconditionally, we have that $(GX'Y')$ is tangent to $\overline{MK}$. Proof. Suppose $\overline{NN'} \cap \overline{MK} = X_1$ and $\overline{LL'} \cap \overline{SMK} = Y_1$. Then by shooting lemma from $E$, the quadrilateral $X'X_1Y_1Y'$ is cyclic. Also, by the dual of Desargues involution theorem (DDIT) on $N'NLL'$, there is an involution swapping $M \leftrightarrow K$, $X_1 \leftrightarrow Y_1$, $G \leftrightarrow G$. Combined with power of a point from $V$, we are able to get \[ VX_1 \cdot VY_1 = VX' \cdot VY' = VM \cdot VK \implies VX' \cdot VY' = VG^2. \]This proves the desired tangency. $\blacksquare$ Thus, the tangency takes place exactly when $F' \in (GX'Y')$. To prove the result, we now need: Claim: Unconditionally, $\overline{GF'}$ and $\overline{GW}$ are isogonal. Proof. Since $\triangle GN'N \overset{-}{\sim} \triangle GLL'$, we get $\triangle GN'X' \overset{-}{\sim} \triangle GLY'$. $\blacksquare$ Now, $WX'F'Y'$ is always a parallelogram, because these four points are the midpoints of $NN'L'L$. So we focus entirely on $\triangle GX'Y'$ now, and consider the following four statements: $\overline{GW}$ and $\overline{GF'}$ are isogonal wrt $\angle G$ (and they are distinct, provided $X' \neq Y' \iff X \neq Y$, which I think should be assumed for the problem to make sense). $W$ and $F'$ are reflections are the midpoint of $\overline{X'Y'}$. $F'$ lies on $(GX'Y')$. $\overline{GW}$ and $\overline{GF'}$ are altitude and diameter respectively. The first two statements are always known; under these two assumptions, one can then show the third and fourth statements are equivalent. Finally, the last bullet is equivalent to $\overline{GF'} \perp \overline{MK}$, because $\overline{MK}$ is a tangent, and hence equivalent to $\overline{LN} \perp \overline{MK}$.

v_Enhance wrote: Thus, the tangency takes place exactly when $F' \in (GX'Y')$. This part is wrong because there are two such circles and $F'$ does not necessarily lie on $(GX'Y') $. Also as shown in post #6 there are another cases and the conditions are not equivalent to $KM\perp LN$.

I think this solution is much simpler than the official one, which uses Casey's theorem and trigonometry. My solution. It is well known that $AC,BD,KM,LN$ are concurrent at $P'$. Let $P=IP'\cap EF$. We also easily prove that $(XPY)$ is tangent to $(EI)$ at $P$. Case 1: $I\notin (XFY)$. Inverting about $(I)$, we have $(X'F'Y')$ is tangent to $KM$. But $(X'P'Y')$ is also tangent to $KM$, thus $X',F',Y',P'$ are concyclic or $X,Y,F,P$ are concyclic. So $(YP,YX)\equiv (FP,FX)\equiv (IP,IL) \pmod{\pi}\implies I,L,P,Y$ are concyclic. Let $W=IL\cap KM$, we have $P'Y.P'W=P'P.P'I=P'L.P'N\implies W,L,Y,N$ are concyclic. Hence $(YW,YN)\equiv (LI,LN)\equiv (NP',NI) \pmod{\pi}$, but $IN\perp YN$ so $KM\perp NL$. Hence $ABCD$ is bicentric quadrilateral and $P$ is the Miquel point of it. One can proves $(TEZ)$ is tangent to $(FI)$ at $P$ by just angle chasing. Case 2: $I\in (XFY)$. Clearly $(XFY)$ and $(EI)$ are tangent at $I$, so $IX=IY$. Let $H$ be the midpoint of $KM$. By Menelaus' theorem and noting that $XN=YL,FL=FN$, we have $L,H,N$ are collinear, thus $L\in AC,BD$. Since $HI.HE=HM.HK=HN.HL\implies E\in (FI)$. Finally, $(FE,FI)+(ZE,ZT)\equiv (LE,LI)+(IK,IF)\equiv (HL,HI)+(IK,IF)\equiv (IF,KM)+(IK,IF)\equiv (IK,KM)\equiv (EI,EC) \pmod{\pi}$ $\implies (FI)$ is tangent to $(TEZ)$ at $E$. So we are done. $\quad \square$

5 We contend that $(XFY),(EI)$ tangent $\iff\overline{KM}\perp\overline{LN}$, which is just another expression of `$ABCD$ bicentric'. Call the incircle $\omega$. Define $S=\overline{KM}\cap\overline{LN}$ and $Q$ as the Miquel point of $KLMN$ aka the inverse of $S$ wrt $\omega$, which obviously lies on the polar $\overline{EF}$ of $S$ wrt $\omega$. It follows that $\angle SQE=\angle SQF=90^\circ$. Let $F'=\overline{QF}\cap\overline{KM}$ and $U$ be the midpoint of $\overline{SF'}$. Claim: $\overline{QF}$ always bisects $\angle XQY$, and $UM\cdot UK=US^2=UX\cdot UY$, so $U$ lies on the radical axis of $(EIMK),(XFY)$. Proof. By degenerate Brianchon, $S=\overline{AB}\cap\overline{CD}$ as well. Apply DIT to $\overline{KM}$ and quadrilateral $ABCD$ and project to $Q$, to obtain an involutive pairing $i:Q(XY;SS;KM)$. The last two pairs imply that $i$ is just reflection in $\overline{QS}$, so $\overline{QS}$ bisects $\angle XQY$. As $\overline{QF}\perp\overline{QS}$, it must also bisect the same angle: $i:Q(F'F')$. By these right angles and angle bisections, $(SF';MK)=(SF';XY)=-1$, so the last result follows by midpoints of harmonic bundles lemma. $\qquad\qquad\square$ Now because $\angle SQF'=90^\circ$, we have $UQ=US=UF'\Rightarrow(EI),(XQY)$ tangent at $Q$; this means that the desired is equivalent to $FQXY$ cyclic.

Note

Next, we show that this happens iff $FX=FY$. ``If'' direction Since $\angle FQX=\angle YQF$, triangles $FQX,FQY$ have equal circumradii, so their circumcircles either coincide or are reflections of each other in $\overline{QF}$. If they were to be reflections, we'd obtain two possibilities, each absurd in the context of the problem: (i) $Q,X,Y$ collinear $\Rightarrow\overline{XY},\overline{EF}$ coincide; and (ii) $X,Y$ reflections in $\overline{QF}\Rightarrow\overline{XY},\overline{SQ}\perp\overline{EF},\overline{XY}\parallel\overline{SQ}$. Thus $(FQX)=(FQY)$ as required. ``Only if'' direction $\angle YQF=\angle FQX\Rightarrow \widehat{YF}=\widehat{FX}$, trivial; To finish the problem, observe that $FX=FY$ is equivalent to $\overline{MK}$ making equal angles with $\overline{AB},\overline{CD}$. As $FN=FL$ as well, $\overline{LN}$ (always) makes equal angles with the same two lines. Since $K,L,M,N$ form a convex polygon, this is in turn equivalent to $\overline{KM}\perp\overline{LN}$, completing the proof.