Let $O$ be the radical center of $\Gamma_1,\Gamma_2,\Gamma_3$ and $O_1,O_2$ be the centers of $\Gamma_1,\Gamma_2$ ,so $O$ is the circumcenter of $\Delta BST$ now $\angle CST=90+\angle BST$ $=90+\frac{\angle BOT}{2}$ $=180-\frac{\angle BO_1T}{2}$ $=180-\angle CAT$ so $A,C,S,T$ is cyclic ,now if $ST\cap AB=X$ since $T,S$ are antihomologous points in $\Gamma_1,\Gamma_2$ so $X$ is the exscimilicenter so $X,D,E$ are collinear so $(XD)(XE)=(XS)(XT)=(XA)(XC)$ so $A,C,D,E$ are also cyclic $\blacksquare$ Alternatively:Invert at $B$ then the problem becomes very trivial.

enhanced wrote: Let $O$ be the radical center of $\Gamma_1,\Gamma_2,\Gamma_3$ and $O_1,O_2$ be the centers of $\Gamma_1,\Gamma_2$ ,so $O$ is the circumcenter of $\Delta BST$ now $\angle CST=90+\angle BST$ $=90+\frac{\angle BOT}{2}$ $=180-\frac{\angle BO_1T}{2}$ $=180-\angle CAT$ so $A,C,S,T$ is cyclic ,now if $ST\cap AB=X$ since $T,S$ are antihomologous points in $\Gamma_1,\Gamma_2$ so $X$ is the exscimilicenter so $X,D,E$ are collinear so $(XD)(XE)=(XS)(XT)=(XA)(XC)$ so $A,C,D,E$ are also cyclic $\blacksquare$ Alternatively:Invert at $B$ then the problem becomes very trivial. What is an antihomologous point @ enhanced ?

Electron_Madnesss wrote: enhanced wrote: Let $O$ be the radical center of $\Gamma_1,\Gamma_2,\Gamma_3$ and $O_1,O_2$ be the centers of $\Gamma_1,\Gamma_2$ ,so $O$ is the circumcenter of $\Delta BST$ now $\angle CST=90+\angle BST$ $=90+\frac{\angle BOT}{2}$ $=180-\frac{\angle BO_1T}{2}$ $=180-\angle CAT$ so $A,C,S,T$ is cyclic ,now if $ST\cap AB=X$ since $T,S$ are antihomologous points in $\Gamma_1,\Gamma_2$ so $X$ is the exscimilicenter so $X,D,E$ are collinear so $(XD)(XE)=(XS)(XT)=(XA)(XC)$ so $A,C,D,E$ are also cyclic $\blacksquare$ Alternatively:Invert at $B$ then the problem becomes very trivial. What is an antihomologous point dear enhanced ? see here

Thanks.$ $

Part (1): Notice that $\triangle TSB $ is tangent to $AC$. Extend $AT, CS$, let $Q=AT\cap CS$. Hence, $\angle TAB+\angle TSC=90^{\circ}-\angle TBA+\angle TBA + 90^{\circ}=180^{\circ}$. Part (2): By the Radical Lemma, it suffices to prove that $T, S, H$ are colinear. We apply Menelaus' Theorem to $\triangle O_1O_2O_3$( where $O_1, O_2, O_3$ are centers of the circles). Hence, the collinearity condition is equivalent to proving $\frac{O_1H}{O_2H}=\frac{r_1}{r_2}$, which is immediate if we consider the positive homothety that maps $\Gamma_2$ to $\Gamma_1$ with ratio $\frac{r_1}{r_2}$. Hence, $A, C, D, E$ are concyclic.

