Tricky Triangles

Define the areas $X=\Delta PLB,\,Y=\Delta PNB,\,Z=\Delta PNA$

$\alpha=\Delta PMA,\,\beta=\Delta PMC,\,\gamma=\Delta PLC,\,S=\Delta ABC$

From the figure, $S=X+Y+Z+\alpha+\beta+\gamma$ , the goal is to find $X+Y+Z$

We know that the areas of triangles with equal altitudes are proportional to the bases of the triangles.

So, $\dfrac{CL}{BL}=\dfrac{\gamma}{X}=\dfrac{\alpha+\beta+\gamma}{X+Y+Z}$

and $\dfrac{AM}{CM}=\dfrac{\alpha}{\beta}=\dfrac{\alpha+Z+Y}{\beta+\gamma+X}=\dfrac{Y+Z}{X+\gamma}$

where the last equality follows because if $\dfrac{a}{b}=\dfrac{c}{d}$ , then $\dfrac{a}{b}=\dfrac{a\pm c}{b\pm d}$

The two expressions imply

$X+Y+Z=\left(\dfrac{\alpha+\beta+\gamma}{\gamma}\right)X$

and $Y+Z=\dfrac{\alpha(X+\gamma)}{\beta}$

solving these equations gives $X=\dfrac{\alpha\gamma^2}{\beta(\alpha+\beta)-\alpha\gamma}$

$\implies X+Y+Z=\dfrac{\alpha\gamma(\alpha+\beta+\gamma)}{\beta(\alpha+\beta)-\alpha\gamma}$

Finally, $S=X+Y+Z+\alpha+\beta+\gamma=\boxed{\dfrac{\beta(\alpha+\beta)(\alpha+\beta+\gamma)}{\beta(\alpha+\beta)-\alpha\gamma}}$

Putting $\alpha=\gamma=1$ , we get

$S=\dfrac{\beta(\beta+1)(\beta+2)}{\beta(\beta+1)-1}$

which is minimum when $\beta\approx 1.6917395$

Use Menelaus' theorem to represent the area of the whole triangle by the area of yellow region. Define it as X.
Use Menelaus's theorem on ${\displaystyle \triangle ABD}$ , so we get:

${\displaystyle \frac{BC\cdot DG\cdot AF}{CD\cdot GA\cdot FB} =1}$

Introduce area relationships, we get:

${\displaystyle \frac{S\cdot 1\cdot 1}{[ S-( 2+X)] \cdot ( 1+X) \cdot X} =1}$

We can figure out S

${\displaystyle S=\frac{X( 1+X)( 2+X)}{X^{2} +X-1}}$

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