There are 24 equally spaced points lying on the circumference of a circle. What is the maximum number of equilateral triangles that can be drawn by taking sets of three points as the vertices?

(a) 4
(b) 6
(c) 8 ✓ UPSC's answer
(d) 12

Why the answer is (c)

• The 24 points are equally spaced, so the central angle between adjacent points is 360°/24 = 15°.

• An equilateral triangle inscribed in a circle must have its vertices separated by arcs of 120° each (since 360°/3 = 120°).

• In terms of point intervals, 120° corresponds to 120°/15° = 8 intervals between vertices.

• To form such a triangle, we select a starting point, then the second point is 8 steps away, and the third is 16 steps away (or 8 steps from the second).

• Since the pattern repeats every 8 steps, there are exactly 8 distinct starting positions (0, 8, 16, 24... but 24 is 0, so effectively 8 unique sets based on the first vertex modulo 8).

• Specifically, the vertices for the k-th triangle are at positions k, k+8, and k+16 for k = 0 to 7. This yields 8 distinct equilateral triangles.

Why the other options are wrong

(a) 4: Option (a) is incorrect because it underestimates the number of distinct starting positions; there are 8, not 4, valid starting points for the 120° separation pattern.
(b) 6: Option (b) is incorrect because it does not account for all 8 possible starting vertices that generate unique equilateral triangles with 120° arc separations.
(d) 12: Option (d) is incorrect because it overcounts; while there are 24 points, the symmetry of the 120° spacing means only 8 distinct triangles can be formed, not 12.

Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2018, held on 3 June 2018. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.