Let $A_1A_2$ and $A_3A_4$ meet at $Q$, $A_1A_4$ and $A_2A_3$ at $P$. We will assume that $P$, $T_1$ and $T_3$ are collinear and prove that so are $Q$, $T_2$ and $T_4$. Consider a circle $\omega$ tangent to $QT_1$ at $T_1$ and tangent to $QT_3$. Then $T_1$ is the insimilicenter of $\omega_1$ and $\omega$ and $P$ is the exsimilicenter of $\omega_1$ and $\omega_3$, so according to Monge d'Alembert's Theorem, the insimilicenter of $\omega$ and $\omega_3$ lies on $PT_1$. Furthermore, it lies on the common tangent $A_3A_4$, but the intersection of these lines is $T_3$ because of our assumption. Thus $T_3$ is the insimilicenter of $\omega$ and $\omega_3$, hence $\omega$ is tangent to $A_3A_4$ at $T_3$. In particular, $QT_1 = QT_3$. Now some calculations that would require some boring notation and use only the fact that the two tangents from a point to a circle are equal will show that $PT_2 = PT_4$. So there is a circle $\omega'$ tangent to $PT_2$ and $PT_4$ at $T_2$ and $T_4$, respectively. Applying Monge d'Alembert's Theorem as we did before, we have that $Q$, $T_2$ and $T_4$ are collinear.

Let $X=A_1A_4\cap A_2A_3$ and $Y=A_1A_2\cap A_3A_4$. By the law of sines or area ratios, $X,T_1,T_3$ are collinear iff \[\frac{XA_1/A_1T_1}{XA_2/A_2T_1} = \frac{XA_4/A_4T_3}{XA_3/A_3T_3} \Longleftrightarrow \frac{XA_1\cdot XA_3}{XA_2\cdot XA_4} = \frac{A_1T_1\cdot A_3T_3}{A_2T_1\cdot A_4T_3}\]and $Y,T_2,T_4$ are collinear iff \[\frac{YA_1\cdot YA_3}{YA_2\cdot YA_4} = \frac{A_1T_4\cdot A_3T_2}{A_2T_2\cdot A_4T_4}.\]However, letting $r_i$ denote the radius of $\omega_i$, right triangle similarities trivially yield that $A_2T_1/A_2T_2=r_1/r_2$, etc., so \[\frac{A_1T_1\cdot A_3T_3}{A_2T_1\cdot A_4T_3} \cdot \frac{A_1T_4\cdot A_3T_2}{A_2T_2\cdot A_4T_4} = 1,\]and again by the law of sines, \[\frac{XA_1/YA_1}{XA_2/YA_2} = \frac{XA_3/YA_3}{XA_4/YA_4}\implies \frac{XA_1\cdot XA_3}{XA_2\cdot XA_4} \cdot \frac{YA_1\cdot YA_3}{YA_2\cdot YA_4} = 1.\]Thus the two conditions are obviously equivalent.

Let $A_1 A_4$ and $A_2 A_3$ intersect at $P$, and $A_1 A_2$ and $A_3 A_4$ intersect at $Q$. It is clearly sufficient to show that if $P$, $T_1$, and $T_3$ are collinear, then $Q$, $T_4$, and $T_2$ are collinear. Let $PT_1$ intersect $\omega_1$ at $U$, let $\omega_1$ be tangent to $PA_2$ and $PA_1$ at $W$ and $X$, respectively, and let $\omega_4$ be tangent to $QA_1$ and $QA_4$ at $Y$ and $Z$, respectively. Because $P$, $T_1$, and $T_3$ are collinear, the homothety centered at $P$ mapping $\omega_3$ to $\omega_1$ maps $T_3$ to $U$, so the tangent to $\omega_1$ through $U$ is parallel to $A_4 A_3$. It follows that the homothety mapping $\omega_1$ to $\omega_4$ sends $U$ to $Z$. Furthermore, we note that this homothety maps $T_1$ to $Y$ and $X$ to $T$. Suppose this homothety also maps $W$ to some point $W'$ on $\omega_4$. Then the tangent to $\omega_4$ through $W'$ is parallel to $A_2A_3$. Furthermore, because $XUWT_1$ is harmonic, $T_4 Z W' Y$ is harmonic, so $Q$, $W'$, and $T_4$ are collinear. The homothety centered at $Q$ mapping $\omega_4$ to $\omega_2$ must therefore send $W'$ to $T_2$ (since the tangent to $\omega_4$ at $W'$ is parallel to $A_2 A_3$), so $Q$, $T_4$, and $T_2$ are collinear, as desired.

Assume that $~$ $N$, $L$, $J$ be collinear. $\Delta MLJ$, $\Delta MOP$, $\Delta MTS$ are isosceles, which are similar to one another. For $\Delta MLJ$, $\angle XLD=\angle XLU+\angle ULD=\angle JWR+\angle WJR=\angle WJC$, where homothety is used. Thus $VI=OL=PJ=IQ$, $~$ $UK=LT=JS=KR$. Therefore $M$, $I$, $K$ are on the radical axis of ($H$),($F$)

Let $A_1A_2,A_3A_4$ meet at $Q$ and the other pair of opposite sides meet at $P$. We only need to show $P,T_1,T_3$ collinear implies $Q,T_2,T_4$. Draw the line parallel to $A_3A_4$ meeting $\omega _1$ at $X$, and line parallel to $A_1A_2$ meeting $\omega _3$ at $Y$. Using homothety and condition we have $P,T_1,T_3,T_1',T_3'$ collinear. Now note that homthety gives us $A_1T_3'$ passes through the tangent point between $\omega _4$ and $A_3A_4$ And etc.,so $A_1T_3', A_4T_1'$ meet at gergonne point (Call it $K$) of $A_1A_4Q$ (maybe the ex-gergonne point)$\Rightarrow Q,K,T_4$ collinear, and similarly $A_2T_3', A_3T_1'$ meet at gergonne point (Call it $L$) of $A_2A_3Q$ $\Rightarrow Q,L,T_2$ collinear. It remains to show $Q,K,L$ collinear, which is true by Desargues Theorem on $\triangle A_1A_2T_3'$ and $\triangle A_3A_4T_1'$, hence we are done.

