Chapter 4: Exploring Some Geometric Themes (Geometry)

Each remaining square gives 8 smaller remaining squares and 1 new hole at the next step:

$$R_{n+1}=8R_n,H_{n+1}=H_n+R_n$$

Rₙ = 8ⁿ and Hₙ = 1 + 8 + … + 8ⁿ⁻¹ = (8ⁿ − 1)/7 (0, 1, 9, 73, …).

Let the side be $s$. At each corner, the small corner triangle has two sides of $\frac{s}{2}$ (half-sides) with a $60°$ angle between them. So it is isosceles with base angles $\frac{180°-60°}{2}=60°$, which makes it equilateral with side $\frac{s}{2}$.

So all three segments joining the midpoints have length $\frac{s}{2}$. The middle triangle therefore has three sides of $\frac{s}{2}$ and is equilateral too. All four triangles have sides $\frac{s}{2}$, so they are congruent (SSS).

Each corner triangle is isosceles with a 60° apex, hence equilateral with side s/2; so all 4 triangles have sides s/2 and are identical.

Step 1: remove the middle of the 4 triangles; Step 2: do the same in each of the 3 that remain (9 small triangles left).

Each remaining triangle becomes 3 remaining triangles and 1 new hole: $T_{n+1}=3T_n$ and $H_{n+1}=H_n+T_n$.

Tₙ = 3ⁿ and Hₙ = (3ⁿ − 1)/2 (0, 1, 4, 13, 40, …).

• Carpet: each step keeps 8 of the 9 equal parts, so the area is multiplied by $\frac{8}{9}$. At step $n$ the area is ${\left(\frac{8}{9}\right)}^n$. (Or: $8^n$ squares, each of area $\frac{1}{9^n}$.)
• Triangle: each step keeps 3 of the 4 equal parts, so at step $n$ the area is ${\left(\frac{3}{4}\right)}^n$. (Or: $3^n$ triangles, each of area $\frac{1}{4^n}$.)

Both areas get smaller and smaller at every step, closer and closer to 0.

Carpet: (8/9)ⁿ sq. units; Triangle: (3/4)ⁿ sq. units.

Step 1 is a six-pointed star (12 sides); Step 2 adds a bump to each of its 12 sides (48 sides).

Each side is replaced by 4 sides, so the number of sides is multiplied by 4 at every step: 3, 12, 48, 192, …

3 × 4ⁿ sides.

At step $n$, there are $3\times 4^n$ sides, each of length ${\left(\frac{1}{3}\right)}^n$:

$$\text{Perimeter}=3\times 4^n\times \frac{1}{3^n}=3{\left(\frac{4}{3}\right)}^n$$

That gives 3, 4, $5\frac{1}{3}$, $7\frac{1}{9}$, …. The perimeter keeps growing without limit, even though the snowflake always fits inside a fixed circle.

3 × (4/3)ⁿ units.

(2) The piece left is a square standing on its corner, with half the area. The four corners are isosceles right triangles. Put their right-angle corners together at one point, with the long sides outwards, and they form another square of the same size as the one left over.

(3) Each corner cut removes a small equilateral triangle of side $\frac{1}{3}$. What remains has 6 equal sides (each $\frac{1}{3}$ of the original side) and 6 angles of $120°$: a regular hexagon.

(4) You get an octagon with all angles $135°$. Its sides are not all equal: the middle thirds of the square's sides have length $\frac{1}{3}$, while the cut edges are longer, $\frac{\sqrt{2}}{3}$. So it is not a regular octagon.

(2) A tilted square of half the area; the 4 corners also make a square. (3) A regular hexagon. (4) An octagon with equal angles but unequal sides.

• (5) Square: a cube seen face-on; or a cylinder whose height equals its diameter, seen from the side; or a square pyramid seen from above.
• (6) Circle: a sphere from any direction; or a cylinder or cone seen from directly above.
• (7) Triangle: a cone or a square pyramid seen from the side; or a triangular prism seen end-on.

Square: cube face-on. Circle: sphere, or a cylinder or cone from above. Triangle: cone or pyramid from the side.

