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

\begin{document}
\title{Cross Ops}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 131}
\maketitle

\section*{Problem}
Given $A$, $B$, $C$, $D$, $E$ on a line, define $f(P) = (A, B; C, P)$.
Construct points $X$ and $Y$ with a straight edge such that $f(X) = f(D)f(E)$
and $f(Y) = f(D) + f(E)$.

\section*{Video}
\href{https://www.youtube.com/watch?v=gjQyGxUu4YY&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/gjQyGxUu4YY}}



\newpage

\section*{Solution}
Impose a coordinate system where $A = \infty$, $B = 0$, $C = c$, $D = d$, $E = e$.
In that case, if $P$ has coordinate $p$, then
\[ f(P) = \frac{CA}{CB} \div \frac{DA}{DB} = \frac{p}{c}. \]
We progressively give the following operations.

\begin{claim*}
  Given $x$ and $y$, one can construct $\frac{x+y}{2}$, $2y-x$, and $y^2/x$.
\end{claim*}
\begin{proof}
  Start by drawing any two other lines through $A = \infty$.
  See the following picture.
  \begin{center}
  \begin{asy}
    size(12cm);
    pair I1 = (-1,0);
    pair J1 = (9,0);
    pair I2 = (-1,2);
    pair J2 = (9,2);
    pair I3 = (-1,3);
    pair J3 = (9,3);
    draw(I2--J2);
    draw(I3--J3);
    pair X = (3,0);
    pair Y = (5,0);
    pair P = (4.2,3);
    pair R = extension(X,P,I2,J2);
    pair Q = extension(Y,P,I2,J2);
    pair S = extension(X,Q,Y,R);
    pair M = extension(P,S,X,Y);
    draw(I1--J1, black+1.5);
    draw(X--P--Y, lightred);
    draw(X--Q, lightred);
    draw(Y--R, lightred);
    draw(M--P, lightred);
    pair T = extension(X,Q,I3,J3);
    pair U = extension(Y,T,I2,J2);
    pair N = extension(P,U,I1,J1);
    draw(Q--T, green);
    draw(Y--T, green);
    draw(P--N, green);
    pair V = extension(P, origin, I2, J2);
    pair W = extension(origin, P, Y, R);
    pair Z = extension(W, Q, I1, J1);
    draw(P--origin, lightblue);
    draw(W--Z, lightblue);
    draw(W--R, lightblue);
    dot("$0$", origin, dir(-90));
    dot("$x$", X, dir(-90));
    dot("$y$", Y, dir(-90));
    dot("$\frac{x+y}{2}$", M, dir(-90), red);
    dot("$2x-y$", N, dir(-90), deepgreen);
    dot("$y^2/x$", Z, dir(-90), blue);
    dot(P);
    dot(Q);
    dot(R);
    dot(S);
    dot(T);
    dot(U);
    dot(V);
    dot(W);
  \end{asy}
  \end{center}
  Follow red, green, blue in order.
\end{proof}
We are initially given points $D$ and $E$ such that $f(D) = d/c$ and $f(E) = e/c$.
The point
\[ X = 2 \frac{d+e}{2} - 0 = d+e \]
thus has $f(X) = \frac{d+e}{c} = f(D)+f(E)$.
Meanwhile, the point
\[ Y = \frac{\frac{(d+e)^2}{c} - \frac{d^2}{c} - \frac{e^2}{c}}{2} = \frac{de}{c} \]
satisfies $f(Y) = \frac{de}{c^2} = f(D)f(E)$. We're done.

\end{document}
