shields@h-sc1.UUCP (thomas shields) (03/04/86)
Here is a hint to help you on your way to finding solutions to
the previously posed puzzle: any chord of a circle can be represented by
its midpoint. Therefore this helps in two ways: it is possible to generate
random chords by generating random points, and a simple check for
intersection with the inner circle is to check if the midpoint of the chord
is in the inner circle. This should help considerably, and make evident
yet another solution to the puzzle. However, we found more solutions based
on this property also, see if you can find them.
Happy puzzling,
- Tom Shields
Harvard Universityins_ampm@jhunix.UUCP (Michael P McKenna) (03/04/86)
In article <965@h-sc1.UUCP> shields@h-sc1.UUCP writes: > > Here is a hint to help you on your way to finding solutions to >the previously posed puzzle: any chord of a circle can be represented by >its midpoint. Therefore this helps in two ways: it is possible to generate >random chords by generating random points, and a simple check for >intersection with the inner circle is to check if the midpoint of the chord >is in the inner circle. This should help considerably, and make evident >yet another solution to the puzzle. However, we found more solutions based >on this property also, see if you can find them. > > Happy puzzling, > > - Tom Shields > Harvard University I don't seem to have received the puzzle, but if it is what I think it is you want to be careful about random generation. This is discussed in one of Martin Gardner's columns, reprinted in _The 2nd Scientific American Mathematical Puzzles & Diversions_ The problem (which sounds like it might be your puzzle), is to determine the probability that a random chord drawn in a circle is longer than the side of an equilateral triangle inscribed in the circle. It is easy to show that this probability depends on the method of generating the random chord. The first suggested method is to pick a point A on the circle. This is one endpoint of the chord. The other endpoint is can range uniformly over the rest of the circle. By symmetry it doesn't matter where A is for computing the probability, so assume that A is at the one of the vertices of the triangle (or just inscribe the triangle so that one of its vertices is at A if this makes you feel uncomfortable). Anyway clearly the chord is longer than the side of the triangle, if and only if it cuts across the triangle. The angle at the vertex is 60 degrees out of a possible 180 degrees so the probability is 1/3. 2nd method: Any chord must be perpendicular to one of the diameters. Pick a diameter, position the triangle so that a vertex is at one of the endpoints. Now the set of chords ranges over the length of the diameter. The diameter intersects the triangle at the midpoint of one side, label this point A. Label the analagous point on the opposite side of the diameter B. Clearly only the chords between A and B are longer than the triangle's side. This length is equal to 1/2 the diameter. Therefore the probability is 1/2 3rd method: The one suggested above. Let the midpoint of the chord range uniformly over the area of the circle. It is easy to show that chords longer that the triangle side must have their midpoints inside a circle with same center, and 1/2 the radius of the original circle. This inner circle therefore has area 1/4 the original circle, therfore the probability of the midpoint being inside it is 1/4. So depending upon which method we choose to generate random chords we get an answer of 1/2, 1/3, or 1/4. When stating that something is to be chosen at random, it is important to state HOW it is to be chosen at random. Interestingly the articly states that humans do not seem to use any of the above methods when asked to draw a random chord. The probability of a subject drawing a chord longer than the side of the triangle is much better than 1/2. Dwight S. Wilson Disclaimer: Why should I disclaim these views? They're mathematically correct. "WARNING: The above contents contain letters and words. USE WITH CAUTION. Improper use may lead to ideas and thoughts. If you find yourself forming opinions, contact your physician at once."
shields@h-sc1.UUCP (thomas shields) (03/06/86)
The original puzzle: draw random chords of a circle of radius 2r, what percentage of them will intersect the concentric circle of radius r? (This is how it was posed to us) > So depending upon which method we choose to generate random chords > we get an answer of 1/2, 1/3, or 1/4. When stating that something > is to be chosen at random, it is important to state HOW it is to > be chosen at random. > > Dwight S. Wilson I stand corrected. The problem was posed to us as intentionally ambiguous, but my real question for you puzzlers is to find as many different methods of drawing random chords that generate different percentages as you can. I found the above three solutions in the space of a few minutes, but after several hours of playing with the problem, I came up with two more, and another person in our seminar came up with still a sixth. It would be very interesting to see what others can come up with. - Tom Shields Harvard University
tan@ihlpg.UUCP (Bill Tanenbaum) (03/08/86)
> The original puzzle: draw random chords of a circle of radius 2r, > what percentage of them will intersect the concentric circle of > radius r? (This is how it was posed to us) > > > So depending upon which method we choose to generate random chords > > we get an answer of 1/2, 1/3, or 1/4. When stating that something > > is to be chosen at random, it is important to state HOW it is to > > be chosen at random. > > > > Dwight S. Wilson > > I stand corrected. The problem was posed to us as intentionally ambiguous, > but my real question for you puzzlers is to find as many different methods > of drawing random chords that generate different percentages as you can. > I found the above three solutions in the space of a few minutes, > but after several hours of playing with the problem, I came up with two more, > and another person in our seminar came up with still a sixth. It would be very > interesting to see what others can come up with. > > - Tom Shields > Harvard University ---------------- All methods of choosing chords are random. There are obviously an uncountable infinity of percentages (all real numbers between zero and one inclusive). Perhaps you'd better place restrictions on what you allow as a "good" random chord generator. Remember, even a random number generator which returns the same number every time is still a random number generator, even if it not a very good one. -- Bill Tanenbaum - AT&T Bell Labs - Naperville IL ihnp4!ihlpg!tan
bhuber@sjuvax.UUCP (03/08/86)
In article <974@h-sc1.UUCP> shields@h-sc1.UUCP writes: >The original puzzle: draw random chords of a circle of radius 2r, >what percentage of them will intersect the concentric circle of >radius r? (This is how it was posed to us) > >> So depending upon which method we choose to generate random chords >> we get an answer of 1/2, 1/3, or 1/4. When stating that something >> is to be chosen at random, it is important to state HOW it is to >> be chosen at random. >> >> Dwight S. Wilson > >I stand corrected. The problem was posed to us as intentionally ambiguous, >but my real question for you puzzlers is to find as many different methods >of drawing random chords that generate different percentages as you can. > It would be very >interesting to see what others can come up with. > > - Tom Shields > Harvard University The problem is quite uninteresting, unless one assumes that the probabilities involved are invariant under rotations of the circle (at least). Even so, it's easy to construct such distributions of chords which yield any answer you please, between 0 and 1. Let a be an angular coordinate on the circle, which we orient. Then almost every chord (the diameters excepted) is uniquely determined by, and uniquely determines, the angle b that it makes with a radius drawn to its first endpoint, where b ranges between 0 and pi/2, excluding pi/2. (To avoid to much pedantry, I let you define 'first endpoint'. This is why one orients the circle.) "Drawing random chords" in a way that respects the symmetry of the circle means choosing a normalized product measure on [0, 2pi[ x [0, pi/2[ (I'm fudging with the diameters) which is uniformly distributed on the first parameter. So choose the uniform measure da/(2pi) for the endpoint of the chord, and let m be any normalized measure for the angle to the radius. The measure da X m is normalized and respects the symmetry of the circle. By choosing m to have support in [pi/6, pi/2], the answer is 0. By choosing m to have support in [0, pi/6] the answer is 1. A linear combination of such m's will give any desired answer between 0 and 1. Either more clarification of this problem needs to be given, or else it isn't a problem at all. bill huber