• (8) A cylinder: rectangle from the side, circle from the top.
• (9) A cone: triangle from the side, circle from the top.
• (10) A triangular prism lying on a face: rectangle from the front, triangle from the end. (A tent shape.)
• (11) A frustum (a cone with its tip cut off flat, like a bucket or a lampshade): trapezium from the side, circle from the top.
• (12) A pentagonal prism: pentagon from the end, rectangle from the side.

Not unique. For example, a square pyramid also has triangular side profiles, and many solids have circular top views. Usually several solids fit.

Cylinder, cone, triangular prism, frustum of a cone, pentagonal prism. The solids are not unique; often several work.

Prism: 2 ends + $n$ side faces; $n$ edges on each end + $n$ joining edges; $n$ vertices on each end.

Pyramid: the base + $n$ triangles; $n$ base edges + $n$ edges to the apex; $n$ base vertices + the apex.

In every case, F − E + V = 2.

Prism: n + 2 faces, 3n edges, 2n vertices (12, 30, 20 for n = 10). Pyramid: n + 1 faces, 2n edges, n + 1 vertices (11, 20, 11).

Each answer was checked by "rolling" a cube over the squares to see which face each square covers.

(ii), (iii), (iv) and (vi) are nets of a cube; (i) and (v) are not.

There are 35 ways to join 6 squares edge to edge (hexominoes). Exactly 11 of them fold into a cube:

• 6 nets with a row of 4 (one square above it and one below it)
• 3 nets with a row of 3
• 1 net with steps of two ("2-2-2")
• 1 net with two rows of three

There are 11 nets: 6 with a row of four, 3 with a row of three, plus the 2-2-2 staircase and the 3-3 shape.

Lay four faces in a strip: length, height, length, height. Attach the two remaining faces (the ones with the two shorter dimensions as sides) on either side of one long face.

(i) Make a strip of four rectangles, each 5 cm long: 5 × 3, 5 × 1, 5 × 3, 5 × 1. Total strip: 5 cm by $3+1+3+1=8$ cm. Attach two 3 cm × 1 cm rectangles to the two 3 cm sides of one 5 × 3 face.

(ii) Make a strip of 6 × 3, 6 × 2, 6 × 3, 6 × 2 (a 6 cm × 10 cm strip). Attach two 3 cm × 2 cm rectangles at the ends of one 6 × 3 face.

A check: the total area must be $2(lb+bh+hl)$, which is 46 cm² for (i) and 72 cm² for (ii).

(i) Faces 5×3 (two), 5×1 (two), 3×1 (two): area 46 cm². (ii) Faces 6×3, 6×2, 3×2 (two of each): area 72 cm².

• Figure 1 (a big triangle cut into 4) is a net.
• Figure 3 (a strip of 4 triangles forming a parallelogram) is a net.
• Figure 2 (three triangles in a row with one below the first) is not a net: when folded, its last triangle lands on the same face as the one below.
• Figure 4 has 5 triangles, but a tetrahedron has only 4 faces, so it is not a net either.

These two shapes are the only nets of a regular tetrahedron.

Tetrahedron net: draw an equilateral triangle of side 10 cm and join the midpoints of its sides. You get 4 equilateral triangles of side 5 cm. Fold the three corner triangles up so their tips meet.

Square pyramid net: draw a square of side 5 cm. On each side, draw an isosceles triangle with base 5 cm and two equal sides of 6 cm. Fold the four triangles up to meet at the apex.

Figures 1 and 3 are nets (the only two); figures 2 and 4 are not.

Cylinder: the rectangle is the curved surface opened out. One side is the height $h$. The other side is the circumference of the base, $2\pi r$, because it wraps exactly once around the circle. The net is this rectangle with a circle of radius $r$ attached to each long side.

Cone: slit along a slant line and unroll it. You get a sector of a circle: its radius is the slant length $l$, and its arc length is the base circumference $2\pi r$. Add the base circle of radius $r$.

If O is not the centre of the boundary arc, the edge of the rolled-up surface is not at the same distance from the tip all the way round. The result is a cone-like surface whose rim is slanted and wavy, not a flat circle. It will not stand evenly on a table; it looks like a cone cut off at a slant.

Cylinder: a rectangle h × 2πr plus two circles. Cone: a sector of radius l with arc 2πr, plus a circle. With O off-centre you get a cone with a slanted, uneven rim.

