PrepShorts · Study sheet · Class 7 Mathematics · Chapter 7, A Tale of Three Intersecting Lines
Chapter 7 · A Tale of Three Intersecting Lines
The triangle inequality: which three lengths can close
This video could not be loaded. Reload the page to try again.
Sign in with Google10 min.
Keep your place in this chapter — sign in, it’s free.Sign in
Try to build a triangle with sides 3, 4 and 8 and the two arcs sweep past each other without ever meeting.
The idea
Two different arguments are needed here, and the chapter is unusually careful to keep them apart. The walking-to-the-tree picture can only ever say no: if one side beats the other two put together, the figure is impossible, and you know it without lifting a compass. It cannot say yes — passing that test leaves you still not knowing whether the arcs will cross. What settles the yes is a second, quite separate look at two circles, which turns out to have exactly three possible outcomes. Only when both halves are in place does a rule about three numbers become a statement about whether a shape exists.
What you should be able to do
- Decide, for a given triple of lengths, whether a triangle with those sidelengths exists, and justify the decision without constructing it
- Explain why a straight path between two points cannot be longer than a path that detours via a third
- Reconstruct the contradiction argument that rules out 10, 15 and 30
- Show that only one of the three comparisons can ever fail, and identify in advance which one that is
- State the triangle inequality in the chapter's own terms and apply it to a list of triples
- Explain why satisfying the inequality does not, on the path argument alone, prove that the triangle exists
- Relate each of the three circle cases to a comparison between the two smaller lengths and the largest
- Given two lengths, describe the whole range of third lengths that would close a triangle
Words to know
| Term | Definition in one line | First introduced |
|---|---|---|
| triangle inequality | the condition that each of three lengths falls short of the other two added together | printed in bold in §7.2, p.154, and again in SUMMARY, p.171 |
| sidelength | the length of a side, written by this book as one word | printed throughout §7.2, from p.148 |
| direct path | the straight-line route between two of the three points | printed on p.152 and used as a measured quantity on p.153 |
| roundabout path | the two-leg route between two points that goes via the third | printed on p.152 and used as a measured quantity on p.153 |
| rough diagram | a sketch drawn to record the stated lengths, not to scale | printed on p.152 |
| radius / radii | the fixed distance from a centre to its circle; plural radii | printed on p.155 and p.157 |
| circle | every point at one fixed distance from the centre | printed throughout §7.2, from p.149 |
| centre | the point a circle is drawn about | printed on p.149 and p.155 |
| base | the side drawn first; here always chosen to be the longest of the three | printed on p.155 and p.157 |
| equilateral triangle | a triangle whose three sides are of equal length | printed as the §7.1 heading, p.146 |
| degenerate case | three lengths where the two smaller ones total exactly the largest, so the "triangle" flattens onto a line | an added term; the phrase is not printed in this chapter, though the situation is drawn as Case 1 on p.157 |
One caution: The chapter never uses "≤" — but it is not symbol-free, and an explanation that avoids notation altogether is not following it. It writes its comparisons with <, > and = between named lengths: three of them for 10, 15 and 30 on p.154, one for 4, 5 and 8 on p.155, the bolded generalisation on p.156, and each of the three circle cases on pp.157–159 as a relation between the two radii added together and AB. What it never does is put algebraic letters in place of the lengths. Keep both halves: relational symbols on named lengths are the book's own notation; a lettered inequality, and "≤", are not.
Where people slip up
- "You have to check all three comparisons." You never do. Order the three lengths and test the largest against the other two; the other two comparisons cannot fail. The chapter walks the student to this on p.154 with a hint rather than announcing it.
- "If the inequality holds, the triangle obviously exists." This is the error the chapter goes out of its way to block on p.155. The path argument is a one-way test. The circle analysis on pp.157–159 is what supplies the other direction, and it is a genuinely different argument, not a restatement.
- "Equal is fine — 3, 6, 9 makes a very thin triangle." It makes no triangle. The two circles touch at exactly one point, which lies on AB itself, so the three points are in a line and there are no angles at all. This is the picture the chapter's opening question on p.146 was pointing at.
- "Fig. 7.4 is drawn wrong, so the argument is unfair." It is drawn wrong on purpose. You cannot draw an impossible triangle correctly; supposing it exists and following the consequences is the only way to reach the contradiction.
- "The rule is about the two shortest sides." It is about every length against the other two. The shortcut — check the longest — is a consequence, not the definition, and stating the shortcut as the rule leaves a student unable to say why it works.
- "Longer sides make a triangle more likely." Scale has nothing to do with it: 10, 15, 30 fails and 10, 15, 20 succeeds; multiply any working triple by a thousand and it still works. The condition is about proportion, not size.
Ask your teacher a person
Your teacher reads this and writes back, usually within a day. For an instant answer, use Ask the video in the sidebar.
Your class sees the question and the answer. Only your teacher sees that it was you.
No questions on this topic yet.
Worked answers to this chapter’s exercises · this video explains Figure it Out · 2 Q1, Figure it Out · 2 Q2, Figure it Out · 2 Q3, Figure it Out · 3 Q1, Figure it Out · 4 Q1, Figure it Out · 4 Q2, Figure it Out · 4 Q3
Transcript1,399 words
Try to build a triangle with sides three, four and eight. Base of eight, drawn with the ruler. Then an arc of three from one end, and an arc of four from the other. Watch them. They sweep out, and they do not meet. Three and four make seven. The base is eight. The arcs finish a whole centimetre short of each other. Try another. Two, three and six. Two and three make five, and the base is six. Short again.
Nothing about your drawing was wrong. There is no such triangle. So which three lengths can close, and which cannot? That is the whole question. Here is a picture that answers half of it. A tent. A tree some way off. And between them, a tall pole. You are at the tent, and you want to get to the tree. You could walk straight there. One leg. Or you could walk to the pole first, and then on to the tree. Two legs.