Suppose $X,Y,Z$ are the centers of $\Gamma_3, \Gamma_1, \Gamma_2$ respectively. Notice that $\triangle BST$ is the contact triangle of $\triangle XYZ$. Let $B'$ be the antipode of $B$ in $\odot(BST)$; then $\angle BSC=\angle BSB'=90^{\circ}$ hence $C,S,B'$ are collinear. Similarly $A,T,B'$ are collinear. Now $$B'S \cdot B'C=B'B^2=B'T \cdot B'A$$hence $A,C,S,T$ are concyclic, proving part (i). Let $L=\overline{PQ} \cap \overline{AC}$. Then $L$ is the ex-similicenter of $\Gamma_1, \Gamma_2$ so $\tfrac{LY}{LZ}=\tfrac{YP}{ZQ}=\tfrac{YB}{ZB}$ hence $(ZY; BL)=-1$. Thus, $L$ lies on line $\overline{ST}$. Now $ACST$ is cyclic so $$LA \cdot LC=LS \cdot LT$$while $DEST$ is cyclic so $$LD \cdot LE=LS \cdot LT$$proving $LA \cdot LC=LD \cdot LE$ hence $A,C,D,E$ concyclic, proving part (ii). $\blacksquare$

Quite similar to the above solution but still posting. Let $O_1, O_2, O_3$ be the circumcenters of $\Gamma_1, \Gamma_2, \Gamma_3$. Also let $PQ \cap \ell = K$. As $B$ and $K$ are the insimilicenter and the exsimilicenter respectively of $\Gamma_1$ and $\Gamma_2 \Rightarrow (O_1,O_2;K,B) = -1$. Also, $B, S, T$ are the points where the incircle of $\triangle O_1O_2O_3$ touches its sides $\Rightarrow (O_1,O_2;ST \cap \ell,B) = -1 \Rightarrow ST \cap \ell = K$ 1. $\frac{KO_1}{BO_1} = \frac {KO_2}{BO_2} \Rightarrow \frac{KO_1-BO_1}{KO_1+BO_1} = \frac{KO_2-BO_2}{KO_2+BO_2} \Rightarrow \frac{KB}{KA} = \frac{KC}{KB} \Rightarrow KB^2 = KC \cdot KA$ Also,$KB$ is tangent to $\odot (BST) \Rightarrow KS \cdot KT = KB^2 = KA \cdot KC \Rightarrow A,C,S,T$ are concyclic. 2. $KD \cdot KE = KS \cdot KT = KA \cdot KC \Rightarrow A,C,D,E$ are concyclic.

Very nice

Is the condition that $DE$ is diameter really required? Invert about $B$ with arbitrary radius and the rephrased question is: Inverted problem wrote: Let $A,B,C$ are collinear points in that order on a line $\ell$. Let $\ell_1$ be line through $A$ perpendicular to $\ell$ and $\ell_2$ be line through $C$ perpendicular to $\ell$. Let $\odot(PQ)$ be a circle with diameter $PQ$ passing through $B$ and tangent to $\ell_1$ at $P$ and $\ell_2$ at $Q$. Also let $\odot(TS)$ be a circle with diameter $TS$ and tangent to $\ell_1$ at $T$ and $\ell_2$ at $S$. Let $\odot(TS)\cap \odot(PQ) = D,E$. Then show that $A,C,S,T$ are concyclic and $A,C,D,E$ are concyclic. Firstly $ACST$ is a rectangle which is cyclic. By symmetry, $DE||TS||AC$ which gives $A,C,D,E$ are concyclic and we are done.

^no, the $DE$ diameter is not required; your solution is right.

The original figure:

Attachments:

And here's the inverted figure (Inversion at B with arbitrary radius) :

Attachments:

The proof of concyclicity of ACTS is that as the quad A'C'T'S' is cyclic(rectangle) and as A'C'T'S' is a circle not passing through the centre of inversion, it's pre-image, i.e- ACTS is also a circle (not passing through the centre of inversion). The proof of concyclicity of ACDE is- First, note that P'Q' and T'S' are diameters of $\odot(B'P'Q')$ and $\odot(T'S'D'E')$. As T'S'||P'Q'||A'C', we can say that the line joining the centres is perpendicular to these three lines. As D'E' is the radical axis of $\odot(B'P'Q')$ and $\odot(T'S'D'E')$, implies that it is perpendicular to the line joining the centres, i.e- D'E'||T'S'||P'Q'||A'C'. This, along with the fact that D'A'=E'C'(Symmetry), gives that D'E'A'C' is an isosceles trapezoid, which in turn implies that it is cyclic. As A'C'D'E' is a circle not passing through the centre of inversion, it's pre-image, i.e- ACDE is also a circle (not passing through the centre of inversion).

