Shapes and spaces.

The Scene at this step

Solid

Cubefixed

Unfold

Solidfixed

Once the prediction is locked in, the Scene changes

Unfold

Cornerfixed

The Scene then shows: 90 ° · The gap left at one corner pressed flat: 360° − the corner’s angles/nodes/solid/outputs/defect

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Hint

Worked solution

The Scene at this step

Solid

Icosahedronfixed

Unfold

Solidfixed

Accepted within

± 0.5 °

Computed from the Scene at this step’s setting

The gap left at one corner pressed flat: 360° − the corner’s angles/nodes/solid/outputs/defect

Any unit of the same quantity is accepted; it is converted before checking.

Wrong answers it recognises

• 300 ° ± 0.5 °

  Misconception: Gives the face angles at a corner added up when asked for the gap they leave pressed flat (300° for the icosahedron instead of 60°).

  Feedback from the catalogue

  The angle defect is what is missing from a full turn, not what is there. Add up the face angles at the corner, then take that from $360^\circ$⁠. In the icosahedron: $5 \cdot 60^\circ = 300^\circ$$5 \cdot 60^\circ =$$300^\circ$⁠, and the defect is $360^\circ - 300^\circ = 60^\circ$$360^\circ - 300^\circ =$$60^\circ$⁠.

Hint

Worked solution

The Scene at this step

Solid

Cube

Unfold

Corner

Choices

Shown in a random order.

• Correct Each angle of a hexagon is $120^\circ$⁠, so three hexagons at a corner make exactly $360^\circ$⁠: they lie flat, as in a honeycomb, and never close into a point

• Wrong A hexagon has too many sides to join its neighbours edge to edge

  Feedback

  Hexagons join edge to edge without any trouble: they tile a whole floor. The trouble is at the corner.

• Wrong There is one: two hexagons at each corner close a solid

  Feedback

  Two faces do not close a corner; they just lie on each other. At least three faces must meet at every corner.

• Wrong Nobody has found it yet, and it may turn up one day

  Feedback

  Euclid proved there is no sixth solid: the rule at the corner closes the list, not just the search.

Hint

Worked solution

The Scene at this step

Solid

Icosahedronfixed

Unfold

Solidfixed

Accepted within

± 0.1

Computed from the Scene at this step’s setting

Edges cut to open the net, V − 1/nodes/solid/outputs/cuts

Hint

Worked solution

The Scene at this step

Solid

Dodecahedronfixed

Unfold

Count

Accepted within

± 0

Computed from the Scene at this step’s setting

Faces, F/nodes/solid/outputs/faces

Wrong answers it recognises

• 48 ± 0

  Misconception: Adds the edges in Euler’s formula, or mixes up which terms take a minus sign.

  Feedback from the catalogue

  In Euler’s formula, $V - E + F = 2$⁠, the corners and faces have a plus and the edges a minus. To find $F$⁠, rearrange: $F = 2 - V + E$⁠.

Hint

Worked solution

The Scene at this step

Solid

Cube

Unfold

Corner

Read from the Scene at

Sides of each face, p/nodes/solid/outputs/faceSides

Read from the Scene at

Faces meeting at each corner, q/nodes/solid/outputs/cornerFaces

One setting that reaches it

Solid

Octahedron

Hint

Worked solution

The Scene at this step

Solid

Dodecahedronfixed

Unfold

Corner

Accepted within

± 0.1 °

Computed from the Scene at this step’s setting

Each angle of a face, 180°(p − 2)/p/nodes/solid/outputs/faceAngle

Any unit of the same quantity is accepted; it is converted before checking.

Wrong answers it recognises

• 72 ° ± 0.1 °

  Misconception: Gives the turn at each corner of a regular polygon, 360° ÷ p, as the angle inside it.

  Feedback from the catalogue

  Walking round a regular polygon, at each corner you turn by $\dfrac{360^\circ}{p}$⁠: the exterior angle. The angle inside the polygon is what is left of a straight line: $180^\circ - \dfrac{360^\circ}{p}$⁠, $108^\circ$⁠ for a pentagon.

Hint

Worked solution

The Scene at this step

Solid

Cube

Unfold

Solid

Choices

• Wrong Less than $720^\circ$⁠, because each gap is smaller

  No feedback of its own; the learner sees the standard message

  Not yet. Look again and try once more.

