Various Geometrical Puzzles

The last of the geometry before the puzzles of points and lines is a miscellany: boxes, ropes, estates, gardens, flags, cones and wheels. Most of them turn on a handful of old theorems, Pythagoras above all, and on a few facts about greatest and least quantities. Several are Dudeney catching out an accepted answer, and in a few the accepted answer can be improved further still.

The units are the old ones: twelve inches to the foot, three feet to the yard, 4,840 square yards to the acre, and 5,280 feet to the mile.

No. 178. The Cardboard Box

No. 179. Stealing the Bell-Ropes

No. 180. The Four Sons

No. 181. The Three Railway Stations

No. 182. The Garden Puzzle

No. 183. Drawing a Spiral

No. 184. How to Draw an Oval

No. 185. St George’s Banner

No. 186. The Clothes Line Puzzle

No. 187. The Milkmaid Puzzle

No. 188. The Ball Problem

No. 189. The Yorkshire Estates

No. 190. Farmer Wurzel’s Estate

No. 191. The Crescent Puzzle

No. 192. The Puzzle Wall

No. 193. The Sheepfold

No. 194. The Garden Walls

No. 195. Lady Belinda’s Garden

No. 196. The Tethered Goat

No. 197. The Compasses Puzzle

No. 198. The Eight Sticks

No. 199. Papa’s Puzzle

No. 200. A Kite-Flying Puzzle

No. 201. How to Make Cisterns

No. 202. The Cone Puzzle

No. 203. Concerning Wheels

No. 204. A New Match Puzzle

No. 205. The Six Sheep-Pens

The Cardboard Box

If the box is $l$ long, $b$ broad and $d$ deep, the faces are $lb = 120$, $ld = 96$ and $bd = 80$. Multiply the first two and divide by the third: $lb \cdot ld / bd = l^2 = 144$, so $l = 12$, and then $b = 10$ and $d = 8$. The rule is Dudeney’s: the product of any two faces divided by the third is the square of the edge they share. The three faces multiplied together give the square of the volume, here $960^2$.

Answer 12 by 10 by 8 inches

Stealing the Bell-Ropes

The rope hangs straight down to the floor from a point on the ceiling. Pulled out to the wall, 48 inches away, it reaches 3 inches up, so it forms the long side of a right-angled triangle whose other sides are 48 and the rope’s length less 3. With $L$ for the length, $$L^2 = 48^2 + (L - 3)^2, \qquad 6L = 2313, \qquad L = 385\tfrac12 .$$ The rope, and the height of the ceiling, is 32 feet $1\tfrac12$ inches. Both thieves had climbed to the top, so each fell practically that whole height. Dudeney’s general rule is the same sum: with one side $a$ of a right-angled triangle and the difference $b$ between the other side and the long side, the long side is $a^2/2b + b/2$.

Answer 32 feet $1\tfrac12$ inches

The Four Sons

Take the estate as 4 on a side, so that the widow has the $2 \times 2$ corner and each son is owed an area of 3. Cut each of the other three quarters along a diagonal, and cut one of the halves of each again into two small triangles, as shown.

Each son gets one large triangle (half of a quarter, area 2) and one small one (a quarter of a quarter, area 1), placed apart so that they cannot be taken for one piece. All four sons’ land is therefore the same in shape and area, and every large triangle has a corner at the well in the middle. Dudeney notes that the conditions do not ask for each son’s land to be in one piece; they do ask for the same shape, which is why the two parts must not touch.

Answer Each son a large and a small half-square triangle, the large one at the well

The Three Railway Stations

The point the same distance from all three stations is the centre of the circle through them, so the answer is the radius of that circle for a triangle with sides 13, 14 and 15 miles. With 14 as the base the height is 12, since $13^2 - 5^2 = 15^2 - 9^2 = 144$, and the area is 84. For any triangle the radius of the circle through its corners is the product of the sides divided by four times the area: $$R = \frac{13 \times 14 \times 15}{4 \times 84} = \frac{2730}{336} = 8\tfrac18 .$$ The triangle is acute, so the centre, and the squire’s house, lies inside it.

