Professor Stewart's Hoard of Mathematical Treasures (59 page)

This shape is rep-9.

A torus can be represented as a rectangle with opposite faces identified - that is, ‘wrapped round’ so that anything that disappears off one edge reappears at the opposite edge. A Möbius band can be drawn as a rectangle with the left- and right-hand edges identified, but with a half twist. Drawn that way, here are possible solutions. Remember, if you draw it on a Möbius band made from paper, then the lines are deemed to ‘soak through’ the sheet of paper.

Connecting utilities on a torus . . .

... and a Möbius band.

Lots of examples prove that in general you can’t bisect three shapes with a single straight line. Here’s one with three circles. It’s easy to show that the only line bisecting the lower two is the one illustrated. But this doesn’t bisect the third one.

Only one line bisects the two lower circles, and this line does not bisect the top one.

In three dimensions, exactly the same idea works with four spheres. The centres of three of them lie on some plane, and provided those centres are not in a straight line, there is exactly one such
plane. Now put the centre of the fourth sphere at a point that is not on the plane.

The Grumpians are septimists, and use base-7 arithmetic. In their system, 100 works out as
So they get very excited instead of being disappointed: the batsthing with decimal 49 runs has just scored a Grumpian century!

Innumeratus’s solution.

Well, it does look pretty convincing ... but something must be wrong, because the area of Innumeratus’s ‘square’ must be less than that of the original ‘square’. In fact, neither shape is a perfect square. The original one bulges slightly outwards in the middle; the second bulges slightly inwards. For example, the two different sized triangles have horizontal and vertical sides in the ratios 8:3 and 5:2 respectively. If the figure were square, these ratios would be equal. But they are 2.67 and 2.5, which are different.

The bosun placed one coin on the table, and then put two on top of it so that they touched at its centre. To do this, hold them in place while you fit the other two coins almost on edge, leaning together to touch at the top. Again, all coins touch, so in particular they are equidistant.

Place the first three coins as on the left, then add the other two.

The next number in the sequence is 46.

Holmes’s point is: don’t look at what’s there, look at what’s missing. The missing numbers are:

These are the multiples of 3, the multiples of 5, anything containing a digit 3, and anything containing a digit 5. The next number in the sequence is therefore 46 (because 45 is a multiple of 5).

Lagrange’s interpolation formula states that the polynomial
satisfies
P
(
x
_j
) =
y
_j
for
j
= 1, ... , n. Remember: Linderholm’s book is based on the premise that mathematics should be made as complicated as possible to enhance the prestige of the mathematician. Actually, the basic idea here is simple. In less compact notation, the formula becomes
When
x
=
x
_j
, all terms except the jth vanish because of the factor (
x
-
x
_j
). The jth term doesn’t have that factor, and it is an apparently complicated fraction times
y
_j
. However, the numerator and denominator of the fraction are identical, so the fraction is 1. And 1 times
y
_j
is
y
_j
. Cunning!

For example, to justify the sequence 1, 2, 3, 4, 5, 19, we take
x
₁
= 1,
x
₂
= 2,
x
₃
= 3,
x
₄
= 4,
x
₅
= 5,
x
₆
= 6 and
y
₁
= 1,
y
₂
= 2,
y
₃
= 3,
y
₄
= 4,
y
₅
= 5,
y
₆
= 19. Then a calculation gives
and
P
(1) = 1,
P
(2) = 2,
P
(3) = 3,
P
(4) = 4,
P
(5) = 5,
P
(6) = 19.

Edward Waring first published the formula in 1779. Euler rediscovered it in 1783, and Lagrange discovered it again in 1795. So it is named after the third person who found it, which is fairly typical when it comes to naming mathematical ideas after people.