A Tale of Three Intersecting Lines · Lesson 4 of 10
Triangle Inequality
“Explain exactly when three lengths make a triangle and how to find the possible third side.”
• Apply the longest-side check to positive lengths in a common unit. • Explain the separate, touching, and intersecting circle cases. • Connect the strict inequality to a successful triangle construction. • Find all possible lengths of a third side when two are given. • Distinguish a very narrow triangle from an impossible straight arrangement.
A Complete Test for Three Lengths
The shortest-path reasoning explains why triangles cannot have a side as long as, or longer than, the other two together. To complete our test, we must also explain the successful case. If the sum of the shorter sides exceeds the longest, the compass circles meet at points away from the base and produce a genuine third vertex.
Each of three positive side lengths is less than the sum of the other two. Equivalently, the longest is less than the sum of the two shorter lengths.
Choose the longest length as AB. Draw circles from A and B with the two smaller lengths as their radii. Their sizes are no greater than the distance AB between their centres, so neither circle can enclose the other. The relevant possibilities are separation, a single touching point on AB, or two intersections away from AB.
Visualising the Construction of Circles
To see the difference, keep a base of 8 cm. Circles of radii 3 cm and 4 cm cannot bridge it: there is a 1 cm gap between the nearest points. Circles of radii 4 cm and 4 cm only touch. Radii 4 cm and 5 cm reach past each other, so the circles cross at two points. A crossing point gives a vertex that is not on the base line.
| Comparison | What the circles do | Does a triangle exist? |
|---|---|---|
| Shorter + shorter < longest | Remain separate | No |
| Shorter + shorter = longest | Touch once on the base line | No: the vertices are collinear |
| Shorter + shorter > longest | Intersect at two points away from the base line | Yes |
For the 4, 5, 8 case, start at A and move 4 cm along AB to a point X. Since AB is 8 cm, BX is also 4 cm. X lies inside B’s circle of radius 5 cm. The point on A’s circle on the opposite side of A is 12 cm from B, so it lies outside B’s circle. The circles therefore cross as the first circle passes from inside to outside the second. Either crossing point constructs the required triangle.
This completes both directions of the rule. If the inequality fails, a triangle does not exist. If it holds for three positive lengths, a triangle exists. In the successful case, the third vertex can be on either side of the base.
The test uses “less than”, not “less than or equal to”. Three lengths 3, 6, 9 fail because 3 + 6 = 9. A very thin triangle can still be genuine, but a completely straight arrangement is not one.
Problem
Which of 2, 2, 5; 3, 4, 6; and 5, 5, 8 can be side lengths?
- 1.For 2, 2, 5: the longest is 5 and 2 + 2 = 4. Since 5 > 4, no triangle exists.
- 2.For 3, 4, 6: the longest is 6 and 3 + 4 = 7. Since 6 < 7, a triangle exists.
- 3.For 5, 5, 8: the longest is 8 and 5 + 5 = 10. Since 8 < 10, a triangle exists.
Problem
Compare the sets 1, 100, 100 and 3, 6, 9.
- 1.For 1, 100, 100, the longest is 100 and the sum of the other two is 101. Since 100 < 101, the triangle exists, despite the large difference among its side lengths.
- 2.For 3, 6, 9, the longest equals 3 + 6. The circles touch on the base line, so there is no triangle.
- 3.It is the strict inequality, not how balanced the lengths look, that decides existence.
Finding the Third Side
Suppose two sides are 3 cm and 7 cm, and the unknown length is x cm. First, x must be smaller than 3 + 7 = 10. But that is not the only restriction: the 7 cm side must also be shorter than 3 + x. Thus x must be greater than 4. The remaining comparison, 3 < 7 + x, automatically holds for any positive x.
The same reasoning works for any two positive lengths. The third length must be greater than their difference and smaller than their sum. If the two lengths are equal, their difference is zero; the third side still has to be positive. Whole-number lengths are not required: suitable decimal lengths work too.
Problem
Find the range of the third side when the other sides are 1 cm and 100 cm.
- 1.The third length x must be less than the sum: x < 101.
- 2.The 100 cm side must be less than 1 + x, so x > 99.
- 3.Therefore 99 < x < 101. Values 99.4, 99.5, 100, 100.5, and 100.7 cm work. Exactly 99 cm and exactly 101 cm do not.
Problem
Two sides are each 5 cm. What third lengths are possible?
- 1.The third length x must be positive.
- 2.The sum of the known sides is 10 cm, so x must be less than 10 cm.
- 3.All lengths satisfying 0 < x < 10 work, including 1, 3, 4, 4.9, and 5 cm. The special value 5 cm gives an equilateral triangle.
For sides s, s, s with s > 0, the longest-side check is s < s + s, which is true. Therefore an equilateral triangle exists for every positive side length, including 50 units.
A useful procedure is now available: express all lengths in one unit, identify the longest, add the other two, and check that their sum is strictly larger. If so, construct with the longest as the base and the other lengths as radii. If not, explain whether the circles remain separate or merely touch.
Quiz
Which set forms a triangle?
What happens when the two smaller lengths add to the longest?
If two sides are 3 cm and 7 cm, which can be the third side?
Which range gives every possible third side x for known sides 5 cm and 5 cm?
Why do 1, 100, and 100 form a triangle?
Practice Problems
- Decide whether 2, 4, 8; 10, 20, 25; 10, 20, 35; and 24, 26, 28 form triangles. Give the decisive comparison.
- Check 1, 1, 5 and 5, 10, 12, explaining what the compass circles would do with the longest side as base.
- For known sides 1 and 100, give five decimal third lengths and explain why 99 and 101 are excluded.
- For known sides 3 and 7, describe all possible third lengths and give five examples.
- Create three sets where the circles touch and three sets where they remain separate.
- Justify why an equilateral triangle of side 50 units exists and why the same argument works for every positive side length.
- Two known sides are 4 cm and 9 cm. Find the complete range of the third side and test x = 5, 5.2, 12, and 13 cm.
Key Takeaways
• Three positive lengths form a triangle exactly when the longest is smaller than the sum of the other two. • Separate circles mean the two smaller lengths do not reach across the base. • Touching circles give collinear vertices and no triangle. • Two off-line circle intersections provide a valid third vertex. • A third side must be strictly between the difference and the sum of the two known sides.