\documentclass[11pt]{scrartcl}
\usepackage{evan}
\begin{document}
\title{Math Prize 2022/2}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 108}
\maketitle
\section*{Problem}
Determine, with proof,
whether or not there exists a \emph{non-isosceles} trapezoid $ABCD$
such that the lengths $AC$ and $BD$ both lie in the set
$\{DA+AB, AB+BC, BC+CD, CD+DA, AB+CD, BC+DA\}$.
\section*{Video}
\href{https://www.youtube.com/watch?v=C9WhZldyMuc&t=2259s}{\texttt{https://youtu.be/C9WhZldyMuc}}
\section*{External Link}
\url{https://aops.com/community/p26628697}
\newpage
\section*{Solution}
The answer is yes, such a trapezoid exists.
We present two possible direct constructions and one indirect one.
\paragraph{First construction}
Let $A=(0, 0)$, $B=(1, 0)$, $C=(2\sqrt2, 2\sqrt2)$, $D=(1, 2\sqrt2)$,
as shown below.
\begin{center}
\begin{asy}
size(7cm);
pair A = (0,0);
pair B = (1, 0);
pair C = (2.828, 2.828);
pair D = (1, 2.828);
filldraw(A--B--C--D--cycle, opacity(0.1)+yellow, red);
draw(B--D, dotted+brown);
dot("$A$", A, dir(225));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
label("$1$", midpoint(A--B), dir(-90), red);
label("$3$", midpoint(A--D), dir(130), red);
label("$2\sqrt2-1$", midpoint(C--D), dir(90), red);
label("$2\sqrt2$", midpoint(B--D), dir(0), brown);
\end{asy}
\end{center}
Then $AC = 4 = 1+3 = AB+AD$, and $BD = 2\sqrt2 = 1 + (2\sqrt2-1) = AB + CD$.
Also, we clearly have $AB \parallel CD$
(since $\angle DBA = \angle CDB = 90\dg$) and $3 = AD \neq BC$,
so this construction is valid.
\paragraph{Second construction}
Construct two similar right triangles $AOM$ and $CON$,
where $OM = 33$, $MA = 44$, $OA = 55$
and $ON = 105$, $NC = 140$, $CO = 175$.
Situate these triangles such that $AOC$ and $MON$ are collinear.
Finally, let $B$ and $D$ be the reflections
of $O$ over $M$ and $N$, respectively.
The resulting figure is depicted below.
\begin{center}
\begin{asy}
size(13cm);
pair A = (-33, 44);
pair B = (-66, 0);
pair C = (105, -140);
pair D = (210, 0);
pair M = (B.x/2,0);
pair N = (D.x/2,0);
label("$55$", midpoint(A--B), dir(A+B), blue);
label("$221$", midpoint(B--C), dir(B+C-N), blue);
label("$175$", midpoint(C--D), dir(C+D), blue);
filldraw(A--B--C--D--cycle, opacity(0.1)+lightcyan, blue);
draw("$33$", origin--M, dir(90), deepgreen);
draw("$33$", M--B, dir(90), deepgreen);
draw("$105$", origin--N, dir(-90), deepgreen);
draw("$105$", N--D, dir(-90), deepgreen);
draw("$44$", A--M, lightred);
draw("$140$", C--N, lightred);
draw("$55$", A--origin, dir(45), blue);
draw("$175$", origin--C, dir(45), blue);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$O$", origin, dir(225));
dot("$M$", M, dir(-90));
dot("$N$", N, dir(90));
\end{asy}
\end{center}
Then because $\triangle AOB$ and $\triangle COD$ are similar,
it follows that $AB$ and $CD$ are parallel.
In that case, we have
\begin{align*}
AB &= AO = 55 \\
CD &= CO = 175 \\
BC &= \sqrt{BN^2 + NC^2} = \sqrt{171^2+140^2} = 221 \\
AC &= AO + CO = 55 + 175 = 230 = AB + CD \\
BD &= BO + DO = 2(33+105) = 276 = BC + CD.
\end{align*}
(The length of $AD$ is not relevant for this solution.)
This completes the construction.
\begin{remark*}
The numbers selected here may seem magical in nature.
Really, the underlying idea is to construct
two similar isosceles triangles $AOB$ and $COD$ as above,
so that $AB \parallel CD$ and $AC = AB+CD$ are automatically true.
In that case, the only condition that needs to
hold is for $BD=AB+BC$ to be true.
Because we have a choice of three numbers
(the length $AO$, $BO$, $CO$ determine the figure),
it should be possible to make this equation true,
and one simply needs to exhibit one solution.
The lengths here were chosen after some calculation
to yield a construction in which all lengths were integers,
but this is neither necessary nor important for the solution to work.
\end{remark*}
\paragraph{Indirect construction using continuity}
We develop the ideas mentioned in the preceding remark
by showing how one can indirectly prove the existence of a valid trapezoid,
without having to actually find all the necessary constants.
Indeed, we again construct two similar right triangles $AOM$ and $CON$,
but this time we set
where $OM = 3$, $MA = 4$, $OA = 5$ (say)
and $ON = 3r$, $NC = 4r$, $CO = 5r$, for some $r > 0$.
Then let $B$ and $D$ be the reflections
of $O$ over $M$ and $N$, respectively.
\begin{center}
\begin{asy}
size(13cm);
pair A = (-33, 44);
pair B = (-66, 0);
pair C = (60, -80);
pair D = (120, 0);
pair M = (B.x/2,0);
pair N = (D.x/2,0);
label("$5$", midpoint(A--B), dir(A+B), blue);
draw("$5$", A--origin, dir(45), blue);
label("$5r$", midpoint(C--D), dir(C+D), blue);
filldraw(A--B--C--D--cycle, opacity(0.1)+lightcyan, blue);
draw("$3$", origin--M, dir(90), deepgreen);
draw("$3$", M--B, dir(90), deepgreen);
draw("$3r$", origin--N, dir(-90), deepgreen);
draw("$3r$", N--D, dir(-90), deepgreen);
draw("$4$", A--M, lightred);
draw("$4r$", C--N, lightred);
draw("$5r$", origin--C, dir(45), blue);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$O$", origin, dir(225));
dot("$M$", M, dir(-90));
dot("$N$", N, dir(90));
\end{asy}
\end{center}
Because $BD = 6+6r$ and $AC = 5+5r$, we have $BD > AC$,
and this trapezoid is not isosceles for any value of $r$.
Now, we vary the parameter $r$ and consider the function
\[ f(r) \coloneqq AB + BC - BD = 5 + \sqrt{(3r+6)^2+(4r)^2} - (6+6r). \]
Note that this function is continuous and
\begin{align*}
f(0.001) &= 5 + \sqrt{3.006^2+4^2} - 6.006 > 0 \\
f(1000) &= 5 + \sqrt{3006^2+4000^2} - 6006 < 0.
\end{align*}
so by the intermediate value theorem,
there must be some $0.001 < r < 1000$ for which $f(r) = 0$.
That value of $r$ gives a valid construction.
\begin{remark*}
As we saw in the previous solution, $r = \frac{35}{11}$ works.
In fact, it is the unique value of $r$ for which $f(r) = 0$.
The choice of a $3$-$4$-$5$ triangle in this construction is just for
concreteness; many other dimensions would work as well.
\end{remark*}
\end{document}