Where should you place the line segment to increase the chance of getting an obtuse triangle

Question

You have a line segment AB (with endpoints A and B) with length 1. You want to place it inside a unit circle. After you've placed it inside the circle a randomly generated point is being placed iniside the circle such that it does not coincide with the line segment. Now, you can connect the three points to form a triangle. Here's the question:

Question:

Where do want to place the line segment inside the unit circle such that it increases the probability that it forms an obtuse triangle?

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I thought I was close to an answer, but just removed it because it was wrong. Will soon upload an image of my progress...

Answer 1

Because of the rotationnal symmetry of the problem, you can suppose that AB is included in the X-axis and write $A=(s,0), B=(s+1,0)$, with $-1 \le s \le 0$. Remark that the triangle is obtuse if and only if the third point, call it $C$ has his $x$ coordinate outside $[s,s+1]$. The probability of this event is proportionnal to the area of the set of such points, which can be computed as: $$P(s) = 2\int_{-1}^s \sqrt{1-x^2} dx + 2\int_{s+1}^1 \sqrt{1-x^2} dx$$ Then the fundamental theorem of analysis gives $$P'(s) = \sqrt{1-s^2} - \sqrt{1-(s+1)^2}$$ and you can study the variations of $P$...

Answer 2

Given a line segment, the only regions in which a placed point does not form an obtuse triangle, is the region bordered by two parallel lines, perpendicular to the segment, and intersecting the segment at its end points, excluding the circular region formed by rotating the line around. I tried to make an illustration of this below, the region that creates acute triangles is marked with the letter $R$

So then what we want to do is minimise the area of intersection or overlap between this region $R$ and that of the big circle. Deriving an expression for the area of this overlap is probably a bit involved, as it will need to depend on $2$ parameters (a third parameter is redundant due to rotational symmetry of the circle) But I think it should be intuitively clear that this overlap will be minimised when the line segment falls on a radial line, with its one end on the circumference. We can maybe give some justification for this, but it doesn't constitute a proof.

If we imagine our 2 degrees of freedom, the distance of the centre of the line segment to the centre of the circle,denoted by $d$, and the angle of rotation of the line segment, denoted by $\theta$, ($0^{\circ}$ and $180^{\circ}$ corresponding to radial lines) then we expect the following to be true: (Where $A(d,\theta)$ represents the area of the overlap)

• $ A(d,90^{\circ}) = c, \mbox{ for } d \in [0,\frac{1}{2}] \mbox{ and } A(d,90^{\circ}) \geq c \mbox{ for } d \geq \frac{1}{2} $

• $ A(0,\theta) = c $

• $A(d,\theta)$ is monotone increasing in $\theta$ when $\theta \in [0,90^{\circ}]$ and $A(d,\theta) = A(d,-\theta)$

• And lastly, $A(d,0^{\circ})$ is monotone decreasing when $d\in[0,\frac{1}{2}]$

The first two claims are not necessary to prove the location of the minimum, it is just there to help you visualise the setup and help with the intuition of the next two claims.

If we take the last two claims to be true, then we can prove the result:

Assume that $A(d_1,\theta_1)$ is a minimum. Then by the third assumption, $A(d_1,\theta_1)\geq A(d_1,0^{\circ})$ so $A(d_1,0^{\circ})$ is also a minimum. But then by claim $4$, $A(\frac{1}{2},0^{\circ}) \leq A(d_1,0^{\circ})$. So $A(\frac{1}{2},0^{\circ})$ Is also a minimum, thus completing the proof.

(The possible values that $d$ can take on depend the value of $\theta$ but the point is that the overlap must attain its minimum when $\theta = 0^{\circ}$, and when $\theta = 0^{\circ}$, $d$ must be less than $\frac{1}{2}$ for the segment to be contained in the circle. We also implicitly assumed that this function does in fact attain a minimum value, which can be justified by the extreme value theorem if we assume that the function is continous in both its parameters, since its domain is compact, but this seems excessive to go into)