Tiling the plane with quadrilaterals · 3 marks

Note two interesting things about the second step: (1) each new copy can be obtained by rotating any one of its neighbours, and both ways give the same result. (2) The new copies fit perfectly. Can you explain these facts by reasoning? This is needed to prove that the method works!

Answer: (1) Two half-turns about different points make a slide by twice the distance between the points. Reaching a corner copy through either neighbour gives the same slide (by the diagonal of SOME from the start tile’s vertex to the corner), so both ways give the same copy. (2) That copy brings the missing angle to the corner, so the four angles there add to 360°, and its two edges at the corner coincide with the neighbours’ edges, so it fits perfectly.

Step-by-step solution

Idea: Use the rule “half-turn about P, then half-turn about Q = slide by 2 × (Q − P)”. Work out the corner copy at M both ways.

Take the corner M of the starting tile T = SOME. Its step-1 neighbours there are U (T turned about the midpoint of OM) and V (T turned about the midpoint of ME). U has angle 2 at M, V has angle 4 at M, T has angle 3; angle 1 is missing.½ mark
Turning U through 180° about the midpoint of its free edge at M, (M + (O − S)/2), gives T slid by 2[(M + (O − S)/2) − (O + M)/2] = M − S. Turning V about the midpoint of its free edge at M, (M + (E − S)/2), gives T slid by 2[(M + (E − S)/2) − (M + E)/2] = M − S. The same slide, so both ways give the same copy: fact (1).1½ marks
This copy is T slid by SM, so its vertex S lands on M, bringing angle 1 there. Angles 1 + 2 + 3 + 4 = 360° fill the corner, and each of its edges at M is the edge shared with U or with V (it was made by a half-turn about that edge’s midpoint). So it fits perfectly: fact (2). The same reasoning works at every corner.1 mark

Both routes are the same slide of the starting tile (along a diagonal), so they give the same copy, and that copy supplies the one missing angle at the corner while sharing its edges with both neighbours, so it fits perfectly.

Check: With S(0, 3), O(4, 3.6), M(5, 0.6), E(−0.6, 0): M − S = (5, −2.4). Via U: 2[(5 + 2, 0.6 + 0.3) − (4.5, 2.1)] = 2(2.5, −1.2) = (5, −2.4) ✓. Via V: 2[(5 − 0.3, 0.6 − 1.5) − (2.2, 0.3)] = 2(2.5, −1.2) = (5, −2.4) ✓.

Answer to write in the exam

Half-turn about P then about Q = slide by 2(Q − P)

Via U: slide = 2[(M + ½(O − S)) − ½(O + M)] = M − S

Via V: slide = 2[(M + ½(E − S)) − ½(M + E)] = M − S ⇒ same copy (fact 1)

New copy = T slid by SM ⇒ angle 1 at M; 1 + 2 + 3 + 4 = 360°, edges shared with U and V

∴ The new copy fits perfectly (fact 2).

Common mistakes that cost marks

Explaining only with the picture. The question asks for a reason why the two routes must agree.
Thinking two half-turns give another half-turn. Two half-turns about different points give a slide.
Forgetting to check the angle sum at the corner where the new copy goes.

How this can come in the exam

Short answer (2 marks)

Point X(3, 0) is turned through 180° about A(1, 1), and the image is turned through 180° about B(2, 4). Find the final position and the slide that takes X there.

Show answer

First image: 2A − X = (2 − 3, 2 − 0) = (−1, 2) (1 mark). Then 2B − (−1, 2) = (4 + 1, 8 − 2) = (5, 6). The slide is (5, 6) − (3, 0) = (2, 6) = 2(B − A) (1 mark).

Try one yourself

In the Method 1 tiling of SOME, which copies are just slides of SOME (not turned)?