Triangular 2,19,53.

I did it in a very easy and straight-forward way.

Join P and A, P and B and P and C.

Therefore, now we have 3 triangles namely PAC, PBC, and PAB inscribed inside ABC.

Now, Area of the small triangles will be equal to the area of triangle ABC.

Hence,

1/2 * 2014* P1 + 1/2 2014 P2 + 1/2 2014 P3 = sqrt(3)/4 * (2014)^2

(P1, P2, P3 denote the perpendiculars)

P1 + P2 + P3 = sqrt(3)/2 * 2014

=1007 sqrt(3)

Hence, 2(a + b) = 2*1010 = 2020

Well I think Its easy to use the formula that for an Equilateral $\triangle$ the sum of perpendicular distances from a random point in the circle to its sides is konstant and the value is , say s,

$s=\frac{\sqrt{3}}{2} \times a$

where $a$ is the side of the $\triangle$ . [Here $a=2014$ ]

After that it is $\dots$ $.$

I really do not know if the sum always stays the same but I'll trust Krishna.

WLOG, we can assume that the chosen point is the point of intersection of all the three angle bisectors

Now, exploiting the obvious properties of triangles, you have: $\frac{EF}{FB}=\tan{30^\circ}$

Imgur

Because $\tan{30^\circ}$ is $\frac{1}{\sqrt{3}}$ , you have $EF=\frac{1007}{\sqrt{3}}$

You've three perpendiculars, so the answer is three times that,i.e, $1007\sqrt{3}$

perpendicular to all sides would be the radius of inscribed circle hence r=\frac { a }{ 6 } \sqrt { 3 }
=\left[ \frac { \sqrt { 3 } }{ 6 } 1007 \right] 3\ =2020