Various Dissection Puzzles

After the Greek cross Dudeney gives a mixed bag of cutting-out puzzles, some very easy and some hard. The rules are the ones he laid down before: cuts must be exact, the fewest pieces is the aim, and pieces may be turned over unless the puzzle says otherwise.

Two tools do most of the work. The first is counting area, which fixes the size of the square or cross to be made and often shows that a piece must be cut. The second is Pythagoras’s theorem, which turns an area into a length that can be drawn with ruler and compasses. Every dissection below has been checked by a program that cuts the pieces exactly and then, without being told where they go, fits them into the target shape.

No. 146. An Easy Dissection Puzzle

No. 147. An Easy Square Puzzle

No. 148. The Bun Puzzle

No. 149. The Chocolate Squares

No. 150. Dissecting a Mitre

No. 151. The Joiner’s Problem

No. 152. Another Joiner’s Problem

No. 153. A Cutting-Out Puzzle

No. 154. Mrs Hobson’s Hearthrug

No. 155. The Pentagon and Square

No. 156. The Dissected Triangle

No. 157. The Table-Top and Stools

No. 158. The Great Monad

No. 159. The Square of Veneer

No. 160. The Two Horseshoes

No. 161. The Betsy Ross Puzzle

No. 162. The Cardboard Chain

No. 163. The Paper Box

No. 164. The Potato Puzzle

No. 165. The Seven Pigs

No. 166. The Landowner’s Fences

No. 167. The Wizard’s Cats

No. 168. The Christmas Pudding

No. 169. A Tangram Paradox

An Easy Dissection Puzzle

The figure has area 6 if the square has side 2, so each piece has area $\tfrac32$. Dudeney’s hint is to divide it into twelve equal right-angled triangles of area $\tfrac12$: draw both diagonals and both midlines of the square, and the three midlines of the attached half-square. Each piece is then three of these triangles.

Each piece is a right-angled trapezium with parallel sides 1 and 2 and height 1: a unit square with half of another attached. The four are copies of the whole figure at half the size, which is why the puzzle works. A search of every way to split the twelve triangles into four connected pieces of the same shape finds exactly one, this one.

Answer Four right-angled trapeziums, each a half-size copy of the figure

An Easy Square Puzzle

With a triangle of legs 1 and 2, each has area 1, so the square has area 5 and side $\sqrt5$, the long side of a triangle. That is the crux: put four triangles round the edge of the square, each with its long side along one side of the square, turned like the sails of a windmill. Their right angles leave a square hole in the middle of side $2 - 1 = 1$.

The fifth triangle, of area 1, must fill the unit hole. Cut it across the middle of its longer leg, at right angles to that leg, and turn the small triangle round: the two pieces make a unit square. Some piece must be cut, because a program that tries every way of fitting pieces into the square finds no arrangement of the five whole triangles.

Answer Four whole triangles round the edge; the fifth cut in two fills the middle

The Bun Puzzle

The diameters are in the proportion 3, 4 and 5, and that is the secret: a triangle with those sides is right-angled, so by Pythagoras the squares on its sides satisfy $9 + 16 = 25$, and so do any similar figures drawn on them: the semicircles B and C in the figure together equal A, and the circles, whose areas go as the squares of their diameters, likewise. The two small buns together equal the large one.

In units where the buns have areas 9, 16 and 25, the whole is 50 and each boy’s share is $12\tfrac12$. Cut the large bun in half, D and E: two boys are served. From the middle bun cut a round piece F with the same centre and half the area of the large bun, $12\tfrac12$; its diameter is the large bun’s diameter divided by $\sqrt2$, the side of the square inscribed in the large bun. F serves the third boy. The ring G left over has area $16 - 12\tfrac12 = 3\tfrac12$, and with the small bun H, area 9, it serves the fourth.

Five pieces is the fewest possible. The large bun and the middle bun are each bigger than a share, so each must be cut into at least two pieces, and the small bun is at least one.

Answer Five pieces: the large bun halved, a disc of half its area cut from the middle bun, the ring with the small bun

The Chocolate Squares

The slab has twenty squares, so each of the four squares has area 5 and side $\sqrt5$: a square tilted so that its sides run two squares along and one across. Put one such square A in the middle, with its corners at the middles of squares of the slab, and extend its four sides right across the slab.

The four lines cut the slab into nine pieces, like a noughts-and-crosses board. A is one square. The two pieces marked B lie on opposite sides of A, and fitted together they make a second square; the two C pieces make a third. The four corner pieces D, each including one of the projecting squares, make the fourth. Each group has area 5, and the fitting program assembles each into a square.

