Ian Stewart

by Professor Stewart's Hoard of Mathematical Treasures

  Aelfred is in a rather different position. He sees two blobs, one on Benedict and one on Cyril. He wonders whether he, Aelfred, also has a blob. If so, then all three of them have blobs, and he knows from the previous version of the puzzle that they will all wait for ring 3 and then raise their hands. So he does not, and should not, raise his hand after ring 1 or 2. Then the other two do raise their hands: at that point he knows that he does not have a blob.

  Again, a complete proof can be given by induction, thinking about the general case of n monks, m of whom have blobs. I’ll spare you the details.

  Pickled Onion Puzzle

  There were 31 pickled onions.

  Suppose there are a pickled onions to start with, b after the first traveller has eaten, c after the second traveller has eaten, and d after the third traveller has eaten. Then

  b = 2(a - 1)/3, c = 2(b - 2)/3, d = 2(c - 3)/3

  which we rewrite as

  a = 3b/2 + 1, b = 3c/2 + 2, c = 3d/2 + 3

  We are told that d = 6. Working backwards, c = 12, b = 20, a = 31.

  Guess the Card

  At each stage, when Whodunni picks up the cards he makes sure that the selected pile is sandwiched between the other two. As a result, the chosen card works its way to the middle of the stack of cards. So on the final deal it is the middle card in the chosen pile.

  And Now with a Complete Pack

  The first part of the trick is equivalent to asking which column the chosen card is in on the second deal. Knowing the column, and now being told the row, Whodunni can easily spot the card.

  This trick is a bit transparent, but it can be dressed up to make it less obvious. It may be better performed with 30 cards, first dealing 6 rows of 5, then 5 rows of 6. It works with ab cards dealt in a rows of b, then b rows of a, for any whole numbers a and b.

  Halloween = Christmas

  Because 31 Oct = 25 Dec. That is, 31 in base-8 numerals (octal) = 25 in base-10 numerals (decimal). In base 8 notation, 31 means 3 × 8 + 1, and this is 25.

  Rectangling the Square

  I know at least two distinct solution, not counting rotations or reflections as distinct. The first was found by M. den Hertog, the second by Bertle Smith. In the first rectangles have sides 1 × 6, 2 × 10, 3 × 9, 4 × 7 and 5 × 8. In the second, the sides are 1 × 9, 2 × 8,3 × 6, 4 × 7 and 5 ×10.

  Solutions found by den Hertog and Smith.

  X Marks the Spot

  From Abandon Hope Point: 113. From Buccaneer Bay: 99. From Cutlass Hill: 85.

  Distances for Redbeard’s map.

  The figure shows the three distances required: a, b, c. Of these, we know b = 99. Let s = 140 be the side of the square. Consider two further lengths x and y as shown. Then Pythagoras’s theorem tells us that

  a2 = x2 + (s - y)2 = x2 + s2 - 2sy + y2

  c2 = y2 + (s - x)2 = y2 + s2 - 2sx + x2

  b2 = x2 + y2

  The first step is to get rid of x and y. Subtract the third equation from the first and second, to obtain

  2sy = s2 + b2 - a2

  2sx = s2 + b2 - c2

  Therefore

  (s2 + b2 - a2)2 + (s2 + b2 - c2)2 = 4s2(x2 + y2) = 4s2b2

  This is the fundamental relation between a, b, c and s.

  Substitute known values s = 140 and b = 99, to get

  (29,401 - a2)2 + (29,401 - c2)2 = 27,7202 = (23 × 32 × 5 × 7 × 11)2

  The hint now tells us that 29,401 - a2 and 29,401 - c2 are both multiples of 7. (The corresponding statement is false for the factors 2 and 5, but true for 3 and 11.) Considering the factor 7 (a similar trick works for 3 and 11) we observe that

  29,401 = 4,200×7 + 1

  so 1 - a2 and 1 - c2 are both multiples of 7. That is, a2 and c2 are of the form 7k + 1 or 7k - 1 for suitable integers k.

  Now it is a matter of trying the possible values for a, and seeing first whether

  23,8002 - (29,401 - a2)2

  is a perfect square, and, if so, whether the corresponding c is a whole number. The business about multiples of 7 shortens the work because the only values for a that we need check are

  1, 6, 8, 13, 15, 20, 22

  and so on. We can stop when c becomes less than a, because then we’ll be trying the same calculations but with a and c interchanged.

  For the 7k + 1 case, we find a = 85, c = 113 when k = 12; there is also the solution a = 113, c = 85, interchanging a and c, when k = 16. There is no solution for the 7k - 1 case.

  Since the instructions on the back of the map say that the nearest marker is C, we want c < a, so a = 113 and c = 85.

  That’s one way to get the answer, but the mathematical story goes further.

