Problem: Let $ABC$ be a triangle with incenter $I$.
Points $K$ and $L$ are chosen on segment $BC$
such that the incircles of $\triangle ABK$ and $\triangle ABL$ are tangent at $P$,
and the incircles of $\triangle ACK$ and $\triangle ACL$ are tangent at $Q$.
Prove that $IP = IQ$.

\end{enumerate}

Solution: We present two solutions.

\paragraph{First solution, mostly elementary (original).}
Let $I_B$, $J_B$, $I_C$, $J_C$ be the incenters
of $\triangle ABK$, $\triangle ABL$,
$\triangle ACK$, $\triangle ACL$ respectively.

\begin{center}
\begin{asy}
unitsize(15);
real r1, r2;
path w1, w2;

pair J1 = (-0.5, 2.7);
pair J2 = (1, 3);
real r1 = abs(J1.y);
real r2 = abs(J2.y);
path w1 = CR(J1, r1);
path w2 = CR(J2, r2);

pair X1 = intersectionpoints(w1, w2)[0];
pair X2 = reflect(J1, J2)*X1;
pair H = extension((1.9, 0), (2, 1), X1, X2);
pair P = intersectionpoints(CR(H, sqrt(abs(H-X1)*abs(H-X2))), w1)[1];
pair Q = intersectionpoints(CR(H, sqrt(abs(H-X1)*abs(H-X2))), w2)[0];
pair I = extension(J1, P, J2, Q);
pair B = extension(J1, P, (0, 0), (1, 0));
pair C = extension(J2, Q, (0, 0), (1, 0));
pair A = extension(B, reflect(B, I) * C, C, reflect(C, I) * B);
pair K = extension(B, C, A, reflect(A, J2) * C);
pair L = extension(B, C, A, reflect(A, J1) * B);
pair I_B = incenter(A, B, K);
pair J_B = incenter(A, B, L);
pair I_C = incenter(A, C, K);
pair J_C = incenter(A, C, L);

pair R = extension(P, Q, B, C);

filldraw(A--B--C--cycle, opacity(0.1)+lightcyan, blue);
draw(incircle(A, B, K)^^incircle(A, C, L), deepcyan);
draw(L--A--K, blue);
draw(w1^^w2, deepcyan);
draw(I--P^^I--Q, red);
draw(B--P^^C--Q, orange);
draw(Q--R--B, deepgreen);
draw(J_C--R, deepgreen+dashed);
draw(I_C--R, deepgreen+dashed);

dot("$A$", A, dir(A-I));
dot("$B$", B, dir(230));
dot("$C$", C, dir(310));
dot("$I$", I, dir(2*I-P-Q));
dot("$K$", K, dir(270));
dot("$L$", L, dir(270));
dot("$P$", P, dir(80));
dot("$Q$", Q, dir(20));
dot("$R$", R, dir(-90));

dot("$I_B$", I_B, dir(-90));
dot("$I_C$", I_C, dir(-90));
dot("$J_B$", J_B, dir(-90));
dot("$J_C$", J_C, dir(-90));
\end{asy}
\end{center}

We begin with the following claim which does not depend
on the existence of tangency points $P$ and $Q$.

\begin{claim*}
  Lines $BC$, $I_BJ_C$, $J_BI_C$ meet at a point $R$
  (possibly at infinity).
\end{claim*}
\begin{proof}
  By rotating by $\half \angle A$ we have the equality
  \[ A(BI; I_B J_B) = A(IC;I_C J_C). \]
  It follows $(BI; I_B J_B) = (IC; I_C J_C) = (CI; J_C I_C)$.
  (One could also check directly that both cross ratios equal
  $\frac{\sin \angle BAK/2}{\sin \angle CAK/2}
  \div \frac{\sin \angle BAL/2}{\sin \angle CAL/2}$,
  rather than using rotation.)

