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This problem is taken from MATHEMATICS CIRCLE book

Can anyone tell me how the figure of this problem look like Pls elaborate with simplest explanation

Naman Vyas
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1 Answers1

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Well, the answer is no, since for every intersection point you would complete the requirement for exactly 2 of the segments, which are, however, 9, an odd number.

George K
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  • Can't understand Pls elaborate as simple as you can – Naman Vyas May 14 '18 at 06:37
  • Each time a segment crosses another, well, that's it for them - they can't cross anything ever again, since each segment crosses exactly one other. So, for each crossing point, two segments go "out of action" - they are no more included in the possible "crossing mates" of other segments. Then the segments are obviously partitioned to couples that cross, each with its own crossing point. So, the segments must be even. Here, however, they are nine, thus odd. So the answer to this question is negative. – George K May 14 '18 at 15:03
  • Excellent explaination sir – Naman Vyas May 15 '18 at 19:01
  • Then you should think about accepting it. – George K May 15 '18 at 19:28