किसी वृत्त की परिधि पर समान दूरी पर 24 बिन्दुएँ हैं। इनमें से शीर्ष के रूप में तीन बिन्दुओं के समुच्चयों से अधिकतम कितने समबाहु त्रिभुज बन सकते हैं?
Read this question in English
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 · (d) 12
उत्तर और व्याख्या · explanation in English
Answer: (c) 8
An equilateral triangle inscribed in a circle has its vertices 120° apart — every 8th point of 24. Each point belongs to exactly one such triangle, so there are 24 ÷ 3 = 8.
- Adjacent points are 360° ÷ 24 = 15° apart.
- An inscribed equilateral triangle needs vertices 120° apart: 120 ÷ 15 = 8 steps.
- Starting at points 1 to 8 gives different triangles (1-9-17, 2-10-18, …, 8-16-24); starting at 9 repeats the first.
- Number of triangles = 24 ÷ 3 = 8.
Remember · On n equally spaced points, inscribed regular k-gons exist only if k divides n, and there are n/k of them.
Question and answer: UPSC's official GS Paper II (2018, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·