Chapter 6: Constructions and Tilings (Geometry)

AB passes through the midpoint O of XY and makes a right angle (90°) with it: AB is the perpendicular bisector of XY.

AB meets XY at its midpoint, at 90°.

Let P be any point with PX = PY, and let A be a point on the other side of XY with AX = AY. We showed that the line joining two such points (one on each side) is the perpendicular bisector. If P is on the same side as A, use the point B below XY instead. Either way P lies on a line through two points equidistant from X and Y, which is the perpendicular bisector; and since XY has only one perpendicular bisector, P lies on it.

Joining P to an equidistant point on the other side gives the perpendicular bisector, and there is only one, so P lies on it.

No. Point A (arcs of radius r above) is equidistant from X and Y, and point B (arcs of a different radius s below) is also equidistant from X and Y. Both lie on the perpendicular bisector, and two points fix a line, so AB is still the perpendicular bisector.

No: A and B are each equidistant from X and Y, so both lie on the perpendicular bisector.

Both pairs can be on the same side, as long as they use different radii (so that they give two different points). Both meeting points are equidistant from X and Y, so the line through them is the perpendicular bisector. (On the same side the two points may be close together, so the line is drawn less accurately; opposite sides give a better line.)

Yes, on the same side with two different radii; the two meeting points still determine the perpendicular bisector.

Yes. The meeting point must be at the same distance from X and from Y. With radius r from X and a different radius s from Y, the meeting point is r from X and s from Y, so it does not lie on the perpendicular bisector.

Yes; otherwise the meeting point is not equidistant from X and Y.

Draw two perpendicular lines through a point O (construct a perpendicular bisector). On each of the four arms mark the same length OX with the compass. Each petal is an "eye" on the segment from O to the end of an arm: construct the perpendicular bisector of that segment, pick two points on it at equal distances from the segment, and draw the two arcs with these as centres (as in the Eyes construction). Repeat for all four arms.

Draw two perpendicular lines, mark equal arms, and draw an eye-shaped petal on each arm using arcs centred on the arm's perpendicular bisector.

With the compass centred at O, mark X and Y on the line with OX = OY. With a larger radius, draw arcs from X and Y meeting at A. Join OA: $\angle AOY=90°$.

Only one pair of arcs is needed, because O is already a point on the perpendicular bisector of XY.

Mark OX = OY, draw equal arcs from X and Y meeting at A; OA ⊥ the line. One pair of arcs is enough.

The rope's midpoint divides it into two equal halves, so when stretched, A is at the same distance (half the rope) from X and from Y: AX = AY. The same is true for B: BX = BY. Points equidistant from X and Y lie on the perpendicular bisector, so AB is the perpendicular bisector.

AX = AY = BX = BY = half the rope, so A and B are equidistant from X and Y.

Method 1 (Pythagorean triple 3, 4, 5): take a rope divided into 12 equal parts by knots. Peg it at the point O, stretch 3 parts along the line to P, and bring the remaining rope so that 4 parts go from O and 5 parts come back to P; the corner where the 4 and 5 parts meet gives Q with $\angle QOP=90°$ (a 3-4-5 triangle is right-angled). The Śulba-Sūtras use such rope triangles.

Method 2 (equal distances): peg the rope at O and mark points X and Y on the line at equal rope-lengths on both sides; then use a longer rope with its ends at X and Y and pull its midpoint out, as in Fig. 6.4.

E.g. a knotted rope forming a 3-4-5 triangle with the 3-part on the line, or marking OX = OY with the rope and then using the midpoint method.

In ΔOBC and ΔOAC: OB = OA, BC = AC and OC is common. By SSS, ΔOBC ≅ ΔOAC, so $\angle BOC=\angle AOC$.

ΔOBC ≅ ΔOAC (SSS), so ∠BOC = ∠AOC.

Draw, for example, angles of about 40°, 75°, 110° and 150° freehand with a ruler (no measuring is needed). Bisect each: arc from O cutting both arms at A and B; equal arcs from A and B meeting at C; draw OC. Check with a protractor that the two halves are equal (e.g. 110° → 55° and 55°).

