Eight equidistant points lie on a circle. Using these points as vertices, right angled triangles are drawn such that one side of each triangle is diameter of the circle. The number of such possible right angled triangles is :

Correct — option (4), 24.

Step 1 — Count the diameters.
8 equally spaced points: each point has one point exactly opposite it.
Pairs of opposite points: 8 ÷ 2 = 4 diameters.

Step 2 — Apply Thales' theorem.
An angle inscribed in a semicircle is a right angle.
So any third point on the circle, joined to both ends of a diameter, gives a right angle at that point.

Step 3 — Count the third points for one diameter.
A diameter uses 2 of the 8 points: 8 − 2 = 6 points remain, and each gives a right triangle.

Step 4 — Multiply.
4 diameters × 6 third points = 24.

Check for double counting: a right triangle inscribed in a circle has exactly one side that is a diameter (its hypotenuse), so no triangle is counted under two diameters.

The idea to remember: a triangle inscribed in a circle is right-angled exactly when one of its sides is a diameter.

Why the others are wrong

• (1)
  8 — 8 would mean only 2 right triangles per diameter (4 × 2). Each diameter has 6 remaining points, and Thales' theorem gives a right angle at every one of them.

  8 is in fact the number of isosceles right triangles among the 24: for each diameter, the two points midway round each semicircle give 45°–45°–90° triangles, and 4 × 2 = 8.

• (2)
  16 — 16 would mean only 4 right triangles per diameter (4 × 4), leaving out 2 of the 6 remaining points.

  16 is the number of right triangles that are not isosceles: for each diameter, the four points next to its ends give 22.5°–67.5°–90° triangles, and 4 × 4 = 16. Adding the 8 isosceles ones gives 24.

• (3)
  20 — 20 would need 5 right triangles per diameter (4 × 5), but each diameter has 6 remaining points, all usable.

  20 is not any complete count here: the right triangles split into 8 isosceles and 16 non-isosceles, a total of 24. For reference, 20 is C(6, 3), the number of triangles from 6 points, which is a different question.

Thales' theorem says an angle inscribed in a semicircle is a right angle. Its converse also holds: if a triangle inscribed in a circle is right-angled, its hypotenuse is a diameter.

So right triangles on a circle can be counted by counting diameters. With n equally spaced points, n even, there are n ÷ 2 diameters, and each one pairs with the n − 2 other points. The count is (n ÷ 2) × (n − 2).

With n odd, no point is exactly opposite another, so no diameter joins two of the points and no right triangle can be formed from them.

Key facts

• Thales' theorem: an angle inscribed in a semicircle is a right angle.
• Converse: a right triangle inscribed in a circle has its hypotenuse as a diameter.
• n equally spaced points on a circle (n even) give n ÷ 2 diameters; each pairs with n − 2 third points.
• Right triangles from 8 equally spaced points: 4 × 6 = 24, of which 8 are isosceles and 16 are not.
• Of the C(8, 3) = 56 triangles from 8 such points, 24 are right-angled and 32 are not.

Common traps

• Count diameters as pairs: 8 points give 4 diameters, not 8. Counting each end as its own diameter doubles the answer to 48.
• Every remaining point works, including the ones next to a diameter's ends. Those triangles are thin, but still right-angled.
• The angle is right at the third point, not at the ends of the diameter. The diameter is always the hypotenuse.

UnlockIAS will link similar questions from RAS Pre 2023 and 2021 here once those papers are published on this site.

• practice — not a real PYQ

  Twelve equally spaced points lie on a circle. How many right-angled triangles can be drawn using these points as vertices?

  • (a)30
  • (b)60
  • (c)66
  • (d)120
  Answer(2) — 12 ÷ 2 = 6 diameters; each leaves 12 − 2 = 10 third points; 6 × 10 = 60. Option (1), 30, uses only 5 third points per diameter; option (3), 66, is C(12, 2), the number of chords; option (4), 120, counts each diameter twice (12 × 10).
• practice — not a real PYQ

  Six equally spaced points lie on a circle. How many triangles with vertices among these points are NOT right-angled?

  • (a)8
  • (b)12
  • (c)14
  • (d)20
  Answer(1) — All triangles: C(6, 3) = 20. Right triangles: 3 diameters × 4 third points = 12. Not right-angled: 20 − 12 = 8 (the 2 equilateral triangles plus the 6 made of three neighbouring points). Option (2), 12, is the right-angled count; option (4), 20, is the total; option (3), 14, matches neither.