Answer $8\tfrac18$ miles

The Garden Puzzle

A tree the same distance from all four corners is the centre of a circle through them, so the garden is a quadrilateral inscribed in a circle. For such a quadrilateral a formula of the Indian mathematician Brahmagupta (seventh century) gives the area from the sides alone; it is Heron’s formula for a triangle with a fourth side added, and reduces to Heron’s when that side shrinks to nothing. With $s$ half the perimeter, $$\text{area}^2 = (s-a)(s-b)(s-c)(s-d) = 64 \times 99 \times 44 \times 81 = 4752^2 .$$ The garden contains 4,752 square yards. The order of the walls round the garden does not matter, and building the inscribed quadrilateral from the four walls confirms the figure.

Answer 4,752 square yards

Drawing a Spiral

Fold the paper to make a straight crease through the middle, and pick two points A and B on it. Draw a half circle on one side of the crease about B, starting midway between A and B, then a half circle on the other side about A, starting where the first ended, and so on alternately. Each half circle ends on the crease exactly where the next one begins, and each is larger than the last by the distance AB, so the curve winds steadily outwards. As Dudeney says, it is not a true spiral, but it is the spiral in the picture.

Answer Half circles drawn alternately about two points on a fold

How to Draw an Oval

Wrap the paper round a cylindrical bottle or tin and draw a circle on it with the compasses. Unwrapped, the curve is an oval, symmetrical both ways, wider across the wrap than along the tin. It is worth adding that it is not an ellipse: on a tin of radius 1 with the compasses opened to 1.2, the flattened curve differs from the ellipse with the same length and breadth by up to 0.03, or three per cent.

Dudeney asked only for an oval, so his answer stands.

Answer Draw the circle on paper wrapped round a cylinder

St George’s Banner

Let the arms be $w$ wide. The red cross covers $4w + 3w - w^2$, the square where the arms cross being counted once, and it must be half the flag, 6 square feet: $$7w - w^2 = 6, \qquad (w - 1)(w - 6) = 0 .$$ So the arms are 1 foot wide. Dudeney’s rule, half the perimeter halved less half the diagonal, $3\tfrac12 - 2\tfrac12$, is the same root in general: for a flag $a$ by $b$ the width is $\tfrac12\bigl(a + b - \sqrt{a^2 + b^2}\bigr)$.

Answer 1 foot

The Clothes Line Puzzle

Let the poles be $a = 7$ and $b = 5$ feet high and $d$ apart, and let the lines cross at height $h$, a distance $x$ from the taller pole. By similar triangles $h/(d - x) = a/d$ and $h/x = b/d$; adding $x$ and $d - x$ gives $h(1/a + 1/b) = 1$, so $$h = \frac{ab}{a + b} = \frac{35}{12} \text{ feet} = 2 \text{ feet } 11 \text{ inches}.$$ The distance $d$ has cancelled: that is why the puzzle need not give it.

Answer 2 feet 11 inches

The Milkmaid Puzzle

Reflect the stool in the near bank of the river to the point A, as far beyond the bank as the stool is in front of it. Every point of the bank is as far from A as from the stool, so a route stool, bank, door is as long as A, bank, door, and that is shortest when it is straight. Join A to the door; where the line crosses the bank, at B, is where she should go.

Answer Aim at the point where the line from the stool’s reflection to the door meets the bank

The Ball Problem

Six. Equal balls resting on the ground have their centres at the same height, so this is the question of how many equal circles can touch one circle in a plane. Their centres lie on a circle of twice the radius, and two neighbours, which may not overlap, have centres at least two radii apart, so they are at least $60^\circ$ apart round that circle, the angle of an equilateral triangle; six fit exactly, as with the pennies of No. 28. For the boy’s question: a sphere’s surface is $\pi d^2$ and its volume $\tfrac16\pi d^3$, so they are equal in number when $d^2 = \tfrac16 d^3$, that is $d = 6$. The ball is 6 feet across, with $36\pi$ square feet of surface and $36\pi$ cubic feet of stone.

Answer Six balls; 6 feet

The Yorkshire Estates

The sides of the triangle are the sides of the three squares, so their squares are 370, 116 and 74. Each of these is a sum of two squares: $370 = 19^2 + 3^2$, $116 = 10^2 + 4^2$ and $74 = 7^2 + 5^2$. So the triangle can be drawn with its corners on the points of a square grid, as Dudeney does, and its area found by counting off right-angled triangles from an enclosing rectangle. Or use Heron’s formula, rewritten in terms of the squares $a$, $b$, $c$ of the sides (multiply out $s(s-x)(s-y)(s-z)$ and only squares of the sides survive): $$\tfrac14\sqrt{4ab - (a + b - c)^2} = \tfrac14\sqrt{1936} = 11 .$$

Answer 11 acres

Farmer Wurzel’s Estate

The middle triangle has area $$\tfrac14\sqrt{4 \cdot 18 \cdot 20 - (18 + 20 - 26)^2} = \tfrac14\sqrt{1296} = 9 \text{ acres}.$$ The other three triangles have the same area, and the reason is a quarter turn: each outer triangle has two sides equal to two sides of the middle one, meeting at an angle that makes up two right angles with the middle triangle’s angle, and triangles with two sides equal and those angles adding to $180^\circ$ have equal areas, since the sine of an angle equals the sine of its supplement.

