by Alex Bellos
But if l = m, the area of the other triangle with base l must be equal to the area of the other triangle with base m, since these two triangles also share the same height.
In other words, A + 3 = B.
In diagram [3] one of the triangles has base n and area 3, and the other has base o and area 7. Since these two triangles have the same height, n = o.
And if n = o, then the area of the other triangle with base n must be of the other triangle with base o, since these two triangles also share the same height.
In other words, A = (B + 7); or 7A = 3B + 21
By substituting the value for B in this equation, we get:
7A = 3(A + 3) + 21
7A = 3A + 9 + 21
4A = 30
A = 7.5
So B = 10.5
And A + B = 18
66 CATRIONA’S ARBELOS
The area is π.
The first thing to notice is that the puzzle doesn’t tell us anything about the location of the vertical line within the large semicircle. We can assume that this means it doesn’t matter where we place it, so let’s move it to somewhere convenient, such as the centre of the figure.
Now the problem is easy. The shaded area is equal to a semicircle with radius 2, minus two semicircles with diameter 2, or radius 1, which is a circle with radius 1. Which is
If you’re not convinced that we can move the line to the middle, then let’s solve the problem another (less elegant) way. Label the diameters of the two semicircles a and b, and add in lines x and y:
We have three right-angled triangles here, which means we can use Pythagoras’ theorem on each one. (Pythagoras’ theorem states that for right-angled triangles, the square of the hypotenuse is equal to the sum of the squares of the other two sides.) One right-angled triangle has hypotenuse x, one has hypotenuse y, and one has hypotenuse (a + b), since the angle on a circle subtended by a diameter is a right angle. So:
x2 = a2 + 22
y2 = b2 + 22
(a + b)2 = x2 + y2
If we combine these three equations to get rid of x and y we get:
(a + b)2 – a2 – b2 = 8
The shaded area is the area of the large semicircle (radius ), minus the areas of the two smaller semicircles (which have radiuses of a⁄2 and b⁄2). This equals:
(a + b)2 – a2 – b2
which is [(a + b)2 – a2 – b2] = = π
So the area is π, regardless of the values of a and b.
67 CATRIONA’S CROSS
The equilateral triangles cover 2⁄3 of the rectangle.
There are many ways to solve this problem. The method below uses no algebra. In tribute to Catriona Shearer, the puzzle’s author, I’ll use the ‘shearing’ property of triangles: two triangles with the same base and height have equal areas. (The height is the perpendicular distance from the base to the opposite vertex.) In other words, if you ‘shear’ a triangle by moving the vertex opposite the base in a direction parallel to the base, the area does not change, because neither the base nor the height changes.
*
Step 1. An equilateral triangle has three internal angles, each of 60 degrees. We know that the four shaded triangles are equilateral. The other angles at the intersection of the four shaded triangles must therefore each be 30 degrees, as shown below, since they must be equal, and all the angles around the intersection must add up to 360 degrees. Let the smallest equilateral triangle have side length a. Look at the top left triangle, marked in bold. The side it shares with the smallest equilateral triangle has length a. Its angle at the top left corner is 30 degrees, since together with the 60 degrees from the medium-sized equilateral triangle it makes a right angle of 90 degrees. If a triangle has two equal angles, the sides opposite those angles have the same length, so the horizontal side of the top left triangle also has length a. Equally, the top right triangle also has horizontal side length a and so the width of the rectangle is 3a. Likewise, considering the bottom right triangle (also in bold) the height of the rectangle is 2b, where b is the length of the side of the medium-sized equilateral triangle.
Step 2. If we let the equilateral triangle with side length a have area 1, then the two isosceles triangles next to it must also have area 1, since they have the same base and height. (As mentioned above, triangles with the same base and height have equal areas.) Now consider the bigger triangle that consists of these three triangles of area 1 (marked in bold in the diagram below). It has area 3. The other triangle in bold must also have area 3, since it has the same base, b, and height.
Step 3. Since we know that the medium equilateral triangle has area 3, the other triangle marked in bold below must also have area 3, since it has the same base, b, and height. We can now fill in all the other areas. If you triple the base of an equilateral triangle, the area multiplies by nine. The triangles therefore cover an area of 9 + 3 + 3 + 1 = 16 in a rectangle of area 24, which means they cover 2⁄3 of the total area.
68 CUBE ANGLE
Join the two ends of the bold line to make a triangle. All three sides have the same length, so it must be an equilateral triangle, and therefore the angle is 60 degrees.
69 THE MENGER SLICE
The holes turn into six-pointed stars.
70 THE PECULIAR PEG
There are an infinite number of solutions.
Finding an object that fits through two of the holes, such as the circle and the square, is straightforward. A cylinder of height 1 and diameter 1 unit does the job. If the cylinder is sat on a table, its horizontal cross section is a circle of diameter 1. Were you to slice it vertically down a diameter, the cross-section would be a square of side-length 1. This cylinder would fit in the circular hole when inserted circular end first, and in the square hole when inserted sideways.
