Tiling the plane with quadrilaterals · 5 marks

Validity of tiling methods. Show using reasoning that each of the three tiling methods we saw produces a tiling of the plane using the given 4-gon. You have to prove that when we place new copies using any of the three procedures, the entire plane is covered with no gaps and no overlaps. First, think carefully about what it means to prove this. Then find a proof for each procedure.

Answer: What to prove: every point of the plane lies in some copy (no gaps), and no two copies share inside points (no overlaps). Method 1: SOME and its half-turn about the midpoint of OM form a hexagon whose opposite sides are equal and parallel; such a hexagon tiles by slides (along the diagonals SM and OE) row by row, exactly like a parallelogram, and its copies are the copies placed by Method 1. Method 2: the Varignon cells form a parallelogram grid that covers the plane; each placed copy covers its own cell plus four corner triangles, and around every grid vertex-cell the four corner triangles of the four surrounding copies fill that cell exactly. Method 3: it places the same slid copies as Method 1 (slides along the diagonals), with the half-turned copies in the gaps, so it gives the same tiling.

Step-by-step solution

Idea: Turn the question into a known fact: parallelograms (and hexagons with opposite sides equal and parallel) tile the plane by slides. Then show each method produces exactly such a pattern.

What it means. A tiling needs two things: (a) no gaps: every point of the plane is inside or on the edge of some copy; (b) no overlaps: no point is inside two copies. Checking corners (angles add to 360°) is necessary but not enough on its own; we must also see that the pattern spreads over the whole plane consistently.1 mark
Method 1 (half-turns). Let H be SOME together with its half-turn about the midpoint of OM. H is a hexagon with vertices S, O, O + M − E, O + M − S, M, E, and its opposite sides are equal and parallel (vectors SO and (O + M − S)M are opposite, and so on). Slide H along vector SM to get a row with no gaps or overlaps (opposite sides match); slide the row along vector EO to stack rows the same way. This covers the plane without overlap. Two half-turns about edge midpoints equal one of these slides, so these are exactly the copies Method 1 places.1½ marks
Method 2 (Varignon grid). Method 2 places exactly the Method 1 copies: the copies on the shaded cells are SOME slid along SM and OE, and the gap copies are its half-turns. So it is valid for the same reason as Method 1. (Picture check: each copy covers its own Varignon cell plus four corner triangles. The cells of placed copies and of gap copies alternate diagonally. Each remaining cell receives four corner triangles, two from placed copies and two from gap copies, one from each corner S, O, M, E, and together they fill it exactly: their total area is ½ ar(SOME), the area of one cell.)1½ marks
Method 3 (slides along diagonals). It places copies of SOME slid by whole-number combinations of the diagonal vectors SM and OE, and the gaps are filled by half-turned copies. These are exactly the copies of Method 1 (two half-turns make such a slide), so Method 3 gives the same tiling and is valid for the same reason.1 mark

Each method is valid: Method 1 and Method 3 place SOME and its half-turn so that their union (a hexagon with opposite sides equal and parallel) repeats by slides along the diagonals, and Method 2 covers each cell of the Varignon parallelogram grid exactly once.

Check: Area check for Method 1: the hexagon H has area 2 × ar(SOME), and the slide parallelogram spanned by SM and OE has area |SM × OE| = 2 × ar(SOME) as well (the area of a quadrilateral is half the product of its diagonals times the sine of the angle between them). Equal areas are consistent with one hexagon per slide-cell, no gaps and no overlaps.

Answer to write in the exam

To prove: (a) no gaps, every point is in some copy; (b) no overlaps, no point inside two copies

Method 1: SOME + half-turn = hexagon with opposite sides equal and parallel ⇒ tiles by slides along SM and OE (rows, then stacked rows); these are the half-turn copies

Method 2: copies on shaded cells = slides of SOME along SM, OE; gap copies = half-turns ⇒ same copies as Method 1

Method 3: same copies as Method 1 (slides along the diagonals + half-turned copies in gaps)

∴ All three methods tile the plane with no gaps and no overlaps.

Common mistakes that cost marks

Checking only that angles add to 360° at one corner and calling it a proof.
Forgetting the “no overlap” half of what must be proved.
Assuming the three methods are unrelated; Methods 1 and 3 give the same set of copies.

How this can come in the exam

Short answer (2 marks)

A hexagon has its opposite sides equal and parallel. Explain why copies of it slid along one direction fit together in a row with no gaps or overlaps.

Show answer

Sliding by the vector that carries one side onto its opposite side makes the copy share that whole side with the original (1 mark). Since the two sides are equal and parallel, they coincide exactly, so neighbouring copies meet along a full side with no gap or overlap, and repeating gives a row (1 mark).

Try one yourself

In Method 1, what single movement is the same as a half-turn about the midpoint of OM followed by a half-turn about the midpoint of SO?