Here is a sketch of my extremely cheap solution that looks to be equivalent to the one above. Note that through easy trig computations (using ratio lemma to compute $\frac {A_1T_1} {A_2T_1}$ and the corresponding ratio $\frac {A_4T_3} {A_3T_3}$ and equating angles), then it is clear this property depends only on the angles of the quadrilateral. Therefore, it suffices to assume that the quadrilateral is circumscribed, in which case each collinearity is true iff the inscribed quadrilateral is harmonic. It helps to have done 2009 G6, which I solved with a similar technique, and just as cheaply.

I dislike the notation, so I write $ABCD$ quadrilateral, and $P=AB \cap CD$, $Q=AD \cap BC$, and $E,F$ are the points of tangency of the incircles of $QBA$ and $PBC$ with $BA$ and $BC$ respectively, and $G,H$ the points of tangency of the excircles (w.r.t $Q$ and $P$, respectively) of $QCD$ and $PDA$ with $CD$ and $DA$, respectively. I prove $QEG \text{ collinear} \Leftrightarrow PFH \text{ collinear}$. Assume $QEG \text{ collinear}$. Let $X$ be the intersection of $QE$ with the incircle of $QAB$. Then the homothety from $Q$ mapping $A$ to $D$ maps the incircle of $QAB$ to the incircle of $QDC$, and so it maps $X$ to $G$. Therefore, the tangent line to the incircle of $QEG$ at $X$ is parallel to $PD$. Let this line cut $PA$ at $Y$. Notice $YX=YE$ by tangency, and therefore $\angle PEG = \angle XEY = \angle YXE = \angle EGP$, and thus $\boxed{PE=PG}$. Notice $2PE=2PB+2BE=2PB+(QB+AB-QA)=PB+PA+QB-QA$ and $2PG=PC+PD+QD-QC$. But $2QF=QC+PB-PC+QB$ and $2QH=QD+PD-PA+QA$. Thus, $2(QH+PE)=QD+PD+B+QB=2(QF+PG)$, and so $\boxed{QH=QF}$. Using the same arguments in paragraph 1, we obtain that $PFH \text{ collinear} \blacksquare$.

Let $X = A_1A_4 \cap A_2A_3$ and $Y=A_1A_2 \cap A_3A_4$ following math154's solution because it's the same one lol Now using this lemma, we have $X, T_1, T_3$ colinear iff $$\frac{\sin \angle A_2XT_1}{\sin \angle A_1XT_1} = \frac{\sin \angle A_3XT_3}{\sin \angle A_4XT_3}$$ Now using standard area formulas, we have $$\frac{\sin \angle A_2XT_1}{\sin \angle A_1XT_1}= \frac{A_2T_1/XA_2}{A_1T_1/XA_1} \text{ and }\frac{\sin \angle A_3XT_3}{\sin \angle A_4XT_3}=\frac{A_3T_3/XA_3}{A_4T_3/XA_4}$$ Therefore, we have $X, T_1, T_3$ colinear if and only if $$\frac{A_2T_1 \cdot A_4T_3}{A_1T_1 \cdot A_3T_3} = \frac{XA_2 \cdot XA_4}{XA_1 \cdot XA_3}$$ Similarly, we can see that $Y, T_2, T_4$ are colinear if and only if $$\frac{A_2T_2 \cdot A_4T_4}{A_3T_2 \cdot A_1T_4} = \frac{YA_2 \cdot YA_4}{YA_1 \cdot YA_3}$$ Let the radius of $\omega_i$ be $r_i$. WLOG, order the configuration so that 1. $\omega_1$ is the incircle of $\triangle XA_1A_2$ 2. $\omega_2$ is the incircle of $\triangle YA_2A_3$ 3. $\omega_3$ is the excircle of $\triangle XA_3A_4$ 4. $\omega_4$ is the excircle of $\triangle YA_4A_1$ Now by simple AA similarities with the centers of $w_i$, we have $\frac{A_{i+1}T_i}{A_{i+1}T_{i+1}} = \frac{r_i}{r_{i+1}}$ for all $i$. This gives us that $$\frac{A_2T_1 \cdot A_4T_3}{A_1T_1 \cdot A_3T_3} = \frac{A_2T_2 \cdot A_4T_4}{A_3T_2 \cdot A_1T_4}$$ Now it suffices to show that $$\frac{XA_2 \cdot XA_4}{XA_1 \cdot XA_3} = \frac{YA_2 \cdot YA_4}{YA_1 \cdot YA_3} \text{ or } \frac{XA_2}{YA_2} \cdot \frac{XA_4}{YA_4} = \frac{XA_1}{YA_1} \cdot \frac{XA_3}{YA_3}$$ By the sine law, we have $$\frac{XA_1}{YA_1} = \frac{\sin \angle XYA_1}{\sin \angle YXA_1} \text{ and } \frac{XA_3}{YA_3} = \frac{\sin \angle XYA_3}{\sin \angle YXA_3}$$$$\frac{XA_2}{YA_2} = \frac{\sin \angle XYA_2}{\sin \angle YXA_2} \text{ and } \frac{XA_4}{YA_4} = \frac{\sin \angle XYA_4}{\sin \angle YXA_4}$$