To get the object through the triangular hole we need to carve out a triangular cross-section from this cylinder that is perpendicular to both the circular and the square cross-sections. One way to do this is to make two diagonal cuts, as shown below left. The object remaining will fit through the triangle when inserted in the direction of the top edge, will fit through the square when inserted perpendicular to the top edge, and through the circle when inserted base-first.
This ‘peg’, which has a perfectly triangular cross-section at only one point, is the object with the largest volume that can pass through the three holes. The object with the smallest volume that can pass through the three holes is shown below right. It has the same top edge as the other shape, but every vertical cross-section perpendicular to that edge is a triangle. Since the second object can be carved out of the first, it is possible to create an infinite number of different sized pegs that are trimmed from the first object but not yet reduced to the second one.
The second one looks a bit like an Afghan karakul hat.
71 THE TWO PYRAMIDS
The solid has five sides, because when you put the pyramids together you get a shape like a slice of cheese. Two of the faces on the triangular-based pyramid are co-planar with two of the faces on the square-based pyramid.
The diagram of the two square-based pyramids helps us visualise the solution. A line drawn between the tops of the two pyramids, as below, creates the outline of a triangular-based pyramid (a tetrahedron) slotted perfectly between the them. One face of this tetrahedron coincides perfectly with the pyramid on the left, and another face coincides perfectly with the pyramid on the right. The sides of the two pyramids and the tetrahedron that face the reader (and that face away from the reader) all lie on the same plane.
72 THE ROD AND THE STRING
Imagine that the rod is a cylinder – a cardboard kitchen-roll cylinder, say. Cut a straight line from one end of the roll to the other, between the points where the ends of the string are. When you unroll the cylinder and place it on a flat surface, you will get a rectangle that is 12cm by 4 cm, as illustrated here, which can be divided into four equally sized, 3cm x 4cm rectangles according to the points where the string crosses the cut edge.
The diagonal of one of these 3cm x 4cm rectangles is the hypotenuse of a right-angled triangle. Here we use the only piece of technical maths needed, Pythagoras’ theorem, which states that the square of the hypotenuse of a right-angled triangle is equal to the sum of the squares of the other two sides. Using Pythagoras, the hypotenuse has length √(32 + 42) = √(9 + 16) = √25 = 5. The total length of the string is equal to four hypotenuses, which is 4 x 5 = 20cm.
73 WHAT COLOUR IS THE BEARD?
The beard is white. Everyone knows that Father Christmas lives at the North Pole! (For those who do not believe in Santa, or claim he resides at an alternative address, any bearded person traversing the extreme north will also likely have a beard coloured white by snow and frost.)
There are an infinite number of places in the world from which a path 10 miles due north, then 10 miles due west, then 10 miles due south takes you back to your starting point. One is the South Pole, where this path maps out a triangle. The question, however, ruled this out as an answer.
The other places are all very close to the North Pole.
Consider the circle of latitude with a circumference of exactly 10 miles (which is only a few miles south of the North Pole). Let your starting point be any point 10 miles south of this circle, as illustrated below. You start your trek, and after 10 miles due north, you reach the circle at, say, point A. The 10 miles due west that you now travel is, in fact, one circumnavigation of the circle, which returns you to A. Travelling due south now for the final leg returns you to your starting position. Equally, consider the circle of latitude that is exactly 5 miles, long, or 2 miles long, or indeed any distance that divides 10 miles into a whole number. If you were to start 10 miles south of any one of these circles, all of which are very close to the North Pole, you would also return back to your starting point, because a 10 mile walk due west on any of these circles is a circumnavigation of that circle that returns you to the point you arrived on it.
74 AROUND THE WORLD IN 18 DAYS
Phileas Fogg arrives back on October 19th.
If you travel around the world eastwards, the sun sets and rises sooner, so you experience shorter days. After 18 sunsets and sunrises (which you experience as 18 days), only 17 days will have passed at your point of origin.
Indeed, Jules Verne used this quirk of circumnavigation in the plot of his novel Around the World in Eighty Days. Phileas Fogg bets that he can travel around the world in 80 days. He counts 80 days on his journey, but due to a missed train arrives back in London five minutes after the deadline. He thinks he has lost the wager. In fact, he is a day early, and he wins it.
75 A WHISKY PROBLEM
This is a beautiful puzzle because it can be solved with a simple insight. When I suggested that you might want to drink some more, but not too much, I was giving you a clue.
When the upright bottle contains whisky to a height of 14cm, there is 13cm above it with no whisky.
If you were to drink 3cm of the whisky, leaving 11cm, there would now be 16cm above it with no whisky.
The upturned bottle would now contain 19 – 3 = 16cm of whisky. In other words, the volume of whisky in the bottle is equal to the volume of the bottle that is empty, meaning that the bottle is half-full.
