In $\vartriangle ABC$ the incircle is tangent to $AB$ at $D$. Let $P$ be a point on $BC$ different from $B$ and $C$, and let $K$ and $L$ be incenters of $\vartriangle ABP$ and $\vartriangle ACP$ respectively. Suppose that the circumcircle of $\vartriangle KP L$ cuts $AP$ again at $Q$. Prove that $AD = AQ$.
Problem
Source: Thailand Mathematical Olympiad 2018 day 2 p9
Tags: geometry, circumcircle, incenter, equal segments
MP8148
21.01.2020 07:46
[asy][asy]
size(8cm);
defaultpen(fontsize(10pt));
pair A = dir(120), B = dir(210), C = dir(330), I = incenter(A,B,C), D = foot(I,A,B), P = B+dir(B--C)*abs(B-C)*0.65 , K = incenter(A,B,P), L = incenter(A,C,P), Q = A+dir(A--P)*abs(A-D), E = foot(I,B,C), M = (K+L)/2, N = (E+P)/2, R = foot(K,B,C), S = foot(L,B,C);
dot("$A$", A, dir(120));
dot("$B$", B, dir(210));
dot("$C$", C, dir(330));
dot("$D$", D, dir(150));
dot("$P$", P, dir(285));
dot("$K$", K, dir(180));
dot("$L$", L, dir(345));
dot("$Q$", Q, dir(60));
dot("$E$", E, dir(255));
dot("$M$", M, dir(90));
dot("$N$", N, dir(270));
dot("$R$", R, dir(270));
dot("$S$", S, dir(270));
draw(A--B--C--A--P--K--L--P);
draw(circumcircle(K,P,L));
draw(Q--K--R^^M--N^^L--S^^D--K--E, dotted);
draw(A--K--B, dashed);
[/asy][/asy]
Let $E$ be the point where the incircle touches $\overline{BC}$; WLOG assume $E$ lies on segment $\overline{BP}$ (the other case is analogous). Define $M$, $N$ as midpoints of $\overline{KL}$, $\overline{EP}$ respectively; $R$, $S$ as projections of $K$, $L$ on $\overline{BC}$ respectively.
Claim: $E$ lies on $(KPL)$.
Proof. First, observe that $\angle KPL = \tfrac{1}{2}(\angle APB+\angle APC) = 90^\circ$, so $M$ is the circumcenter of $\triangle KLP$. This means that we just need to prove $\overline{MN} \perp \overline{BC}$, or equivalently, $RE = SP$. Indeed, we have \begin{align*}RE &= BE-BR = \dfrac{1}{2}(AB+BC-CA)-\dfrac{1}{2}(AB+BP-AP) \\ &= \dfrac{1}{2}(BC-CA-BP+AP) = \dfrac{1}{2}(AP+CP-CA) = SP.\end{align*}$\square$
Notice that $BE = BD$, so $\triangle BKD \cong \triangle BKE$ by SAS. Now $$\angle ADK = 180^\circ - \angle KDB = 180^\circ - \angle KEB = \angle KEP = \angle AQK,$$so $\triangle ADK \cong \triangle AQK$ by AAS and the conclusion follows. $\blacksquare$
yayups
24.08.2020 04:16
Note that \[\angle KPL = \angle KPA+\angle LPA = \frac{1}{2}\angle APB + \frac{1}{2}\angle APC = \pi/2,\]so $(KPL)$ has diameter $KL$. Let $M$ be the midpoint of $KL$, and let $\phi$ be the projection map onto line $AP$. We see that $(KPL)$ has center $M$, so \[Q=2\phi(M)-P.\]Since $\phi$ is a linear map, we have $2\phi(M)=\phi(K)+\phi(L)$, so \[Q=\phi(K)+\phi(L)-P.\]We see that \[|P\phi(K)|=\frac{AP+BP-AB}{2}\]and \[|P\phi(L)|=\frac{AP+CP-AC}{2},\]so \[PQ = |P\phi(K)| + |P\phi(L)| = AP-\frac{AB+BC-AB}{2} = AP-AD,\]so $AQ=AD$, as desired. This completes the solution.