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22 Feb 2021, 06:15
Kudos
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A
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Difficulty:
95%
(hard)
Question Stats:
33% (02:07) correct
67%
(01:54)
wrong
based on 49
sessions
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Let there be 9 fixed point on the circumference of a circle. Each of these points is joined to every one of the remaining 8 points by a straight line and the points are so positioned on the circumference that at most 2 straight lines meet in any interior point of the circle. The number of such interior intersection points is:
(A) 126
(B) 351
(C) 756
(D) 775
(E) 810
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
(A) 126
(B) 351
(C) 756
(D) 775
(E) 810
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
22 Feb 2021, 06:34
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Bunuel
If we take three points on circumference and join each with each other point there there will be no point of intersection inside teh circle
i.e. for one intersection inside the circle there must be 4 points (Intersection will be the intersection of diagonals of quadrilateral formed by those 4 points)
i.e. Total intersections = Total group of 4 points
Total group of 4 points out of 9 points = \(^9C_4 = 126\)
Answer: Option A
22 Feb 2021, 07:16
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Bunuel
We want to find how many times 2 of the lines will intersect in the circle.
In every set of 4 points, ie. In a quadrilateral, the lines formed by the 4 points will intersect only once - the 2 diagonals intersect once.
Thus, if we can determine the number of quadrilaterals inside the polygon, we will have the answer
= 9C4 = 126
Note that there won't be any overlap of any 2 such quadrilaterals; i.e. the 126 points are all distinct.
Answer A
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