So, the whisky in the bottle is equal to half the bottle’s volume plus 3cm of whisky.
The full bottle is 750cc, so half the bottle is 375cc. The only calculation we need to do is to find the volume of 3cm of whisky, which is πr2 x 3cm, where r is the radius of the bottle. Since the diameter is 7cm, the radius is 3.5cm, so the volume of 3cm is:
3.14 x (3.5)2 x 3 = 115 cubic cm (approximately).
The whisky left in the bottle = 375 + 115 = 490 cubic cm (approximately).
Cheers!
Tasty teasers
Pencils and utensils
1) Step 1: rotate your right hand so that your four right fingers are between the thumb and index finger of your left hand.
Step 2: place your right index finger on the pencil and twist it round in a clockwise direction, whilst bringing your left thumb around your right thumb.
2) A magnetic rod will have positive charge at one end, negative charge at the other, and no charge in the middle. So, move one of the rods so that its end is touching the middle of the other rod, as if to make a T. If there is no attraction between the rods, the ‘stem’ is not the magnet, but if there is an attraction, the stem is the magnet.
3) The dentist puts on both pairs of gloves to treat the first patient, and then takes the outer pair off. She treats the second patient with the gloves that are still on. The assistant turns the gloves that the dentist is no longer wearing inside out, and puts them back on the dentist, who is still wearing the first pair of gloves. The two contaminated sides are in contact, leaving the sterile side on the outside, so the dentist can treat the third patient.
4)
5)
If you cut the postcard in this zigzag pattern, it will open up to become a loop large enough for a person to step through.
A wry plod
WORD PROBLEMS
76 THE SACRED VOWELS
[1] Ellen the jezebel chews sweet preserves then belches beer
[2] Orthodox monk Otto wolfs down two bowls of pork wonton
[3] An Arab lass gnaws at a lamb shank and has jam tarts as a snack
[4] Bud’s drunk chum glugs rum punch, upchucks hummus brunch, slumps
[5] This rich dish is figs in icing, which I finish whilst swigging gin
77 WINTER REIGNS
The first letter of each new word is the same as the last letter of the preceding word.
78 FIVE DEFT SENTENCES
[1] All of these words contain three consecutive letters of the alphabet, and the final one has four: Deft Afghans hijack somnolent understudies. Only a few words in English have this property.
[2] The sentence is a palindrome, meaning that the letters read the same forwards as they do backwards.
[3] The first sentence is formed exclusively from letters on the top row of a keyboard. The second uses only letters from the middle row, and the third was only letters from the bottom row.
[4] The first word has one letter, the second word has two, the third three, and so on, with the eleventh word having eleven letters. This constraint is what’s known as a ‘snowball sentence’. This example, from Language on Vacation, by Dmitri Borgmann, has 20 words, although I didn’t use the second part since it rather gives the game away:
I do not know where family doctors acquired illegibly perplexing handwriting; nevertheless, extraordinary pharmaceutical intellectuality, counterbalancing indecipherability, transcendentalizes intercommunications’ incomprehensibleness.
[5] This obeys Oulipo’s ‘prisoner’s constraint’, which forces you to use only letters without ascenders or descenders. In other words, only the letters a, c, e, i, m, n, o, r, s, u, v, w, x and z. The other letters all extend either above or below the height of these letters. It’s called the prisoner’s constraint because if you were in prison with only a single piece of paper, you would be able to fit more lines of text on the page.
79 THE CONSONANT GARDENER
80 KANGAROO WORDS
hen fiction part sated
taint noble me
dead tutor rain
81 THE TEN-LETTER WORDS
Each word contains four different letters, of which one appears once, one twice, one three times and one four times. Words in which one letter occurs once, a second letter occurs twice, a third letter occurs three times, and so on, are called ‘pyramid words’.
82 TEN NOTABLE NUMBERS
[1] Four
[2] Eighty four
[3] Eighty eight
[4] Forty
[5] One
[6] One thousand and eighty four
[7] Twenty
[8] One hundred and one
[9] One billion
[10] One octillion
83 THE QUESTIONS THAT COUNT THEMSELVES
[1] 73
[2] 79
[3] 75
[4] 93
[5] 168
84 THE SEQUENCE THAT DESCRIBES ITSELF
SIX, ONE, EIGHT, FIVE, TWO, SEVEN, FOUR, TEN, ONE, TWO, …
Here’s how we get there. We started with:
We know that the next E must appear at the end of a chunk of five:
We can eliminate numbers with an E before the fifth term, such as ONE, THREE, FIVE, SEVEN, EIGHT, NINE, TEN, and so on. This leaves us with TWO, FOUR and SIX. Let’s try TWO, the lowest number, since the question tells us we must choose the lowest number that fits. The TWO gives us the length of the next chunk:
The only number that fits here is SEVEN:
Again let’s choose the lowest number that fits, which is TWO:
