Find the locus of incenters of $\Delta PBC$ Consider a fixed point $P$ on the plane and a line $m$. Find the locus of the incenters of $\Delta PBC$ where $B, C \in m$ and $\Delta PBC$ is acute-angled $\Delta$. 
I don't really know what the locus will be like. But I can understand that if $B$ is fixed then the locus will be a line along $BI$ where $I$ is the Incenter. But for the figure in the question, i.e., $B,C$ not fixed, I can understand that a lot of arms will be formed, but what exactly is the locus k can't really decipher. I have tried it using a Geometry software called $\text {Geogebra}$, which I hope is pretty common among the geometers. I have traced the locus and it was somewhat weird: a set of arms (line-segments) and not some solid locus.
 A: We may assume that the line $m$ is horizontal and $A$ lies above it, at a fixed distance $h$ from it.
Let $H_A$ be the projection of $A$ on $m$, $N$ the midpoint of $AH_A$ and $n$ the parallel to $m$ through $N$.
With a bit of lateral thinking, we may just pick some point $I$ in the upper half plane delimited by $m$ and ask ourself if this point is the incenter of a triangle $ABC$ with $B,C\in m$. This is easily solved: there is a unique circle $\Gamma_I$ centered at $I$ which is tangent to $m$, and the tangents from $A$ to $\Gamma_I$ intersect $m$ at $B,C$.

Clearly $I$ has to lie between $m$ and $n$ ($h>2r$). Now we may deal with the acute-angled constraint. Let $r$ be the radius of $\Gamma_I$ and $I_A$ the projection of $I$ on $m$. Since $\widehat{A}\leq 90^\circ$, $AI\geq \sqrt{2}\,II_A$, hence $I$ has to lie below a hyperbola:

Now it just remains to ensure that also $\widehat{B}$ and $\widehat{C}$ are $\leq 90^\circ$. This is simple to deal with: it just means that $I$ lies above the lines departing from $H_A$ and forming angles equal to $45^\circ$ with $m$.

I leave to you to check that the mentioned conditions are not only necessary but also sufficient to grant that $ABC$ given by $I$ is acute-angled.
Thus the wanted locus is the shaded region above, i.e. a right and isosceles triangle whose hypotenuse has been replaced by an arc of hyperbola. A vertex of such hyperbola is the point $V$ on the segment $AH_A$ such that $\frac{AV}{VH_A}=\sqrt{2}$.
