Problem: Let $ABC$ be a scalene triangle.
Points $A_1$, $B_1$ and $C_1$ are chosen
on segments $BC$, $CA$, and $AB$, respectively,
such that $\triangle A_1B_1C_1$ and $\triangle ABC$
are similar.
Let $A_2$ be the unique point on line $B_1C_1$
such that $AA_2 = A_1A_2$.
Points $B_2$ and $C_2$ are defined similarly.
Prove that $\triangle A_2B_2C_2$ and $\triangle ABC$ are similar.

Solution: We give three solutions.
\paragraph{Solution 1 (author).}
We'll use the following lemma.

\begin{lemma*}
  Suppose that $PQRS$ is a convex quadrilateral with $\angle P = \angle R$. Then
  there is a point $T$ on $QS$ such that $\angle QPT = \angle SRP$, $\angle
  TRQ = \angle RPS$, and $PT=RT$.
\end{lemma*}

Before proving the lemma, we will show how it solves the problem. The lemma
applied for the quadrilateral $AB_1A_1C_1$ with $\angle A = \angle A_1$ shows
that $\angle B_1A_1A_2 = \angle C_1AA_1$. This implies that the point $A_2$ in
$\triangle A_1B_1C_1$ corresponds to the point $A_1$ in $\triangle ABC$. Then
$\triangle A_2B_2C_2 \sim \triangle A_1B_1C_1 \sim \triangle ABC$, as desired.

Additionally, $PT=RT$ is a corollary of the angle conditions because
\[\measuredangle PRT = \measuredangle SRQ - \measuredangle TRQ - \measuredangle SRP = \measuredangle QPS - \measuredangle RPS -
\measuredangle QPT = \measuredangle TPR.\]
Therefore we only need to prove the angle conditions.

\subparagraph{Proof 1 of lemma}
  Denote $X = PQ \cap RS$ and $Y = PS \cap RQ$. Note that $\angle XPY = \angle
  XRY$, so $PRXY$ is cyclic. Let $T$ be the point of intersection of tangents to this circle at $P$ and $R$. By Pascal's theorem for the degenerate hexagon $PPXRRY$, we have $T \in QS$ (alternatively, $Q$, $S$, and $T$ are collinear on the pole of $PR \cap XY$ with respect to the circle). Also, $\measuredangle QPT = \measuredangle XRP = \measuredangle SRP$ and similarly $\measuredangle TRQ = \measuredangle RPY = \measuredangle RPS$, so we're done.

\begin{center}
  \begin{asy}
    size(8cm);

    pair P, Q, Rp, R, S, X, Y, T;

    P = dir(50);
    Q = dir(200);
    S = dir(340);
    Rp = dir(40);
    R = 2*foot(Rp, Q, S) - Rp;
    X = extension(P, Q, R, S);
    Y = extension(P, S, R, Q);
    T = extension(Q, S, (P+R)/2, (P+R)/2+dir(90)*(P-R));

    draw(P--Q--R--S--cycle);
    draw(P--X--S--Y--R--cycle);
    draw(Q--S);
    draw(P--T--R, red);

    //draw(anglemark(Q, P, T, 10));
    //draw(anglemark(S, R, P, 10));
    //draw(anglemark(T, R, Q, 9, 11));
    //draw(anglemark(R, P, S, 9, 11));

    draw(circumcircle(P, R, X));

    string[] names={"$P$","$Q$","$R$","$S$","$X$","$Y$","$T$"};
    pair[] pts={P, Q, R, S, X, Y, T};
    pair[] labels={dir(110), Q, dir(220), S, dir(120), dir(240), T};
    for(int i=0;i<names.length;++i){
      dot(names[i], pts[i], dir(labels[i]));
    }
  \end{asy}
\end{center}


\subparagraph{Proof 2 of lemma}
Let $P'$ and $R'$ be the reflections of $P$ and $R$ in $QS$. Note that $PR'$ and
$RP'$ intersect at a point $X$ on $QS$. Let $T$ be the second intersection of
the circumcircle of $\triangle PRX$ with $QS$. Note that
\begin{align*}
  \measuredangle PXT &= \measuredangle R'PQ + \measuredangle PQS \\
                       &= \measuredangle R'SQ + \measuredangle PQS  \\
                       &= \measuredangle QSR + \measuredangle PQS \\
                       &= \measuredangle (PQ, SR) \\
                       &= \measuredangle QPR + \measuredangle PRS.
\end{align*}
This means that
\begin{align*}
  \measuredangle QPT &= \measuredangle QPR - \measuredangle TPR \\
                       &= \measuredangle QPR - \measuredangle TXR \\
                       &= \measuredangle QPR - \measuredangle PXT \\
                       &= \measuredangle QPR - \measuredangle QPR -
                       \measuredangle PRS \\
                       &= \measuredangle SRP.
\end{align*}
Similarly, $\measuredangle QRT = \measuredangle SPR$, so we're done.

