We use EGMO 2019 P4 to solve this. I have changed the labellings a little and we find $E'F'B'C'$ as the desired bicentral quadrilateral. Let the midpoints of the arcs be $P,Q,R$ respectively. We first draw $\odot(PQR)$ and its center $I$. Now let $\ell_1, \ell_2,\ell_3$ denote the tangents to $\odot(PQR)$ at $P,Q,R$ respectively. Now let $A=\ell_1 \cap \ell_3, B=\ell_1 \cap \ell_2, C=\ell_2 \cap \ell_3$. Now, let the line perpendicular to $AI$ at $I$ and the perpendicular bisector of $BI$ meet at $O_1$. Now we draw a circle $\odot(O_1)$ with centre $O_1$ and radius $r=O_1B=O_1I$. Now let $E=\odot(O_1) \cap AB$. We first observe that as the centre of $\odot(EBI)$ is $O_1$ and $\angle AIO_1 = 90^{\circ} \implies AI$ is tangent to $\odot(EBI)$. So, we just constructed the circle through $B$ tangent to $AI$ at $I$ which meets side $AB$ again at $E$. Similarly, we construct the point $F$ on $AC$ such that the circle through $C$ tangent to $AI$ at $I$ meets side $AC$ again at $F$. Using EGMO 2019 P4, we know that $EF$ is tangent to the incircle of $\triangle ABC$ which is $\odot(PQR)$. We also know from the same that $EFCB$ is bicentral. Now $X,Y,Z$ denote the midpoints of the arcs of $EB,BC,CF$ of $\odot(EFCB)$. Note that as $XB=XE \implies$ tangent to $\odot(EBCF)$ at $X \parallel BE$. Similarly, the tangents to $\odot(EBCF\}$ at $Y,Z$ are $\parallel BC,CF$ respectively. So, let $K,L,M=\cap$ of tangents at $\{X,Z\},\{X,Y\},\{Y,Z\}$ respectively. So, now we know that $KL\parallel AB,LM \parallel BC,MK\parallel CA \implies \{KA,LB,MC\}$ are concurrent at the homothetic center of $\{\triangle ABC, \triangle KLM\}$. Let $T=KA\cap LB \cap MC$. Thus we can deduce that there exists a homothety $\Psi$ centred at $T$ that maps $\triangle KLM \mapsto \triangle ABC$. We have the following maps for $\Psi$: \begin{align*} \text{incircle of } \triangle KLM \mapsto \text{incircle of } \triangle ABC &\implies \odot(EBCF) \mapsto \odot(PQR)\\ KL-\text{intouch point} \mapsto AB-\text{intouch point}&\implies X \mapsto P\\ LM-\text{intouch point} \mapsto BC-\text{intouch point}&\implies Y \mapsto Q\\ MK-\text{intouch point} \mapsto CA-\text{intouch point}&\implies Z \mapsto R \end{align*} Now we move onto the final part. Let $E'=TE \cap \odot(PQR),B'=TB \cap \odot(PQR),C'=TC \cap \odot(PQR),F'=TF \cap \odot(PQR)$. As, $\Psi$ maps $\odot(EBCF) \mapsto \odot(PQR)$, we also get that $E\mapsto E',B\mapsto B',C\mapsto C',F\mapsto F'$. I claim that the quadrilateral $E'B'C'F'$ is our desired bicentral quadrilateral. Firstly note that as $EBCF$ is bicentral, $E'B'C'F'$ also becomes bicentral. Thus it is enough to show that $P,Q,R$ are the midpoints of the arcs $E'B',B'C',C'F'$ of $\odot(E'B'C'F')$. Firstly see that as $X$ is the midpoint of arc $BE$ of $\odot(EBCF)$, and $E\mapsto E', F\mapsto F',X\mapsto P$, we get $P$ is the midpoint of the arc $B'E'$ of $\odot(E'B'C'F')$. Similarly, $Q,R$ are the midpoints of the arcs $B'C',C'F'$ and we are done.

Okay, look I might get cancelled for this solution, but it's all fair game. No one said you can't bash lengths using constructions.

Attachments:
Sharygin-Problem-18.pdf (137kb)

consider the labellings as in the diagram below . We are given $Q,R,S$ and we need to restore $ABCD$. It is well known that $UW \perp VX \iff ABCD$ is bicentric . $T$ is midpoint of arc $AD$ . Note that $QRST$ and $VWXU$ are homothetic so $RT \perp SQ$ so we construct $T$ from here . Now drawing tangents at $Q,R,S,T$ and marking their intersections we have $A'B'C'D'$ and $ABCD$ are homothetic . Their circumcenters be $O',O$ respectively . construct $H$ the external homothetic center of $(O),(O')$ which is the homothetic center of $ABCD,A'B'C'D$ so we mark intersections $HA',HB',HC',HD'$ with $(O)$ to restore $ABCD$.

Attachments:

GeoKing wrote: consider the labellings as in the diagram below . We are given $Q,R,S$ and we need to restore $ABCD$. It is well known that $UW \perp VX \iff ABCD$ is bicentric . $T$ is midpoint of arc $AD$ . Note that $QRST$ and $VWXU$ are homothetic so $RT \perp SQ$ so we construct $T$ from here . Now drawing tangents at $Q,R,S,T$ and marking their intersections we have $A'B'C'D'$ and $ABCD$ are homothetic . Their circumcenters be $O',O$ respectively . construct $H$ the external homothetic center of $(O),(O')$ which is the homothetic center of $ABCD,A'B'C'D$ so we mark intersections $HA',HB',HC',HD'$ with $(O)$ to restore $ABCD$. exactly what I did to solve it.