Now this implies $\frac{XA_1}{YA_1} \cdot \frac{XA_3}{YA_3} = \frac{XA_2}{YA_2} \cdot \frac{XA_4}{YA_4}$, as desired. $\blacksquare$

Really nice! Let $X=A_1A_4 \cap A_2A_3$ and $Y=A_1A_2 \cap A_3A_4$ and suppose that the circles $\omega_4$ and $\omega_2$ touch the lines $A_1A_2$ and $A_3A_4$ at points $U_1,V_1$ and $U_2,V_2$ respectively. We may assume without loss of generality that $Y,A_1,A_2$ and $Y,A_4,A_3$ are collinear in this order. Let $\omega$ be a circle tangent to the lines $A_1A_4$ and $A_2A_3$ at points $T_4$ and $T_2'$, respectively. Notice that the insimilicenter of circles $\omega$ and $\omega_2$ is a point, say $T_2"$, on the segment $T_2T_2'$. We see that the exsimilicenter of circles $\omega_2$ and $\omega_4$ is $Y$ and the insimilicenter of circles $\omega$ and $\omega_4$ is the point $T_4$. Applying Monge D'Alembert's Theorem, we see that the points $Y,T_4,T_2"$ are collinear. However, by our initial assumption, $T,T_4,T_2$ are collinear and so $T_2=T_2'=T_2"$ and we see that $XT_4=XT_2$ being tangents to the circle $\omega$. Therefore, $X$ lies on the radical axis of the circles $\omega_2$ and $\omega_4$. Notice that by lengths chasing, we get that $$0=XT_4-XT_2=(A_1X+A_1T_4)-(A_2X+A_2T_2)=(A_1X-A_2X)+(A_1U_1-A_2V_1)=(A_1T_1-A_2T_1)+(A_1U_1-A_2V_1)=T_1U_1-T_1V_1$$and we get that $T_1$ also lies on the radical axis of circles $\omega_2$ and $\omega_4$. A similar length chase for side $A_3A_4$ yields that $T_3$ lies on the radical axis of the circles $\omega_2$ and $\omega_4$. Thus, we conclude that the points $X,T_1,T_3$ are collinear. The other direction can be proven in an analogous manner.

Suppose that $A_1A_2$, $A_3A_4$, and $T_2T_4$ are concurrent at $X$, and let $A_2A_3$ and $A_4A_1$ meet at $Y$. Consider the circle $\omega$ through $T_2$ externally tangent to $\omega_2$ and $\omega_4$ at $T$. By Monge on $\omega$, $\omega_2$, and $\omega_4$, $X$, $T_2$, and $T$ are collinear, so $T$ is either $T_4$ or the second intersection of $XT_2T_4$ with $\omega_4$. It is easy to see that $T$ cannot be the second intersection as then it would be the image of $T_2$ under the homothety at $X$ mapping $\omega_2$ to $\omega_4$. Thus, $T=T_4$. Notably, the $YT_2$ and $YT_4$ are the tangents from $Y$ to $\omega$, so $YT_2=YT_4$. Now, let $A_2A_3$ and $A_4A_1$ be tangent to $\omega_1$ at $P_2$ and $P_4$, respectively, and let $A_1A_2$ be tangent to $\omega_2$ and $\omega_4$ at $Q_2$ and $Q_4$, respectively. Note that as $XT_2=XT_4$ and $XP_2=XP_4$, $P_2T_2=P_4T_4$. But $T_2P_2$ and $T_1Q_2$ are the common internal tangents of $\omega_1$ and $\omega_2$, and $T_4P_4$ and $T_1Q_4$ are the common internal tangents of $\omega_1$ and $\omega_4$, so $T_1Q_2=T_1Q_4$. Thus, $T_1$ has equal tangents to $\omega_2$ and $\omega_4$, so it is on their radical axis. The same holds true for $Y$ and $T_3$, so $T_1T_3$ is the radical axis of $\omega_2$ and $\omega_4$ that passes through $Y$, as desired.

THERES ACTUALLY NO WAY (DID NOT DESERVE TO FIND THIS) Let $P=A_1A_2\cap A_3A_4$ and let $Q=A_1A_4\cap A_2A_3$. Consider a homothety at $P$ which takes $\omega_4$ to $\omega_2$. Then $A_1A_4$ goes to $\ell$ and $T_4$ goes to $T_4'\in \ell$. Notice that the angle bisector of $\ell$ and $A_2A_3$ is just perpendicular to $T_2T_4'$; since $\ell\parallel A_1A_4$ we find that $T_2T_4$ is perpendicular to the angle bisector of $\angle T_4QT_2$. As a result, $QT_4=QT_2$. In a stroke of pure genius (luck? how did I even manage to find this), we get: Claim: $T_1T_3$ is the radical axis of $\omega_2$ and $\omega_4$. Proof: I defined the following points in clockwise order in my diagram (so it may be confusing to readers): $W$ is the tangency point of $\omega_4$ with $A_1A_2$. $X$ is the tangency point of $\omega_1$ with $A_1A_4$. $Y$ is the tangency point of $\omega_1$ with $A_2A_3$. $Z$ is the tangency point of $\omega_2$ with $A_1A_2$. Now we get \[WT_1=T_4X=T_4Q-XQ=T_2Q-YQ=T_2Y=ZT_1\]and similarly for $T_3$, so the claim is proven. Finally $PT_1=PT_3$, and by reversing the first part of this solution, we get $QT_1T_3$ collinear.

