A Tale of Three Intersecting Lines · Lesson 3 of 10
Are Triangles Possible for any Lengths?
“Discover why the three lengths must satisfy a shortest-path relationship before a triangle can exist.”
• Recognise that some sets of three positive lengths cannot form a triangle. • Compare a direct path with a path through a third non-collinear point. • Use all three side comparisons to reject impossible triangles. • Explain why changing the order of impossible side lengths does not help. • Identify the longest-side comparison as the decisive check.
When a Construction Fails
Try to construct a triangle with side lengths 3 cm, 4 cm, and 8 cm. Draw the 8 cm base and make arcs of radii 3 cm and 4 cm from its endpoints. Even carefully drawn arcs cannot reach each other. This failure is caused by the lengths themselves, not by a poor choice of pencil position.
The same difficulty occurs for lengths 2 cm, 3 cm, and 6 cm. It is tempting to try another side as the base, but a different arrangement cannot remove a contradiction among the three distances. To decide whether a triangle is possible, we need to understand a relationship that every genuine triangle must obey.
Triangle Inequality
Imagine a tent, a tree, and a pole at three points that are not all on one line. To go from the tent to the tree, you can take the direct straight path, or travel to the pole and then to the tree. The direct path is shorter. The detour has a bend, so it cannot be the shortest connection between the same endpoints.
Apply this idea to a triangle. Between A and B, the direct side AB must be shorter than AC + CB. Between B and C, the direct side BC must be shorter than BA + AC. Between C and A, the direct side CA must be shorter than CB + BA. Each side is tested against the sum of the other two.
These strict comparisons are called the triangle inequality. “Strict” means that equality is excluded. If the third point lies along the direct route, the two pieces can add to exactly the direct distance, but then the three points are on one straight line and do not enclose a triangle.
Problem
Could a triangle have AB = 15 cm, BC = 10 cm, and CA = 30 cm?
- 1.For AB, compare 15 with 10 + 30 = 40. The required 15 < 40 is true.
- 2.For BC, compare 10 with 15 + 30 = 45. The required 10 < 45 is true.
- 3.For CA, compare 30 with 10 + 15 = 25. The required 30 < 25 is false. A direct 30 cm route cannot be longer than a 25 cm detour between the same endpoints.
- 4.The triangle cannot exist, even though two comparisons worked.
Two successful comparisons are not enough. A rough sketch can show a triangle-like shape with impossible labels, but that does not make the measurements possible. Trust the distance relationships rather than the appearance of an unscaled sketch.
Problem
Decide whether 3 cm, 4 cm, and 8 cm can be side lengths.
- 1.The longest given length is 8 cm. The sum of the other two is 3 + 4 = 7 cm.
- 2.A side of 8 cm would need to be shorter than 7 cm, which is impossible.
- 3.No triangle can have these lengths. Moving the 8 cm label to another side still leaves the same failed comparison.
Which Comparison Should We Check First?
Put three positive lengths in increasing order: shortest, middle, longest. The shortest is automatically less than the middle plus the longest. The middle is also less than the longest plus the positive shortest. Therefore the only comparison that might fail is the longest against the sum of the other two.
For example, for 10, 15, and 30, the comparisons involving 10 and 15 must succeed because their right-hand sums already contain a length at least as large as themselves and another positive length. The comparison 30 < 10 + 15 is the useful one. This shortcut avoids unnecessary arithmetic without ignoring any requirement.
Problem
Check 5 mm, 10 mm, and 20 mm.
- 1.All three are in millimetres, so they can be compared directly.
- 2.The longest is 20 mm, and the sum of the shorter two is 5 + 10 = 15 mm.
- 3.Since 20 > 15, no triangle exists. The same reasoning applies to centimetres or kilometres when the lengths use a common unit.
Make three positive lengths of your own and order them. Explain why the shortest and middle comparisons always succeed. Then create a set in which the longest comparison fails. This turns a pattern seen in examples into a general explanation.
We now know that failing even one required comparison makes a triangle impossible. A further question remains: if the longest side is shorter than the other two together, can we be sure that the construction will succeed? The next lesson uses circles to answer that question.
Quiz
Which comparison must hold for side BC of a triangle?
Why can 10 cm, 15 cm, and 30 cm not form a triangle?
What does equality between one length and the sum of the other two describe?
For positive lengths 7, 10, and 15, which comparison is the decisive one?
Can rearranging side labels make the lengths 2, 3, and 6 form a triangle?
Practice Problems
- Attempt the 2, 3, 6 cm construction using 6 cm as the base, then explain the failure with a distance comparison.
- Use the longest-side check for lengths 10, 10, 25 km; 12, 20, 40 cm; and 3, 3, 7 cm.
- For 7, 10, and 15, write all three required comparisons. Explain why two were guaranteed before calculating.
- Create three different positive-length sets that cannot form triangles. State the failed comparison in each.
- A rough drawing is labelled with sides 5 cm, 7 cm, and 15 cm. Explain why a triangle-shaped sketch is not evidence that these lengths work.
- Compare a direct walk from home to a shop with a route through a third point. Explain when the detour is strictly longer and when the points would lie on one line.
Key Takeaways
• A triangle’s side must be shorter than the route along its other two sides. • Every side must be less than the sum of the other two. • One failed comparison rules out a triangle. • For positive lengths, checking the longest against the sum of the shorter two is sufficient to test all three comparisons. • Rearranging labels cannot repair an impossible set of lengths.