Answer Nine pieces: A, two B, two C and four D, as in the figure

Dissecting a Mitre

With the square of side 2, the mitre has area 3, so the new square has side $\sqrt3$, and the difficulty is to find the two points C and F. Dudeney’s construction: extend the base BD to A with AB half of BD, raise AE parallel to BH, and swing the arc from B through H to meet it at E. Then $AE^2 = BE^2 - AB^2 = 4 - 1 = 3$, so AE is $\sqrt3$, and C is placed on the base with BC equal to AE. F lies on the line across the mitre at the height of the notch, with FG equal to BC less AB, that is $\sqrt3 - 1$.

Cut off the two triangles 1 and 2 above that line, cut from its left end down to C, and cut down from F to meet that cut.

The five pieces make the square of side $\sqrt3$, as the fitting program confirms.

Dudeney’s objection to the four-piece “step” answer is right, and his arithmetic checks. With the long sides 84 inches, packing the two horns into the notch leaves an 84 by 63 rectangle, area 5292. Steps $10\tfrac12$ inches high and 12 wide turn it into 72 by $73\tfrac12$, still a rectangle. The step method needs the sides in a special ratio: with the short side $61\tfrac57$ it would work, since $61\tfrac57 \times 84 = 5184 = 72^2$. Whether a four-piece dissection exists at all Dudeney left open, saying he did not believe one possible; nothing here settles it.

Answer Five pieces, with BC $= \sqrt3$ and FG $= \sqrt3 - 1$ when the side is 2

The Joiner’s Problem

Let the square BCDF have side 2. The triangle BEF, a quarter of it, has area 1 and its point E is 1 above the middle of BF, so the board has area 5 and the table-top must have side $\sqrt5$. Find A, the middle of BC, and cut from A to D and from A to E: three pieces.

The crux is that the two cuts are exactly right for a corner of the square. Each is $\sqrt5$ long, since $AD^2 = 2^2 + 1^2$ and $AE^2 = 1^2 + 2^2$, and they meet at A at a right angle, because AD goes 2 across for every 1 down and AE goes 1 across for every 2 up. So A becomes one corner of the table, AD and AE two of its sides, and pieces 2 and 3 swing round to fill the rest, as the right-hand figure shows. No piece is turned over.

Dudeney says three is the fewest, and the solution here meets it. A proof that two pieces could never do cannot come from a search, since a single cut could wander anywhere, and none is given.

Answer Three pieces: cut from the middle of BC to D and to E

Another Joiner’s Problem

Let the square ACLF have side $a$, and put the half-square CED against its side, with its long side CD, of length $c$, along CL. Mark B on AC with AB half of CD, and make two cuts right through both boards, from B to F and from B to E.

The same idea as in the last puzzle does all the work. BF and BE are both $\sqrt{a^2 + \tfrac14 c^2}$ long, since BF runs $a$ down and $\tfrac12 c$ across, and BE runs $a$ across and $\tfrac12 c$ down; and for the same reason they are at right angles. The area to be made is $a^2 + \tfrac14 c^2$, the square plus the half-square, so B is a corner of the new square and BF and BE two of its sides. The two cuts and the join between the boards make five pieces: the triangle cut off at A, the corner of the square at C above BE, the two parts of the half-square, and the rest. The small pieces move round to complete the square BEKF, none of them turned over.

The fitting program confirms it for half-squares of many sizes relative to the square, up to nearly the same area, which is Dudeney’s point that the proportions do not matter.

He adds that when the two areas are equal, cutting the square along its diagonal gives three pieces, and that when the half-square is the larger, six are needed; those two remarks are his and are not checked here.

Answer Five pieces: cut from B, with AB half of CD, to F and to E

A Cutting-Out Puzzle

The square has area 5, so its side is $\sqrt5$, the mean proportional between the strip’s length 5 and breadth 1. Cut off the end of the strip $\sqrt5$ long: it will lie along the bottom of the square as the band 4, one unit high. The rest of the strip, $5 - \sqrt5$ long and 1 high, must become the top of the square, $\sqrt5$ wide and $\sqrt5 - 1$ high.

That is a slide. Cut the remaining piece from its top left corner to the point $\sqrt5$ along its bottom edge, and slide the upper piece 3 down along the cut until it is $\sqrt5$ wide. It then sticks out on the left by a small triangle and leaves a gap of exactly the same triangle at the bottom right, so cut that triangle, piece 1, off beforehand and put it in the gap. Four pieces.