  This puzzle is a special case of the four distance problem: does there exist a square whose side is a whole number, and a point whose distances from the four corners of the square are all whole numbers? No one knows the answer. For a long time no one even knew whether three of those distances could be whole numbers.

  We’ve already derived a relation between s, a, b and c:

  (s2 + b2 - a2)2 + (s2 + b2 - c2)2 = (2bs)2

  The fourth distance d (shown dotted on my diagram) must satisfy

  a2 + c2 = b2 + d2 J. A. H. Hunter discovered a formula giving some (but not all) solutions of the first equation:

  a = m2 - 2mn + 2n2

  b = m2 + 2n2

  c = m2 + 2mn + 2n2

  s2 = 2m2 (m2 + 4n2)

  and observed that s is a whole number provided that we take

  m = 2(u2 + 2uv - v2)

  n = u2 - 2uv - v2

  for whole numbers u and v.

  The choice u = 2, v = 1 leads to s = 280, a = 170, b = 198, c = 226, and we can remove the factor 2 to get s = 140, a = 85, b = 99, c = 113. The fourth side here is d = which is not a whole number, indeed not rational. In fact, it is known that in Hunter’s formula the fourth distance d can never be rational, so this formula alone won’t solve the four distance problem. However, there are solutions of the three distance problem that don’t arise from Hunter’s formula.

  This tantalising problem has deep connections with Kummer surfaces in algebraic geometry: see Richard K. Guy, Unsolved Problems in Number Theory.

  Whatever’s the Antimatter?

  Dirac’s electron equation looks like this:

  where

  ψ is the quantum wave-function of the electron

  A0 is the scalar potential of the electromagnetic field

  A is the vector potential of the electromagnetic field

  ρ = (ρ0, ρ1, ρ2, ρ3) is the momentum vector of the electron

  σ = (σ1, σ2, σ3) is a list of three 4×4 spin matrices

  r = (r1, r2, r3) is a list of three 4×4 matrices which anticommute

  with σ1, σ2, σ3

  e is the charge on the electron

  m is the mass of the electron

  c is the speed of light

  Got that? I include it only to make the point that the equation is far from obvious, and because it seemed a cheat to leave it out. Everyone writes out E = mc2, even Stephen Hawking,55 and it’s wrong to discriminate against formulas on grounds of complexity. Dirac spends nearly four pages of his book The Principles of Quantum Mechanics explaining how he derived it, and most of the previous 250 pages setting up the ideas that go into it.

  Leaning Tower of Pizza

  With five boxes, the maximal overhang is 1.30455. With six boxes, it is 1.4367. The stacks look like this:

  Maximal overhangs with five and six boxes.

  The paper by Paterson and Zwick is ‘Overhang’, American Mathematical Monthly, vol. 116, January 2009, pp. 19-44.

  PieThagoras’s World-Famous Mince πs

  The areas of the three pies are (mini) 9π, (midi) 16π, and (maxi) 25π, so the maxi-pie has the same area as the other two put together. If the maxi-pie is split in half, Alvin and Brenda can each have a piece (25π/2). That leaves the other two pies to share between Casimir and Desdemona. This can be done by slicing 7π/2 off the midi-pie and giving that piece plus the mini-pie to Casimir
(9π + 7π/2 = 25π/2). Desdemona gets the larger piece of the midi-pie (16π - 7π/2 = 25π/2).

  There are lots of ways to slice the midi-pie. The traditional one is to place the mini-pie over the middle of the midi-pie, trace round half of its circumference, and then join the ends to the edge of the midi-pie. But you could slice off any shape with area 7π/2. And the cut that divides the maxi-pie in half need not be a diameter - it could be curved.

  Cut the three pies as on the left, and share as on the right.

  Diamond Frame

  There are ten basically different answers, where, for example, swapping the ace and seven in the right-hand side doesn’t count as different since it’s a simple transformation that obviously keeps the sums the same. There are two with a sum of 18, four with 19, two with 20, and two with 22. Here’s one of them.

  One of the ten answers: sums are all 18.

  Pour Relations

  You can solve this by trial and error, or by listing all the possible states and moves and finding a path from the initial state to the desired final one. Here is one solution, which takes nine moves (two are combined in the second picture). There’s a shorter solution, which I’ll derive in a moment.

  One way to divide the water.

  However, there is a more systematic method, which seems to have first been published by M. C. K. Tweedie in 1939. It uses a grid of equilateral triangles, which engineers call isometric paper and mathematicians call trilinear coordinates.

  Representing the possible states of the jugs.

  Here the triples of numbers indicate how much each jar holds, in the order (3-litre jug, 5-litre jug, 8-litre jug). So for example 251 means that the 3-litre jug holds 2 litres, the 5-litre jug holds 5 litres, and the 8-litre jug holds 1 litre. If you look at the first number, then the lowest line in my diagram always starts 0, the line above starts 1, and so on. Similarly, the second number reads 0, 1, 2, 3, ... from left to right in each row. So the two arrows in the diagram are ‘coordinate axes’ for the amount in the 3-litre jug and the amount in the 5-litre jug.

