We create the unit equilateral triangle and put one vertex on each side the of the equilateral triangle and then connect them. What is the expected value of the triangle formed by the connection of these points?


You can write the area of a triangle with edge of length $a$ and $b$ and angle $\theta$ between them as $\frac{\sin(\theta)}{2}ab$.

Suppose the points you pick are $a$, $b$ units from one corner of the triangle. The third point is $c$ units away from one of the remaining edges. I'm assuming $a,b,c \sim U[0,1]$, and independently drawn.

The total area of the triangles other than the one in the middle is given by: $A=\frac{\sin(\theta)}{2} (ab+(1-a)c+(1-b)(1-c))=\frac{\sin(\theta)}{2} (ab-ac+bc +1-b)$, where $\theta=\pi/3$. Taking the expectation, we obtain $A= \frac{\sin(\theta)}{2} (1/4 +1-1/2)=\frac{3}{4}\frac{\sin(\theta)}{2} $.

The area of the triangle we started with is $\frac{\sin(\theta)}{2}$. Thus, the answer is $\frac{1}{4}\frac{\sin(\theta)}{2}=\frac{\sqrt{3}}{16}$.

  • $\begingroup$ In fact, the same argument shows that for an arbitrary starting triangle, each corner we remove is, in expectation, $\frac14$ of the area of the original triangle, so the area remaining is, in expectation, $\frac14$ of the original area as well. (The area of an equilateral triangle of side length 1 is $\frac{\sqrt3}{4}$, and $\frac14$ of that is $\frac{\sqrt3}{16}$.) $\endgroup$ – Misha Lavrov Sep 7 '17 at 23:09

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