Triangular prism:

• Draw three rectangles of 3 cm × 8 cm side by side, making a 9 cm × 8 cm strip.
• Attach an equilateral triangle of side 3 cm to the top and to the bottom of the middle rectangle.

Octahedron: 8 equilateral triangles (say of side 4 cm) arranged as in the book's net. Folded, they make two square pyramids joined base to base.

Sphere: no flat net can wrap a ball exactly. A sphere curves in every direction at once, but paper can bend in only one direction at a time without stretching. Any wrapping leaves wrinkles, gaps or overlaps. That is why world maps always distort shapes or areas.

Prism: 3 equal rectangles + 2 triangles. A sphere has no net: paper cannot cover it without wrinkles, gaps or overlaps.

Unfold the cuboid so that the two faces lie flat, side by side. On the flat net, the shortest route is the straight line between the ant and the laddu. Folding the net back turns it into the shortest route on the surface.

• Laddu on the top face: the straight line on the net goes straight up the side face and straight across the top. The red path is shortest.
• Laddu at the centre of an edge: the bent blue path is not straight on the net, so it is not the shortest. The straight segment from the ant to the laddu on the net is.

This is how we can be sure: every path on the surface becomes a path of the same length on a net, and on a flat sheet a straight line beats every other path. But a straight line must stay inside the net, and we must check all the ways of unfolding the faces in between.

Unfold into a net and join the two points by a straight line; it must lie inside the net, and every possible unfolding must be compared.

In the first unfolding, the straight segment goes outside the net, so it is not a real path. Unfold instead so that the end face is attached directly to the face next to the laddu.

In the flat picture, the ant is 2 cm from the shared edge (the centre of a 4 cm face) and 2 cm from the front edge. The laddu is 2 cm on the other side of the shared edge, on the front edge. So:

$$d=\sqrt{(2+2{)}^2+2^2}=\sqrt{20}\approx 4.5\text{ cm}$$

Use the unfolding that joins the ant's face directly to the laddu's face; the path is then a straight segment of about √20 ≈ 4.5 cm.

Compare the unfoldings:

• Straight over the top: $1+30+11=42$ cm. (From the ant up 1 cm, along the top, then down 11 cm to the laddu.)
• Unfolding across the top, a side and the bottom (the book's second net): a right triangle with legs 24 and 32, so $d=\sqrt{24^2+32^2}=\sqrt{1600}=40$ cm.
• Other unfoldings (for example, going round one side wall: √(42² + 10²) ≈ 43.2 cm) are all longer than 40 cm.

The shortest is 40 cm. Surprisingly, it crosses five of the six faces.

40 cm, shorter than the obvious 42 cm route.

In Fig. 4.3, AECD is a rectangle (it has right angles at C and D, and AE ⊥ BC), so $AE=DC=p$.

Triangle AEB has a right angle at E, so the hypotenuse is the longest side: $l=AB\ge AE=p$.

They are equal only when E = B, i.e. when the line is parallel to the plane. The more the line tilts, the shorter its projection; a line perpendicular to the plane projects to a single point.

p ≤ l always; p = l exactly when the line is parallel to the plane.

Square: a square (parallel to the plane), a rectangle (tilted about an axis parallel to one side), a rhombus or a general parallelogram (tilted about other directions), or a line segment (seen edge-on).

Parallelogram: always another parallelogram, or a segment. Parallel lines project to parallel lines, so its two pairs of parallel sides stay parallel. It can never become a quadrilateral that is not a parallelogram, such as a trapezium or kite.

Regular n-gon: the projection is the polygon "squashed" in one direction. It has the same number of sides, and sides that were parallel stay parallel. So a regular hexagon gives a hexagon with three pairs of parallel sides. Edge-on, it becomes a line segment.

Square → square, rectangle, parallelogram (rhombus) or segment. Parallelogram → always a parallelogram (or segment), never a non-parallelogram. Regular n-gon → a squashed n-gon keeping parallel sides parallel.

• Cube: a square, a rectangle or a hexagon, depending on how it is turned.
• Cone: a circle (looking along its axis, with the apex at the centre) or a triangle (seen from the side).