Dudeney states the general rule: a strip more than $n^2$ and not more than $(n+1)^2$ times as long as it is broad can be cut into $n + 2$ pieces to make a square, $n - 1$ of them plain rectangles. The same construction proves it: cut off $n - 1$ bands of length $\sqrt L$, and the rest, between $\sqrt L$ and $2\sqrt L$ long, slides in three pieces into the last band. The program builds it for strips 5, 10 and 24 long, in 4, 5 and 6 pieces.

Answer Four pieces: a $\sqrt5$ band cut off, and the rest slid along a diagonal

Mrs Hobson’s Hearthrug

The whole rug was $36 \times 27 = 972$, and the two triangles cut off make an oblong $12 \times 6$ together, 72. So the square rug has area 900 and side 30. The top edge of the rug, from its left end to the cut corner, is 30 long, and so is the bottom edge from the other cut corner to its right end: both can stay as edges of the square, one top and one bottom. The rug is 27 high, so the part with the bottom edge has to move 3 down, and 6 to the left to bring its right edge in line with the square.

Two pieces do it, if the cut between them repeats every 6 across and 3 up, for then moving one piece by one step of the pattern, 6 left and 3 down, fits it back against the other. The cut is a staircase of teeth: each tooth runs 6 across and 12 down, parallel to the cut corners, and then rises 15 straight up.

Placed so that one rise runs up the line 30 from the left edge to the top corner, the staircase cuts the rug into A and B; move B one tooth down and they make the 30 by 30 square. A search of every placing of such a staircase finds that this is the only one that works.

Dudeney mentions easy answers in four and three pieces; two is the fewest possible, since the rug is not itself a square.

Answer Two pieces, cut by a staircase and slid one tooth

The Pentagon and Square

First the drawing.

With the circle of radius 1, A is $\tfrac12$ from the centre, AD is $\sqrt{1 + \tfrac14} = \tfrac12\sqrt5$, and E lies $\tfrac12\sqrt5 - \tfrac12$ beyond the centre, so $DE^2 = 1 + (\tfrac12\sqrt5 - \tfrac12)^2 = \tfrac{5 - \sqrt5}{2}$. That is exactly the square of the side of the regular pentagon in the circle, $(2\sin 36^\circ)^2$, so the construction is exact.

Now the dissection, which Dudeney does in six pieces, improving, he says, on a seven-piece answer by Paul Busschop. Take the pentagon ABCDE with side 1 and the diagonal AC level. Cut off the top triangle ABC along AC, and cut it again from F, the middle of AC, to M on AB with AM equal to AF. The two pieces, turned without being turned over, fit against the side AE of the trapezium left below and make a parallelogram GHDC of the same height.

A parallelogram becomes a square by the standard two cuts. The square’s side is the mean proportional between the base HD and the height of the parallelogram; mark it off from C to K on the base line, cut along CK, and cut from G at right angles to CK as far as CK. The pieces of the parallelogram then slide into the square. Traced back into the pentagon these cuts give six pieces in all, which the fitting program assembles into both the pentagon and the square without turning any over. Of the two ways the first three pieces can make the parallelogram, only this one gives six; the other gives seven.

The square’s side is $\sqrt{1.720477\ldots} = 1.311670\ldots$ times the pentagon’s side, an irrational number, as Dudeney says, so the cuts must be found geometrically. He mentions a correspondent’s five-piece answer resting on the idea that half the diagonal plus half the side equals the side of the square. That sum is $\tfrac12(1.618034 + 1) = 1.309017$, short of $1.311670$ by a fifth of one per cent: close enough to fool the eye, and wrong.

Answer Six pieces: triangle ABC cut into three, the trapezium into three

The Dissected Triangle

Measure the triangle in triangles of side 1: one of side 5 holds 25, and the smaller triangles must use the same 25. The crux is to find two ways of writing 25 as a sum of squares that share a part: $25 = 9 + 16$ and $25 = 9 + 12 + 4$. The 9 is a triangle of side 3, which a cut parallel to the base takes off the top whole, piece 1. From the ends of that cut drop two perpendiculars to the base, cutting off the small triangles 2 and 3, and cut the rectangle between them along a diagonal into 4 and 5.

Pieces 2 and 3 are each half of an equilateral triangle of side 2, cut down its middle, and together they make one. Pieces 4 and 5 are each half of an equilateral triangle of side $2\sqrt3$, whose area is 12 small triangles, and together they make that. With piece 1 that is three triangles; and 2, 3, 4 and 5 also fit together as one triangle of side 4, which with piece 1 makes two.