Nothing much different but follows a slight different approach, India TST #1 2018 P2 wrote: Let $A,B,C$ be three points in that order on a line $\ell$ in the plane, and suppose $AB>BC$. Draw semicircles $\Gamma_1$ and $\Gamma_2$ respectively with $AB$ and $BC$ as diameters, both on the same side of $\ell$. Let the common tangent to $\Gamma_1$ and $\Gamma_2$ touch them respectively at $P$ and $Q$, $P\ne Q$. Let $D$ and $E$ be points on the segment $PQ$ such that the semicircle $\Gamma_3$ with $DE$ as diameter touches $\Gamma_2$ in $S$ and $\Gamma_1$ in $T$. Prove that $A,C,S,T$ are concyclic. Prove that $A,C,D,E$ are concyclic. Solution: Let $PQ \cap AC=F$ $\implies$ $\Delta QBC \stackrel{F}{\mapsto} \Delta PAB$ $\implies$ $\angle PBQ=90^{\circ}$ $\implies$ $APQC$ cyclic. Let $K.L.M$ be the midpoints of $AB$,$BC$,$DE$ $\implies$ $M$ $-$ $K$ $-$ $T$ and $M$ $-$ $S$ $-$ $L$. Also, $\Delta BTS$ is contact triangle WRT $\Delta KLM$. Let $I$ be incenter of $\Delta KLM$ $$\begin{cases} \angle ATS=90^{\circ}+\angle BTS=90^{\circ}+\frac{1}{2}\angle BIS=180^{\circ}-\frac{1}{2}\angle SLB=180^{\circ} -\angle TSC \\\ \angle TPQ=\angle TAP=\angle BAP-\angle TAC=\angle CBQ-(180^{\circ}-\angle TSC)=\angle CSQ+\angle TSC-180^{\circ}=180^{\circ}-\angle TSQ \end{cases} \Longrightarrow ATSC, PQST \text{ cyclic}$$Apply Radical Axes Theorem $\implies$ $TS$ passes through $F$ and Apply POP to conclude $ADEC$ cyclic

My first time using inversion on an actual olympiad problem (Yay) so I might have messed up. Invert at $B$ with arbitrary radius. Then, the semicircles with diameters $AB,BC$ invert to parallel lines and $A',C'$ are the feet of perpendiculars from $B$ onto the lines. The line $PQ$ inverts to a circle passing through $B$ and tangent to the two lines at $P',Q'$. The circle $DEST$ inverts to another circle tangent to the parallel lines at $S',T'$. We see that $A'C'S'T"$ is a rectangle and so is cyclic. So, $ACST$ must be cyclic as well. Similarly, we get that $PQCA$ and $PQST$ are cyclic. So, by radical axis theorem, $PQ,ST,AC$ are concurrent at a point $Z$. Then, by PoP, we have that $ZA.ZC = ZS.ZT = ZD.ZE$ and so $ADEC$ is cyclic as well

Ankoganit wrote: Let $A,B,C$ be three points in that order on a line $\ell$ in the plane, and suppose $AB>BC$. Draw semicircles $\Gamma_1$ and $\Gamma_2$ respectively with $AB$ and $BC$ as diameters, both on the same side of $\ell$. Let the common tangent to $\Gamma_1$ and $\Gamma_2$ touch them respectively at $P$ and $Q$, $P\ne Q$. Let $D$ and $E$ be points on the segment $PQ$ such that the semicircle $\Gamma_3$ with $DE$ as diameter touches $\Gamma_2$ in $S$ and $\Gamma_1$ in $T$. Prove that $A,C,S,T$ are concyclic. Prove that $A,C,D,E$ are concyclic. Invert around $B$. $ACST$ becomes a rectangle and $ACDE$ becomes an isosceles trapezoid,hence done.

For (a): Let $O,O_1,O_2$ the midpoints of $DE,AB,BC$ respectively. $O_1,T,O$ and $O_2,S,O$ are collinear. Let $K$ be the intersection of $AT$ and $SC$. $\angle BTK= \angle BSK =90^{\circ}$ so $KSBT$ is cyclic. $\angle TBK= \angle KAC= \angle TSK$. So, $TSCA$ is cyclic.