Cal Poly Department of Mathematics
Puzzle of the Week
Nov 19 - Dec 2, 2015
Consider the following construction: Begin with a unit circle and its circumscribed equilateral triangle; then circumscribe about the triangle a
new circle, and circumscribe about it a square; continue in this way, so
that after constructing a regular n-gon you circumscribe it with a circle
and circumscribe the new circle with a regular n + 1-gon. If r2 = 1 denotes
the radius of the first circle, and in general we let rn denote the radius of
the circle which circumscribed the regular n-gon, then what is a formula
for rn ? Correct to four decimal places, what is r1000 ? Do you think rn
tends to a finite value, or infinity, as n → ∞? (No proof required)
Solutions should be submitted to Morgan Sherman:
Dept. of Mathematics, Cal Poly
Email: sherman1 -AT- calpoly.edu
Office: bldg 25 room 329
before the due date above. Those with correct and complete solutions will have
their names listed on the puzzle’s web site (see below) as well as in the next
email announcement. Anybody associated to Cal Poly is welcome to make a
submission.
http://www.calpoly.edu/˜sherman1/puzzleoftheweek
Solution:
A formula for rn is given by:
rn =
n
Y
k=3
sec
π
π
π
π
= (sec )(sec ) · . . . · (sec )
k
3
4
n
from which we compute r1000 ≈ 8.6572. In
fact the sequence {rn }∞
n=3 converges.
The formula for rn follows from the below picture:
rn−1
π/n
rn
In general we have rn = rn−1 sec πn , from which we get the formula given above.
To see the sequence converges first note that the terms rn are all greater than 1 and they are
monotonically increasing. So either the sequence {rn } converges or it diverges to infinity. Now note
that
π
π
π
π
log rn = log sec + log sec + log sec + . . . + log sec .
3
4
5
n
P∞
π
Hence the sequence {rn } converges if and
if the sum n=3 log sec n converges. We do this by
P only
1
which we know converges:
a limit comparison test with the series
n2
− xπ2 tan πx
1
x→∞
x→∞
− x23
x2
π 2 tan πx
π2
= lim
=
π
x→∞ 2
2
x
P
π
The first equality is due to L’Hôpital’s Rule. Hence the sum ∞
n=3 log sec n converges, and hence
the sequence {rn } also converges.
lim
log sec πx
= lim
To get a better idea of the limiting value we numerically compute
r1000 ≈ 8.6572
(I used the computer algebra system Sage). To see how good an approximation this is we can use
the integral test, as depicted below:
y
y = log sec πx
0
... n − 1 n n + 1
...
R∞
P∞
π
π
π
k=n log sec k ≤ log sec n + n log sec x dx
x
First we obtain the more precice bounds
1
1
1 2
u ≤ log sec u ≤ u2 + u4
2
2
4
which holds, for example, for 0 ≤ u ≤ π/3. (This can be obtained by integrating the corresponding
bounds θ ≤ tan θ ≤ θ + θ3 which are easy to verify for positive θ < π/3.) Then for any n ≥ 3 we
have
π
π
π
En := log sec
+ log sec
+ log sec
+ ...
n+1
n+2
n+3
Z ∞
π
log sec dx
≤
x
n
Z π/n
1
π
=π
log sec u du
(u = )
2
u
x
0
Z π/n
1 1
≤π
( + u2 ) du
2 4
0
π3
π
+
=π
2n 12n3
We also compute (again using the integral test)
Z ∞
π
En ≥
log sec dx
x
n+1
Z π/(n+1)
1
=π
log sec u du
u2
0
Z π/(n+1)
1
π2
≥π
du =
2
2n + 2
0
Hence for every n we find
π2
π4
π2
≤ En ≤
+
.
2n + 2
2n 12n3
Note that for any n we also have
r := lim rk = rn · exp(En ).
k→∞
2
2
π
π
Using the above bounds for En for n = 1000 we find that 8.6996 · exp( 2002
) ≤ r ≤ 8.6996 · exp( 2000
+
π4
) which comes to
1.2×1010
8.700015 ≤ r ≤ 8.700058
In fact the value of r is known to be approximately 8.700036625 . . .. It is known as the polygon
circumscribing constant; you can find more information from the On-Line Encyclopedia of Integer
Sequences where the digits of this constant are recorded in entry A051762.