Constructions and Tilings: Drawing with Logic, Not Luck

Take a line segment XY. Bisection means dividing something into two identical parts, and the perpendicular bisector of XY is the line that cuts XY exactly in half at a right angle. Here's how to construct it using only an unmarked ruler and a compass: open the compass to some fixed radius (long enough to reach past the midpoint of XY), and from X, draw an arc above XY and another arc below XY. Without changing the radius, repeat from Y — draw an arc above and another below. The two arcs above XY cross at a point; call it A. The two arcs below XY cross at a point; call it B. Draw the line through A and B — this line AB is exactly the perpendicular bisector of XY.

Why does this work? Every point on arc A is the same distance from X as it is from... no — more precisely, A was drawn using the same radius from both X and Y, so A is equidistant from X and Y (AX = AY). The same is true for B (BX = BY). It turns out that any point equidistant from X and Y must lie on the perpendicular bisector of XY — which is exactly why joining any two such points (like A and B) always gives that perpendicular bisector, regardless of which radius was chosen for the arcs.

This fact can be proved using congruent triangles. In triangles ABX and ABY: AX = AY (both equal the first radius), BX = BY (both equal the second radius), and AB is a side shared by both triangles. By the SSS congruence criterion (all three corresponding sides equal), triangle ABX is congruent to triangle ABY. This congruence can then be used to show that the point where AB crosses XY is exactly the midpoint of XY, and that AB meets XY at a right angle — confirming AB really is the perpendicular bisector.

To construct a 90-degree angle at a point O on a line, first extend the line on both sides of O, then use the compass to mark two points X and Y on the line at equal distances from O (so O becomes the midpoint of segment XY). Now construct the perpendicular bisector of XY exactly as before — since O already lies on segment XY as its midpoint, and the perpendicular bisector of XY must pass through that midpoint, the resulting perpendicular bisector automatically passes through O and meets the line at exactly 90°.

To bisect any given angle ∠XOY exactly in half: mark two points A and B on the two arms of the angle such that OA = OB (one compass setting from O works for both). Then, keeping the compass open at any convenient radius, draw an arc from A and another arc of the same radius from B, so that the two arcs cross at a point C. The ray OC is the angle bisector — it splits ∠XOY into two exactly equal angles, ∠AOC and ∠BOC.

This works because triangles OAC and OBC are congruent: OA = OB (by construction), AC = BC (both equal the second radius used), and OC is a shared side — matching all three sides, so by SSS congruence, triangle OAC ≅ triangle OBC, which means ∠AOC = ∠BOC exactly.

Bisecting repeatedly gives a whole family of angles from just one starting construction. Starting from 90°, bisecting once gives 45° (since 90 ÷ 2 = 45), and bisecting again gives 22.5° (since 45 ÷ 2 = 22.5). Starting instead from 60°, bisecting once gives 30°, and bisecting again gives 15°.

To construct a 60-degree angle at a point A on a line, draw an arc of any radius centred at A, meeting the line at a point B. Keeping the same radius, draw another arc centred at B that crosses the first arc at a point C. Then angle CAB is exactly 60°. This works because AB, AC and BC are all equal (all three were drawn using the same radius), making triangle ABC equilateral — and every angle in an equilateral triangle measures exactly 60°, since the three equal angles must add up to 180° and 180 ÷ 3 = 60.

This single 60° construction unlocks a regular hexagon. A full angle around a point measures 360°, and since 6 × 60° = 360° exactly, six equilateral triangles can be arranged around one shared centre point with absolutely no gap and no overlap. The outer boundary of these six triangles together forms a regular hexagon, because every outer side has the same length (each triangle's side) and every outer angle is 60° + 60° = 120° (formed by two adjacent triangles meeting at each outer vertex).

This matches the general formula for a regular polygon's interior angle: for a polygon with n sides, each interior angle measures (n − 2) × 180° ÷ n. For a hexagon, n = 6, giving (6 − 2) × 180° ÷ 6 = 4 × 180° ÷ 6 = 720° ÷ 6 = 120° — exactly the angle produced by combining two 60° triangle-angles, confirming the hexagon construction is correct.

To copy a given angle ∠A onto a new ray starting at a point X: draw an arc centred at A, crossing both arms of the angle, to get two points — call the distances along the arms B and C (so AB and AC are both the same known radius). Then, without changing that radius, draw an arc of the same radius centred at X on the new ray, marking a point Z on it. Open the compass to the length BC (measuring the distance between the two original arm-points), and from Z, draw an arc of that length crossing the first new arc — mark that crossing point Y. The angle at X between the ray through Z and the ray through Y is now an exact copy of the original angle A, because triangle ABC and the new triangle XZY are congruent by the SSS criterion (all three corresponding sides equal by construction), which forces the angles at A and at X to be equal too.

This angle-copying trick is exactly the tool needed to construct a line parallel to a given line m. Draw any line l that crosses m — line l becomes a transversal. Pick a point B on l, away from where l meets m, and copy the angle that l makes with m at that new point B, on the same side, so the new angle at B matches the original angle exactly (this is the corresponding-angles idea). The new line drawn through B at that matching angle is parallel to m, because when a transversal cuts two lines and the corresponding angles are equal, the two lines must be parallel.

Tiling is covering a flat region with copies of one or more shapes so that there are no gaps between them and no overlapping. A square tiles a flat surface perfectly on its own, and so does an equilateral triangle, and so does a regular hexagon — because in each case, a whole number of copies of that shape's interior angle divides evenly into 360° at every point where corners meet (4 × 90° = 360° for squares, 6 × 60° = 360° for triangles, 3 × 120° = 360° for hexagons). A regular pentagon, by contrast, cannot tile a flat surface alone: its interior angle is (5 − 2) × 180° ÷ 5 = 108°, and 360° ÷ 108° is not a whole number, so there's no way to fit a whole number of pentagon-corners around a single point with no gap and no overlap.

A related question: can a rectangular grid be exactly covered using 2×1 domino tiles? A 4×6 grid has 4 × 6 = 24 unit squares — an even number — and it can indeed be tiled, for instance by covering each column with vertical dominoes (since the number of rows, 4, is even). But a 5×7 grid has 5 × 7 = 35 unit squares — an odd number — and it can never be tiled with 2×1 dominoes, because every single domino covers exactly 2 unit squares, so any number of dominoes together must cover an even total. Since 35 is odd, no combination of dominoes can ever add up to covering it exactly.

A clever trick for trickier regions is to colour the grid like a checkerboard (alternating two colours) and count each colour. A tiled region must always have an equal number of each colour, since every domino covers exactly one square of each colour. A 5×3 grid has 15 unit squares total; checkerboard-colouring it gives 8 squares of one colour and 7 of the other. If you remove one square of the majority colour (one of the 8), the remaining region has 7 of each colour — balanced, and potentially tileable. But if you remove one square of the minority colour (one of the 7) instead, the remaining region has 8 of one colour and 6 of the other — an imbalance of 2, which can never be fixed by dominoes, so that region can never be tiled.