Another object with the same projection as a cone:

• Triangular profile: a triangular prism (seen end-on), a square pyramid, or a flat triangle.
• Circular profile: a cylinder, a sphere or a disc.

That is why we use three views and not just one.

The cone projects to a circle or a triangle; e.g. a triangular prism or square pyramid gives the same triangle, and a cylinder the same circle.

Let a line's extent be $a$ along the length, $b$ along the depth and $c$ along the height. Its true length satisfies $l^2=a^2+b^2+c^2$, and the three views have lengths:

• front view $\sqrt{a^2+c^2}$
• top view $\sqrt{a^2+b^2}$
• side view $\sqrt{b^2+c^2}$

So:

• Each view is at most the true length.
• $(\text{front}{)}^2+(\text{top}{)}^2+(\text{side}{)}^2=2l^2$.
• A line parallel to a plane shows its full length in that view, and a line perpendicular to it shows only as a point. In the first row of the figure, the side view is a single dot.

Each view is no longer than the line, and front² + top² + side² = 2 × (true length)².

Standing each solid on the horizontal plane in the usual way:

See the table; e.g. an upright cylinder gives rectangle, circle, rectangle, and a cone gives triangle, circle, triangle.

The rows of projections are listed from top to bottom:

Jug–8, funnel–6, hammer–7, car–1, slide–3, chair–4, fan–5, bucket–2 (rows counted from the top).

When the torch is closer, the shadow is bigger. When it moves further away, the shadow shrinks towards the true size of the projection.

A torch sends out rays that spread out from a point. Near the torch the rays spread steeply, so the shadow is enlarged. From far away, the rays reaching the object are almost parallel, so the shadow is almost the same as the projection. The Sun is so far away that its rays are parallel for all practical purposes.

Closer torch → larger shadow, because the rays diverge from one point; a very distant source (like the Sun) gives parallel rays and true projections.

Each solid is drawn below as unit squares. In each top view, the number is the height of the stack of cubes standing there.

In a neat drawing, also mark the edges where two visible faces at different depths meet, as the book does.

See the figures: each view shows the squares seen from that direction (front: width × height, top: width × depth, side: depth × height).

(i) The C is one cube thick:

• Side: a vertical bar of 4 squares, like the letter I.
• Top: a row of 3 squares, like a dash —.

(ii) Keep the C as the front layer. Add 7 cubes behind its top row, all at the top level:

• 2 cubes behind the left and right ends
• then 3 cubes in a full row
• then 2 cubes at the ends again

From the top, this reads as an A (3 wide, 4 deep: ### / #.# / ### / #.#). From the front, the new cubes hide behind the C's top row, so the front still shows C. Total: 15 cubes.

(iii) From the side, the shape now shows a full column (the C) with a top bar (the A layer): it looks like Γ. To make an F, add a middle bar.

Glue 2 cubes behind the C's left cube on the second level from the bottom. They are hidden from the front behind the C, and hidden from the top under the A. From the side, they add a shorter bar of 3.

The side view becomes F (top bar of 4, middle bar of 3). Total: 17 cubes.

(iv) Many combinations work. For example, a row of 3 with a column of 2 below its centre looks like T from the front and I from the side.

(i) From the side 'I', from the top '—'. (ii) Add 7 cubes at the top level behind the C (15 cubes). (iii) Add 2 more at the second level behind the left column (17 cubes).

Reading the views:

• Top view: a back row of two parts, a full middle row, and a front row only at the left.
• Front view: the left column is taller than the right.
• Side view: low at the front and tall at the back.

So the solid has:

• a back wall whose left part is higher than its right part (a step down to the right),
• a full-width middle strip one cube high,
• a front block only at the left.

Only solid (ii) has all three. The others have a U-shaped notch at the front ((i), (iii), (v), (vi)), a flat-topped back wall ((i), (v)), or a different shape altogether ((iv), (vii)).

Solid (ii).

Read the front and side views to get the heights, and the top view for where the columns stand:

(i)–(iii):

• Back row: an empty corner at the left, then two stacks of 2.
• Middle row: three single cubes.
• Front row: single cubes at the left and right only.

(iv)–(vi):

• Back row: two stacks of 2.
• Middle row: two single cubes.
• Front row: one single cube at the left.