  What of the 8-litre jug? Because the total amount of water is always 8 litres, the third number is always determined by the first two. Just add them and subtract from 8. But there is a nice pattern here. Thanks to the geometry of isometric paper, the third number is constant along lines that slope up and to the left - that is, the third system of lines in the picture. For example, look at the line through 035, 125, 215, 305.

  If we represent the possible ‘states’ of the jugs - how much each contains - in this way, then the allowable moves from any state to any other take on a simple geometric form, as I’ll now explain.

  First, observe that the states in which some jug is either full or empty - which are the states allowed by the moves - are precisely those on the boundary of the picture.

  The allowable moves, starting from some state (necessarily on the boundary) amount to moving along a line until you next hit the boundary. If you start at a corner, and move along the boundary (say from 008 to 053), then you can’t stop part way along, but have to go all the way to the next corner.

  One solution.

  So we can solve the puzzle by starting from 008 (lower left corner) and bouncing around the parallelogram like a billiard ball. The arrowed path shows what happens: we visit the states

  008, 305, 035, 332, 152, 107, 017, 314, 044

  and notice that this is the final state we want, so we stop.

  This is precisely the solution given above. But there’s another one:

  An alternative.

  Now the sequence is

  008, 053, 323, 026, 206, 251, 341, 044

  which uses seven moves instead of nine. You may have found that one instead.

  The Sacred Principle of Mat

  How to ensure the greatest mat.

  Remember, the avatars ‘watch over’ every other cushion. Otherwise you could make the mat bigger.

  Target Practice

  Tuck hit the outer ring (light grey), while Robin hit the three inner rings (dark grey).

  The rings Robin and Tuck hit.

  The areas of successive circles are πr2, where r = 1, 2, 3, 4, 5, respectively:

  π, 4π, 9π, 16π, 25π

  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 πr2. 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 πrs = πr2, 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.

  There is a corresponding position at 5/6 of a cycle, obtained by reflecting the diagram in the line ES.

  How Dudeney Cooked Loyd

  • The children’s puzzle seems impossible on grounds of parity (odd/ even). All the numbers are odd, and six odd numbers must add to an even number, not 21. Gardner’s answer was to cook his own puzzle, turn the page upside down, and circle the three 6’s and the three 1’s that then appear. But a reader, Howard Wilkerson, circled each of the 3’s, one of the 1’s, and then drew a big circle round the other two 1’s (giving 11). I think this is a more elegant quibble-cook.

  • Loyd’s construction leads to a rectangle which, while almost square, has sides of differing lengths. If the mitre is made from a square of side 1, giving it an area of , then Loyd’s ‘square’ measures 6/7 horizontally by 7/8 vertically.

  • Dudeney’s five-piece solution, which is exact if the lengths are chosen correctly, looks like this:

  Dudeney’s solution.

  No one knows a four-piece dissection from the mitre to the square, and there probably isn’t one, but its existence has not been ruled out.

  Cooking with Water

  What I really should have said, to be super-cautious, was ‘you are not allowed to pass the connections through a house or a utility company building’. As David Uphill pointed out, there is a way to solve the problem if the words are interpreted literally, even with pipes forbidden to pass through houses. I’ve modified his suggestion slightly to fit my question more closely. It uses two big water tanks to pass the water connections through two of the houses. No connecting pipe even enters a house, let alone passes through i
t.

  The utilities puzzle quibble-cooked.

  Hmm . . . If you feel that a tank used like this is just a large pipe by another name, which was my first reaction, then this layout doesn’t fit the conditions. That’s why I consider it a quibble-cook. But it’s an ingenious one and deserves to be better known.

  Calculator Curiosity 2

  When you multiply 0588235294117647 by 2, 3, 4, 5, ... , 16, the same sequence of digits arises, in the same cyclic order. That is, you have to start at a different place, and when you get to the end, you continue from the beginning. Specifically,

  0588235294117647 × 2 = 1176470588235294

  0588235294117647 × 3 = 1764705882352941

  0588235294117647 × 4 = 2352941176470588

  0588235294117647 × 5 = 2941176470588235

  0588235294117647 × 6 = 3529411764705882

  0588235294117647 × 7 = 4117647058823529

  0588235294117647 × 8 = 4705882352941176

  0588235294117647 × 9 = 5294117647058823

  0588235294117647 × 10 = 5882352941176470

  0588235294117647 × 11 = 6470588235294117

  0588235294117647 × 12 = 7058823529411764

  0588235294117647 × 13 = 7647058823529411

  0588235294117647 × 14 = 8235294117647058

  0588235294117647 × 15 = 8823529411764705