Dudeney turns piece 5 over in the two-triangle and three-triangle forms, and he could not have done otherwise. Pieces 4 and 5 are the same right-angled triangle turned half round, not mirror images, while the two halves of an equilateral triangle cut down its middle are mirror images; and the fitting program finds no way to make either form without turning a piece over.

Answer Cut off the top triangle of side 3, then two half-triangles and a rectangle cut diagonally

The Table-Top and Stools

Dudeney’s point is that “oval” need not mean ellipse, and his stools are the pointed oval that architects call the vesica piscis:

the lens where two equal circles overlap. The crux is to use arcs of the table’s own radius everywhere. Take the table’s radius as 1. In the middle mark a curved square whose corners are at the points $(\pm a, \pm a)$, each side an arc of radius 1 bulging outwards, and quarter it along its diagonals: pieces 5, 6, 7 and 8. From its four corners cut straight up and down to the rim, giving the side pieces 1 and 2 and the top and bottom pieces 3 and 4.

Pieces 1 and 2 join along their straight edges to make one stool: its outline is two arcs of the table’s circle and its hole is two of the curved square’s arcs, a smaller vesica. For the second stool, turn 3 and 4 through a right angle and close them together, so their rims make the sides and their inner arcs the hole, and put the quarters at the two ends.

The quarters fit only if the ends of the stool are the same shape as a quarter: two straight edges at a right angle, as long as half the diagonal of the curved square, and an arc of radius 1. That fixes $a$. The end of the stool has straight edges $\sqrt{1 - a^2} - a$ long, and they must equal $a\sqrt2$, so $a = \sin 22\tfrac12^\circ$. With that value the program finds the two stools identical, and with any other, for instance $a = 0.3$, they are not. Each hand-hole is $2\sin 22\tfrac12^\circ = 0.77$ table-radii long and only $0.15$ wide. As Dudeney remarks, 5 and 6 never part company and could be one piece, and so could 7 and 8, making six pieces in all; he kept eight to match the story.

Answer Arcs of the table’s radius throughout, with the inner corners at $\pm\sin 22\tfrac12^\circ$

The Great Monad

(I) They are equal. In the old drawings the outer circle passes through the corners of the square round the inner circle, so its diameter is that square’s diagonal CD. A square has half the area of the square on its diagonal, and circles go as the squares of their diameters, so the outer circle has twice the area of the inner one, and the ring left over equals the inner circle.

(II) Cut along a second S, the first turned through a right angle. The circle falls into four pieces, each a quarter turn of the next, so all four are the same size and shape.

(III) Cut straight through the centre at $45^\circ$ to the line through the centres of the two small half-circles. Each of the Yin and the Yan falls into two pieces, of different shapes, and Dudeney shows they have the same area. The circle drawn on half the diameter, K in his figure, has a quarter of the area of the whole, and so does each piece of the second question; comparing them, what one piece of the straight cut loses on one side of the S it gains on the other. The program confirms the halves are equal to within rounding, and a search through every whole number of degrees finds that $45^\circ$ is the only direction that halves the Yin.

Answer (I) Equal; (II) a second S at right angles; (III) a straight cut at $45^\circ$

The Square of Veneer

The nails sit at the whole-inch points, so the trick is to cut along a grid that never meets them. Rule the square into $13 \times 13$ small squares, each $\tfrac5{13}$ of an inch: a line of that grid lies at a multiple of $\tfrac5{13}$, which is a whole number of inches only at the edges. Every cut along it misses every nail.

Now $13^2 = 12^2 + 5^2$, so the square can become squares of 12 and 5 small units. Cut out a $5 \times 5$ square A whole from one corner, and cut the rest into three pieces B, C and D along the grid, as shown, which fit together as a $12 \times 12$ square.

The two squares measure $\tfrac{60}{13}$ and $\tfrac{25}{13}$ inches, and indeed $(\tfrac{60}{13})^2 + (\tfrac{25}{13})^2 = 25$. The program checks that the four pieces fill the veneer, that B, C and D make the larger square, and that no cut passes through a nail. Dudeney’s general formula for splitting a square of side $a$ into two, $x = 2pqa/(p^2+q^2)$ and $y = a(p^2 - q^2)/(p^2+q^2)$, gives these sizes with $p = 3$, $q = 2$ and $a = 5$.