• Correct Exactly $720^\circ$⁠

• Wrong More than $720^\circ$⁠, because there are many more corners

  No feedback of its own; the learner sees the standard message

  Not yet. Look again and try once more.

Once the prediction is locked in, the Scene changes

Solid

Dodecahedron

The Scene then shows: 720 ° · The gaps at every corner added up: 720° for every convex solid (Descartes)/nodes/solid/outputs/totalDefect

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Hint

Worked solution

The Scene at this step

Solid

Tetrahedronfixed

Unfold

Solid

Accepted within

± 0.2 °

Computed from the Scene at this step’s setting

The angle between two faces that share an edge/nodes/solid/outputs/dihedral

Any unit of the same quantity is accepted; it is converted before checking.

Wrong answers it recognises

• 60 ° ± 0.2 °

  Misconception: Takes an angle inside a face for the angle between two faces.

  Feedback from the catalogue

  The angle inside a face is measured between two of its sides, on the face itself. The angle between two faces is measured in a plane cutting their shared edge square on, like how far a book is opened. In the tetrahedron the first is $60^\circ$⁠ and the second about $70.5^\circ$⁠.

Hint

Worked solution

The Scene at this step

Turn in 4D

0 °fixed

Cut height

0 edges

Pose

Square onfixed

Shadow to slice

Shadowfixed

Choices

Shown in a random order.

• Correct A cubic cell of the tesseract, as big as the near cell, that is farther from the lamp along $w$⁠

• Wrong A smaller cell of the tesseract, sitting inside the big one

  Misconception: Takes the smaller square or cube in a shadow for a smaller part of the shape, rather than an equal part farther from the lamp.

  Feedback from the catalogue

  A lamp’s shadow grows the nearer a thing is to the lamp. In a cube’s shadow, the near face casts the big square and the far face, just as big, casts the small one. The tesseract’s shadow is the same one dimension up: the small cube is a whole cell, as big as the large one, that is simply farther from the lamp along $w$⁠.

• Wrong An empty space at the tesseract’s centre

  Feedback

  The tesseract is not hollow; the small cube is a real cell of it, just as the small square in a cube’s shadow is a real face.

Hint

Worked solution

The Scene at this step

Turn in 4D

0 °

Cut height

0 edges

Pose

Square on

Shadow to slice

Shadow

Accepted within

± 0.1

A fixed value in the question

Hint

Worked solution

The Scene at this step

Turn in 4D

0 °fixed

Cut height

0.5 edges

Pose

Corner firstfixed

Shadow to slice

Slicefixed

Choices

• Correct An octahedron: a tetrahedron with its corners cut off until the cuts meet

• Wrong A bigger tetrahedron

  Feedback

  In Flatland, the cube passing corner first gives a triangle near the tip, but in the middle its corners are cut off and a hexagon is left. The same happens here, one dimension up.

• Wrong A cube

  Feedback

  A cube comes when the tesseract passes Square on, a cell first. Corner first, the slices are tilted to every cell.

Once the prediction is locked in, the Scene changes

Cut height

0 edges

The Scene then shows: 8 · Faces of the tesseract’s slice, one per cell it cuts/nodes/hyper/outputs/sliceFaces

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Hint

Worked solution

The Scene at this step

Turn in 4D

45 °fixed

Cut height

0 edges

Pose

Square onfixed

Shadow to slice

Shadowfixed

Accepted within

± 0.1

Computed from the Scene at this step’s setting

Tesseract corners nearest the lamp: 8 a cell leads, 4 a square face, 2 an edge, 1 a corner/nodes/hyper/outputs/leading

Hint

Worked solution

The Scene at this step

Crossings

2fixed

Draw and untangle

Drawing

Choices

• Correct The pushed side will slide back and both crossings will vanish: it is a plain loop

• Wrong Both crossings will stay, because a loop with crossings is a knot

  Misconception: Thinks a drawing with crossings must be knotted, or that more crossings make a loop more surely knotted.

  Feedback from the catalogue

  Crossings in a drawing are only what you see from above, and you can add as many as you like even to a plain loop: a twist adds one crossing, and pushing one side over the other adds two. To know a loop is knotted you need an invariant, a property no move changes, such as a colouring in three colours.

• Wrong One crossing will vanish and one will stay

  Feedback

  In move II, the push, one side passes wholly over the other, so its crossings appear and vanish in pairs.