For each angle: mark OA = OB, draw equal arcs from A and B meeting at C, and join OC.

The 8 supporting lines make $360°÷8=45°$ with each other. Construct two perpendicular lines through O (90°), bisect each of the four right angles to get the 45° lines, mark equal lengths on all 8 arms, and draw an eye-shaped petal on each arm as in the four-petal design.

Make 90° angles at O, bisect them to get 8 lines at 45°, and draw a petal on each arm.

Yes. The meeting point C is still at equal distances from A and B (CA = CB), and OA = OB, so O and C both lie on the perpendicular bisector of AB. So does the bisector of the angle (since ΔOAB is isosceles). Hence the line OC (extended through O) is the angle bisector; C just lies on the extension of the bisector behind O.

Yes; O and C are both equidistant from A and B, so the line OC is still the bisector (extended through O).

From 90°: 45°, 22.5°, 11.25°, … From 60°: 30°, 15°, 7.5°, … Adding and subtracting these gives many more, e.g. $90°+45°=135°$, $60°+45°=105°$, $45°+30°=75°$.

65.5° cannot be constructed exactly this way. Every angle we can make by halving 60° and 90° and adding or subtracting is a whole multiple of 15° divided by a power of 2 (such as 7.5°, 3.75°, 1.875°, …), and 65.5° = 131/2 degrees is not of that form. We can only get close, e.g. 60° + 3.75° + 1.875° = 65.625°.

45°, 22.5°, 30°, 15°, 75°, 105°, 135°, …; 65.5° cannot be made exactly this way (only approximately).

Peg the rope at the vertex O and mark points A and B on the two arms with the same rope length (OA = OB). Then take a rope with loops at its ends, put the loops on pegs at A and B, and pull its midpoint out between the arms until both halves are tight: that point C satisfies CA = CB. The line OC is the bisector.

Mark OA = OB with the rope, then pull a doubled rope from A and B by its midpoint to C (CA = CB); OC bisects the angle.

Each petal joins the centre O of the square to a corner. Mark the midpoints of the four sides (by perpendicular bisection). With each side midpoint as centre and radius half the side, draw a semicircle inside the square: it passes through the two corners of that side and through the centre O (all three are half a side away from the midpoint). The four semicircles overlap in pairs to form the four petals.

Draw semicircles inside the square with the side midpoints as centres and half the side as radius; each pair of neighbouring semicircles forms a petal from O to a corner. This radius gives the largest petals that fit.

1. Draw a ray from the new vertex X.
2. With centre A and any radius, draw an arc cutting the arms of ∠A at B and C.
3. With the same radius and centre X, draw an arc cutting the ray at Z.
4. Set the compass to the length BC; with centre Z, cut the arc at Y.
5. Draw XY. Then ΔABC ≅ ΔXYZ (SSS), so ∠YXZ = ∠BAC.

Fig. 6.6: draw the base line; construct one sector (two equal arms at the given angle with an arc joining their ends); copy the angle at the next points, alternately pointing up and down, keeping all arm lengths equal with the compass.

Same-radius arcs at A and X, then copy the chord BC from Z to get Y; ΔABC ≅ ΔXYZ by SSS. Repeat to build Fig. 6.6.

For each pair: draw line m and any line l crossing it at A; choose B on l; copy the angle between l and m at B (in the corresponding position); the new line through B is parallel to m.

The 8-pointed figure: draw 8 lines through the centre at 45° to each other (bisect right angles) and mark the inner points A to H at equal distances. Each outer point (S, T, U, …) is then found by drawing lines parallel to the neighbouring spokes through the inner points, using the parallel-line construction; shade alternate triangles as shown.

Copy the angle made with a transversal to draw each parallel line; the star is built from 8 congruent rhombuses on lines at 45°.

Draw AD. Construct equal angles at A and D (e.g. 90° each, or copy one angle to the other end). Mark B and C with AB = DC using the compass. Then draw the three lobes: one arc on BC (centre at the midpoint of BC) and two smaller equal arcs at the sides (centres on AB and DC extended), adjusting the radii until the arch looks balanced.

Equal angles at A and D, AB = DC by compass; then a middle arc on BC and two equal side arcs.