Answer Four pieces: squares of $\tfrac{60}{13}$ and $\tfrac{25}{13}$ inches

The Two Horseshoes

The key is the Great Monad of No. 158.

Take a circle of radius 2 and divide it into the Yin and the Yan by the usual S, two semicircles of radius 1 whose centres $P_1$ and $P_2$ lie on a diameter, one unit either side of the centre $O$. Each shoe is one half of the Monad, cut in two and reassembled. In the circle below, the Yin is $A$ and $B$ together, and the Yan is $C$ and $D$.

The piece $B$ is cut from the Yin by a circle of radius 1 that touches the circle about $P_2$ and runs out to the rim. Give $B$ a quarter turn about the point one unit below $O$, and the arc along which it was cut falls exactly on the arc of the circle about $P_2$ that bounds $A$. The two fit into shoe 1, the long side of the shoe being part of the rim of the circle and the notch at the heel the arc that $B$ left behind.

The second shoe must be the same shape, and its pieces different from $A$ and $B$. Dudeney’s device is the curvilinear square: the region round the point one unit above $O$ that lies outside four circles of radius 1, centred on $P_1$, $P_2$ and the two points two units above them. Neighbouring circles touch, so the square has four concave sides and a quarter turn about its centre carries it on to itself. $D$ is the Yan’s copy of $B$ with this square added, and $C$ is what remains of the Yan. A quarter turn of $D$ about the centre of the square moves the $B$-like part exactly as $B$ moved, and leaves the square where it was, so shoe 2 comes out the same shape as shoe 1, turned half round. The four pieces all differ: their areas are $\pi + 2$, $\pi - 2$, $2\pi - 2$ and 2, the square itself being $4 - \pi$, and each shoe has area $2\pi$, half the circle.

Dudeney gave the principle and the drawing but no construction, and his drawing is too rough to measure. The construction above is the one his words force. The curvilinear square must have $P_1$ and $P_2$ among its four centres for the quarter turn to carry the Monad’s circles into each other, which fixes it; and the cut for $B$ must touch the circle about $P_2$, as it does in his drawing. A program built the pieces as exact shapes and confirmed each claim. The four pieces tile the circle, each shoe is two pieces meeting without overlap, and the two shoes are the same symmetrical shape, bounded by exactly the four arcs described in the puzzle. It also shows why $B$ must run out to the rim: cutting off only the lens inside the circle would leave $A$ in two parts.

Answer Cut each shoe as shown: $A$ and $B$, $C$ and $D$; the four make the Monad’s circle

The Betsy Ross Puzzle

Fold the circle in half along a diameter, and fold the half into five equal sectors of $36^\circ$, like a fan. The paper now lies in ten layers, all in one narrow wedge. Cut straight across the wedge, from a point A on one edge to a point on the other edge nearer the centre.

Unfolding reflects the cut in every fold line. The folds lie along the five lines of symmetry of a regular pentagon, whose ten symmetries turn the one cut into ten, joined end to end into a closed star: five points where A was, $72^\circ$ apart, and five notches between. The nearer to the centre the cut reaches on the second edge, the longer the points. For the regular star, the outline of a pentagram, the notch must be $\sin 18^\circ / \sin 54^\circ = 0.382$ of the way out, which is $1/\phi^2$ with $\phi$ the golden ratio.

Answer Fold in half, then into five $36^\circ$ sectors; one straight cut across the wedge

The Cardboard Chain

This is a craft puzzle, and the secret is to cut partly through the card from each side. Take a card 8 inches by $2\tfrac12$. Rule lines BB and CC half an inch from the long edges, and short cross lines half an inch apart between them, the same on both sides of the card, pricking through with a needle so that they match.

Split the card edgeways, through its thickness, from the edge AA down to BB and from DD up to CC. Cut right through along all the short cross lines, and half through along the stretches of BB and CC shown solid; then turn the card over and cut half through along the other stretches, those that lie under the dotted parts. With a little care the card now separates into two ladders that interlace, and cutting away the waste leaves the chain in one piece, every link solid. There is nothing to calculate; the method is Dudeney’s, and cutting two keys on a ring the same way is a good variation.

Answer Split the card edgeways from both edges and cut half through from each side

The Paper Box

There is nothing to solve here, and Dudeney printed no answer. His diagram of creases and cuts is not reproduced; the description above, with a square of paper, some patience and his second drawing of the box half folded, is how he left it.