Once the prediction is locked in, the Scene changes

Draw and untangle

Try to untangle

The Scene then shows: 0 · Crossings in the drawing once untangling is done: the curl and the pushed side come out, a knot keeps all of them/nodes/knot/outputs/crossingsLeft

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Hint

Worked solution

The Scene at this step

Crossings

3fixed

Draw and untangle

Try to untangle

Choices

Shown in a random order.

• Correct Its arcs can be coloured in three colours by the rule, no move spoils such a colouring, and a plain loop has none

• Wrong People have tried many times to untangle it and failed

  Misconception: Takes failing to untangle a loop as proof that it is knotted.

  Feedback from the catalogue

  If you could not untangle it, perhaps you just did not find the right order of moves; some plain loops look very tangled. A proof needs a property that every move keeps and the plain loop does not have: for instance, a colouring of the arcs in three colours by the rule.

• Wrong It has three crossings, and a loop with three crossings is always a knot

  Misconception: Thinks a drawing with crossings must be knotted, or that more crossings make a loop more surely knotted.

  Feedback from the catalogue

  Crossings in a drawing are only what you see from above, and you can add as many as you like even to a plain loop: a twist adds one crossing, and pushing one side over the other adds two. To know a loop is knotted you need an invariant, a property no move changes, such as a colouring in three colours.

• Wrong The rope is tied too tight to move

  Feedback

  In topology the rope is perfectly flexible: you may stretch and move it as you like, only not cut it.

Hint

Worked solution

The Scene at this step

Crossings

4fixed

Draw and untangle

Drawingfixed

Accepted within

± 0.1

Computed from the Scene at this step’s setting

Arcs of the drawing as tied: one per under pass, or 1 with none/nodes/knot/outputs/arcs

Hint

Worked solution

The Scene at this step

Crossings

6

Draw and untangle

Try to untangle

Choices

Shown in a random order.

• Correct Nothing yet: it may be a knot or a tangled plain loop. You need an invariant to know

• Wrong It is surely a knot, because it has more than three crossings

  Misconception: Thinks a drawing with crossings must be knotted, or that more crossings make a loop more surely knotted.

  Feedback from the catalogue

  Crossings in a drawing are only what you see from above, and you can add as many as you like even to a plain loop: a twist adds one crossing, and pushing one side over the other adds two. To know a loop is knotted you need an invariant, a property no move changes, such as a colouring in three colours.

• Wrong It is surely a knot whose fewest crossings are six

  Feedback

  A drawing with six crossings may belong to a simpler knot, such as the trefoil, drawn with extra crossings. A knot’s crossing number is the smallest count over all its drawings.

Hint

Worked solution

The Scene at this step

Crossings

2fixed

Draw and untangle

Try to untangle

Accepted within

± 0

Computed from the Scene at this step’s setting

The fewest crossings any drawing of this loop can have: 0 for the unknot/nodes/knot/outputs/crossingNumber

Wrong answers it recognises

• 2 ± 0

  Misconception: Thinks a drawing with crossings must be knotted, or that more crossings make a loop more surely knotted.

  Feedback from the catalogue

  Crossings in a drawing are only what you see from above, and you can add as many as you like even to a plain loop: a twist adds one crossing, and pushing one side over the other adds two. To know a loop is knotted you need an invariant, a property no move changes, such as a colouring in three colours.

Hint

Worked solution

The Scene at this step

Crossings

3

Draw and untangle

Drawing

Read from the Scene at

How many colours the colouring rule counts with: the smallest prime dividing the determinant, 0 for the unknot/nodes/knot/outputs/colours

One setting that reaches it

Crossings

4

Hint

Worked solution

The Scene at this step

Half-twists

0

Squeeze the mug

0default

Cut down the middle

Whole loopfixed

Choices

• Correct The loop will have one side: the ant will cover all the paper and come back to where it started

• Wrong Still two sides: a half-twist only turns the paper

  Feedback

  After one lap the paper reaches the join upside down: the top side is glued to the bottom side, and the two sides become one.

• Wrong Three sides: the twist makes a new side

  Feedback

  A twist adds no paper, so no sides either. It only changes how the sides join at the glue.

Once the prediction is locked in, the Scene changes

Half-twists

1

The Scene then shows: 1 · Sides of the paper loop: 1 with an odd number of half-twists, else 2/nodes/strip/outputs/sides

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Hint

Worked solution

The Scene at this step

Half-twists

1fixed

Squeeze the mug

0default

Cut down the middle

Whole loop

Choices

• Correct One loop, twice as long

• Wrong Two separate loops, as when you cut a plain loop

  Misconception: Expects cutting any loop of paper down its middle to leave two loops.