So the estate is $18 + 20 + 26 + 4 \times 9 = 100$ acres. On a grid, with $26 = 5^2 + 1^2$, $20 = 4^2 + 2^2$ and $18 = 3^2 + 3^2$, all four triangles come out at 9 acres, as the program confirms.

Answer 100 acres

The Crescent Puzzle

Let the large circle have radius $x$. The small circle touches it at A, and across the diameter BA the crescent is 9 wide, so the small circle’s radius is $x - 4\tfrac12$ and its centre lies $4\tfrac12$ from C. On the radius CF the small circle crosses at E, at height $x - 5$ above C; that point is on the small circle, so by Pythagoras $$(x - 5)^2 + (4\tfrac12)^2 = (x - 4\tfrac12)^2,$$ which reduces to $(x - 5)^2 = x(x - 9)$. That is Dudeney’s statement that $x - 5$ is the mean proportional between $x - 9$ and $x$. So $x = 25$: the diameters are 50 inches and 41 inches.

Answer 50 inches and 41 inches

The Puzzle Wall

Each cottage stands between the lake and a mansion, so a single wall that keeps the cottages out and lets every mansion reach the lake must wind round them like a pinwheel. The old books drew a curving wall; most people now draw straight lines from mansion to cottage to mansion; Dudeney’s better wall runs straight from the far end of each mansion to the near end of the next, and cuts inwards round the cottages only where it must.

Measured on his own drawings, the straight-line wall is about 4.1 times the distance between two opposite mansions and his is about 3.8, some eight per cent shorter.

Dudeney made the same comparison by measurement, and nothing here proves his wall the shortest possible.

Answer The pinwheel wall of the second drawing, about 8 per cent shorter than the straight-line one

The Sheepfold

The accepted answer supposes the pen is 24 hurdles by 1, and widens it to 24 by 2 with one extra hurdle at each end. Dudeney’s objection is that nothing fixes the shape, and even granting it the answer fails. The same fifty hurdles set out 13 by 12 enclose 156 squares instead of 24, room for 650 sheep. A pen of exactly twice the area, 48, needs only twenty-eight hurdles, 8 by 6. And if all fifty must be used for exactly 48, set them as a parallelogram with sides of 12 and 13, leaning over until its height is 4, when the sine of its angle is $\tfrac4{13}$.

His point goes further than his figures. If the pen may be any shape, a ring of $n$ equal hurdles encloses most when it is a regular $n$-sided figure, with area $n/(4\tan(180^\circ/n))$. Twenty-five hurdles enclose up to 49.5 and so can be set to enclose exactly 48; twenty-four enclose at most 45.6. So 25 hurdles are enough to double the fold, not 28, and all fifty could enclose nearly 199 squares rather than 156.

Answer None are needed; the same fifty can hold far more, and 25 can hold exactly double

The Garden Walls

Dudeney’s answer turns on two words. The walls are brick, so they have thickness and the solution must work whatever it is; and they must be three walls, not walls that cross or join, since two walls built into each other become one. That rules out the obvious answers: two diameters crossing make one wall or four, and any wall that divides the field by a curve drawn with no thickness comes out only approximately right.

His answer is a straight wall across one diameter and, on either side of it, a wall that runs from the boundary wall to the first along the other diameter, the line of the cottages. Each side wall dips in a half circle under its outer cottage and arches in a half circle over its inner one, so the outer cottage goes with the garden above and the inner one with the garden below. The three walls touch but are not built into one another.