Answer A folding recipe, not a puzzle

The Potato Puzzle

Twenty-two. Think about what one more cut does. If a new cut crosses $k$ earlier cuts inside the slice, at $k$ different points, those points divide it into $k + 1$ stretches, and each stretch splits one piece into two. So the new cut adds $k + 1$ pieces, and to add as many as possible it must cross every earlier cut, at a point where no other cut passes. With cuts $1, 2, \dots, n$ that gives $$1 + (1 + 2 + \dots + n) = 1 + \tfrac12 n(n+1)$$ pieces: 2, 4, 7, 11, 16 and 22 for up to six cuts. Six cuts all touching a small circle in the middle, at equal angles, do it: no two are parallel, no three meet at a point, and all the crossings fall inside the slice.

Loyd’s cheese is the same argument one dimension up. A new plane cut meets the $k$ earlier planes in $k$ lines, which divide it into at most $1 + \tfrac12 k(k+1)$ regions, and each adds a piece; adding these up gives Dudeney’s formula $\tfrac16 (n-1)n(n+1) + n + 1$, which the program confirms up to ten cuts.

Answer 22 pieces; in general $\tfrac12 n(n+1) + 1$

The Seven Pigs

Three straight lines divide a square into at most seven parts, by the rule of the last puzzle, so with seven pigs every part must hold exactly one: the fences must cross each other inside the pen, no two parallel and not all through one point. Within that the answer is found by trial, and Dudeney’s fences are these.

The program cuts the pen along his three fences, finds seven sties with one pig in each, and checks that every fence passes clear of every pig.

Answer Three fences crossing inside the pen, as shown

The Landowner’s Fences

Four fences are needed, and the reason is the count from the potato puzzle: $n$ straight lines divide a field into at most $\tfrac12 n(n+1) + 1$ parts, so three fences give at most 7 enclosures, too few for eleven trees, and four give at most 11. Eleven is then exactly the most four can make, so every two fences must cross inside the field, no three at one point, and every enclosure must hold one tree. Dudeney’s four fences do it.

The program cuts the field along his fences, as measured from his diagram, and finds eleven enclosures with one tree in each and every fence clear of the trees.

Answer Four fences, each crossing the other three inside the field

The Wizard’s Cats

Three circles that overlap in the usual way, like a diagram of three sets, divide the plane into eight parts:

seven inside the circles and one outside them all. That is two short of ten. The extra two come from the magic circle itself: make each of the three circles touch it. The space inside the magic circle but outside the three is then pinched off at the three touching points into three separate parts, giving $7 + 3 = 10$ enclosures, and the cats sit one in each.

In Dudeney’s drawing the three circles come within a hair’s breadth of the magic circle, and drawn to touch it they give ten enclosures with one cat in each, none on a line.

Answer Three overlapping circles, each touching the magic circle

The Christmas Pudding

The principle is symmetry about the centre. Any cut from edge to edge that looks the same after a half turn about the centre divides the disc into two parts, each of which is the other turned half round: the same size and shape. So the cut must pass through the centre, and each stretch on one side must be matched by the same stretch turned half round on the other. That leaves the job of threading such a cut between the plums.

Dudeney’s cut runs from the edge down to A, across to B, through the centre C, on to D, across to E and down to the edge, and the stretch from E outwards is the half-turn of the stretch from A outwards. The program checks that it gives two halves, each the other turned half round, and that with A and B placed to within the accuracy of his drawing it misses every plum.

As he says, the stretches beyond the plums can be varied endlessly, straight or crooked, provided each is matched by its half-turn.

Answer A cut through the centre, the same after a half turn, threaded between the plums

A Tangram Paradox

The foot is not created from nothing: it is paid for by a thin strip along the side of the first figure’s body, too thin for the eye. Counting areas shows it. Take the square of the Tangram as 4 on a side: the two large triangles have area 4 each, the medium one 2, the two small ones 1 each, and the square and the parallelogram 2 each, 16 in all. In both figures the head is the square, the hat a small triangle and the arm the parallelogram, 5 of the 16. That leaves 11 for the body and the foot.

In the first figure the body uses the other four pieces, both large triangles, the medium one and the other small one, and has area 11. In the second the body uses only the two large triangles and the medium one, area 10, and the small triangle becomes the foot. Both bodies are 2 wide at the bottom, but the first is 6 tall and the second only $4\sqrt2 = 5.66$. The dotted line AB in the first figure is the outline of the second body laid over it: the first body is larger by the thin strip between AB and its own edge, a strip of area exactly 1, the area of the foot. Spread along the whole side it is too thin for the eye to notice.

Answer The foot is the strip along the side of the first body, rearranged