Read Professor Stewart's Hoard of Mathematical Treasures Online

Authors: Ian Stewart

Tags: #Mathematics, #General

Professor Stewart's Hoard of Mathematical Treasures (56 page)

BOOK: Professor Stewart's Hoard of Mathematical Treasures
8.24Mb size Format: txt, pdf, ePub
ads
The areas of the rings are the differences between these:
π, 3π, 5π, 7π, 9π
These are π multiplied by consecutive odd integers. Robin’s integers are less than or equal to Tuck’s, since Robin’s arrows are closer to the centre. The two sets of odd integers must have the same sum. The only possibility is 1 + 3 + 5 = 9.
Bonus point:
Six rings. The sixth ring has area 11π, so 1 + 3 + 7 = 11 is a second solution.
Further bonus point:
Eight rings. Robin’s odd numbers must be consecutive, and so must Tuck’s. The next two rings have areas 13π, 15π, and 3 + 5 + 7 = 15 is a second solution with consecutive odd numbers.
Just a Phase I’m Going Through
AC is exactly
AB. The times at which such crescents appear are
and
through the lunar cycle, starting from the new moon. (Not
and
!)
When the area of the crescent is one-quarter of the area of the disc.
The inner edge of the crescent is half of an ellipse (Cabinet, page 283). The full ellipse is shown. The white crescent has
the area of the circle when the ellipse has
the area of the circle. Let AB =
r
, AC =
s
. The area of the circle is π
r
2
. The area of an ellipse is πab where a and b are the ‘semiaxes’ - half of the widths in the widest and narrowest directions. Here
a
=
r
and
b
=
s
. So we want π
r
s =
π
r
2
, so
s
=
r
.
For the timing of these crescents, view the Moon from ‘above’. The centre of the Earth is at E; its orbit is shown as the larger circle, but not to scale. The light from the Sun S illuminates half the Moon, leaving the other half dark (here shown in grey). New moon occurs when the centre of the Moon is at point O.
Geometry of lunar orbit.
The points A, C, B correspond to those in the question, and we want to choose angle SEA to make C the midpoint of AB, where P is the edge of the dark area of the Moon and FPC is parallel to EA (the assumption of parallel projection). Then BP = AP (since triangle APB is isosceles), but AP = AB since both are radii of the Moon. Therefore triangle APB is actually equilateral, so angle PAB = 60°. Therefore angle PAE = 30° and angle SEA = 60°, one-sixth of a full circle. The Moon is thus 1/6 of a cycle round from the new moon.
BOOK: Professor Stewart's Hoard of Mathematical Treasures
8.24Mb size Format: txt, pdf, ePub
ads

Other books

Lake People by Abi Maxwell
Pnin by Vladimir Nabokov
Her Ladyship's Girl by Anwyn Moyle
Lost at Running Brook Trail by Sheryl A. Keen
Shutterspeed by Erwin Mortier
The Marriage Market by Spencer, Cathy
The Convenient Bride by Teresa McCarthy
Crockett's Seduction by Tina Leonard