Find the area of ​region triangular BPM For reference: In triangle $ABC$, the external angle bisector $BE$ and the median $AM$ intersect at $P$. Determine the area of ​​triangular region $BPM$; if $AB =3BC$ and $S_{ABC}=40\ \mathrm{m^2}$.

My progress:
$S_{ABM}=S_{AMC}=\frac{40}{2} = 20\\
\frac{S_{ABP}}{S_{APE}}=\frac{BP}{PE}\\
\frac{S_{BPM}}{S_{BCE}}=\frac{BM.BP}{BC.BE}=\frac{BP}{2BE}\\
\frac{S_{ABM}}{S_{ABP}}=\frac{AB.AM}{AB.AP}\implies \frac{20}{S_{ABP}} = \frac{AM}{AP}\\
\frac{S_{ABC}}{S_{BCE}}=\frac{AC}{CE}$
...?

The only data of this problem is $AB=3BM$ and area $S=40$ in the triangle $\triangle {ABC}$ in which $M$ is the midpoint of $BC$. There are a lot of such triangles whose shape can be quite different so either the problem is undetermined or there is a family of distinct triangles having area as a very nice invariant. We choose the angle $\beta=90^{\circ}$ and we must find the value $4$ according Math Lover's answer.
Put  $\angle {ABC}=\beta=90^{\circ}, AB=6k, BC=2k$.
$$S_{\triangle{ABC}}=6k^2\sin(90^{\circ})=6k^2=40\Rightarrow k=\sqrt{\frac{20}{3}}\\S_{\triangle{BMP}}=\frac12k\cdot BP\sin45^{\circ}=\frac{k\cdot BP\sqrt2}{4}$$
In $\triangle{ABP} :  \dfrac{BP}{\sin(\alpha_1)}=\dfrac{6k}{\sin(135^{\circ}+\alpha_1)}=\dfrac{12k}{\sqrt2(\cos(\alpha_1)-\sin(\alpha_1))}=\dfrac{12AM}{5\sqrt2}=\dfrac{6\sqrt2(AM)}{5}$
Then $BP=\sin(\alpha_1)\dfrac{6\sqrt2(AM)}{5}=\dfrac{k}{AM}\cdot\dfrac{6\sqrt2(AM)}{5}=\dfrac{6\sqrt2k}{5}$.
Consequently $$\\S_{\triangle{BMP}}=\frac{k\sqrt2}{4}\cdot\dfrac{6\sqrt2k}{5}=\frac{3k^2}{5}=\dfrac35\cdot\dfrac{20}{3}=\color{red}{\boxed4}$$

Thus, in the great variety of possible shapes for the triangle $\triangle{BMP}$ (we can change the angle $\beta$) the area of all is a pretty invariant.