The supports are two equal vertical segments. Mark their midpoints. For a pointed arch, draw an arc from the top of the left support with centre at the top of the right support (radius = the gap between them), and the mirror arc from the right; they meet at a point above the middle. Using a radius larger than the gap gives a taller, sharper arch; a smaller radius (but more than half the gap) gives a flatter one.

Draw two equal arcs, each centred on the opposite support's top; their meeting point is the tip. Changing the radius changes the height of the arch.

Gap: $40°+60°+50°+30°+40°+90°=310°$, so the gap is $360°-310°=50°$. The 70° angle does not fit.

Straight lines: at O, three equilateral triangles lie between OA and OD on each side: $\angle AOD=60°+60°+60°=180°$, a straight angle. The same holds for BOE and COF.

Hexagon of side 4 cm: draw a circle of radius 4 cm. With the compass still at 4 cm, start at any point on the circle and mark off six points around it; join them in order. (Each side equals the radius because each triangle with the centre is equilateral.)

Gap = 50°, so 70° does not fit; ∠AOD = 3 × 60° = 180°. Hexagon: step the radius 4 cm six times round a circle of radius 4 cm.

AB = AC (radius of the first arc) and BC = AB (same radius used from B). So ΔABC is equilateral, and each of its angles is 60°: $\angle CAX=\angle CAB=60°$.

Hexagon of side 5 cm:

(Or: draw a 5 cm side, construct 120° at each end by making 60° twice, mark 5 cm on each new arm, and continue.)

ΔABC is equilateral (all sides = the compass radius), so ∠CAX = 60°. Step a 5 cm radius six times round a circle of radius 5 cm for the hexagon.

Construct 60° (equilateral triangle) and bisect it: 30°. Bisect the 30° angle: 15°.

Bisect 60° to get 30°, then bisect 30° to get 15°.

Construct a regular hexagon GHIJKL and draw an equilateral triangle outward on each side (60° at both ends of each side, using the 60° construction), or extend alternate sides.

Yes, they are equilateral. Each interior angle of the regular hexagon is 120°, so the angles of a point triangle next to the hexagon are $180°-120°=60°$ each (they lie on straight lines through the hexagon's sides). A triangle with all angles 60° is equilateral.

Yes: each base angle is 180° − 120° = 60°, so all three angles are 60°.

(a) Inflexed arc: two vertical supports; at the top of each, draw an arc curving outward and then inward to meet at a point above the middle (two pairs of arcs with centres outside and inside the arch).

(b) Six-petal flower: draw a circle; step its radius six times round it to get 6 points. With each point as centre and the same radius, draw an arc outside the circle between the neighbouring points: these arcs are the petals. (This uses only the compass.)

(c) Hexagon in a circle: as in the regular hexagon construction, join the 6 points stepped around the circle.

(d) Six circles: find the 6 points around a circle (as in (c)); draw a circle of half the side length at each point; neighbouring circles touch each other.

(e) Triangle pattern: draw the regular hexagon, join opposite vertices (6 equilateral triangles), then divide each side into halves (perpendicular bisectors) and join the midpoints with lines parallel to the sides to make the smaller triangles and stars.

Each figure starts from 6 points stepped round a circle with its own radius; petals and circles are drawn with those points as centres.

We "see" two white triangles, one pointing up over the other, even though no triangle is drawn. Our brain joins the edges of the cut-out shapes into continuous lines. To recreate it: draw an equilateral triangle lightly, draw small circles at its vertices and erase the parts inside the triangle; draw another (upside-down) triangle's corners as V-shapes; then erase all the guide lines.

A white triangle appears though none is drawn; the brain completes the edges. Construct it with faint guide triangles and erase them.

The star's 6 points are the vertices of the hexagon. Join every second vertex to get two overlapping equilateral triangles (ACE and BDF): each angle at the star's points is 60°, the angles of the small triangles at the hexagon's sides are 30°, 30°, 120°, and the inner hexagon has angles of 120°. Construct the regular hexagon (stepping the radius six times round a circle), then join alternate vertices.