  Feedback from the catalogue

  In a plain loop the cut separates the left half from the right half, so two are left. In a loop with an odd number of half-twists there is no “left half” and “right half”: after one lap, the left side of the line runs on into the right side. So what is left is one piece, twice as long.

• Wrong Two loops, linked together

  Feedback

  Two linked loops come from a loop with two half-twists. The Möbius strip leaves only one piece after the cut.

Once the prediction is locked in, the Scene changes

Cut down the middle

Cut

The Scene then shows: 1 · Loops left after one cut down the middle/nodes/strip/outputs/pieces

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Hint

Worked solution

The Scene at this step

Half-twists

3

Squeeze the mug

0default

Cut down the middle

Whole loop

Choices

• Correct One long loop, tied in a trefoil knot

• Wrong One long plain loop, as with the Möbius strip

  Feedback

  Right that there is one piece, as the number of half-twists is odd. But three half-twists also wind it round itself, and it comes out knotted.

• Wrong Two separate loops

  Misconception: Expects cutting any loop of paper down its middle to leave two loops.

  Feedback from the catalogue

  In a plain loop the cut separates the left half from the right half, so two are left. In a loop with an odd number of half-twists there is no “left half” and “right half”: after one lap, the left side of the line runs on into the right side. So what is left is one piece, twice as long.

Once the prediction is locked in, the Scene changes

Cut down the middle

Cut

The Scene then shows: 1 · 1 if the single loop after the cut is tied in a knot (a trefoil), else 0/nodes/strip/outputs/knotted

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Hint

Worked solution

The Scene at this step

Half-twists

1

Squeeze the mug

0default

Cut down the middle

Whole loop

Choices

Shown in a random order.

• Correct None: the inside of the glass is a dent, and the glass can be squashed into a ball without tearing

• Wrong One: the space inside it

  Misconception: Counts a dent or hollow, such as the inside of a cup, as a hole.

  Feedback from the catalogue

  In topology a hole is something you can thread a string through and tie it, like a mug’s handle. The inside of a cup is only a dent: you can fill it with the walls’ material and squash the cup into a ball, without tearing or gluing.

• Wrong Two: the opening at the top and the space inside

  Misconception: Counts a dent or hollow, such as the inside of a cup, as a hole.

  Feedback from the catalogue

  In topology a hole is something you can thread a string through and tie it, like a mug’s handle. The inside of a cup is only a dent: you can fill it with the walls’ material and squash the cup into a ball, without tearing or gluing.

Hint

Worked solution

The Scene at this step

Half-twists

1fixed

Squeeze the mug

0default

Cut down the middle

Whole loop

Accepted within

± 0

Computed from the Scene at this step’s setting

Edges of the paper loop: 1 with an odd number of half-twists, else 2/nodes/strip/outputs/edges

Wrong answers it recognises

• 2 ± 0

  Misconception: Thinks a loop of paper always has two edges, a top and a bottom, however it is twisted.

  Feedback from the catalogue

  A half-twist joins the top edge to the bottom one, so they become one long edge. With an odd number of half-twists there is one edge, with an even number two.

Hint

Worked solution

The Scene at this step

Half-twists

2fixed

Squeeze the mug

0default

Cut down the middle

Whole loop

Choices

• Wrong Two separate loops, as with an ordinary loop

  Misconception: Thinks that once a loop has two edges, its twists make no difference to what a cut leaves.

  Feedback from the catalogue

  With two edges the cut really does leave two loops. But the twists stay in them, and each full twist winds them round each other once. With one full twist they are linked and cannot be pulled apart without cutting.

• Wrong One loop twice as long, as with the Möbius strip

  No feedback of its own; the learner sees the standard message

  Not yet. Look again and try once more.

• Correct Two loops linked through each other

Once the prediction is locked in, the Scene changes

Cut down the middle

Cut

The Scene then shows: 1 · Times the two loops after the cut wind round each other, 0 for a single loop/nodes/strip/outputs/linking

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Hint

Worked solution

The Scene at this step

Half-twists

2fixed

Squeeze the mug

0default

Cut down the middle

Cut

Accepted within

± 0.01

Computed from the Scene at this step’s setting

Length of each loop after the cut, times the original loop/nodes/strip/outputs/pieceLength

Wrong answers it recognises

• 2 ± 0.01

  Misconception: Thinks cutting a twisted loop down the middle always gives a loop twice as long, as with the Möbius strip.