Let $P=\overline{A_1A_2} \cap \overline{A_3A_4}$ and $Q=\overline{A_1A_4} \cap \overline{A_2A_3}$. Suppose that $P$, $T_1$, and $T_3$ are collinear. The hardest part of the problem is the following claim. Claim: We have $QT_2=QT_4$. Proof. Let $W$ denote the point on $\omega_4$ whose tangent line is parallel to $\overline{A_1A_3}$. By homothety, $W$ lies on $\overline{PT_4T_2}$. Note that this line is parallel to the line through the tangencies of $\omega_1$ with $\overline{A_4A_1}$ and $\overline{A_2A_3}$, which implies the desired result. The rest of the proof is relatively simple. By the claim, we have that the length of the internal tangents between $(\omega_4, \omega_1)$ and $(\omega_1, \omega_2)$ are equal, which implies that $T_1$ and $T_3$ are the midpoints of the external tangents to $\omega_4$ and $\omega_2$. In particular, $\overline{T_1T_3}$ is the radical axis of $\omega_2$ and $\omega_4$. Since $Q$ lies on the radical axis by the claim, we have that $Q$, $T_1$, and $T_3$ are collinear, as desired.

[asy][asy] unitsize(1cm); pair A1, A2, A3, A4, T1, T3; A1=(1,6); A2=(4,7); A3=(8,0); A4=(0,0); pair X=extension(A1,A4,A2,A3); T1=A1+A2-foot(incenter(X,A1,A2),A1,A2); T3=foot(incenter(X,A3,A4),A3,A4); draw(X--A3--A4--X--A1--A2); draw(incircle(X,A3,A4)); draw(circle(extension(X,incenter(X,A1,A2),T1,T1+(A1-A2)*(0,1)),abs(extension(X,incenter(X,A1,A2),T1,T1+(A1-A2)*(0,1))-T1))); label("$A_1$", A1, W); label("$A_2$", A2, NE); label("$A_3$", A3, SE); label("$A_4$", A4, SW); label("$T_1$", T1, NW); label("$T_3$", T3, S); [/asy][/asy] We claim $A_2A_3$, $A_4A_1$, and $T_1T_3$ concur if and only if $\cos^2\frac{A_1}2\cos^2\frac{A_3}2=\cos^2\frac{A_2}2\cos^2\frac{A_4}2$. By the Ratio Lemma, this occurs if and only if $\frac{\sin A_1}{\sin A_2}\frac{A_1T_1}{T_1A_2}=\frac{\sin A_4}{\sin A_3}\frac{A_4T_3}{T_3A_3}$. Since $\frac{A_1T_1}{T_1A_2}=\frac{\tan\frac{A_2}2}{\tan\frac{A_1}2}$, the left hand side is equal to $$\frac{\sin A_1}{\sin A_2}\frac{\tan\frac{A_2}2}{\tan\frac{A_1}2}=\frac{\cos^2\frac{A_1}2}{\cos^2\frac{A_2}2}.$$Similarly, the right hand side is equal to $$\frac{\cos^2\frac{A_4}2}{\cos^2\frac{A_3}2},$$so both sides are equal if and only if $\cos^2\frac{A_1}2\cos^2\frac{A_3}2=\cos^2\frac{A_2}2\cos^2\frac{A_4}2$. Since this is symmetric, this happens if and only if $A_1A_2$, $A_3A_4$, and $T_2T_4$ concur, so both concurrencies are equivalent.