The proof needs no measuring. Take the left-hand side wall. Its curved part, the dip and the arch with the straight piece between them, is carried to itself by a half turn about the point midway along the radius: the dip becomes the arch and the arch the dip. That half turn changes above into below. So the ground the dip adds to the upper garden is exactly the ground the arch gives the lower, and the upper face of the curved wall, which is the outer face of the arch and the inner face of the dip, has exactly the same length as the lower face, the inner face of the arch and the outer face of the dip. Thicker bricks make the outer faces longer and the inner shorter, but each garden has one of each. The straight ends, at the boundary wall and at the first wall, are the same above and below, and so is the rest of each half of the field. The two gardens on the left are therefore equal in area and in length of wall, and the right-hand side is their mirror image. A program built the walls with bricks of three different thicknesses and measured the four gardens: equal areas and equal lengths of wall in every case, and one cottage in each.

Answer A diameter and two side walls, each dipping and arching in half circles, touching but not built into one another

Lady Belinda’s Garden

With the garden $a$ by $b$ and the path $w$ wide, the bed is $(a - 2w)(b - 2w)$, and it must be half of $ab$. That is a quadratic whose useful root is $$w = \tfrac14\bigl(a + b - \sqrt{a^2 + b^2}\bigr).$$ The tape does it without numbers. From the corner A mark E a quarter of the way along one side (fold the tape in four) and F a quarter of the way along the other. Then EF is a quarter of the diagonal. Lay off EG along the side equal to AF, so that AG is a quarter of $a + b$, and then GH back towards A equal to EF: AH is the width of the path. For a garden 12 by 5, where the diagonal is 13, the path is $\tfrac14(12 + 5 - 13) = 1$.

Answer AH, found as a quarter of the sides less a quarter of the diagonal

The Tethered Goat

The goat grazes a sector of a circle with its angle at the corner, $60^\circ$, a sixth of the whole circle. It must be a quarter of an acre, so the whole circle would be $1\tfrac12$ acres, or 9,408,960 square inches. The radius is $\sqrt{9{,}408{,}960/\pi} = 1730.6$ inches, so the tether is 1,731 inches, 48 yards 3 inches, to the nearest inch. The sector lies wholly inside the field, since the tether is shorter than the height of the triangle, about 2,331 inches.

Answer 48 yards 3 inches

The Compasses Puzzle

Dudeney’s construction, with AB the given length. Draw circles of radius AB about A and B, meeting at D. Round the circle about B mark E and F with DE and EF equal to AB; F is then as far beyond B as A is before it. With centres A and F and radius DF draw arcs meeting at G. With radius BG draw an arc about B through G, cutting the circle about A at H, and an arc about A cutting the circle about B at N; mark K on the arc about B with HK equal to AB, and L on the circle about B with HL equal to HB. Arcs of radius AB about K and L meet at I. Mark M on the circle about A with BM equal to BI. Finally the arc about M through B cuts the line at C, the middle of AB.

Why it works: with AB as 1, DF is $\sqrt3$, so G is at height $\sqrt2$ above B and $BG^2 = 2$, twice $AB^2$. That makes H, A, B, N the corners of a square, and K and L are placed so that I is its centre. Then $BM^2 = BI^2 = \tfrac12$, and the circle about M through B meets AB again exactly halfway along. The program follows every step with coordinates and lands on $(\tfrac12, 0)$. Dudeney remarks that this is not the shortest construction; the problem belongs to the geometry of Mascheroni, who showed that every construction with ruler and compasses can be done with compasses alone.

Answer The construction above; C is the middle of AB

The Eight Sticks

The answer nearly everyone gives, three squares tilted like diamonds, fails the conditions: two of the sticks would rest on the table at one end only, and every stick must lie on the table. The honest answer needs sticks of the right thickness. With long sticks of 2 feet and short ones of 1 foot, lay the long ones end to end in two rows, 4 feet long, and stand the four short ones between them as uprights. The spaces are 1 foot high, and $(48 - 4t)/3$ inches wide if the sticks are $t$ inches thick; they are squares when $t = 3$.

With matches, about twenty-one times as long as they are thick, the spaces would be oblongs, which is why Dudeney said sticks.

Answer Sticks of 2 feet and 1 foot, 3 inches thick, laid flat as shown

Papa’s Puzzle

The card, cut from A on the top edge to the bottom corner D, hangs with its top level when its centre of gravity lies straight below A. Many people take BA as a third of BC, to make the areas on either side equal; but the triangle’s point sticking out to D pulls harder, so balance turning effects, not areas. Let the card be $L$ long and $h$ high, and BA $= t$. Below A, the piece is a rectangle $t$ by $h$ to the left of the thread and a right-angled triangle with legs $L - t$ and $h$ to the right. The rectangle’s weight $th$ acts $\tfrac12 t$ to the left of the thread; the triangle’s weight $\tfrac12 (L-t)h$ acts at its centroid, a third of the way along its base, $\tfrac13 (L-t)$ to the right. Balance needs $$th \cdot \tfrac12 t = \tfrac12 (L-t)h \cdot \tfrac13 (L-t), \qquad\text{that is}\qquad 3t^2 = (L-t)^2,$$ so $L - t = \sqrt3\, t$ and $BA : AC = 1 : \sqrt3$. BA is $(\sqrt3 - 1)/2 = 0.366$ of the length of the card.

