Cool Geometry

Let the pentagon be ABCDE, and the perpendiculars from P to sides AB, BC, CD, DE, and EA have lengths $p_1, p_2, p_3, p_4, and \ p_5$ , respectively.

Since P is inside the pentagon, we have that the area of the pentagon is the sum of the areas of the triangles ABP, BCP, CDP, DEP, and EAP. This sum is simply equal to

$[ABCDE]=\dfrac{1}{2}(AB\cdot p_1+BC\cdot p_2+CD\cdot p_3+DE\cdot p_4+EA\cdot p_5)$ .

Now, since ABCDE is a regular pentagon, we have that all of its sides are equal to some positive s, and therefore

$p_1+p_2+p_3+p_4+p_5=2[ABCDE]/s$ .

Let O be the center of the pentagon. We know that $\angle AOB=72°$ , so $[AOB]=(1/2)r^2\sin 72°$ . We were given that ABCDE is a regular pentagon, so it follows that

$p_1+p_2+p_3+p_4+p_5=(5r^2/s)\sin 72°$ .

We now find s in terms of r. We know that $\angle OAB=54°, so [AOB]=(1/2)rs \sin 54°$ . Therefore $s=r(\sin 72°/ \sin 54° )$ . This shows that

$p_1+p_2+p_3+p_4+p_5=5r\sin 54^{\circ}.$

$5 \times 5 \times \dfrac{\sqrt{5}+1}{4}$

Therefore $m=25 , n=5 , o=25 , p=4$ ; Hence $m+n+o+p =59$ .

Each perpendicular to a side, and the side forms a triangle.
The perpendicular is the altitude of a triangle the side is the base.
Five such triangle cover full area of the pentagon.
However the same 5 sides with their apothem also cover the full area of the pentagon.
So the sum of the five apothem = sum of the perpendiculars.
Sum of 5 Apothem $=5 * R * cos36 = 5*5*\dfrac{\sqrt5 + 1} 4=\dfrac{25\sqrt5 + 25} 4=\dfrac{m\sqrt n + o} p.$
m+n+o+p=25+5+25+4=59.

since P is any point, hence take p to be the centre of the circle and hence ans=5Rcos(pi/5)