This can't be this simple can it? [asy][asy] import geometry; size(10cm); defaultpen(fontsize(9pt)); pen pri; pri=RGB(24, 105, 174); pen sec; sec=RGB(217, 165, 179); pen tri; tri=RGB(126, 123, 235); pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair O, X, A_2, A_3, O_2, U_2, U_3, T_2, O_4, U_1, U_4, K, T_4, A_1, A_4, Y, O_3, T_3, T_1,O_1, V_1, V_2; O=(0, 0); X=dir(140); A_2=dir(220); A_3=dir(340); O_2=incenter(X,A_2,A_3); U_2=foot(O_2,X,A_2); U_3=foot(O_2,X,A_3); T_2=foot(O_2,A_2,A_3); O_4=8O_2-7X; U_1=8U_2-7X; U_4=8U_3-7X; K=intersectionpoints(circle(U_2,U_3,T_2), line(X,T_2))[1]; T_4=8K-7X; A_1=extension(O_4,((U_1+T_4)/2),X,A_2); A_4=extension(A_1,T_4,X,A_3); Y=extension(A_2,A_3,A_1,A_4); O_3=incenter(Y,A_3,A_4); T_3=foot(O_3,A_3,A_4); T_1=extension(Y,T_3,X,A_2); O_1=extension(O_4,A_1,Y,O_3); V_1=foot(O_1,Y,A_4); V_2=foot(O_1,Y,A_3); filldraw((path)(A_1--A_2--A_3--A_4--cycle), white+0.1*pri, pri); draw(A_2--X, pri); draw(A_3--X, pri); draw(A_3--Y, pri); draw(A_4--Y, pri); draw(A_2--V_2, pri); draw(A_1--V_1, pri); draw(A_1--U_1, pri); draw(A_4--U_4, pri); draw(X--O_4,sec+dashed); draw(Y--O_1,sec+dashed); draw(X--T_4,sec); draw(Y--T_1,sec); draw(O_2--T_2,sec+dashed); draw(O_4--T_4,sec+dashed); filldraw(circumcircle(U_2,U_3,T_2), tfil, tri); filldraw(incircle(Y,A_3,A_4),tfil,tri); clipdraw(circle(O_4,abs(O_4-T_4)),tri); clipdraw(circle(O_1,abs(O_1-T_1)),tri); dot("$A_1$",A_1,dir(260)); dot("$A_2$",A_2,dir(140)); dot("$A_3$",A_3,dir(65)); dot("$A_4$",A_4,dir(0)); dot("$T_1$",T_1,dir(180)); dot("$T_2$",T_2,dir(260)); dot("$T_3$",T_3,dir(260)); dot("$T_4$",T_4,dir(290)); dot("$U_1$",U_1,dir(180)); dot("$U_2$",U_2,dir(140)); dot("$U_3$",U_3,dir(65)); dot("$U_4$",U_4,dir(20)); dot("$O_1$",O_1,dir(260)); dot("$O_2$",O_2,dir(50)); dot("$O_3$",O_3,dir(80)); dot("$O_4$",O_4,dir(50)); dot("$X$",X,dir(140)); dot("$Y$",Y,dir(65)); [/asy][/asy] Without loss of generality assume that $\overline{A_1A_2}$, $\overline{A_3A_4}$, $\overline{T_2T_4}$ concur at $X$, and define $Y $ as the intersection of $\overline{A_2A_3}$ and $\overline{A_1A_4}$. Define the center of $\omega_i$ to be $O_i$, and let $U_2$, $U_1$ and $U_3$, $U_4$ to be the feet from $O_2$ and $O_4$ to $\overline{A_1A_2}$ and $\overline{A_3A_4}$ respectively. Claim: The concurrency at $X$ is equivalent to $\overline{YT_2} = \overline{YT_4}$. Proof. Observe that $180 - \angle A_1T_4X = \angle A_2T_2X$ as from homothety both intercept the same arc measure in $\omega_2$ and $\omega_4$ respectively. Then it follows that, \begin{align*} \angle T_2T_4A_4 = \angle XT_2A_2 = \angle YT_2T_4 \end{align*}proving the claim. $\square$ Note then that $A_3T_2 + A_3T_3 = A_4T_3 + A_4T_4$. Claim: $A_1T_1 + A_1T_4 = A_2T_1 + A_2T_2$. Proof. The claim simply follows from noting the tangents from $Y$ to $\omega_1$ are equal in length, combined with the previous claim. $\square$ Now let $A_3T_2 + A_3T_3 = A_4T_3 + A_4T_4 = x$ and $A_1T_1 + A_1T_4 = A_2T_1 + A_2T_2 = y$. Claim: $x = y$. Proof. Note that, \begin{align*} 2x = U_3A_3 + A_3T_3 + T_3A_4 + A_4U_4 = U_3U_3 &= U_1U_2\\ &= U_2A_2 + A_2T_1 + T_1A_1 + A_1U_1\\ &= 2y \end{align*}which proves the claim. $\square$ However then it follows that $XT_1 = XT_3$ which we have previously shown to be equivalent to $\overline{A_2A_3}$, $\overline{A_1A_4}$, $\overline{T_1T_3}$ being concurrent. This finishes the problem (hopefully).

