Assuming indifference over all the angular positions, it follows that there is a probability of 1/3 that a chord is longer than the side of the triangle.
| HPS 0628 | Paradox | |
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Bertrand's Geometrical Paradoxes: Proofs
John D. Norton
Department of History and Philosophy of Science
University of Pittsburgh
http://www.pitt.edu/~jdnorton
| This first case arises when we consider chords diverging from the apex of the equilateral triangle. Then there are three, equal angles of 60o at the apex, over which the chords can be distributed. Chords in the two outer angles will be shorter than the side of the triangle. Only chords in the middle angle will be longer than the side of the triangle. Assuming indifference over all the angular positions, it follows that there is a probability of 1/3 that a chord is longer than the side of the triangle. |
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The figure above appears to deal with a special case of chords emanating from the apex of the triangle. It is easy to see that the result is more general. The chords of this case could emanate from any point on the circumference of the circle. The same result would still follow: there is a probability of 1/3 that a chord has a length greater than the side of the triangle. |
| To recover this next case, it is helpful to add a second, inverted triangle; and then add the smaller shaded triangles that are 1/3rd the height of the larger triangles. We add chords that lie paralled to one side of the large, equilateral triangle. These chords intersect the shaded smaller triangles in two regions. One is the inner region in which the smaller triangles lie within a hexagon formed by the intersection of the two triangles. The other, outer region is defined by the two extreme, smaller triangles. The height of the inner region is that of two of the smaller triangles; and the combined height of the two-part, outer region is the same: that of two smaller triangles. If we are indifferent over all possible of these parallel chords, it follows that there is an equal chance of the two cases: a chord is in the inner region, and has a length greater than the side of the triangle; and a chord is in the outer region and has a length shorter than the side of the triangle. That is, the probability that a chord has a length greater than the side of the triangle is 1/2. |
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Once again, the argument does not hold only in a special case in which the chords are parallel to the side of the triangle. The same analysis can be made if the set of parallel chords are not parallel to the side of the triangle. If you are uncertain that this is so, imagine that we add another triangle to the figure, so that its base is parallel to the new set of chords depicted at right. The analysis then proceeds as above and, once again, returns a probability of 1/2. |
| In this next case, we inscribe a circle within the equilateral triangle. As a first step, we must establish that the radius of the inscribed circle is half that of the radius of the large cirlce. To do this, we draw the right angle triangle shown. Its two sides are the radius of the small circle and the radius of the large circle. Since the radius of the large circle bisects the angle of the equilateral triangle, the new triangle has an angle of 30o at its apex. The triangle overall is a 30o-60o-90o triangle. Its sides are known from elementary geometry to stand in the ratio √3 to 1 to 2. It now follows that the radius of the small circle is half that of the large circle. |
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The next result we need is that, if we pick a point anywhere in the circle, that point is the midpoint of just one chord. To see that only one chord has that property, draw a second chord through that point, as shown in the figure. In comparison to the original chord, the new chord is lengthened on the left and shortened on the right. As a result, its midpoint is also shifted to the left. More generally, any other chord drawn through the first chord's midpoint will have its midpoint dislocated in the same way. There is only one chord that has that original point as its midpoint. |
| We now apply indifference to the points in the circle such that we are indifferent over the location of points that lie in equal areas of the figure. Since the small circle has half the radius of the large circle, the small circle has 1/4th of its area. Thus there is probability of 1/4 that a point is in the small circle. Each point in the circle identifies a unique chord that has the point as its midpoint. Those chords whose midpoints lie in the small circle are the ones with a length greater than the side of the equilateral triangle. It follows that there is a probability of 1/4 that a chord is longer than the side the equilateral triangle. |
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The construction of the last case gives us another way to recover the probability 1/2 case. In it, we consider chord midpoints that lie on a single radius of the larger cirlce. we now assume that we are indifferent over locations of points along the length of the radius. Since the radius of the small circle is 1/2 that of the large cirlce, it follows that there is a probability of 1/2 that a chord midpoint lies within the smaller circle. Chords with such a midpoint are longer than the side of the equilateral triangle. That is, the probability of a chord being longer than the side of the equilateral triangle is 1/2. |
January 31, 2025.
Copyright, John D. Norton