Advanced level (June 2003)
This month's problem is a variation on the Buffon Needle Problem.
A needle of length 2 is bent at its midpoint forming a right angle. It is then dropped onto a floor on
which a family of parallel lines spaced Sqrt[2] units apart have been drawn. What is the probability
that the needle lands on one of the lines? Assume that where the midpoint of the needle lands and
with what orientation are both uniformly distributed. What if the lines are spaced 2 units apart? 1
unit?
Solution
Let O be the midpoint and A, B the needle ends, with the assumption that OA rotates 90 degrees
counterclockwise to overlap OB. For sake of generality, let d be the distance between two parallel
lines on the floor, and l/2 = OA = OB. We can assume point O to fall within a strip from one line to
half the distance from the next one (other configurations are symmetrical). Let y be the distance
from O to the line, n the normal to the line, and x the angle measured counterclockwise between n
and the segment OA. It is easy to show that
- for 0 < x   / 2, there will be intersection if y  l/2 sin x or y  d – l/2 cos x
- for  / 2 < x  , there will be intersection if y  l/2 sin x or y  – l/2 cos x
- for  < x  3 / 2, there will be intersection if y  – l/2 cos x or y  d + l/2 sin x
- for 3 /2 < x  2, there will be intersection if y  d + l/2 sin x or y  d – l/2 cos x
Let’s define the following functions:
y = f1(x): y = l/2 sin x for 0 < x  , y = 0 elsewhere
y = f2(x): y = – l/2 cos x for  / 2 < x  3 / 2, y = 0 elsewhere
y = f3(x): y = d + l/2 sin x for  < x  2, y = 0 elsewhere
y = f4(x): y = d – l/2 cos x for 0 < x   / 2 and for 3 /2 < x  2, y = 0 elsewhere
(the graphs of these four function are given in the Excel file Graphs.xls for the three different values
of distance d).
Let’s also define the following sets:
S1 = (x,y): 0  x  2, y  f1(x), S2 = (x,y): 0  x  2, y  f2(x), S3 = (x,y): 0  x  2, y 
f3(x), S4 = (x,y): 0  x  2, y  f4(x), S = (x,y): 0  x  2, 0  y  d/2.
Let A = d be the area of the rectangle S, and let A’ be the area of the set (S1S2S3S4)S; the
probability that the needle lands on the line is then given by A’/A.
A’ can be now computed by integrals.
Case 1: l = 2, d = sqrt[2]
Functions f1, f4 and f2, f3 are tangent at x = /4, y = sqrt[2]/2 and x = 5/4, y = sqrt[2]/2 respectively,
while functions f1, f2 and f3, f4 intersect at x = 3/4, y = sqrt[2]/2 and x = 7/4, y = sqrt[2]/2
respectively (see sheet “d = sqrt(2)” in Graphs.xls).
Instead of A’, it is simpler to compute A – A’, that is the area of “no intersection”. Taking account of

7

7
A  A'  2 4 ( f 4 ( x)  f 1 ( x))dx 23 4 f 3 ( x)dx  2 4 ( 2  cos x  sin x)dx 23 4( 2  sin x)dx 
0
2
0
2
  2  2sin ( / 4)  2sin (0)  2 cos( / 4)  2 cos(0)  2 cos(7 / 4)  2 cos(3 / 2) 
 2  2  0  2  2  2  0  2  2  2
the symmetries, we find
Being A =  sqrt[2], it follows A’ = 2 + sqrt[2], and the probability is A’/A = (2 + sqrt[2])/ sqrt[2],
that is approximately 0,7685.
Case 2: l = 2, d = 2
It is easy to see that in this case the only possible intersections are given by the points in S1 and S2
(see sheet “d = 2” in Graphs.xls). Area A’ is
A'  2
3
0
4
f1 ( x)dx 2
3
0
4
sin xdx   2 cos(3 / 4)  2 cos(0)  2  2
A’ has the same value as in Case 1; however, A is greater now, and the probability is A’/A = (2
+sqrt[2])/2, that is approximately 0,5434.
Case 3: l = 2, d = 1
In this case, the only region where there is no intersection is for 3 /2 < x  2 and y < f3, f4 (see
sheet “d = 1” in Graphs.xls). We find immediately
7
7
A  A'  23 4 f 3 ( x)dx 23 4(1  sin x)dx   / 2  2 cos(7 / 4)  2 cos(3 / 2)   / 2  2
2
2
It is A = , and A’ =  – /2 + sqrt[2] = /2 + sqrt[2]. The probability is (/2 + sqrt[2])/, that is
approximately 0,9502.
Solution by:
Marcello Cammarata
Cassina de’ Pecchi (Italy)
marcello.cammarata@tiscali.it