RMM 2010 p3 Here is a sketch [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(11.965728128960736cm); 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 = -4.545333464299365, xmax = 7.420394664661371, ymin = -4.1846818179915415, ymax = 2.737815733102022; /* image dimensions */ pen wwzzff = rgb(0.4,0.6,1.); pen ffqqtt = rgb(1.,0.,0.2); pen afeeee = rgb(0.6862745098039216,0.9333333333333333,0.9333333333333333); pen ffttww = rgb(1.,0.2,0.4); /* draw figures */ draw(circle((1.917232250314305,-2.7793903492715946), 1.244997039315295), linewidth(1.) + wwzzff); draw(circle((3.4995020041907616,-0.9475668498855881), 0.7665921011284942), linewidth(1.) + wwzzff); draw(circle((1.8502026731016514,-0.008308267626357484), 0.5743850769864233), linewidth(1.) + wwzzff); draw(circle((0.765261505294918,-1.1451546125070666), 0.2360795505494103), linewidth(1.) + wwzzff); draw((2.069500180477472,-1.5437398706379015)--(1.7927912473734944,2.365148044238439), linewidth(1.) + ffqqtt); draw((-0.4514830700572181,-1.2330817127101155)--(2.7603854386101982,-1.1509633886884703), linewidth(1.) + ffqqtt); draw((0.765261505294918,-1.1451546125070666)--(3.4995020041907616,-0.9475668498855881), linewidth(1.) + afeeee); draw((1.7927912473734944,2.365148044238439)--(1.917232250314305,-2.7793903492715946), linewidth(1.) + afeeee); draw((1.2896834569685265,0.11713664064100313)--(2.404001133584057,0.14409081974629342), linewidth(1.) + ffttww); draw((0.7022898840773685,-2.5074847137536045)--(3.117607175260799,-2.449060714719149), linewidth(1.) + ffttww); draw((0.7022898840773685,-2.5074847137536045)--(1.7927912473734944,2.365148044238439), linewidth(1.) + blue); draw((1.7927912473734944,2.365148044238439)--(3.117607175260799,-2.449060714719149), linewidth(1.) + blue); draw((-0.4514830700572181,-1.2330817127101155)--(3.4057448394488556,-1.7084039175762735), linewidth(1.) + blue); draw((-0.4514830700572181,-1.2330817127101155)--(3.297327631151047,-0.2081150445604557), linewidth(1.) + blue); /* dots and labels */ dot((0.9488824858649298,-1.405647311444667),dotstyle); label("$A_4$", (0.8270342670707617,-1.562201917500575), NE * labelscalefactor); dot((1.0812930933077354,-0.8140036097061314),dotstyle); label("$A_1$", (1.1243194379766186,-0.7128157149124076), NE * labelscalefactor); dot((2.5566523716662393,-0.41062392361651034),dotstyle); label("$A_2$", (2.6001279649735505,-0.3093572686830281), NE * labelscalefactor); dot((2.896518288088268,-1.645652456462818),dotstyle); label("$A_3$", (2.939882446008815,-1.5409672624358708), NE * labelscalefactor); dot((-0.4514830700572181,-1.2330817127101155),linewidth(4.pt) + dotstyle); label("$P$", (-0.40457572668207353,-1.1481261437388435), NE * labelscalefactor); dot((1.7927912473734944,2.365148044238439),linewidth(4.pt) + dotstyle); label("$Q$", (1.8356803826442043,2.451147889728516), NE * labelscalefactor); dot((3.4995020041907616,-0.9475668498855881),linewidth(4.pt) + dotstyle); label("$G$", (3.545070115352881,-0.8614583003653369), NE * labelscalefactor); dot((1.917232250314305,-2.7793903492715946),linewidth(4.pt) + dotstyle); label("$H$", (1.9630883130324288,-2.6982559634622487), NE * labelscalefactor); dot((0.765261505294918,-1.1451546125070666),linewidth(4.pt) + dotstyle); label("$I$", (0.8057996120060577,-1.0631875234800268), NE * labelscalefactor); dot((1.8502026731016514,-0.008308267626357484),linewidth(4.pt) + dotstyle); label("$J$", (1.8887670203059646,0.07286652248164721), NE * labelscalefactor); dot((2.069500180477472,-1.5437398706379015),linewidth(4.pt) + dotstyle); label("$T_{3}$", (2.111730898485357,-1.456028642177054), NE * labelscalefactor); dot((2.7603854386101982,-1.1509633886884703),linewidth(4.pt) + dotstyle); label("$T_2$", (2.801857188088239,-1.0631875234800268), NE * labelscalefactor); dot((2.001686027355041,-0.5623579134604122),linewidth(4.pt) + dotstyle); label("$T_1$", (2.048026933291245,-0.4792345092006616), NE * labelscalefactor); dot((0.995642011197483,-1.1967140606004063),linewidth(4.pt) + dotstyle); label("$T_4$", (1.1243194379766186,-1.0419528684153225), NE * labelscalefactor); dot((0.7022898840773685,-2.5074847137536045),linewidth(4.pt) + dotstyle); label("$Y$", (0.7420956468119455,-2.4222054476210944), NE * labelscalefactor); dot((3.117607175260799,-2.449060714719149),linewidth(4.pt) + dotstyle); label("$X$", (3.162846324188208,-2.369118809959334), NE * labelscalefactor); dot((2.404001133584057,0.14409081974629342),linewidth(4.pt) + dotstyle); label("$V$", (2.451485379520622,0.2321264354669286), NE * labelscalefactor); dot((1.2896834569685265,0.11713664064100313),linewidth(4.pt) + dotstyle); label("$R$", (1.3366659886236592,0.20027445286987233), NE * labelscalefactor); dot((0.7363880675429864,-1.3794618441069406),linewidth(4.pt) + dotstyle); label("$S$", (0.7739476294090015,-1.2967687291917727), NE * labelscalefactor); dot((3.4057448394488556,-1.7084039175762735),linewidth(4.pt) + dotstyle); label("$N$", (3.4495141675617127,-1.6259058826946875), NE * labelscalefactor); dot((0.7029999257018558,-0.9174331820395142),linewidth(4.pt) + dotstyle); label("$U$", (0.7420956468119455,-0.8296063177682806), NE * labelscalefactor); dot((3.297327631151047,-0.2081150445604557),linewidth(4.pt) + dotstyle); label("$M$", (3.343340892238192,-0.11824537310069044), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $P$ and $Q$ denote $A_1A_2 \cap A_3A_4$ and $A_2A_3 \cap A_1A_4$. Main claim is. Claim: $QT_2=QT_4$. We first show if the above holds implies that we have the desired concurrency. After that we show that $PT_1=PT_3$ iff $QT_2=QT_4$. Which is done by some easy length chase.

