What is the average sum of distances of a random point inside a triangle to its three sides?

Question

Given a non- Equilateral Triangle with following side sizes: $45,60,75$. Find the sum of distances from a random located point inside a triangle to its three sides.

Note 1: Viviani's theorem related only to equilateral triangles.

Note 2: Fermat point is related to the minimization of distances from a random point inside the triangle and its vertices.

As we can see that both notes are not helpful to solve that problem.

I have been given that puzzle during an hour an a half exam. There were only 6 minutes to solve that problem. Afters many hours I still do not have an answer. I will be very glad to get some assistance or maybe the whole solution

Regards, Dany B.

Answer 1

I understand that a point $P$ is chosen at random inside a triangle $ABC$ according to a uniform probability distribution, and you want the expected value of the sum of the distances from $P$ to the sides of the triangle.

The distance from $P = (x,y)$ to one of the sides is a linear function $ax + by + c$ of the coordinates $x, y$. Thus the sum of the distances is also linear. Therefore the average value is the average of the values for $P = A$, $P = B$ and $P = C$, i.e. the average of the three altitudes of the triangle.

In the present case the altitudes are $36, 45, 60$. So the expected value is $47$.

Answer 2

Given a rectangular triangle with sides $a\leq b\leq c$ we will compute the sum of the distances for an arbitrary interior point. Choose orthonormal coordinates such that the hypotenuse goes through the origin and the other sides are horizontal or vertical. Let $a$ be the height of the vertical side. All interior points have positive coordinates so we can avoid absolute value signs.

The distance from $p(x,y)$ to the horizontal side is $y.$ The distance to the vertical side is $b-x.$ The distance to the hypotenuse $H\leftrightarrow \frac{a}{b}x-y=0$ is

$$\frac{\frac{a}{b}x-y}{\sqrt{\left(\frac{a}{b}\right)^2+1}}=\frac{ax-by}{c}$$

where we have used $a^2+b^2=c^2.$

The sum of the distances is

$$s(x,y)=y+4-x+\frac{ax-by}{c}=\left(\frac{a}{c}-1\right)x+\left(1-\frac{b}{c}\right)y+b.$$

Integration gives

$$\int_{x=0}^b\int_{y=0}^{\frac{ax}b}s(x,y)dy\ dx=\frac{ab}6\left(a+b+\frac{ab}c\right).$$

The expected value of the sum of the distances is obtained after dividing this integral by the surface area $ab/2,$ giving

$$\overline{s}=\frac{a+b+\frac{ab}c}3=47.$$

Afterthought. The terms $a,$ $b$ and $ab/c$ in the numerator are the three heights of the triangle so the average sum of the distances is equal to the average of the three heights.

Answer 3

Let $ (x,y)$ be a point in a semi-circle with diameter inclined at $ \sin ^{-1} \frac35 $ to x-axis having sides proportional to Pythagorean triplet (8,6,10) as given.

The three perpendicular distance to sides of a scalene triangle are $ (x,y, 4 x/5 + 3 y/5) , $

totaling to $ \dfrac {9 x + 8 y} {5} $ which is variable , needing to be averaged.

If it were constant, a result like those from Viviani, Fermat would have been in existence now for more than two centuries.