Perpendiculars from P

Draw segments $PA$ , $PB$ , and $PC$ . Notice that $[PAB]+[PBC]+[PAC]=[ABC]$ . Since the area of an equilateral triangle with side length $s$ is $\frac{s^2\sqrt{3}}{4}$ , the area of triangle $ABC$ is $\frac{(20\sqrt{3})^2\sqrt{3}}{4}=300\sqrt{3}$ . Since the area of any triangle is $\frac{1}{2}bh$ , we have $[PAB]=\frac{1}{2}\cdot20\sqrt{3}\cdot PD$ , $[PBC]=\frac{1}{2}\cdot20\sqrt{3}\cdot PE$ , and $[PAC]=\frac{1}{2}\cdot20\sqrt{3}\cdot PF$ . Substituting the values of $PD$ and $PE$ gives $[PAB]=90\sqrt{3}$ and $[PBC]=100\sqrt{3}$ . This means that $90\sqrt{3}+100\sqrt{3}+10PF\sqrt{3}=300\sqrt{3}\Rightarrow \\ 10PF\sqrt{3}=110\sqrt{3}\Rightarrow \boxed{PF=11}$

Let $x$ be the length of $\overline{PF}.$ Draw line segments from $P$ to $A, B,$ and $C.$ We have now created three smaller triangles $\triangle{APB}, \triangle{BPC},$ and $\triangle{APC}$ which make up the whole triangle. The triangles all have base $20\sqrt{3}$ and have heights $9, 10,$ and $x,$ respectively. The area of the entire triangle is $(20\sqrt{3})^2 \sqrt{3}/4 = 300\sqrt{3} ,$ so we have:

$\frac{9(20\sqrt{3})}{2} + \frac{10(20\sqrt{3})}{2} + \frac{x (20\sqrt{3})}{2} = 300\sqrt{3}$

$\frac{9}{2} + \frac{10}{2} + \frac{x}{2} = 15$

$9 + 10 + x = 30$

$x = 11$

So, the length of $\overline{PF}$ is $\boxed{11}.$

Let,PF=h Since area of triangle ABC is= area of triangle PAB+area of triangle PBC
+area of triangle PAC

=>√3/4 X (20√3)(20√3) =1/2 X 9 X20√3 +1/2X10X20√3 +1/2 X hX20√3 => h = 11

area of triangle ABC = (1/2) PD AB + (1/2) PE BC + (1/2) PF AC = (1/2) AB (PD + PE + PF)

Also, area of triangle ABC = (1/2) height AB

height = (√3/2)*20√3 = 30

So, PD + PE + PF = height = 30 => PF = 11

(9 + 10 + x)/3 = (20 3^(1/2))/(2 3^(1/2)) then we get the value of x = 11 So the length of PF = 11

We will get three small triangles within the equilateral triangle.,if we join the point P to all the vertices of equilateral triangle.

Equating the sum of the areas of three small triangles equal to th area of equilateral triangle,

(1/2) 20root(3) [9 + 10 + PF] = {root(3) 400 3}/4, we will obtain PF = 11

Let $[ABC]$ denote the area of the triangle $ABC$ . Since $ABC$ is an equilateral triangle we have $[ABC] = \frac{1}{2} AB^2 \sin 60^\circ = \frac{\sqrt{3}}{4} \left(20\sqrt{3}\right)^2 = 300 \sqrt{3}$ . So, we have $\begin{aligned} [ABC] &= [APB] + [APC] + [BPC] \\ &= \frac{1}{2}AB \cdot PD + \frac{1}{2}BC \cdot PE + \frac{1}{2}AC \cdot PF \\ 300 \sqrt{3} &= \frac{1}{2} \cdot 20\sqrt{3} \cdot 9 + \frac{1}{2} \cdot 20\sqrt{3} \cdot 10 + \frac{1}{2}\cdot 20\sqrt{3} \cdot PF \\ PF &= 30 - 9 - 10 \\ PF &= 11 \\ \end{aligned}$

Note: If $P$ is any point within the triangle, and $D, E, F$ are defined as the foot of the perpendiculars to the sides, then the above argument shows that $PD + PE + PF = 30$ .

P is in triangle. so CD, BF, and AE are perpendicular to AB, AC, and BC respectively. BC= 20 \sqrt{3} and FC= 10\sqrt{3} also we know that BFC=90^\circ, FCB=60^\circ so FBC=30^\circ. Then BF= 20 \sqrt{3} x \cos 30^\circ= 30= CD= AE from ceva theorem: \frac {DP}{\frac {DC}} + \frac {PE}{\frac {AE}} + \frac {PF}{\frac {BF}} = 1 \frac {9}{\frac {30}} + \frac {10}{\frac {30}} + \frac {PF}{\frac {30}} = 1 \frac {19+PF}{\frac {30}} = 1 19+PF=30 PF=11