Find the area of ​region triangular BPM

Solution 1:

$S_{\triangle MBP} = \frac 12 \cdot \frac{BC}{2} \cdot BP \sin{(90^\circ - \frac{\angle B}{2})} = \frac{BC \cdot BP}{4} \cos (\frac{\angle B}{2})$

$S_{\triangle ABP} = \frac1 2 \cdot 3 BC \cdot BP \sin{(90^\circ + \frac{\angle B}{2})} = 6 \cdot S_{\triangle MBP}$

$S_{\triangle ABM} = S_{\triangle ABP} - S_{\triangle MBP} = 5 \cdot S_{\triangle MBP} = 20$

$S_{\triangle MBP} = 4$

Or applying external angle bisector theorem in $\triangle ABM$,

$\frac{AB}{BM} = \frac{AP}{MP} \implies AP = 6 \cdot MP$ or $AM = 5 \cdot MP$

$S_{\triangle ABM} = 5 S_{\triangle MBP} = 20$

$ \therefore S_{\triangle MBP} = 4$

Solution 2:

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.