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CAT 2017 Slot 1 — QA Question 29

Geometry based questionsEasy

Let AB, CD, EF, GH, and JK be five diameters of a circle with center at O. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and O so as to form a triangle?

Answer & solution

Correct answer: 160

Answer: 160

Solution

Easy

There are 1111 points: 1010 on the circle (five diametrically-opposite pairs) plus the centre OO. Count all 33-point selections, then subtract the collinear ones. A triple is collinear only when it is the two ends of a diameter together with OO.

O
1

Total selections of 33 from 1111.

(113)=11⋅10⋅96=165\begin{aligned} &\binom{11}{3} = \frac{11\cdot 10\cdot 9}{6} = 165 \end{aligned}
2

Subtract the collinear (degenerate) triples. Three points are collinear only when they are the two endpoints of one diameter plus OO. There are exactly 55 diameters, hence 55 such bad triples.

collinear triples=5(one per diameter, with O)\begin{aligned} &\text{collinear triples} = 5 \quad\text{(one per diameter, with }O\text{)} \end{aligned}
3

Triangles formed.

165−5=160\begin{aligned} &165 - 5 = 160 \end{aligned}

Split by whether OO is chosen. Without OO: (103)=120\binom{10}{3}=120 (no 33 circle-points are collinear). With OO: pick 22 of the 1010 in (102)=45\binom{10}{2}=45 ways, minus the 55 diameter pairs that line up with OO, giving 4040. Total 120+40=160120+40=160.

160 triangles160 \text{ triangles}

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CAT 2017 Slot 1 QA Q29: Let AB, CD, EF, GH, and JK be five diameters of a circle with center at O. In how many ways can three points b — Solution | TheCATExam