\documentclass[11pt]{scrartcl}
\usepackage{evan}

\begin{document}
\title{Twitch 032.3}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 32}
\maketitle

\section*{Problem}
Square $DEAL$, with side length $2$, is drawn,
sharing a vertex with unit square $DIPS$. $DIPS$ has centroid $R$.
Finally, square $RENT$ is drawn.
If all of these squares were drawn in counterclockwise order of vertices,
and $IAN$ is a line, find the possible values for the side length of $RENT$.

\section*{Video}
\href{https://www.youtube.com/watch?v=Nrj2bc9L_Tw&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/Nrj2bc9L\_Tw}}



\newpage

\section*{Solution}
We delete the squares, basically,
and focus on $45\dg-45\dg-90\dg$ triangles $DRI$, $DEA$ $REN$.
Let us define $X$ as the intersection of $\ol{DR}$ and $(DAE)$.

The spiral similarity between $DEA$ and $REN$ gives a rotation
\[ \triangle RDE \cong \triangle NAE \]
with $RD = NA = 1/\sqrt2$ and $DE = EN = 2$.

By the displayed rotation we have $\ol{NA} \perp \ol{RD}$,
so if we let $X$ be their intersection, we get a right angle.
Hence $IR$ and $XAN$ are parallel.
\begin{center}
\begin{asy}
import graph; size(8.74833cm);
pen zzttqq = rgb(0.6,0.2,0.); pen qqwuqq = rgb(0.,0.39215,0.);
pair D = (-1.,0.), A = (1.,2.), L = (-1.,2.), I = (-0.60097,0.91694), P = (-1.51791,1.31596), R = (-1.25895,0.65798), X = (-1.41332,1.05019);
draw(D--I--R--cycle, linewidth(0.6) + zzttqq);
draw(D--(1.,0.)--A--cycle, linewidth(0.6) + qqwuqq);
draw(R--(1.,0.)--(1.65798,2.25895)--cycle, linewidth(0.6) + zzttqq);
draw(D--I, linewidth(0.6) + zzttqq);
draw(I--R, linewidth(0.6) + zzttqq);
draw(R--D, linewidth(0.6) + zzttqq);
draw(D--(1.,0.), linewidth(0.6) + qqwuqq);
draw((1.,0.)--A, linewidth(0.6) + qqwuqq);
draw(A--D, linewidth(0.6) + qqwuqq);
draw(R--(1.,0.), linewidth(0.6) + zzttqq);
draw((1.,0.)--(1.65798,2.25895), linewidth(0.6) + zzttqq);
draw((1.65798,2.25895)--R, linewidth(0.6) + zzttqq);
// draw((xmin, 0.39356*xmin + 1.60643)--(xmax, 0.39356*xmax + 1.60643), linewidth(0.6)); /* line */
draw(circle((0.,1.), 1.41421), linewidth(0.6));
draw(D--X--(1.65798,2.25895), linewidth(0.6));
dot("$D$", D, dir(225));
dot("$E$", (1.,0.), dir(-45));
dot("$A$", A, dir((1.406, 3.575)));
// dot("$L$", L, dir(135));
dot("$I$", I, dir(-45));
// dot("$P$", P, dir((1.347, 3.716)));
// dot("$S$", (-1.91694,0.39902), dir((1.370, 2.718)));
dot("$R$", R, dir((1.318, 3.049)));
dot("$N$", (1.65798,2.25895), dir((1.355, 2.829)));
dot("$X$", X, dir(X-A));
\end{asy}
\end{center}

Hence if $I$ lies on line $AN$,
we need to actually have $X=R$.
One of two possibilities is shown below
(in the other, $X$ is on the other side of line $AD$).

\begin{center}
\begin{asy}
import graph; size(8.74833cm);
pen zzttqq = rgb(0.6,0.2,0.); pen qqwuqq = rgb(0.,0.39215,0.);
pair D = (-1.,0.), A = (1.,2.), I = (-0.74909,0.96801), R = (-1.35855,0.60945), X = (-1.35985,0.61166);
draw(D--I--R--cycle, linewidth(0.6) + zzttqq);
draw(D--(1.,0.)--A--cycle, linewidth(0.6) + qqwuqq);
draw(R--(1.,0.)--(1.60945,2.35855)--cycle, linewidth(0.6) + zzttqq);
draw(D--I, linewidth(0.6) + zzttqq);
draw(I--R, linewidth(0.6) + zzttqq);
draw(R--D, linewidth(0.6) + zzttqq);
draw(D--(1.,0.), linewidth(0.6) + qqwuqq);
draw((1.,0.)--A, linewidth(0.6) + qqwuqq);
draw(A--D, linewidth(0.6) + qqwuqq);
draw(R--(1.,0.), linewidth(0.6) + zzttqq);
draw((1.,0.)--(1.60945,2.35855), linewidth(0.6) + zzttqq);
draw((1.60945,2.35855)--R, linewidth(0.6) + zzttqq);
draw(circle((0.,1.), 1.41421), linewidth(0.6));
draw(D--X, linewidth(0.6));
draw(X--(1.60945,2.35855), linewidth(0.6));
dot("$D$", D, dir(225));
dot("$E$", (1.,0.), dir(-45));
dot("$A$", A, dir((1.406, 3.575)));
// dot("$L$", L, dir(135));
dot("$I$", I, dir(-45));
// dot("$P$", P, dir((1.347, 3.716)));
// dot("$S$", (-1.91694,0.39902), dir((1.370, 2.718)));
dot("$R$", R, dir((1.318, 3.049)));
dot("$N$", (1.60945,2.35855), dir((1.538, 2.928)));
dot("$X$", X, dir(X-A));
\end{asy}
\end{center}

The extraction is Ptolemy's theorem, left as exercise.
Answer is $\half(\sqrt{15}\pm1)$.

\end{document}
