The Square Cross-over Part (a)

In the diagram above, we have chosen our first two points: $x$ and $y$ . We have also drawn the two limits to triangles – we have drawn the right most, and left most possible triangles with these values of $x$ and $y$ . Thus we have shown the total distance in which our third point could be in. This distance is $AB$ . By rotational symmetry, we can prove that $XY$ is equal to $AB$ . Thus we have that for any given values of $x$ and $y$ where $x>y$ , the total distance in which a third point can be chosen is the shortest distance between $x$ and $y$ along the perimeter of the square.

In this diagram, we have placed the first point of our triangle - point $x$ , randomly on the perimeter of the square and drawn in point $k$ , which is the point directly opposite of point $x$ . Let the perimeter of the square be equal to $p$ .

Notice that the closer we put point $y$ to $k$ , the shortest distance between $x$ and $y$ along the perimeter of the square increases linearly to a maximum of $\frac { p }{ 2 }$ . This is the same no matter if we let point $y$ approach point $k$ clockwise or counter-clockwise from point $c$ . The minimum distance from $x$ to $y$ possible is obviously $0$ .

Thus the mean shortest distance from $x$ to $y$ along the perimeter of the square, is the average of $0$ and $\frac { p }{ 2 }$ , which is equal to $\frac { p }{ 4 }$ . Dividing this by $p$ we get our answer: $\frac { p }{ 4p } =\frac { 1 }{ 4 }$

This is a very nice problem; thanks for sharing it!

If you work this out for a circle, you'll get the same answer. In fact, if you do this for any regular polygon with an even number of sides, you'll get the same answer. Can anyone show this?

For each point the two remaining points must reside on opposite half-perimeters determined by the diameter defined by the first point. This condition is always true for at least one of three random points. The probability of the condition being true for the remaining two random points is 1/2 each giving a total probability of 1/4.

For any two points on the square (side s) of minimal distance x from each other, with distance measured along the outline of the square. Due to symmetry, I find the range for the third point to be also x. The probability for containing the center would be x/4s if the third point is randomly chosen along the outline of the square. We integrate x/4s over the whole possible values of x = 0 to 2s. = (integral [from x = 0 to 2s] of (x/4s) dx) / 2s = 1/4

I got the answer by using the following: 1 - ( (0.5)^3 * 3! ) = 0.25

I got this right by relying on intuitive maths and Vladimir Smith's solution kind of summed up my gut instinct.