A point is reflected in the three sides of a triangle; when do the lines connecting the reflected points to corresponding vertices concur? 
Let $P$ be a point in $\triangle ABC$, and let $P_A$ be its reflection about $\overline{BC}$, and similarly define $P_B$ and $P_C$.  When do $\overleftrightarrow{AP_A}, \overleftrightarrow{BP_B},$ and $\overleftrightarrow{CP_C}$ concur?

They appear to concur when $P$ is the orthocenter, the incenter (no idea why though), and the circumcenter (and some other points) (also not sure why it happens here).
 A: I'll derive a condition based on the distances from $P$ to the edges of $\triangle ABC$. For simplicity, I'll assume that $P$ lies in the interior of the triangle; some tweaks to the intermediate may be required otherwise, but the final condition is (probably) universal.

Let the midpoints of $\overline{PP_A}$, $\overline{PP_B}$, $\overline{PP_C}$ be $D$, $E$, $F$, respectively, and define
$$d := |\overline{PD}| \qquad e := |\overline{PE}| \qquad f := |\overline{PF}|$$
Note that, because $\square AEPF$, $\square BFPD$, $\square CDPE$ each has two right angles, we have 
$$\angle EPF = \pi - A \qquad \angle DPF = \pi - \angle B \qquad \angle DPE = \pi - \angle C$$
Let's coordinatize, placing $P$ at the origin and $D$ along the positive $x$-axis at $D = (d,0)$. Then 
$$\begin{align}
E &= e \left(\; \cos(\phantom{-}\angle DPE), \sin(\phantom{-}\angle DPE) \;\right) = e \left(\;-\cos C, \phantom{-}\sin C\;\right) \\
F &= f \left(\; \cos(-\angle DPF), \sin(-\angle DPF)\;\right) = f\left(-\cos B, -\sin B\;\;\right)
\end{align}$$
The sides of the triangle are perpendicular to $\overline{PD}$, $\overline{PE}$, $\overline{PF}$ at $D$, $E$, $F$, and we derive
$$\overleftrightarrow{BC}: x = d \qquad \overleftrightarrow{CA}: x \cos C - y \sin C = -e \qquad \overleftrightarrow{AB}: x \cos B + y \sin B = -f$$
so that
$$\begin{align}
A &= \frac{1}{\sin A}\left(\;-e \sin B - f \sin C, e \cos B - f \cos C\right) \\[8pt]
B &= \left(d, -d \cot B - f \csc B \right)\\[8pt]
C &= \left(d, \phantom{-}d \cot C + e \csc C \right)
\end{align}$$
Writing $D^\prime$ for the point where $\overleftrightarrow{AP_A}$ meets $\overleftrightarrow{BC}$, we can find
$$D^\prime = \left(\;d\;,\; \frac{d \left( e \cos B - f \cos C \right)}{
  2 d \sin A + e \sin B + f \sin C}\;\right)$$
Also, 
$$\begin{align}
\overrightarrow{BD^\prime} = D^\prime - B = 
\left(\;0\;,\; \frac{(f + 2 d \cos B) ( d \sin A + e \sin B + f \sin C )}{\sin B\;(
 2 d \sin A + e \sin B + f \sin C)}\;\right) \\[8pt]
\overrightarrow{D^\prime C} = C - D^\prime = 
\left(\;0\;,\; \frac{(e + 2 d \cos C) ( d \sin A + e \sin B + f \sin C )}{\sin C\;(
 2 d \sin A + e \sin B + f \sin C)}\;\right) \\[8pt]
\end{align}$$
Conveniently, the $y$-components of these vectors are exactly the signed distances used in the corresponding ratio from Ceva's Theorem:
$$\frac{|BD^\prime|}{|D^\prime C|} = \frac{\sin C\;(f+2d \cos B)}{\sin B\;(e+2d \cos C)}$$
With the appropriate variants of the above for the other sides of the triangle, Ceva's Theorem gives (after canceling sines, clearing fractions, and combining terms) this condition for concurrency:
$$\begin{align}
0 &= d ( e^2 - f^2 ) ( \cos A - 2 \cos B \cos C )  \\
&+ e ( f^2 - d^2 ) ( \cos B - 2 \cos C \cos A )  \tag{$\star$}\\
&+ f ( d^2 - e^2 ) ( \cos C - 2 \cos A \cos B ) 
\end{align}$$
Right away, we see that the incenter satisfied this condition, as $d = e = f = \text{inradius}$. Less-obviously, if $P$ is the circumcenter and $r$ the circumradius, then
$$d = r \cos A \qquad e = r \cos B \qquad f = r \cos C$$
and $(\star)$ holds. Also, for $P$ the orthocenter (and $r$ still the circumradius), 
$$d = 2 r \cos B \cos C \qquad e = 2 r \cos C \cos A \qquad f = 2 r \cos A \cos B$$
satisfying $(\star)$. Exploration of other cases, and/or conversion of $(\star)$ into a better form, is left as an exercise to the reader.

It's worth pointing-out some geometric interpretations of expressions used above. For instance, from the Law of Sines (with $r$ the circumradius), we have
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2 r$$
so that 
$$ d \sin A = \frac{d a}{2r} = \frac{1}{r}|\triangle PBC| \qquad e\sin B = \frac{1}{r}|\triangle APC| \qquad f \sin C = \frac{1}{r} |\triangle ABP|$$ 
and
$$ d \sin A + e \sin B + f \sin C = \frac{1}{r}|\triangle ABC| \qquad 2d\sin A + e \sin B + f \sin C = \frac{1}{r}|\square ABP_AC|$$
Moreover, if $E^{\prime\prime}$ and $F^{\prime\prime}$ are the feet of the perpendiculars from $P_A$ to $\overleftrightarrow{PE}$ and $\overleftrightarrow{PF}$, respectively, then $\angle P_APE^{\prime\prime} = \angle C$ and $\angle P_APF^{\prime\prime} = \angle B$, so that 
$$|EE^{\prime\prime}| = e + 2d \cos C \qquad |FF^{\prime\prime}| = f + 2d \cos B$$
These facts may somehow provide a coordinate-free path to $(\star)$.
