Tiling the plane with quadrilaterals · 4 marks

Let us see a third method to tile the plane using a 4-gon. Focus on only two coloured copies of SOME sharing a vertex. What do you see? It appears that each 4-gon is just a shifted copy of the other. Let us see exactly how. Draw two copies of SOME as shown in the figure so that the diagonals EO and EʹOʹ are collinear with O = Eʹ. Now place a cutout of one copy on top of SOME with diagonal EO drawn on it. Slide this cutout so that segment EO moves along EOʹ until EO matches EʹOʹ. Verify that your cutout exactly matches SʹOʹMʹEʹ.

(i) Prove the exact match of SOME with SʹOʹMʹEʹ using four parallelograms.
(ii) Follow the described procedure along each diagonal of the starting coloured 4-gon in both directions to get 4 new 4-gons. Using these 4 copies, repeat the procedure 4 more times to place 4 new copies, resulting in a 3 by 3 grid of 9 copies exactly like the 9 coloured copies. These 9 copies meet each other only at vertices and once again the blank spaces among them also form 4-gons congruent to SOME!

Answer: (i) S′O′M′E′ is a copy of SOME with diagonal E′O′ on the line EO, same length and same direction, and S′, M′ on the same sides. So ∠SEO = ∠S′E′O′ (corresponding angles) gives ES ‖ E′S′, and ES = E′S′: ESS′E′ is a parallelogram. In the same way EMM′E′, OSS′O′, OMM′O′ are parallelograms. So SS′, MM′, EE′ and OO′ are all equal and parallel to EO: every vertex moves by the same slide, and the cutout matches S′O′M′E′ exactly. (ii) Sliding along both diagonals both ways gives a 3 × 3 grid of copies that meet only at corners; each blank space has sides equal to SO, OM, ME, ES and angles 1, 2, 3, 4, so it is a copy of SOME turned through 180°.

Step-by-step solution

Idea: A slide moves every point the same distance in the same direction. Showing that the four segments joining corresponding vertices are equal and parallel (four parallelograms) shows the second copy is exactly a slide of the first.

(i) Prove the exact match of SOME with SʹOʹMʹEʹ using four parallelograms.

S′O′M′E′ ≅ SOME with E′O′ = EO on the same line, in the same direction, and E′ = O. So the slide vector is EE′ = EO = OO′ (= E′O′).½ mark
∠S′E′O′ = ∠SEO (corresponding parts of congruent copies), and these are corresponding angles on the line EO′, so E′S′ ‖ ES; also E′S′ = ES. So ESS′E′ is a parallelogram (one pair of opposite sides equal and parallel): SS′ = EE′ and SS′ ‖ EE′.1 mark
In the same way: EMM′E′ is a parallelogram (MM′ = EE′, MM′ ‖ EE′); OSS′O′ and OMM′O′ are parallelograms (OO′ = EO = SS′ = MM′). So all four vertices move by the same distance in the same direction as EO: the slid cutout lands exactly on S′O′M′E′.1 mark

ESS′E′, EMM′E′, OSS′O′ and OMM′O′ are parallelograms, so each vertex of SOME moves by the same vector EO onto the matching vertex of S′O′M′E′.

(ii) Follow the described procedure along each diagonal of the starting coloured 4-gon in both directions to get 4 new 4-gons. Using these 4 copies, repeat the procedure 4 more times to place 4 new copies, resulting in a 3 by 3 grid of 9 copies exactly like the 9 coloured copies. These 9 copies meet each other only at vertices and once again the blank spaces among them also form 4-gons congruent to SOME!

The 9 coloured copies are the 9 green copies of SOME in this picture: the darker one in the middle is SOME, and each of the others is a slid copy of it along the two diagonals SM and EO. The 4 brown shapes are the blank spaces between them.
Slide SOME along diagonal EO (forwards and backwards) and along diagonal SM (forwards and backwards): 4 new copies, each touching SOME at one vertex (O, E, M or S). Sliding these again along the other diagonal fills the corners of a 3 × 3 grid of 9 copies, all slid copies of SOME.½ mark
Each blank space is surrounded by four copies, one touching it along each side; its sides equal SO, OM, ME, ES and its corners receive the angles 1, 2, 3, 4. So each blank space is a copy of SOME turned through 180°, and filling them and continuing tiles the plane (this is the same tiling as the half-turn method).1 mark

Slides along the two diagonals give a 3 × 3 grid of copies meeting only at corners; the gaps are half-turned copies of SOME.

(i) Four parallelograms (ESS′E′, EMM′E′, OSS′O′, OMM′O′) show every vertex moves by the same vector EO, so the slid cutout matches S′O′M′E′. (ii) Repeating slides along both diagonals makes a grid of copies meeting at vertices, with half-turned copies of SOME filling the gaps.

Answer to write in the exam

(i)

E′O′ = EO on line EO′, E′ = O ⇒ EE′ = OO′ = EO

∠S′E′O′ = ∠SEO ⇒ E′S′ ‖ ES (corresponding angles); E′S′ = ES ⇒ ESS′E′ parallelogram ⇒ SS′ = EE′, SS′ ‖ EE′

Similarly EMM′E′, OSS′O′, OMM′O′ are parallelograms ⇒ MM′ = OO′ = EE′, all ‖ EO

∴ Sliding SOME by EO carries S, O, M, E onto S′, O′, M′, E′: exact match

(ii)

Slide SOME along EO and SM, both ways ⇒ 4 copies, each meeting SOME at one vertex

Repeat from these ⇒ 3 × 3 grid of 9 slid copies meeting only at vertices

Each gap: sides SO, OM, ME, ES from four neighbours, angles 1, 2, 3, 4 ⇒ copy of SOME turned 180°

∴ Filling the gaps and repeating tiles the plane

Common mistakes that cost marks

Checking only one vertex. All four corresponding vertices must move by the same vector.
Saying ESS′E′ is a parallelogram because SS′ ‖ EE′ “looks” true; it follows from E′S′ ‖ ES and E′S′ = ES.
Sliding along a side instead of a diagonal; then the copies overlap.

How this can come in the exam

Short answer (2 marks)

Quadrilateral WXYZ is slid without turning so that W moves to W′ with WW′ = 5 cm. Show that XX′ = 5 cm and XX′ ‖ WW′.

Show answer

After the slide W′X′ = WX and W′X′ ‖ WX (same direction) (1 mark). So WXX′W′ has a pair of opposite sides equal and parallel: it is a parallelogram, giving XX′ = WW′ = 5 cm and XX′ ‖ WW′ (1 mark).

Try one yourself

SOME is slid once along EO to S′O′M′E′, and then once more by the same vector to S″O″M″E″. Which vertices of S′O′M′E′ and S″O″M″E″ meet?