Which is shorter? And you know this one without measuring anything. The straight walk. Going via the pole cannot possibly save you distance. That is not a fact about tents. It is a fact about straight lines. So let us say it for any three points at all. Call them A, B and C. Anywhere you like. The straight path from A to B is never longer than the path from A to C and then C to B.
Never. Not for most arrangements — for all of them. I put twenty thousand random triples of points through this, and the straight path won every single time. There is one case where it ties. When C sits exactly on the line between A and B. Then going via C is walking the very same route, with a pause in the middle. Move C off that line by a hair and the detour gets longer. Off the line, the straight path strictly wins.
Now watch what that does to a set of lengths. Ten, fifteen and thirty. Suppose — just suppose — that a triangle with those three sides exists. I will even draw it. A B is fifteen. A C is thirty. B C is ten. And yes, this drawing is wrong. It has to be. You cannot draw an impossible triangle correctly. That is rather the point. It is a rough sketch, and all it does is record which length sits on which side.
So take the sketch at its word, and see what follows from it. Three journeys, three comparisons. B to C straight is ten. B to A to C is fifteen and thirty — forty-five. Ten against forty-five. Fine. The straight way is shorter, exactly as it should be. A to B straight is fifteen. A to C to B is thirty and ten — forty. Fifteen against forty. Fine again.
Now the third one. C to A straight is thirty. C to B to A is ten and fifteen. Twenty-five. Thirty against twenty-five. The straight path is five longer than the detour — and that is absurd. The sketch has destroyed itself. Hold on, you might say. Perhaps I put the lengths on the wrong sides. Fair enough. Let us try every arrangement. Three lengths on three sides. Six ways to do it.
Swap fifteen and thirty. Still one length beating the other two. Put ten opposite instead. The same trouble. All six. Every one of them fails, and every one fails in the same place. Because it is always thirty that does it. Ten and fifteen reach twenty-five between them, whichever sides they sit on. Relabelling moves the problem around the triangle. It does not remove it. So let us drop the walking and look at the numbers on their own.
For three lengths to be the sides of a triangle, each one has to fall short of the other two added together. Ten is less than fifteen and thirty. True. Fifteen is less than ten and thirty. True. Thirty is less than ten and fifteen. False. One false is enough. The whole set is out. Now try a set that works. Ten, fifteen and twenty. Ten under thirty-five. Fifteen under thirty. Twenty under twenty-five. All three true, and that set is fine.
Three comparisons, though, is two more than you need. Put the lengths in order. Smallest, middle, largest. Look at the smallest. We are asking whether it falls short of the other two added. But one of those two is already at least as big as it is — and then the other gets added on top. So the smallest can never fail. It never had a chance to. Same for the middle one. It is no bigger than the largest, and the smallest is added on top of that.
Only the largest is left. And the largest genuinely can fail — that is precisely what thirty did. So: sort the three, and make one comparison. Does the largest fall short of the other two added together? This rule has a name. It is called the triangle inequality. And it is worth being careful about what it actually says. It says every length falls short of the other two added together. All three of them.
The shortcut — check the largest — is a consequence of that. It is not the rule. Learn only the shortcut and you can apply it, but you cannot say why it works. Let us test a few. One, a hundred, and a hundred. Largest is a hundred. The other two add to a hundred and one. A hundred is less, so it works. Five, ten, twelve. Twelve against fifteen — works. One, one, five. Five against two — it does not.
Now the part that is easy to skip, and you should not. Four, five and eight. Largest is eight. Four and five add to nine. Eight is less than nine, so it passes. Does the triangle exist? And the honest answer, from everything we have done so far, is: I do not know. Look at what we actually proved. If one length beats the other two, there is no triangle. That is all it says.
It is a test that can only say no. Passing it means we have failed to rule the triangle out — not that we have ruled it in. Going straight from the test passing to the triangle existing skips an argument. So here is the argument. Draw the base. Eight centimetres, from A to B. Now the circle of radius four about A. Not an arc this time — the whole circle.
It cuts the base at a point. Call that point X. X is four from A, so X is four from B as well, because the base is eight. And now compare that four with the second radius, which is five. Four is less than five. So X lies inside the circle about B. But go round to the far side of A's circle. That point is eight and four from B. Twelve. Well outside.
So A's circle has a point inside B's circle and a point outside it. Travelling from inside to outside it has to cross — and by symmetry it crosses twice. Once above the base, once below. One case is not a proof. So let us watch every case at once. Keep the base fixed, and make it the longest of the three — otherwise one circle can swallow the other and there is a fourth picture to worry about.
Now let the two radii grow. Small ones first: the circles sit clear of each other. Nothing crosses, so nothing closes. Grow them. At one exact moment the circles touch, at a single point, and that point lies on the base itself. That is not a very thin triangle. It is no triangle at all — three points in a line, no angles, nothing enclosed. That moment is exactly where the two radii add up to the base. Three, six and nine sits precisely there.
A hair more, and the circles overlap. Two crossings, one above the base and one below, and a triangle at each. So it runs both ways now. Less than the base, and nothing. Equal, and a flat line. More, and the triangle exists. The test that could only say no can finally say yes.
Where this fits
Taken from the notes each video was made from, not from the reading order — these are the ideas this one rests on and the ones that later rest on it.
Builds on
- Why a compass beats trial and error for building a triangleClass 7 · Ch 7, A Tale of Three Intersecting Lines
- The four angles at a crossing: vertically opposite and linear pairsClass 7 · Ch 5, Parallel and Intersecting Lines
Comes up again in
- Constructing from two sides and the angle between themClass 7 · Ch 7, A Tale of Three Intersecting Lines