We will show that the sum of the perpendicular distances from an arbitrary point P inside the equilateral triangle to the three sides of the triangle is a constant.
Draw lines from P to each of the vertices of the triangle. These lines divide the equilateral triangle into three triangles. Each of these triangles has as its base one side of the equilateral triangle and as its height the perpendicular distance from P to that side. Let those perpendicular distances be h1, h2, and h3.
The area of a triangle is equal to ½ × base × perpendicular height.
Hence the area of each internal triangle is, respectively, ½h1, ½h2, and ½h3.
The height of the equilateral triangle is readily found by dropping a perpendicular from a vertex to the opposite side, and applying Pythagoras' Theorem.
We find that the height is /2, and so the area of the equilateral triangle is /4.
The area of the equilateral triangle is equal to the sum of the areas of the three internal triangles.
Hence /4 = ½h1 + ½h2 + ½h3, and so h1 + h2 + h3 = /2.
Since the sum of the perpendicular distances is a constant, the expected value of the sum of the perpendicular distances from P to the three sides of an equilateral triangle of side length 1 is /2.
In general, when considering such a question, we need to specify how the point P is chosen. In other words, we must specify what is meant by choosing a point at random. Of course, the answer to the problem posed above would be the same regardless of how P is chosen!
To illustrate the importance of specifying the randomization method, consider Bertrand's Paradox, in which we are asked to find the probability that a chord drawn at random in a circle is longer than the side of an inscribed equilateral triangle. (Or, sometimes, longer than the radius of the circle.) As explained in the reference above, the answer depends upon how we choose a random chord.
In equilateral triangle ABC of side length d, if P is an internal point with PA = a, PB = b, and PC = c, the following pleasingly symmetrical relationship holds:
3(a4 + b4 + c4 + d4) = (a2 + b2 + c2 + d2)2.