i love strophoids like = more overkill solutions [asy][asy] import graph; size(40cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(15); defaultpen(dps); real xmin=-4.3049852179398,xmax=25.62879315657571,ymin=18.543173353827218,ymax=36.30813843682065; pen ududff=rgb(0.30196078431372547,0.30196078431372547,1.), uuuuuu=rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); pair A_1=(17.817069746412454,26.319909233528513), A_2=(17.4185781329044,19.44159746993304), X=(18.363761621821734,35.75628638733201), Y=(-3.9390713480409496,19.899665748941352), I_2=(20.611157422405057,22.390496287153983), I_3=(19.648933475911473,23.0496345824509), I_1=(21.607997250294876,23.311250173859417), I_4=(20.278982393978048,24.366019114652243), T_2=(20.546472085395727,19.374511950947603), T_3=(23.028053246236436,24.44276391822182), T_4=(19.551359645919717,26.83169868090141), T_1=(17.656028976804365,23.540205514640988), A=(20.628150382374795,23.182801057663777), B=(20.446689235405728,23.378529983286242); draw(circle((20.598601434667856,23.404073317134834),1.0136547975187986),linewidth(2.)); draw(Y--(25.15835047755776,19.275598669316214),linewidth(2.)); draw((25.15835047755776,19.275598669316214)--X,linewidth(2.)); draw(X--A_2,linewidth(2.)); draw((21.77287736967186,27.48726919912399)--Y,linewidth(2.)); draw(Y--T_3,linewidth(2.)); draw(I_1--Y,linewidth(2.)); draw(I_2--X,linewidth(2.)); draw(T_1--I_1,linewidth(2.)); draw(I_3--T_3,linewidth(2.)); draw(A--T_2,linewidth(2.)); draw(A--T_4,linewidth(2.)); draw((14.933624366252051,25.469003673817035)--A_2,linewidth(2.)); dot(A_1,ududff); label("$A_1$",(17.06555417365283,26.65571554817517),NE*lsf,ududff); dot((25.15835047755776,19.275598669316214),ududff); label("$A_3$",(24.295481823708336,19.519683062406138),NE*lsf,ududff); dot((21.77287736967186,27.48726919912399),ududff); label("$A_4$",(20.971593007968533,27.65100428961138),NE*lsf,ududff); dot(A_2,ududff); label("$A_2$",(16.539741253648796,19.707473390979008),NE*lsf,ududff); dot(X,linewidth(4.pt)+uuuuuu); label("$X$",(17.572588060799582,35.76354648395933),NE*lsf,uuuuuu); dot(Y,linewidth(4.pt)+uuuuuu); label("$Y$",(-3.8730674622221986,20.233286310983043),NE*lsf,uuuuuu); dot(I_2,linewidth(4.pt)+uuuuuu); label("$I_2$",(20.633570416537367,22.054852498139873),NE*lsf,uuuuuu); dot(I_3,linewidth(4.pt)+uuuuuu); label("$I_3$",(19.03735262366797,23.12525737100523),NE*lsf,uuuuuu); dot(I_1,linewidth(4.pt)+uuuuuu); label("$I_1$",(21.722754322260013,23.10647833814794),NE*lsf,uuuuuu); dot(I_4,linewidth(4.pt)+uuuuuu); label("$I_4$",(20.314326857963486,24.683917098160045),NE*lsf,uuuuuu); dot(T_2,linewidth(4.pt)+uuuuuu); label("$T_2$",(19.957525233675035,19.669915325264434),NE*lsf,uuuuuu); dot(T_3,linewidth(4.pt)+uuuuuu); label("$T_3$",(22.267346275121337,24.683917098160045),NE*lsf,uuuuuu); dot(T_4,linewidth(4.pt)+uuuuuu); label("$T_4$",(18.94345745981534,27.068854271035484),NE*lsf,uuuuuu); dot(T_1,linewidth(4.pt)+uuuuuu); label("$T_1$",(17.009217075080972,23.91397675101128),NE*lsf,uuuuuu); dot(A,linewidth(4.pt)+uuuuuu); label("$A$",(20.689907515109226,22.86235091100321),NE*lsf,uuuuuu); dot(B,linewidth(4.pt)+uuuuuu); label("$B$",(20.445780087964497,23.594733192437403),NE*lsf,uuuuuu); dot((14.933624366252051,25.469003673817035),linewidth(4.pt)+uuuuuu); label("$A_4'$",(14.229920212202492,25.735542938168113),NE*lsf,uuuuuu); dot((15.854929579108406,23.23432195451809),linewidth(4.pt)+uuuuuu); label("$T_3'$",(15.093755723637694,23.444500929579107),NE*lsf,uuuuuu); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Unlike pretty much everyone else, i was too lazy to draw circles so I completely ignored them and only considered incentres/excentres. Let $X=A_1A_2\cap A_3A_4$. Let $Y=A_1A_4\cap A_2A_3$. Assume that $Y,T_1,T_3$ are collinear. It suffices to show that this implies $X,T_2,T_4$ collinear, as then some relabelling gives the converse. Let $I_i$ be the intersection of the angle bisectors of $\measuredangle A_{i-1}A_iA_{i+1}$ and $\measuredangle A_iA_{i+1}A_{i+2}$. Note that because $I_4$ is the excentre of $XA_1A_4$, that \[\measuredangle A_4I_4A_1=90^\circ-\measuredangle A_1XA_4 = 180^\circ-\measuredangle A_2I_2A_3\]Thus $I_1I_2I_3I_4$ is cyclic. Further, $Y,I_1,I_3$ are collinear, being an angle bisector, and so are $X,I_2,I_4$. We thus proceed onto the crux of the problem: Claim: $XT_1=XT_3$ Proof: It suffices to show that $\measuredangle A_2T_1T_3=\measuredangle T_1T_3A_3$. Let $A_4'$ be the point on $A_1A_4$ such $A_4'A_2\parallel A_3A_4$. If $A_2A_4'\cap T_1T_3=T_3'$, we only need $A_2T_3'=A_2T_1$. Note that by homothety, $T_3'$ is the $Y$-intouch point and $T_1$ the $Y$-extouch point of their respective triangles. Fix $\triangle A_1A_2Y$. We try to characterize where $T_3'$ can be. Move the incentre $I$ of $A_2A_4'Y$ along the angle bisector of $\measuredangle A_2YA_1$. Let $T$ be the in-touch wrt $Y$ of $\triangle A_2A_4'Y$. Claim: $T$ moves on a strophoid. Consider the