\begin{center}
  \begin{asy}
    size(8cm);

    pair P, Q, R, S, Pp, Rp, X, T;

    P = dir(50);
    Q = dir(200);
    S = dir(340);
    Rp = dir(70);
    R = 2*foot(Rp, Q, S) - Rp;
    Pp = 2*foot(P, Q, S) - P;
    X = extension(P, Rp, R, Pp);
    T = 2*foot(circumcenter(P, R, X), Q, S) - X;


    draw(P--Q--R--S--cycle);
    draw(P--R--T--cycle);
    draw(Pp--Q--Rp--S--cycle, dashed);
    draw(Q--X);
    draw(R--X--Rp, red);

    draw(circumcircle(P, Q, S));
    draw(circumcircle(R, Q, S));
    draw(circumcircle(P, R, X), blue);

    string[] names={"$P$","$Q$","$R$","$S$","$P'$","$R'$","$X$","$T$"};
    pair[] pts={P, Q, R, S, Pp, Rp, X, T};
    pair[] labels={P, Q, R, dir(330), Pp, Rp, X, dir(220)};
    for(int i=0;i<names.length;++i){
      dot(names[i], pts[i], dir(labels[i]));
    }

  \end{asy}
\end{center}

\subparagraph{Proof 3 of lemma}
  Let $T$ be the point on $QS$ such that $\angle QPT = \angle SRP$. Then we have
  \[
    \frac{QT}{TS} = \frac{\sin QPT \cdot PT / \sin PQT}{\sin TPS \cdot PT / \sin TSP} = \frac{PQ / \sin PRQ}{PS / \sin SRP} = \frac{R(\triangle PQR)}{R(\triangle PRS)},
  \]
  which is symmetric in $P$ and $R$, so we're done.

\paragraph{Solution 2 (Ankan Bhattacharya).}
We prove the main claim $\tfrac{B_1A_2}{A_2C_1} = \tfrac{BA_1}{A_1C}$.

Let $\triangle A_0B_0C_0$ be the medial triangle of $\triangle ABC$.
In addition, let $A_1'$ be the reflection of $A_1$ over $\ol{B_1C_1}$,
and let $X$ be the point satisfying $\triangle XBC \stackrel{-}{\sim} \triangle AB_1C_1$,
so that we have a compound similarity
\[ \triangle ABC \sqcup X
\stackrel{-}{\sim} \triangle A_1'B_1C_1 \sqcup A. \]
Finally, let $O_A$ be the circumcenter of $\triangle A_1'B_1C_1$,
and let $A_2^*$ be the point on $\ol{B_1C_1}$
satisfying $\tfrac{B_1A_2^*}{A_2^*C_1} = \tfrac{BA_1}{A_1C}$.

Recall that $O$ is the Miquel point of $\triangle A_1B_1C_1$,
as well as its orthocenter.
\begin{claim*}
  $\ol{AA_1'} \parallel \ol{BC}$.
\end{claim*}
\begin{proof}
  We need to verify that the foot from $A_1$ to $\ol{B_1C_1}$
  lies on the $A$-midline.
  This follows from the fact that $O$ is both the Miquel point
  and the orthocenter.
\end{proof}

\begin{claim*}
  $\ol{AX} \parallel \ol{B_1C_1}$.
\end{claim*}
\begin{proof}
  From the compound similarity,
  \[ \dang (\ol{BC}, \ol{AX})
  = \dang (\ol{AA_1'}, \ol{B_1C_1}). \]
  As $\ol{AA_1'} \parallel \ol{BC}$,
  we obtain $\ol{AX} \parallel \ol{B_1C_1}$.
\end{proof}

\begin{claim*}
  $\ol{AX} \perp \ol{A_1O}$.
\end{claim*}
\begin{proof}
  Because $O$ is the orthocenter of $\triangle A_1B_1C_1$.
\end{proof}

\begin{claim*}
  $\ol{AA_1'} \perp \ol{A_2^*O_A}$.
\end{claim*}
\begin{proof}
  Follows from $\ol{AX} \perp \ol{A_1O}$
  after the similarity
  \[ \triangle ABC \sqcup X
  \stackrel{-}{\sim} \triangle A_1'B_1C_1 \sqcup A.\qedhere\]
\end{proof}

\begin{claim*}
  $AA_2^* = A_1'A_2$.
\end{claim*}
\begin{proof}
  Since $\dang C_1AB_1 = \dang C_1A_1'B_1$,
  $AO_A = A_1'O_A$,
  so $\ol{AA_1'} \perp \ol{A_2^*O_A}$
  implies $AA_2^* = A_1'A_2^*$.
\end{proof}

Finally, $A_1'A_2^* = A_1A_2^*$ by reflections,
so $AA_2^* = A_1A_2^*$, and $A_2^* = A_2$.