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An equilateral triangle has one vertex at the centre of a disk.
One side of the triangle lies completely outside the disk and is colored green.
A red line is drawn through two independent, uniformly random points on the disk.

Disk, equilateral triangle, line through two random points on disk

What is the probability that the red line crosses the green side?

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    $\begingroup$ rot13(Vs lbh punatr gur natyr fhograqrq ol terra frtzrag gb nal engvbany senpgvba bs π, gura ol gur fnzr nethzrag, gur nafjre vf gung senpgvba. Ol pbagvahvgl, gur cebonovyvgl sbe nal natyr θ (engvbany be veengvbany zhygvcyr bs π) vf whfg gur engvb bs θ naq π. Vg’f abg pyrne gb zr ubj gur nethzrag va gur npprcgrq nafjre rkgraqf gb guvf ceboyrz. Ohg zl gevpx fbyhgvba qbrf, nygubhtu vg’f abg n cebbs.) $\endgroup$
    – Pranay
    Commented 5 hours ago

2 Answers 2

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1/3

We can add 5 more identical equilateral triangles with a vertex at the center of the circle, and their "green" sides would form a regular hexagon that surrounds the circle. Any line in the circle will almost surely pass through 2 of the triangles' green sides. Since it's equally likely that any 60 degree rotation about the center of that line would be randomly chosen, any particular green line will be intersected one third of the time.

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  • $\begingroup$ I'm not confirming or denying your answer since I haven't worked it through, but the last sentence of your answer is not fully justified, because the method of sampling of the red points doesn't naturally correspond to your last sentence. $\endgroup$ Commented 12 hours ago
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    $\begingroup$ @BenjaminWang You could rephrase the last sentence to say: By symmetry all 6 triangles in the hexagon need to have identical probabilities. $\endgroup$
    – quarague
    Commented 11 hours ago
  • $\begingroup$ The red line intersects at two different lines. There are 6 combinations where the "regions" are adjacent, 6 that skip 1 region, and 3 with opposite regions. If you would include 1 region, there would be 2 regions in the first group, 2 in the second, and 1 in the third. Regardless of the distribution of the probabilities between all these types of combinations, it's always 1/3 like-for-like. $\endgroup$
    – Nautilus
    Commented 6 hours ago
  • $\begingroup$ Why "almost surely", are there any edge cases I'm not seeing? $\endgroup$
    – MaxD
    Commented 2 hours ago
  • $\begingroup$ @MaxD it can pass through a corner (or two corners) of the hexagon $\endgroup$ Commented 1 hour ago
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Here’s a (cheeky) trick solution relying on the fact that

the radius of the disk isn’t given.

So we can

shrink the disk to a point (the apex of the triangle)

while still satisfying the condition that the green segment is entirely outside the disk. Now it’s clear that the red line

is chosen randomly from all the lines passing through the apex. When you rotate that line about the apex by 180° starting from any initial orientation, it goes back to the original orientation, and intersects the green segment exactly for a third of the orientations (because the green segment subtends an angle of 60° about the point of rotation, the apex).

Hence the answer is

1/3.

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    $\begingroup$ Nice answer. As an equivalent alternative from the another perspective that avoids the degeneracy of "two points" into one (and the technical inability to follow the line-drawing procedure using only one point), we can increase the size of the triangle arbitrarily, at which point the line's translation within the circle becomes irrelevant, and only its angle matters. $\endgroup$ Commented 56 mins ago

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