, 2016

We have $\frac{MN}{NP} = \frac{23}{13}$.
Let $I$ and $J$ be the incentres of $\triangle ADE$ and $\triangle BCE$ respectively. Note that $I$ is the intersection point of $AN$ and $DM$, while $J$ is the intersection point of $BN$ and $CM$. Applying Pascal's theorem to $ANBDMC$, we find that $I, J, E$ are collinear.

Since $\angle IAD = \angle JBC$ and $\angle IDA = \angle JCB$, we have $\triangle AID \sim \triangle BJC$. Thus, we have

$\triangle$ MIN $\sim \triangle$ AID $\sim \triangle$ BJC $\sim \triangle$ MJN,

with $\triangle MIN \cong \triangle MJN$. This implies $IJ \perp MN$. Also, the isogonal line of $IJ$ with respect to $\angle MIN$, the isogonal line of $IJ$ with respect to $\angle MJN$ and the line $MN$ are concurrent at some point $P'$ by symmetry. Observe that $IP' \perp AD$ since $\triangle MIN$ and $\triangle AID$ are oppositely similar. Similarly, $JP' \perp BC$. We now claim that $P' = P$.

Let $E_1, P_1, O_1$ be the projection of $E, P', O$ on $AD$, and let $E_2, P_2, O_2$ be the projection of $E, P', O$ on $BC$. Note that $E_1, P_1, O_1$ are the foot of altitude from $E$, the contact point of the incircle with the side $AD$, and the midpoint of $AD$.

in $\triangle EAD$. Similar results hold for $E_2, P_2, O_2$ in $\triangle EBC$. As $\triangle EAD \sim \triangle EBC$, we have $\frac{E_1P_1}{P_1O_1} = \frac{E_2P_2}{P_2O_2}$. Therefore, $E, P', O$ are collinear. This implies $P'$ lies on both $EO$ and $MN$, and hence $P' = P$.

Now, let $F$ be the intersection of $AD$ and $IJ$. Then $\frac{MP}{NP} = \frac{AF}{DF}$ due to the isogonal lines. By the angle bisector theorem, we have $\frac{AF}{DF} = \frac{AE}{DE}$.

Note that $\triangle EAD \sim \triangle EBC$, and the ratio is $\frac{AD}{BC} = 2$. Let $BE = x$ and $CE = y$ so that $AE = 2x$ and $DE = 2y$. It is given that $2x + y = 11$ and $x + 2y = 12$. We easily find that $x = \frac{10}{3}$ and $y = \frac{13}{3}$. Therefore,
$$$$
$\text{MN NP = MP}$ + NP}{NP} = $\frac{AE}{DE}$ + 1 = $\frac{x}{y}$ + 1 = $\frac{23}{13}.$