Mystic cevians 2

To begin with, the ratio $\large{\frac{BB_{a}}{CC_{a}} = \kappa = \frac{BD}{DC}}$

This is from an elementary similiarity relationship between $\Delta BB_{a}D$ and $\Delta CC_{a}D$ . So, doing the same with all other ratios: we end up with the figure on the right with just the cevian incidence ratios on the edges shown.

Apply Menelaus' theorem (magnitudes) on $\Delta ABD$ with transversal $FC$ in order to determine the ratio $\large{\frac{AP_{3}}{P_{3}D}}$ . Hence:

$\large{\frac{AF}{FB}\frac{BC}{CD}\frac{DP_{3}}{P_{3}A} = 1}$

$\implies \large{\frac{AP_{3}}{P_{3}D} = \frac{FB}{AF}\frac{CD}{BC} = \frac{\kappa}{1}\frac{1+\kappa}{1}} = \kappa(\kappa+1)$

So, $AP_{3} = \frac{\kappa(\kappa+1)}{\kappa(\kappa+1)+1}\times AC$

Now, area of $\Delta ADC = \frac{1}{\kappa+1}\times \Delta$ __ (1)

since, $DC = \frac{1}{\kappa+1}\times BC$ (area of triangles with a common height is proportional to the respective bases)

But, area of $\Delta AP_{3}C = \frac{\kappa(\kappa+1)}{\kappa(\kappa+1)+1}\times$ area of $\Delta ADC$ __ (2)

since, $AP_{3} = \frac{\kappa(\kappa+1)}{\kappa(\kappa+1)+1}\times AC$ (area of triangles with a common height is proportional to the respective bases)

Combining (1) and (2),

area of $\Delta AP_{3}C = \frac{\kappa}{\kappa(\kappa+1)+1}\times \Delta$

The same relation holds for the other triangles : $\Delta CP_{1}B$ and $\Delta BP_{2}A$ .

So, we have :

Area of $\Delta P_{1}P_{2}P_{3} = \Delta - (Area\Delta CP_{1}B+Area\Delta BP_{2}A+Area\Delta AP_{3}C)$

$= \Delta - \frac{3 \times \kappa}{\kappa(\kappa+1)+1}\times \Delta$

$= \frac{(\kappa - 1)^2}{\kappa(\kappa+1)+1}\times \Delta$

( Note : The famous case of $\kappa =2$ yields $\frac{\Delta}{7}$ )

So, we need to solve: $\frac{\kappa(\kappa+1)+1}{(\kappa - 1)^2} = 2014$

This becomes a quadratic equation and (as expected) the roots are reciprocals being : $~1.039$ and $~ 0.962$

Now, the percentage deviation from unity (for either number) $^*$ to the nearest integer is $4 \%$ .

Footnote:

$^*$ It is interesting that the percentage deviation is the same for the two reciprocals. Indeed this so because : $\large{\frac{1}{1+\delta}} \approx 1-\delta$ for $\delta \ll 1$