oblique strophoid with pole $A_2$ and fixed point $O$, the foot from $A_2$ to $YI$. Our conic (line) for reference will the the midline wrt $A_2$ of $\triangle A_2A_1Y$. This passes through $O$ because of Iran Lemma. Let $M$ be the midpoint of $A_2Y$ and $A_2A4'\cap MO=N$. It suffices to show that $NT=NO$, for $T$ to lie on the desired strophoid. However, by Iran Lemma, because $MO$ is also a midline in $\triangle A_4'YA_2$, if $OT\cap A_4'Y=S$, then $S$ is another intouch point in $\triangle A_4'YA_2$, or equivalently $A_4'S=A_4'T$. Yet by homothety at $T$, due to $OM\parallel A_4'Y$, we have $NT=NO$. Hence, $T$ moves on this strophoid. Now, suppose we intersect $YT_1$ with this strophoid, say $\mathcal{S}$, which gives $T_3'$ Now, because we have $MY=MA_2=MO$ by Thales theorem (recall that $\measuredangle YOA_2$ is right), by the definition of a strophoid, $YA_2$ is tangent to $\mathcal{S}$ because its other intersection is just $A_2$. Now, we generalize the claim that $A_2T_3'=A_2T_1$. Indeed, we claim that for any point $P$ on $\mathcal{S}$, if $YP\cap \mathcal{S}=Q$, then $A_2P=A_2Q$. We will draw a new diagram Reformulate the statement more generally. For any oblique strophoid $\mathcal{S}$, let the reference line be $\ell$. Let $A_2$ be the pole of $\mathcal{S}$, and $O$ the other fixed point. If the tangent at $A_2$ to $\mathcal{S}$ intersects $\mathcal{S}$ again at $Y$, and some arbitrary line through $Y$ intersects $\mathcal{S}$ at $P,Q$, prove that $A_2P=A_2Q$. Recall that $\measuredangle A_2OY$ is right from our lemmas before. Reconstruct $Y$ as the intersection of $PQ$ and the perpendicular to $A_2O$ through $O$. Note that by the definition of a strophoid, it suffices to show that the midpoint of $A_2Y, F$ lies on $\ell$, or equivalently by Thales that $\triangle FA_2O$ is isosceles. By definition, if $A_2Q\cap \ell=Q'$ and $A_2P\cap \ell=P'$, $P'O=P'P$ and $Q'O=Q'Q$. Thus, by incircle substitution's uniqueness, as $A_2P=A_2Q$ also, $P,Q,O$ are the intouch points thus $(PQO)$ is the incircle of $\triangle A_2P'Q'$. Let $A_2O$ intersect $(PQO)$ again at $O'$, and $PQ$ at $U$. Let the tangent at $O'$ to $(OPQ)$ intersect $\ell$ at $E$. $E$ is the pole of $A_2O$ hence by La-Hire $E$ lies on $PQ$, the polar of $A$. Further, by definition of poles and polars, if $I'$ is the centre of $(OPQ)$, then $I'E\perp A_2O$. Note that as $\triangle EO'O$ is isosceles by ice cream cone theorem, we just want $FA_2\parallel O'E$. Yet $YO\perp A_2O$ thus $I'E\parallel YO$. Hence, \[\frac{UE}{UY}=\frac{UV}{UO}=\frac{UO'}{UA_2}\]which suffices. Note that $\frac{UV}{UO}=\frac{UO'}{UA_2}$ is true because upon rearranging, we want $UO\times UO'=UA_2\times UV$ but this follows from $UO\times UO'=UP\times UQ=UV\times UA_2$ by power of a point on $(OPQ)$ and cyclic $A_2PVI'Q$ (because $\measuredangle I'VA_2$ is right etc.). Hence, $FA_2\parallel O'E$, so our claim is true. Thus, indeed we do have $XT_1=XT_3$. Now, let $T_1I_1\cap T_3I_3=B$. Let $T_4I_4\cap T_1I_1=A$. Because of $XT_1=XT_3$, and right angles (note $\measuredangle XT_1I_1, \measuredangle XT_3I_3$ are right by tangency definition), we have $XT_1BT_3$ is a kite. Thus, $XB$ is the bisector of $\measuredangle A_1XA_4$, hence $X,I_4,B,I_2$ are collinear. (We showed $X,I_4,I_2$ collinear earlier) Claim: $A,B$ are isogonal conjugates in $I_1I_2I_3I_4$ Note that because \[\measuredangle I_4BI_3+\measuredangle I_2BI_1=180^\circ-XBT_3+\measuredangle T_1BX=180^\circ\]$B$ lies on the isoptic cubic of $I_1I_2I_3I_4$ hence has an isogonal conjugate. Note that $A_1, I_1,I_4$ collinear because they lie on a common angle bisector, hence $\measuredangle AI_4I_1=90^\circ-\frac{\measuredangle A_1}{2}=\measuredangle A_4I_4X=\measuredangle I_3I_4I_2$ because of Fact 5 (Bowtie $\measuredangle A_1I_4X$ to half of $\measuredangle A_4A_1X$ and use right angles), and as $A_4,I_4,I_3$ are analogously collinear. Consequently, $I_4A$ is isogonal to $I_4I_2$. But as $I_4,I_2,B$ collinear, $I_4A$ and $I_4B$ are isogonal. Analogously, $I_2A$ and $I_2B$ are isogonal. Hence $A,B$ are isogonal conjugates (2 isogonal lines is enough to uniquely characterize a conjugate). Now, we proceed with another angle chase \[\measuredangle I_2I_3I_1=\measuredangle I_2I_4I_1=\measuredangle XI_4A_1=\measuredangle T_3I_3A_4=\measuredangle BI_3I_4\]Hence $I_1I_3$ and $I_3B$ are isogonal, yet $I_3B$ and $I_3A$ are isogonal, thus $A$ must lie on $I_1I_3$. But $I_1I_3$ also passes through $Y$ ! Consequently, by symmetry, as $YT_4AT_2$ is a kite from $A$ lying on the angle bisector of $\measuredangle A_2YA_1$. Thus, $YT_4=YT_2$. Now, going all the way back to our proof of $XT_1=XT_3$, we actually already showed this converse direction when we assumed $A_2P=A_2Q$ to prove $Y,P,Q$ collinear. Hence, we don't have to worry about this converse direction at all as it is already covered!!! Consequently, $X,T_4,T_2$ are indeed collinear, so we are done.