Buffon's Needle: a Proof
Date: 08/13/97 at 07:51:37 From: Mohammad Shakil Akhtar Subject: Buffon's needle I have known the Buffon's needle problem for the last 28 years and have yet to come across its theoretical proof. I have used this theorem with great enthusiasm with students (secondary) where we use it to calculate the value of pi using statistical methods by throwing some needles (toothpicks) on a ruled paper. The formula we use is pi = 2(total no. of throws)/no. of cuts No. of throws means the total number of needles thrown (like 10 needles thrown 100 times = 1000). Cuts = needles landing with an intersection with any line. Distance between parallel lines = length of the needle. I would be very grateful if I could get its proof. I am amazed by how come pi and needles. What is the connection? I am a teacher and a graduate from London. Thanks in advance.
Date: 08/13/97 at 13:01:12 From: Doctor Anthony Subject: Re: Buffon's needle Let the needles have length L1 and the parallel lines be drawn a distance L2 (L2 > L1) apart. A 'success' occurs when any part of a needle cuts a line. We can think of the centre of the needle being uniformly distributed between 0 and L2/2. Let the smaller of the angles between the direction of a needle and the parallel lines be theta, so that theta is uniformly distributed between 0 and pi/2. If y is the distance of mid-point of the needle from the closest line, then we get an intersection if: y < (L1/2)sin(theta) We now draw two axes with y up the vertical axis varying from 0 to L2/2, and theta along the horizontal axis varying from 0 to pi/2. The sample space is any point within this rectangular area = (pi/2)(L2/2). If you draw the curve y = (L1/2)sin(theta) from 0 to pi/2, then the area under this curve divided by the total area of the rectangle will give the probability of an intersection. The area under the sine curve is INT[(L1/2)sin(theta)] = -(L1/2)cos(theta) from 0 to pi/2 = -L1/2[0 - 1] = L1/2 L1/2 Probability of an intersection = --------- (pi/2)(L2/2) 2L1 Probability = ---------- pi.L2 No.of cuts 2L1 Also Probability = ------------ = -------- No.of throws pi.L2 2L1.Number of throws From this pi = ---------------------- L2.Number of cuts [I note that you have made L1 = L2, but this is not necessary provided; L2 is not less than L1.] -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 08/20/97 at 08:37:29 From: Mohammad Shakil Akhtar Subject: Buffon's needle problem Dear Sir, I was delighted to see your proof. Most importantly, it is simple and elegant against all my expectations. Thank you very much. Sincerely yours M.S.Akhtar
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