The curious point is that this does not depend on the height of the card at all: $h$ cancelled from the balance, and the program finds the balance point for cards of several heights and gets 0.366 of the length every time. Pappus’s two cards were the same length, so the little girl had only to lay the cut card on the other and mark A at the same place.

Answer BA $= 0.366$ of the length; since the cards are equally long, copy A from the first

A Kite-Flying Puzzle

Put the ball in a cylindrical hat-box that just fits it, 24 inches across and 24 high. The ball fills exactly two-thirds of it, a fact known since Archimedes, so the ball holds as much as a cylinder 24 inches across and 16 high. Think of that cylinder as made of wires a hundredth of an inch thick standing side by side with no gaps. Circles go as the squares of their diameters, so there are $(24 \div \tfrac1{100})^2 = 5{,}760{,}000$ wires, each 16 inches long: 92,160,000 inches, which is 1,454 miles 2,880 feet. The text as printed calls the square of $\tfrac1{100}$ one hundred-thousandth, a slip for ten-thousandth; the arithmetic that follows uses the right value.

Answer 1,454 miles 2,880 feet

How to Make Cisterns

Cut squares of side $x$ from an $a$ by $b$ sheet and the cistern holds $x(a - 2x)(b - 2x)$. The volume is greatest where its rate of change is zero, $12x^2 - 4(a + b)x + ab = 0$, and the smaller root is Dudeney’s $$x = \tfrac16\bigl(a + b - \sqrt{a^2 + b^2 - ab}\bigr) = \tfrac16\bigl(11 - \sqrt{49}\bigr) = \tfrac23 \text{ foot}$$ for the 8 by 3 foot sheet: squares of 8 inches, and a cistern of $7\tfrac{11}{27}$ cubic feet. A search over sizes to a thousandth of a foot agrees.

Answer Squares of 8 inches

The Cone Puzzle

A cylinder standing inside a cone of height $H$ and base radius $R$, reaching a height $y$, has radius $R(1 - y/H)$ and volume $\pi R^2 (1 - y/H)^2 y$. Its derivative with respect to $y$ is $\pi R^2 (1 - y/H)(1 - 3y/H)$, which is zero at the tip ($y = H$, no cylinder) and at $y = H/3$, where the volume is greatest: cut at one-third of the altitude.

Answer Cut at one-third of the height

Concerning Wheels

The flange of a railway wheel reaches below the top of the rail, beyond the circle that rolls. A point there traces a looped curve, and at the bottom of each loop it is moving backwards: a point one-tenth of a radius below the rail moves at a tenth of the train’s speed towards London. At every moment some points of every flange are doing this.

The rolling wheel turns twice. A mark at its top is at the bottom, touching the fixed wheel, after half the journey, and so has already made one turn; it makes another in the second half. In general a wheel rolled round another turns $(R + r)/r$ times, the fixed wheel’s radius being $R$.

Answer Points on the flanges below the rail; two turns

A New Match Puzzle

(1) Six matches make a rectangle 2 by 1, area 2, and the other twelve a parallelogram with sides of 4 and 2 matches, leaning until its height is $1\tfrac12$, area 6: three times as large.

(2) On a grid of equilateral triangles of side one match, take a parallelogram 3 by 1 and cut off the small triangle at one of its sharp corners: a five-sided figure of 5 triangles, with a boundary of 7 matches. Do the same with a parallelogram 4 by 2: 15 triangles, a boundary of 11 matches. Seven and eleven make eighteen, and the second figure is three times the first.

Answer (1) a 2 by 1 rectangle and a slanted 4 by 2 parallelogram; (2) two cut parallelograms of triangles

The Six Sheep-Pens

Lay six matches as a regular hexagon and the other six as spokes from its centre to its corners. The spokes are as long as the sides, so the hexagon falls into six equal equilateral triangles: six pens of the same size from twelve hurdles.

Answer A hexagon with six spokes