Professor Stewart's Hoard of Mathematical Treasures (60 page)

These 11 circles require four colours.

Eleven circles are needed. Here’s an arrangement that requires four colours. Suppose for a contradiction that it can be 3-coloured. Colour the top middle circle with colour A, and the two adjacent ones with colours B and C. Then the next circle to the left must have colour A, the one below that colour B, and the left-hand circle of the two lowest ones must have colour A. The colours round the right-hand side of the figure must be similar: here either X = B and Y = C, or X = C and Y = B. Either way, the right-hand circle of the two lowest ones must have colour A. But now two adjacent circles have the same colour, namely A - contradiction.

It can be proved that with ten or fewer circles, at most three colours are needed.

The Earth goes round the Sun exactly once in a year, so it returns to the same point in its orbit on any given date (subject to a bit of drift because the exact period is not an integer number of days, whence our use of leap years). In particular, every 13 April it returns to a position at which the orbit of Apophis crosses that of Earth, a necessary condition for a collision.

Apophis has a period of 323 days, and the distance between Apophis and the Sun ranges from 0.7 astronomical units to 1.1 astronomical units, where an astronomical unit is the average distance between the Sun and the Earth. So sometimes it is inside Earth’s orbit, sometimes outside. If Apophis and the Earth orbited in the same plane, their orbits would cross at two points. However, it’s not quite that simple. The orbits are inclined at a small angle, and Apophis is light enough to be affected significantly by the gravitational pull of the other planets, causing changes to its orbit. So although the orbits don’t necessarily cross, the orbit of Apophis - though not necessarily the asteroid itself - comes close to that of the Earth in two places, and the Earth reaches those positions on two specific dates. The one that counts, for the near future, is where the Earth is every 13 April. Whether there’s a collision depends on precisely where Apophis is in its orbit on that date, and a lot of high-precision
observations are needed to determine that. So the date is easy, but the year is difficult.

Orbit of Apophis relative to that of the Earth.

Actually, there was more to it. There always is. Calculations showed that if Apophis happens to pass through a particular region of space, about 600 metres across, during a near-miss in 2029, then it is guaranteed to return to almost the same spot, hitting the Earth in 2036. Fortunately, the latest observations indicate that the chance of such a collision is at most 1 in 45,000. See
neo.jpl.nasa.gov/news/news146.html
science.nasa.gov/headlines/y2005/13may_2004mn4.htm

Both of these refer to Apophis by its provisional designation 2004 MN
₄
.

No, she’s wrong, the probability is
. And it’s very naughty of her to try to swindle poor Innumeratus like that.

Whichever card he picks first, the remaining cards include two of the opposite colour but only one of the same colour. So the probability of picking a card of the opposite colour is
. Since this holds whichever card he picks first, the probability that his two cards are different is
.

Here’s another way to see it. There are 6 distinct pairs of cards. Of these, precisely 2 (♠ ♣ and ♥ ♦) have the same colours, and 4 have different colours. So the probability of getting one of these four is
.

In analysis, ε is always taken to be a small positive number, to the point of cliché. So this joke is a more intellectual variant on the mechanics question that began ‘An elephant whose mass can be neglected ...’

The cards were the King of Spades, Queen of Spades and Queen of Hearts. The first has to be a Spade and the third a Queen, but the precise sequence is not determined.

The first two statements tell us that the cards must either be KQQ or QKQ.

The last two statements tell us that the cards must either be ♠ ♠ ♥ or ♠ ♥ ♠.

Combining these we find four possible arrangements:

The fourth of these contains the same card twice, so it is ruled out. The other three all use the same three cards, in various orders.

This puzzle was invented by Gerald Kaufman.

Surprisingly, Smith earned more - even though £1,600 per year is greater than Smith’s accumulated £500 + £1,000 over a year. To see why, tabulate their earnings for each six-month period:

Note that Jones’s £1,600 splits into two amounts of £800 for each half-year, so his half-year figures increase by £800 every year. Smith’s half-year figures increase by £500 every half-year. Despite that, Smith is ahead in every period after the first, and gets ever further ahead as time passes. In fact, at the end of year n, Smith has earned a total of 10,000n + 500n(2n - 1) pounds, while Jones has earned a total of 10,000n + 800n(n - 1) pounds. So Smith - Jones = 200
n
²
+ 300n, which is positive and grows with n.

Emperor Frederick II was seeking a rational number x such that x,
x
- 5 and
x
+ 5 are all perfect squares. The simplest solution is
for which
Leonardo explained his solution in 1225 in his Book of Squares. In modern notation, he found a general solution
Here the role of
x
is played by the number
(
m
²
+
n
²
), and we want
mn
(
m
²
-
n
²
) = 5. Choosing
m
= 5,
n
= 4, we get
x
= 3
and
mn
(
m
²
-
n
²
) = 180. This may not seem much help, but 180 = 5 × 6
²
. Dividing x by 6 yields the answer.