CSAT Solved Papers/ 2022/Q56

2022 CSAT — Q56

Quant Counting & combinatorics 2.5 marks Medium

There are eight equidistant points on a circle. How many right-angled triangles can be drawn using these points as vertices and taking the diameter as one side of the triangle?

A 24 Answer
B 16
C 12
D 8

Worked rationale

By Thales’ theorem, an inscribed angle on a diameter is a right angle. So every right triangle with the diameter as a side has its right-angle vertex on the circle.

With $8$ equidistant points, opposite points pair into diameters: $8/2 = 4$ diameters.

For each diameter, the third vertex (the right angle) can be any of the remaining $8 - 2 = 6$ points:

4 \times 6 = 24.

Answer: (a) 24.

Visual solution

The same solve, worked by hand — read it, then trace it.

Hand-drawn worked solution for UPSC 2022 CSAT Q56 — Counting & combinatorics — Tap the drawing to open it full size for the fine detail.

Why the other options miss

B

miscounted the apex choices: uses $4 \times 4$ , undercounting the available third vertices.
C

only one semicircle used: takes just $3$ third vertices per diameter (one semicircle), halving the count.
D

fixed one apex per diameter: counts one triangle per point or per diameter, ignoring the free choice of the right-angle apex.

Specialist insight

The two-step count: diameters first ( $8$ points $\to 4$ diameters), then the right-angle apex ( $6$ remaining points each). Thales guarantees every such apex yields a genuine right angle, so no over- or under-counting — $4 \times 6 = 24$ . The trap is forgetting that points on both semicircles serve as valid apexes (all $6$ , not $3$ ).

The trap, in one line

$4$ diameters $\times\ 6$ apex choices (Thales' right angle) $= 24 \Rightarrow$ (a).

← All 2022 CSAT questions