(vii)–(ix):

• Back row: stacks of 2 and 3.
• Middle row: two single cubes.
• Front row: one single cube at the left.

The dashed lines in the side views (iii) and (ix) are edges hidden behind nearer cubes.

(i)–(iii): 9 cubes; (iv)–(vi): 7 cubes; (vii)–(ix): 8 cubes (heights as in the top views shown).

The stack is a staircase seen from a corner. Its visible top cubes form rows of 4, 3, 2 and 1, rising one level for each row back.

For every cube to rest on something, the column under each visible top cube must be full. So the columns have heights 1 (4 columns), 2 (3 columns), 3 (2 columns) and 4 (1 column):

$$4\times 1+3\times 2+2\times 3+1\times 4=20$$

20 cubes (10 can be seen; 10 are hidden underneath).

• Square: a face parallel to the plane.
• Rectangle: the cube turned about one axis that is parallel to the plane.
• Hexagon: turned in a general direction. It is a regular hexagon when the cube balances on a corner (the isometric view).

The outline can never be a triangle or a pentagon. The cube's edges come in three parallel families, and their projections pair the outline's sides as parallel opposites, so the outline has 4 or 6 sides.

A square, a rectangle or a hexagon (regular when the cube stands on a vertex).

Standing on a corner, with the long diagonal vertical, the three edges at the top corner are tilted equally to the floor. By symmetry, turning the cube by $120°$ about the vertical diagonal swaps these three edges. So their projections are equal, and so are those of all the edges parallel to them.

On isometric paper, the three families of edges (height, length and depth) go in three fixed directions: |, ╱ and ╲. A unit step along any axis has the same length on paper. Parallel edges stay parallel, so one can draw the solid by counting steps along the three directions.

Symmetry about the vertical diagonal makes the three edge directions equally tilted, so all edges project to the same length; this is what isometric grids exploit.

Yes, 3 more. These are the non-flat shapes, which cannot lie flat in one layer:

• Corner (tripod): three cubes around one corner cube, one going in each of the three directions.
• Left screw and right screw: an L of 3 cubes with the 4th cube on top of an end cube. They are mirror images of each other.

Altogether there are 8 ways (7 if mirror images count as the same).

Yes: 3 more non-flat ones (the corner/tripod and the two mirror-image "screws"), making 8 in all.

Draw edge by edge, counting units:

1. Choose a corner on a dot of the grid.
2. Move along | for height, along ╱ for length and along ╲ for depth.
3. Decide for each edge whether it goes up or down, forwards or backwards.
4. Start with the base, then the vertical edges, then the top. Lightly draw hidden edges, and erase them at the end.

Use | for height, ╱ for length and ╲ for depth, counting grid steps for every edge.

Yes. Follow the arrows: the ball moves round a closed loop of blocks and comes back to where it started. Yet along the way it seems to go steadily up (or level) all the time. A real loop must come back to the same height, so no real set of blocks could look like this. It is an impossible figure, like the Penrose staircase.

Each small part of the drawing is a real corner of blocks. The trick is that the depth directions are swapped where two parts meet. On the isometric grid you can draw each piece correctly, but the pieces can only be joined into a loop by breaking the rules at one join.

Yes: the loop seems to keep rising yet returns to its start, which is impossible in 3D; the parts are realisable but their joins switch depth.

(i) No model made of real cubes can look like this from every point of view. Each corner is a real right-angled corner of three bars of cubes, but going round the triangle, the three bars would have to be at the same depth and at different depths at once.

A real object can only look like it from one special viewpoint, where one bar's end lines up with another bar that is actually much further away. From the front, top or side, such a model shows three straight bars that do not close up. The profiles are like an L or a staircase of bars, not a triangle.

(ii) On the isometric grid, draw three bars of cubes along the three axis directions (|, ╱, ╲), each turning by $120°$ on the paper at a corner. Then join the last bar to the first.

(iii) Isometric drawing loses depth: a point further back and higher up can land on the same spot of the paper as a nearer, lower point. Our eyes check each corner, which looks right, but not the whole figure. So the brain accepts a join that is impossible in space.

(i) No real model; only an object seen from one special angle can look like it, and its true profiles do not close up. (iii) Isometric pictures hide depth, so each corner is fine but the whole is inconsistent.