Points on a circle
Three points A, B and C lie on a circle. No three points on a circle are ever in a straight line, so every pair of them gives a different straight line, and every three give a triangle.
That turns counting lines and triangles into counting choices of points.
Three points on a circle: A, B and C.
A line is a pair of points
Choosing the two ends of a line in order gives 3 × 2 = 6 ordered pairs: AB, BA, AC, CA, BC and CB. But the line from A to B is the line from B to A, so each line has been counted twice.
So there are 6 ÷ 2 = 3 lines: AB, AC and BC. The order of the two points does not matter, so this is a combination: 3C2 = 3.
The three lines AB, BC and CA, one for each pair of points.
A triangle is three points
Add a fourth point, D. A triangle needs three corners, and the order they are named in does not matter, since ABC and CAB are the same triangle. So the triangles are the choices of 3 points from 4: .
They are ABC, ABD, ACD and BCD. Another way to see it: each triangle uses all the points but one, so naming the triangle is the same as naming the point it leaves out, and there are 4 points to leave out.
Triangle ABC, which leaves out D. Leaving out A, B or C instead gives the other three triangles.
Six points
With 6 points on a circle, the lines are the choices of 2 from 6: . The triangles are the choices of 3 from 6: .
The counts grow quickly. 10 points give 10C2 = 45 lines and 10C3 = 120 triangles.
Six points on a circle with every pair joined: 6C2 = 15 lines.
Diagonals of a polygon
Join the six points in turn around the circle and they make a hexagon. Of the 15 lines between pairs of corners, 6 are the sides of the hexagon, so 15 − 6 = 9 are diagonals.
For any polygon with n corners, the diagonals number nC2 − n. An octagon has 8C2 − 8 = 28 − 8 = 20 diagonals.
The 9 diagonals of the hexagon ABCDEF: the 15 lines between pairs of corners, without the 6 that join neighbors.
Four corners
Any four points on a circle are the corners of one quadrilateral, so 6 points give 6C4 = 6C2 = 15 quadrilaterals. Choosing the 4 corners is the same as choosing the 2 points left out.
When points are in a line
Points that are not on a circle can line up. If three points lie on one straight line, the 3C2 = 3 pairs among them all give the same line, so they count as 1 line, not 3. And the three of them make no triangle, so 1 must come off the count of triangles.
Count as if no three points were in a line, then correct for the points that are.
The usual mistakes
Counting ordered pairs. With 6 points, 6 × 5 = 30 counts every line twice, once from each end; there are 15 lines.
Joining only neighboring points. 6 points around a circle have 6 sides between neighbors, but every pair of points makes a line, 15 in all.
Leaving out the line that the points in a row lie on. Five points along a fence give 10 pairs that are all one line: take off 10 and count the fence back once.
Taking off one triangle for each point in a row. Every choice of three of those points fails to make a triangle, and five points in a row hold 5C3 = 10 such choices.
Posts in a park
In the application below, twelve posts give 12C2 lines and 12C3 triangles if no three are in a line. Five of the posts stand along a straight fence, and both counts are corrected for them.
Worked example: Ropes and Triangles Between Twelve Posts in a Park, Five of Them Along a Straight Fence
Question A park designer has put up 12 posts. Five of them stand along a straight fence, none of the other seven stands on the fence's line, and no three of the other posts, or any two of them with a fence post, lie in a straight line. A straight rope can be stretched through any two posts. (a) How many different straight lines can the ropes follow? (b) How many triangles have three of the posts as their corners?
1.If no three posts were in a line, every pair would give a different line: 122 = 12 × 112 = 66 lines.
If no three posts were in a line, each of the 122 = 66 pairs would give its own line. 2.The five fence posts give 52 = 5 × 42 = 10 pairs, but all ten lie on one line, the fence.
The 52 = 10 pairs of fence posts all give the same line, the fence. 3.(a) Take the 10 off and count the fence once: 66 − 10 + 1 = 57 different lines.
(a) 66 − 10 + 1 = 57 different lines. 4.Any three posts give a triangle unless they are in a line: 123 = 12 × 11 × 103 × 2 × 1 = 220 choices of three posts.
Three posts make a triangle unless they are in a line; there are 123 = 220 choices of three. 5.(b) The choices of three fence posts, 53 = 10, lie on the fence and make no triangle: 220 − 10 = 210 triangles.
(b) The 53 = 10 threes along the fence make no triangle: 220 − 10 = 210.
Answer: (a) 57 lines; (b) 210 triangles
Common mistakes
- Answering (a) with 66 − 10 = 56. The fence posts do give a line, the fence itself; it must be counted once, not removed.
- Taking off only 5 in (b), one for each fence post. It is every choice of three posts along the fence that fails to make a triangle, and there are 53 = 10 of those.