Consider 8 vertices of a regular octagon and its centre. If T denotes the number of triangles and S denotes the number of straight lines that can be formed with these 9 points then T − S has the value equal to

1. Counting triangles (T)

1. Start naïvely: Pick any 3 of the 9 points.

   $\binom{9}{3} = 84$

2. Remove the degenerate ones (three collinear points):

   • In a regular octagon no three vertices alone are collinear.
   • But the centre and any pair of opposite vertices lie on the same straight line (a diameter of the circumcircle).
   • There are 4 such diameters because an octagon has 4 distinct pairs of opposite vertices.

   Hence the number of forbidden (collinear) triplets is 4.

   $T = 84 - 4 = 80$

2. Counting distinct straight lines (S)

1. Start naïvely: Any 2 points give a line.

   $\binom{9}{2} = 36$

2. Adjust for over-counting:

   • Along each diameter we have 3 collinear points (centre + opposite pair). The 3 pairs on that line were all counted separately above, but they represent just one line.
   • Each diameter therefore contributes
     $\binom{3}{2} = 3\text{ pairs}$ while it should contribute only 1 line, so we over-counted by 2.
   • With 4 diameters, total over-count = $4 \times 2 = 8$.

   $S = 36 - 8 = 28$

3. Final difference

$T - S = 80 - 28 = 52$

What’s the puzzle?

We have 9 dots – eight are the corners of a regular stop-sign shape (an octagon) and one is the dot right in its centre.

We want to know

1. How many different triangles we can draw using any three of those dots.
2. How many different straight lines we can draw using any two of those dots.

After we get those two counts, we simply do (number of triangles) − (number of straight lines).

The trick to watch out for

If three dots are already arranged in one straight line, they cannot make a triangle (it would be flat). Also, when two different pairs of dots sit on the same infinite straight line, we must count that line only once.

So the whole game is spotting those special cases!

Step-by-step Solution

1. Total triangles from 9 points $\binom{9}{3}=84$

2. Subtract collinear triplets

   • 4 diameters ⇒ 4 collinear triplets. $T = 84 - 4 = 80$

3. Total pairs (candidate lines) $\binom{9}{2}=36$

4. Correct duplicate lines

   • Each diameter: 3 pairs ⇒ counted 3 times, should be 1 ⇒ over-count = 2.
   • 4 diameters ⇒ $4 \times 2 = 8$ duplicates. $S = 36 - 8 = 28$

5. Required difference $T - S = 80 - 28 = 52$

Final Answer: $52$