Is there a 4-gon with given side lengths?
- (i) Recall the following fact about triangles and check it by construction. For given positive numbers a, b, c, is there a triangle whose sides have these lengths? The answer is Yes exactly when the sum of any two numbers is greater than the third. If we arrange the numbers in increasing order (suppose a ≤ b ≤ c), then this amounts to requiring a + b > c. (Hint: Start by drawing a segment of length c.)
- (ii) Suppose a 4-gon has 2, 5, 11 as three side lengths. Can the length of the fourth side be 100? Can it be 10? Can it be 1? What are the possible lengths of the fourth side?
- (iii) For given positive numbers a, b, c, d, how will you decide if there is a 4-gon whose sides have these lengths?
Answer: (i) Draw AB = c, then arcs of radius b from A and a from B: they meet (giving a triangle) exactly when a + b > c (with a ≤ b ≤ c). (ii) 100: no; 10: yes; 1: no. Possible fourth sides: 4 < x < 18. (iii) A 4-gon exists exactly when the longest side is less than the sum of the other three.
Step-by-step solution
Idea: The straight segment is the shortest path between two points. So any one side of a polygon is shorter than the path made by the other sides, and conversely, if this holds, two triangles can be built on a suitable diagonal.
(i) Recall the following fact about triangles and check it by construction. For given positive numbers a, b, c, is there a triangle whose sides have these lengths? The answer is Yes exactly when the sum of any two numbers is greater than the third. If we arrange the numbers in increasing order (suppose a ≤ b ≤ c), then this amounts to requiring a + b > c. (Hint: Start by drawing a segment of length c.)
- Draw AB = c. Draw a circle of radius b about A and of radius a about B. A third vertex C must lie on both circles. If a + b > c, the circles cross (above and below AB), giving the triangle. If a + b = c they only touch on AB (no triangle), and if a + b < c they do not meet. Example: 3, 4, 6 works; 2, 3, 6 does not. With a ≤ b ≤ c the other two conditions hold automatically.1 mark
Yes exactly when a + b > c (for a ≤ b ≤ c).
(ii) Suppose a 4-gon has 2, 5, 11 as three side lengths. Can the length of the fourth side be 100? Can it be 10? Can it be 1? What are the possible lengths of the fourth side?
- In a 4-gon each side is shorter than the path along the other three sides (a straight segment is the shortest path).½ mark
- 100: 100 < 2 + 5 + 11 = 18 is false: no. 10: longest side 11 < 2 + 5 + 10 = 17: yes. 1: 11 < 2 + 5 + 1 = 8 is false: no.1 mark
- If x ≥ 11 we need x < 18; if x ≤ 11 we need 11 < 7 + x, i.e. x > 4. So the possible lengths are 4 < x < 18.½ mark
100: no; 10: yes; 1: no. Possible fourth side: any length x with 4 < x < 18.
(iii) For given positive numbers a, b, c, d, how will you decide if there is a 4-gon whose sides have these lengths?
- Test: let d be the largest. A 4-gon exists exactly when d < a + b + c.½ mark
- Needed: side d joins two vertices that are also joined by the path of the other three sides, which is longer than the straight side.½ mark
- Enough: choose a diagonal length e with d − a < e < b + c and e > |b − c| (possible because d − a < b + c, and |b − c| < d + a since d is largest), and also e < a + d. Then triangles with sides a, d, e and b, c, e both exist (by (i)); placed on opposite sides of a common segment of length e they form a 4-gon with sides a, b, c, d.1 mark
Check that the longest of a, b, c, d is less than the sum of the other three.
(i) A triangle exists exactly when the two shorter sides add to more than the longest. (ii) The fourth side x must satisfy 4 < x < 18: 100 no, 10 yes, 1 no. (iii) A 4-gon exists exactly when the longest side is less than the sum of the other three.
Check: (ii) with x = 10, build it as in (iii): d = 11, a = 2, b = 5, c = 10 needs 9 < e < 13. Take e = 10: triangles with sides 2, 11, 10 (2 + 10 > 11 ✓) and 5, 10, 10 (5 + 10 > 10 ✓) both exist, and together they form a 4-gon with sides 2, 5, 10, 11 ✓.
Answer to write in the exam
(i)
Draw AB = c; arcs radius b from A and radius a from B
Arcs meet off AB ⇔ a + b > c
∴ Triangle exists ⇔ a + b > c (a ≤ b ≤ c)
(ii)
Each side < sum of other three
x = 100: 100 < 18 false ⇒ no; x = 10: 11 < 17 ⇒ yes; x = 1: 11 < 8 false ⇒ no
x ≥ 11: x < 18; x ≤ 11: 11 < 7 + x ⇒ x > 4
∴ 4 < x < 18
(iii)
Let d be the largest
Necessary: d < a + b + c (straight side shorter than the path of the other three)
Sufficient: pick diagonal e with triangles (a, d, e) and (b, c, e) both possible; join them along e
∴ A 4-gon exists ⇔ largest side < sum of the other three
Common mistakes that cost marks
- Checking only one side, e.g. 2 + 5 + 11 > 10, and forgetting that the longest side (11) must also be less than the sum of the others.
- Writing 4 ≤ x ≤ 18: at x = 4 or 18 the figure is flat (degenerate), not a 4-gon.
- In (iii), using the triangle test on three of the four sides; the 4-gon test uses all four.
How this can come in the exam
MCQ (1 mark)
Three sides of a quadrilateral are 3 cm, 4 cm and 9 cm. The fourth side could be
- 1 cm
- 2 cm
- 8 cm
- 17 cm
Show answer
(C) 8 cm
Need 9 < 3 + 4 + x ⇒ x > 2, and x < 3 + 4 + 9 = 16. Only 8 cm fits.
Try one yourself
Is there a quadrilateral with sides 1, 2, 3 and 7?
Show answer
No: the longest side 7 is not less than 1 + 2 + 3 = 6.
More questions like this
- Counting diagonals of a polygon.
- Sum of angles of a polygon. What is the sum of angles of a (planar non-self-intersecting) n-gon? We know that the answer is 180° for n = 3 and 360° for n = 4. Find the next few values. Then find a formula in terms of n and prove it.
- Multiple converses to a theorem. Let us see how multiple statements can be considered converses to the Midpoint Theorem and how the converse of the Midpoint Theorem (the line through the midpoint of one side parallel to another side bisects the third side) is one of them. To formulate a converse we should express the original statement in “If … then …” form. For a complex statement, there may be multiple ways to do that. The Midpoint Theorem starts with ∆ABC and points P and Q on sides AB and AC respectively. The theorem has two assumptions and two conclusions, which we have named for further discussion.
Assumptions: (P MID) P is the midpoint of AB, and (Q MID) Q is the midpoint of AC.
Conclusions: (PRLL) PQ ‖ BC, and (HALF) PQ = BC2.
Let us use these four named conditions to discuss various possible statements. - 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.
- What fraction of the square is shaded?
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