Consider 4 points A, B, C, D in the plane with no three collinear. Answer the following questions. Some answers may require you to consider different cases, depending on how the points are positioned in the plane.
- (i) How many different quadrilaterals do they form if the quadrilateral is allowed to be self-intersecting or non-convex?
- (ii) How many of these quadrilaterals are self-intersecting? How many are convex?
Step-by-step solution
Idea: A quadrilateral is decided by which pairs of points are joined as sides; the remaining two pairs are its diagonals. The 3 ways to split 4 points into two pairs give 3 quadrilaterals. Then look at whether one point is inside the triangle of the others.
(i) How many different quadrilaterals do they form if the quadrilateral is allowed to be self-intersecting or non-convex?
- There are 24 orders of A, B, C, D, but each quadrilateral has 8 names (4 starting points × 2 directions). 24 ÷ 8 = 3.½ mark
- Equivalently, a quadrilateral is fixed by its pair of diagonals: AC & BD (quadrilateral ABCD), AD & BC (ABDC), or AB & CD (ACBD). So there are 3 quadrilaterals.½ mark
(ii) How many of these quadrilaterals are self-intersecting? How many are convex?
- Case 1: no point inside the triangle of the other three (the four points are corners of a convex shape). Joining them round the boundary gives one convex quadrilateral. In each of the other two, a pair of opposite “sides” are actually the crossing diagonals of that convex shape, so both are self-intersecting: 1 convex, 2 self-intersecting.1½ marks
- Case 2: one point (say D) inside triangle ABC. Segments from D stay inside the triangle and cannot cross a side of the triangle, and no two of the triangle’s sides cross; so no quadrilateral is self-intersecting. In each, D is inside the triangle of the other three, so the angle at D is reflex: all three are non-convex: 0 convex, 0 self-intersecting.1½ marks
Check: Checked by computer on 20 000 random sets of four points: every set gave either {convex, self-intersecting, self-intersecting} (about 70%) or {non-convex, non-convex, non-convex} (about 30%); no other pattern occurs.
Answer to write in the exam
(i)
Number of orders = 4! = 24; each quadrilateral has 8 names
24 ÷ 8 = 3
∴ 3 quadrilaterals: ABCD, ABDC, ACBD
(ii)
Case 1: points in convex position ⇒ 1 convex (boundary order) + 2 self-intersecting (diagonals used as sides)
Case 2: D inside ∆ABC ⇒ no crossings; D inside triangle of the others ⇒ reflex ∠D in all three
∴ Case 1: 2 self-intersecting, 1 convex; Case 2: 0 self-intersecting, 0 convex
Common mistakes that cost marks
- Answering 24 (all orders) or 6. Different names of the same quadrilateral must not be counted twice.
- Giving one answer for (ii) without separating the two cases.
- Thinking Case 2 can give a convex quadrilateral. A point inside the triangle of the other three always makes a dent.
How this can come in the exam
Points P(0, 0), Q(4, 0), R(4, 4), S(0, 4) are given. How many of the quadrilaterals formed by these four points are self-intersecting?
- 0
- 1
- 2
- 3
Show answer
(C) 2
The points are corners of a square (convex position): PQRS is convex and the other two (PQSR, PRQS) are bow-ties.
Try one yourself
Points A(0, 0), B(6, 0), C(3, 6), D(3, 2). Name the three quadrilaterals and classify them.
Show answer
D is inside ∆ABC, so ABCD, ABDC and ADBC are all non-convex; none is convex or self-intersecting.
More questions like this
- Suppose P is a point on side AB of ∆ABC and the line through P parallel to BC meets AC in point Q. For parts (i) to (iii), assume AP = 1, AQ = √5, and see the figure.
- (i) Suppose ABCD is a parallelogram and M, N are midpoints of AB and CD respectively. Show that segments DM and BN trisect segment AC.
(ii) Use part (i) to find a procedure to trisect any given segment PQ. Find two other ways to trisect PQ, one using the fact that a line parallel to one side of a triangle divides the other two sides in the same ratio and a third using the Centroid Theorem. - Is the Midpoint Theorem for Quadrilaterals true when the quadrilateral is non-convex? How about when it is self-intersecting? Experiment and check. The pictures may look strange, but the theorem appears to be true.
- Let P, Q, R, S be four points on sides AB, BC, CD and DA respectively of a quadrilateral ABCD. Suppose PQRS is a parallelogram. Must P, Q, R and S be midpoints of the respective sides? This can be considered a possible converse question to the Midpoint Theorem for Quadrilaterals.
- (i) Complete the following proof of the Midpoint Theorem. Extend segment PQ beyond Q until point S. By how much should we extend PQ? It would be good to be able to prove that ∆APQ ≅ ∆CSQ. (Why?) Use this as a guide to specify the location of S and then complete this proof.
(ii) Give a similar proof of the converse of the Midpoint Theorem. Start by extending PQ up to a suitable point S.
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