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Stanford
Math
Circle
Sneaky Segments
Sam Vandervelde
February 5, 2006
Today we answered the question, “If two lattice points are chosen at random in the plane, thenwhat is the probability that they determine a sneaky segment?” To find out what a sneaky segmentis, or what a lattice point is, please consult the slides posted along with this handout at the website.The questions below recap and also expand on some of the ideas presented at this math circle. Andplease contact me at <[email protected]> if you have any questions.
1. Let m and n be integers, not both zero. Explain why the segment joining (m,n) to the origin issneaky if and only if m and n are relatively prime. (That is, have no common factors except 1.)
2. Consider a square grid four point wide by three points high. If we choose two distinct points atrandom from among these twelve points, what is the probability that the segment joining them willbe sneaky? (Check: the numerator and denominator of the answer have a difference of 10.)
3. Find the probability that. . . a) a “random integer” is not divisible by 7, b) that this integer isalso divisible by 3, c) that two integers chosen at random are both divisible by 5, and d) that twointegers chosen at random do not have a factor of 5 in common.
4. Work through the questions in the slides to demonstrate that if S is the probability that twointegers chosen “at random” are relatively prime, then
1S
=112
+122
+132
+142
+ · · ·
5. Convince yourself that Q(x) =(1 − x
3
) (1 + x
1
) (1 + x
4
)is a degree three polynomial with roots
x = 3, −1, −4, and also that Q(0) = 1. Build on this idea to find a degree four polynomial Q(x)with roots at x = −3, −2, 2, and 3, with Q(0) = 1 as well.
6. Complete the remaining questions in the slides to explain why
sin(πx)πx
= 1 − (πx)2
6+
(πx)4
120− (πx)6
5040+ · · · =
(1 − x2
12
)(1 − x2
22
)(1 − x2
32
)(1 − x2
42
)· · ·
We now have two different formulas for sin(πx)πx . Compare coefficients of x2 in each to conclude that
112 + 1
22 + 132 + · · · = π2
6 , and conclude that the answer to our original question is S = 6π2 .
Further questions
7. Let us say a segment is “almost sneaky” if it passes through exactly one other lattice point.Prove that the probability that two lattice points chosen at random in the plane determine analmost sneaky segment is 3
2π2 .
8. Suppose we work with a hexagonal instead, so that the points form equilateral triangles insteadof squares. Find the probability that two lattice points chosen at random create a sneaky segment.(Hint: there is no need to start from scratch; just relate this problem to our original question.)
9. The product formula for sin(πx), rewritten as
sin(πx)πx
=(
1 − x
1
)(1 +
x
1
)(1 − x
2
)(1 +
x
2
)· · · ,
can be used to discover some pretty wierd formulas. For example, try substituting x = 12 to confirm
a famous formula for π due to John Wallis: 2π = 1
2 ·32 ·
34 ·
54 ·
56 ·
76 · · ·. What other unusual formulas
can you create by using other strategic values for x?
10. It is possible to compare coefficients of x4 in the identity on the previous page to find the valueof the sum
114
+124
+134
+144
+154
+ · · · ,
although this computation requires a little more care. (Hint: consider squaring both sides of theequality 1
12 + 122 + 1
32 + · · · = π2
6 .)
11. There is a very clever alternative method of finding the above sum. Try replacing x by ix onthe previous page, then strategically combining the new identity with the old one.
12. You are now ready for the following variation on the sneaky segments problem: what is theprobability that two points chosen at random from a four-dimensional square lattice determine asneaky segment? (You can think of these lattice points as having integer coordinates (a, b, c, d),since it is hard to visualize the problem.)
13. What is the probability that three points chosen at random in the plane form a sneaky triangle?(Warning: I don’t even know if it’s possible to write the answer to this question in closed form!)
Sneaky Segments
Sam Vandervelde
Mathematician at large
1
Dubious definitions
A lattice point is a point in the Cartesian plane with integer coordinates,such as (4,−7) or (0, 0). A sneaky segment extends from one lattice pointto another without passing through any other lattice points along the way.
2
Simplify, simplify
1) To begin, we may assume without loss of generality that the first latticepoint chosen is the origin.
2) Observe that the segment joining the point (m, n) to the origin is sneakyif and only if m and n are relatively prime. (This means that m and n
have no common divisors other than 1 and −1. For example, 4 and −7 arerelatively prime, but 4 and −22 are not.)
3
Now wait just a minute
One approach to handling infinitely many points is to employ a limitingprocess. For instance, we could answer the question for all lattice pointswith coordinates from 0 to 99, then try expanding our grid to include co-ordinates from 0 to 999, and so on. If the probabilities seem to approach alimiting value, we could declare this to be the overall answer.
4
Your turn
Have your calculator produce a random integer from 0 to 99 by typing100*rand, then ignoring all the digits past the decimal point. Press Enterrepeatedly to obtain a list of twenty such integers, writing them down asyou go. Group them in pairs to obtain ten lattice points. Now check howmany of them determine sneaky segments when connected to the origin bytesting which pairs are relatively prime. We will tally the results below.
5
Enter number theory
We can also handle the difficulty of choosing a lattice point at random byconsidering the question one prime at a time. To see how this works, answerthe following questions.
1) What is the probability that a “random integer” is not divisible by 7?
2) What is the probability that this integer is also divisible by 3?
3) What is the probability that two integers chosen at random are bothdivisible by 5?
4) What is the probability that these two integers do not have a factor of 5in common?
6
Getting down to business
What is the probability S (for ‘Sneaky’) that two integers chosen at randomare relatively prime?
7
You gotta be kidding
Use the formula for the sum of an infinite geometric series to rewrite 1/S.
1
1 − r= 1 + r + r2 + r3 + r4 + · · · , for − 1 < r < 1.
8
Are you insane!?
Now multiply out these (infinitely many) series, each of which has infinitelymany terms.( 112 + 1
22 + 142 + 1
82 + · · ·) ( 1
12 + 132 + 1
92 + · · ·) ( 1
12 + 152 + · · ·
) ( 112 + 1
72 + · · ·)· · ·
9
The Basel problem
Euler managed to find a simple closed form expression for the sum
1
12 +1
22 +1
32 +1
42 + · · ·
by finding two different formulas for the function
P (x) =sin(πx)
πx.
10
An infinite sum
First, he used the standard fact that
sin x = x − x3
3!+
x5
5!− · · ·
to conclude that
sin x
x=
and hence
sin(πx)
πx=
11
An infinite product
On the other hand, Euler knew all the zeros of P (x) = sin(πx)πx , so he could
write down all the factors of P (x):
12
It all comes together
Comparing coefficients of x2, we find that
1
12 +1
22 +1
32 +1
42 + · · · =
Hence the probability of obtaining a sneaky segment when choosing twolattice points at random is . . .
13