  Therefore, the concurrence follows from the so-called
  \emph{prism lemma} on $\ol{I B I_B J_B}$ and $\ol{I C J_C I_C}$.
\end{proof}
%\[ (B I; I_B J_B)
%    = \frac{\sin \angle I_B A B}{\sin \angle I_B A I}
%    \div \frac{\sin \angle J_B A B}{\sin \angle J_B A I}
%    = \frac{\sin \tfrac12 \angle BAK}{\sin \tfrac12 \angle CAK}
%    \div \frac{\sin \tfrac12 \angle BAL}{\sin \tfrac12 \angle CAL}.
%  \]
%  Similarly
%  \[ (C I; J_C I_C)
%    = \frac{\sin \angle J_C A C}{\sin \angle J_C A I}
%    \div \frac{\sin \angle I_C A C}{\sin \angle I_C A I}
%    = \frac{\sin \tfrac12 \angle CAL}{\sin \tfrac12 \angle BAL}
%    \div \frac{\sin \tfrac12 \angle CAK}{\sin \tfrac12 \angle BAK}.
%  \]
%


%\begin{proof}
%    Note $\angle BAI_B = \angle IAI_C$
%    and $\angle I_BAI = \angle I_CAC$.
%    Then
%    \[\frac{BI_B}{I_BI} \div \frac{II_C}{I_CC}
%    = \left(\frac{BA}{AI} \cdot \frac{\sin \angle BAI_B}{\sin \angle I_BAI}\right)
%    \div \left(\frac{IA}{AC} \cdot \frac{\sin \angle IAI_C}{\sin \angle I_CAC}\right)\\
%    = \frac{BA \cdot AC}{AI^2}.\]
%    In the same way,
%    we obtain
%    \[\frac{BI_B}{I_BI} \div \frac{II_C}{I_CC}
%    = \frac{BJ_B}{J_BI} \div \frac{IJ_C}{J_CC}
%    \implies
%    \frac{BI_B}{I_BI} \cdot \frac{IJ_C}{J_CC}
%    = \frac{BJ_B}{J_BI} \cdot \frac{II_C}{I_CC}\]
%    and we're done by Menelaus.
%\end{proof}

\begin{remark*}[Nikolai Beluhov]
  This result is known;
  it appears as {4.5.32} in Akopyan's \emph{Geometry in Figures}.
  The cross ratio is not necessary to prove this claim:
  it can be proven by length chasing
  with circumscribed quadrilaterals.
  (The generalization mentioned later also admits a trig-free proof
  for the analogous step.)
\end{remark*}

We now bring $P$ and $Q$ into the problem.
\begin{claim*}
  Line $PQ$ also passes through $R$.
\end{claim*}

\begin{proof}
  Note $(BP; I_BJ_B) = -1 = (CQ; J_CI_C)$,
  so the conclusion again follows by prism lemma.
\end{proof}

% \begin{corollary*}
%   The inversion at $R$
%   swapping the incircles
%   of $\triangle ABK$ and $\triangle ACL$
%   also swaps $P$ and $Q$.
% \end{corollary*}

We are now ready to complete the proof.
Point $R$ is the exsimilicenter of
the incircles of $\triangle ABK$ and $\triangle ACL$,
so $\tfrac{PI_B}{RI_B} = \tfrac{QJ_C}{RJ_C}$.
Now by Menelaus,
\[\frac{I_BP}{PI} \cdot \frac{IQ}{QJ_C} \cdot \frac{J_CR}{RI_B} = -1
\implies IP = IQ.\]

\begin{remark*}[Author's comments on drawing the diagram]
  Drawing the diagram directly is quite difficult.
  If one draws $\triangle ABC$ first, they must locate both $K$ and $L$,
  which likely involves some trial and error
  due to the complex interplay between the two points.

  There are alternative simpler ways.
  For example, one may draw $\triangle AKL$ first;
  then the remaining points $B$ and $C$ are not related
  and the task is much simpler (though some trial and error is still required).

