, 2010

Solution (By Gabriel Carroll). Let $\omega_1, \omega_C, \omega_B$, and $\omega$ denote the incircles of triangles $ABC, MNC, PBQ$, and $ARS$, respectively. Denote by $I$ and $I_1$ the incenters of triangles $ABC$ and $ARS$, respectively. Let $\omega_1$ touch sides $AB$ and $AC$ at $R_1$ and $S_1$, respectively.

It is clear that there is a homothety $\mathbf{H}_1$ centered at $C$ sending triangle $CMN$ to $CAB$, and that images of $M, E, N$, and line $EN$ under $\mathbf{H}_1$ are $A, S_1, B$, and line $S_1B$. In particular, $BS_1 \parallel RE$ with $AB/AR = AS_1/AE$. In exactly the same way, we can prove that $CR_1 \parallel SF$ with $AC/AS = AR_1/AF$. By equal tangents, we have $AS_1 = AR_1$. By the given condition, $AE = AF$. It follows that
$$\frac{AB}{AR} = \frac{AS_1}{AE} = \frac{AR_1}{AF} = \frac{AC}{AS},$$
implying that $BC \parallel RS$. Thus, there is a homothety $\mathbf{H}$ centered at $A$ sending triangle $ABC$ to triangle $ARS$. It is clear that the images of $S_1, R_1, \omega_1$, and $I_1$ under $\mathbf{H}$ are $E, F, \omega$, and $I$, respectively. Thus, $I$ lies on $\omega$, which is what we wish to show, if and only if $I_1$ lies on $\omega_1$. But the latter claim holds because the midpoint of minor arc $\widehat{R_1S_1}$ on $\omega_1$ is the incenter of triangle $AR_1S_1$.