  Feedback from the catalogue

  The piece is twice as long only when there is one edge, going twice round the loop: with an odd number of half-twists. With an even number each edge gets its own strip, one lap long.

Hint

Worked solution

The Scene at this step

Curvature K

0.25fixed

Triangle size

1.8

Measure the triangle

Trianglefixed

Choices

• Correct More than $180^\circ$⁠

• Wrong Exactly $180^\circ$⁠, as in any triangle

  Misconception: Assumes the flat plane's facts hold on any surface: a triangle's angles add up to 180°, and lines that start parallel stay the same distance apart.

  Feedback from the catalogue

  A sum of $180^\circ$⁠, and parallel lines that stay the same distance apart, hold only on a flat plane; the proofs of both rest on the parallel postulate. On a ball the sum is more than $180^\circ$⁠ and parallels close in; on a saddle the sum is less and parallels spread apart. The difference from $180^\circ$⁠ equals the curvature times the area.

• Wrong Less than $180^\circ$⁠

  Feedback

  Less than $180^\circ$⁠ comes on a Saddle, whose curvature is negative. On a ball the sides bulge outwards and the angles open up.

Once the prediction is locked in, the Scene changes

Measure the triangle

Anglesfixed

The Scene then shows: 257.3 ° · The three angles added up/nodes/shell/outputs/angleSum

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Hint

Worked solution

The Scene at this step

Curvature K

0.25fixed

Triangle size

1.8fixed

Measure the triangle

Anglesfixed

Accepted within

± 3%

Computed from the Scene at this step’s setting

The area inside, along the surface, square grid units/nodes/shell/outputs/area

Wrong answers it recognises

• 309.3 ± 3%

  Misconception: Divides the angle excess in degrees by K to get the area, instead of converting it to radians first.

  Feedback from the catalogue

  In the Gauss–Bonnet theorem, $\alpha + \beta + \gamma - \pi = K \cdot A$$\alpha + \beta + \gamma - \pi =$$K \cdot A$⁠, the angles are measured in radians: that is why it has $\pi$⁠ and not $180^\circ$⁠. Convert the excess first, $\text{rad} = \text{deg} \cdot \dfrac{\pi}{180}$$\text{rad} =$$\text{deg} \cdot \dfrac{\pi}{180}$⁠, and only then divide by $K$⁠.

Hint

Worked solution

The Scene at this step

Curvature K

0.25fixed

Triangle size

2.2

Measure the triangle

Anglesfixed

Choices

• Correct It will come close to $180^\circ$⁠

• Wrong It will stay the same, because the sum depends only on the curvature

  Feedback

  The excess is $K \cdot A$⁠: curvature times area. A smaller triangle on the same ball gets a smaller excess.

• Wrong It will drop below $180^\circ$⁠

  Feedback

  On a ball the excess is always positive, because $K$⁠ is. The sum can only come close to $180^\circ$⁠ from above.

Once the prediction is locked in, the Scene changes

Triangle size

0.6

The Scene then shows: 186.9 ° · The three angles added up/nodes/shell/outputs/angleSum

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Hint

Worked solution

The Scene at this step

Curvature K

0

Triangle size

1.8

Measure the triangle

Parallelsfixed

Choices

• Correct They will spread apart

• Wrong They will stay the same distance apart, as on the plane

  Misconception: Assumes the flat plane's facts hold on any surface: a triangle's angles add up to 180°, and lines that start parallel stay the same distance apart.

  Feedback from the catalogue

  A sum of $180^\circ$⁠, and parallel lines that stay the same distance apart, hold only on a flat plane; the proofs of both rest on the parallel postulate. On a ball the sum is more than $180^\circ$⁠ and parallels close in; on a saddle the sum is less and parallels spread apart. The difference from $180^\circ$⁠ equals the curvature times the area.

• Wrong They will close in and meet

  Feedback

  That happens on a Ball, whose curvature is positive. On a saddle the curvature is negative, and the opposite happens.

Once the prediction is locked in, the Scene changes

Curvature K

-0.1

The Scene then shows: 1.96 · Gap between the parallels' ends, grid units/nodes/shell/outputs/endGap

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Hint

Worked solution