  In fact, by breaking symmetry, we may only require one application of guesswork.
  Start by drawing $\triangle ABK$ and its incircle;
  then the incircle of $\triangle ABL$ may be constructed, and so point $L$ may be drawn.
  Thus only the location of point $C$ needs to be guessed.
  I would be interested in a method to create a general diagram without any trial and error.
\end{remark*}

\paragraph{Second solution, inversion (Nikolai Beluhov).}
As above, the lines $BC$, $I_BJ_C$, $J_BI_C$ meet at some point $R$
(possibly at infinity).
Let $\omega_1$, $\omega_2$, $\omega_3$, $\omega_4$ be the incircles
of $\triangle ABK$, $\triangle ACL$, $\triangle ABL$, and $\triangle ACK$.
\begin{claim*}
    There exists an inversion $\iota$ at $R$
    swapping $\{\omega_1, \omega_2\}$
    and $\{\omega_3, \omega_4\}$.
\end{claim*}
\begin{proof}
    Consider the inversion at $R$
    swapping $\omega_1$ and $\omega_2$.
    Since $\omega_1$ and $\omega_3$ are tangent,
    the image of $\omega_3$ is tangent to $\omega_2$
    and is also tangent to $BC$.
    The circle $\omega_4$ is on the correct side of $\omega_3$
    to be this image.
\end{proof}

\begin{claim*}
    Circles $\omega_1$, $\omega_2$, $\omega_3$, $\omega_4$
    share a common radical center.
\end{claim*}
\begin{proof}
    Let $\Omega$ be the circle with center $R$
    fixed under $\iota$,
    and let $k$ be the circle through $P$
    centered at the radical center of $\Omega$, $\omega_1$, $\omega_3$.

    Then $k$ is actually orthogonal to $\Omega$, $\omega_1$, $\omega_3$,
    so $k$ is fixed under $\iota$
    and $k$ is also orthogonal to $\omega_2$ and $\omega_4$.
    Thus the center of $k$ is the desired radical center.
\end{proof}

The desired statement immediately follows.
Indeed, letting $S$ be the radical center,
it follows that $\ol{SP}$ and $\ol{SQ}$
are the common internal tangents to $\{\omega_1, \omega_3\}$ and $\{\omega_2, \omega_4\}$.

Since $S$ is the radical center,
$SP = SQ$.
In light of $\angle SPI = \angle SQI = 90^{\circ}$,
it follows that $IP = IQ$, as desired.

\begin{remark*}[Nikolai Beluhov]
  There exists a circle tangent to all four incircles,
  because circle $k$ is orthogonal to all four,
  and line $BC$ is tangent to all four;
  thus the inverse of line $BC$ in $k$ is a circle
  tangent to all four incircles.

  The amusing thing here is that Casey's theorem is completely unhelpful
  for proving this fact:
  all it can tell us is that there is a line or circle tangent to these incircles,
  and line $BC$ already satisfies this property.
\end{remark*}

\begin{remark*}[Generalization by Nikolai Beluhov]
  The following generalization holds:
  \begin{quote}
  Let $ABCD$ be a quadrilateral
  circumscribed about a circle with center $I$.
  A line through $A$ meets $\overrightarrow{BC}$ and $\overrightarrow{DC}$
  at $K$ and $L$;
  another line through $A$ meets $\overrightarrow{BC}$ and $\overrightarrow{DC}$
  at $M$ and $N$.
  Suppose that the incircles of $\triangle ABK$ and $\triangle ABM$ are tangent at $P$,
  and the incircles of $\triangle ACL$ and $\triangle ACN$ are tangent at $Q$.
  Prove that $IP = IQ$.
  \end{quote}

  The first approach can be modified to the generalization.
  There is an extra initial step required:
  by Monge, the exsimilicenter of the incircles of $\triangle ABK$ and $\triangle ADN$
  lies on line $BD$;
  likewise for the incircles of $\triangle ABL$ and $\triangle ADM$.
  Now one may prove using the same trig approach that these pairs of incircles
  have a common exsimilicenter,
  and the rest of the solution plays out similarly.
  The second approach can also be modified in the same way,
  once we obtain that a common exsimilicenter exists.
  (Thus in the generalization, it seems we also get
  there exists a circle tangent to all four incircles.)
\end{remark*}