Construct a regular hexagon and join alternate vertices; the star's points are 60°, the corner triangles 30°-30°-120°.

With centre P and a radius large enough, draw an arc cutting l at X and Y; then PX = PY, so P lies on the perpendicular bisector of XY. With a larger radius, draw arcs from X and Y on the other side of l meeting at Q. The line PQ is perpendicular to l.

Arc from P cuts l at X and Y (PX = PY); equal arcs from X and Y meet at Q; PQ ⊥ l.

Every figure uses all 7 pieces: 2 large triangles, 1 medium triangle, 2 small triangles, 1 square and 1 parallelogram, with no overlaps. A good strategy: first place the two large triangles (they cover half the area) in the biggest parts of the shape, then the medium triangle, the square and the parallelogram, and fill the small corners with the small triangles. For example:

• Fish: the two large triangles make the body (a square turned diagonally), the medium triangle and the parallelogram make the tail, the small triangles make the fins.
• Windmill: the four triangles (2 large, 1 medium and 2 small combined) are placed as blades around the centre.
• Running man: the square is the head, the large triangles form the body, the parallelogram and small triangles make the legs.

Use all 7 pieces each time; start with the two large triangles in the largest parts, then fit the medium triangle, square, parallelogram and the two small triangles.

• 4 × 7: yes: each of the 7 columns has 4 squares, so cover each column with 2 vertical tiles.
• 5 × 7: no: it has 35 squares, and each tile covers 2, so any tiling covers an even number of squares; 35 is odd.
• Both even: yes: cover each column with vertical tiles (the number of rows is even).
• One even, one odd: yes: if the number of rows m is even, cover each column with $m÷2$ vertical tiles; if the number of columns n is even, cover each row with $n÷2$ horizontal tiles.
• Both odd: no: $m\times n$ is odd, but the tiles always cover an even number of squares.

4 × 7 yes; 5 × 7 no (35 is odd). Tileable exactly when at least one side is even; never when both are odd.

Colour the grid like a chessboard with the corners black: a 5 × 3 grid has 8 black and 7 white squares. Each tile covers one black and one white square.

• Corner removed (black): 7 black and 7 white, and it can be tiled.
• Middle of row 2 removed (also black): 7 and 7, and it can be tiled, for example (letters show the tiles; ■ is the removed square):

• Top middle removed (Fig. 6.13): that square is white, leaving 8 black and 6 white, so it cannot be tiled.

Other squares that make the 5 × 3 grid non-tileable when removed: any white square: the middle of rows 1, 3, 5 and the two ends of rows 2 and 4.

(Removing any black ■ square leaves a tileable region; removing any white □ square does not.)

Corner and centre-of-row-2 removals are tileable; Fig. 6.13 is not (8 black vs 6 white). Removing any white square (□ above) makes it non-tileable.

Yes, both ways. Any 2 × 1 tile always covers two neighbouring squares, and neighbouring squares always have different colours, so a tiling of the plain grid automatically puts one black and one white square under each tile; and a black-and-white tiling is also a plain tiling. So the two problems are the same, and the colouring just helps us count.

Yes, both: every domino covers one black and one white square anyway.

(1) Yes. The region has 12 squares, so we need 4 L-tiles. One tiling (letters show the tiles):

(2) No. With chessboard colouring, the two opposite corners have the same colour, so the region has 30 squares of one colour and 32 of the other. Each tile covers one of each, so no tiling is possible.

(1) Yes (4 L-tiles, as shown) (2) No: removing two opposite corners leaves 30 of one colour and 32 of the other.

Squares, rectangles, equilateral triangles, regular hexagons, any triangle (two copies make a parallelogram) and any quadrilateral (rotating copies by 180° around the midpoints of its sides) all tile the plane.

Among regular polygons, only the equilateral triangle, square and regular hexagon work: their angles (60°, 90°, 120°) divide 360° exactly ($6\times 60°$, $4\times 90°$, $3\times 120°$). A regular pentagon's angle is 108°, and 360° ÷ 108° is not a whole number, so pentagons leave gaps.

Equilateral triangles, squares and regular hexagons (their angles divide 360°); also every triangle and every quadrilateral.