In $ \triangle{ABC}$ with $ AB = 12$, $ BC = 13$, and $ AC = 15$, let $ M$ be a point on $ \overline{AC}$ such that the incircles of $ \triangle{ABM}$ and $ \triangle{BCM}$ have equal radii. Let $ p$ and $ q$ be positive relatively prime integers such that $ \tfrac{AM}{CM} = \tfrac{p}{q}$. Find $ p + q$.
It seems that a rather obvious solution in terms of insight is to find a ratio of areas, but I was reading a more synthetic solution that brought up a confusing claim.
First, let the incenters be $I_1$ and $I_2$ for $\triangle ABM$ and $\triangle CBM$ respectively, and point $P$ is the radius from $I_1$ to $AM$. It is then claimed that $MP=(MA+MB-AB)/2$.
I've tried manipulating some relations, but I'm stuck. How can we prove this?
Let the incircle $\Gamma$ of $\Delta ABM$ touch the sides $AB$ and $BM$ at $Q$ and $R$ respectively. It already touches $AM$ at $P$ as the problem mentions.
Then $AP=AQ=x\ (\text{say})$, since they are tangents from the point $A$ to $\Gamma$.
Similarly, we have $BQ=BR=y\ (\text{say})$ and $MP=MR=z \ (\text{say})$.
Then, if you have already marked the points as outlined,
$$x+y=AB,\ y+z=BM, \ z+x=MA \\ \implies 2(x+y+z) = AB+BM+MA \\ \implies MP = z= x+y+z-(x+y) = \dfrac{AB+BM+MA}2-AB = \dfrac{MA+MB-AB}2$$
This is a well-known formula for the distance between the vertex of a triangle and it's intouch point (point of contact of triangle with incirle) on a side. containing that vertex.