Andrei marked \(4n\) points on the plane. He then connected with segments every pair of points which were \(1\) cm apart. He found that amid every \(n+1\) points there were at least two points connected by segments. Prove that Andrei drew at least \(7n\) segments.

## Comments

Sort by:

TopNewestWhat exactly does this statement mean:

"He found that amid every \(n+1\) points there were exactly two points connected by segments" – Raghav Vaidyanathan · 1 year, 5 months ago

Log in to reply

– Sudeep Salgia · 1 year, 5 months ago

I think the word exactly should be replaced by atleast.Log in to reply

@Raghav Vaidyanathan @Sudeep Salgia – Andrei Golovanov · 1 year, 5 months ago

Yes, that is correct. It should say "at least". Thanks for pointing it out.Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

– Vishnu C · 1 year, 4 months ago

But you just said that it is not possible with n=1.Log in to reply

It's like "If \(a, b \) are 2 positive reals such that \( a^2 + b^2 = 0 \), then \( a+b = 0\)" is a true statement. Because under all cases where the condition holds (which is no case), the result is true. Note that even though \( a+b = 0 \) can never be a true statement under the restricted condition that (a,b) are 2 positive reals. However, by adding additional conditions, we can make it a statement about nothing, which is trivially true. – Calvin Lin Staff · 1 year, 4 months ago

Log in to reply

– Andrei Golovanov · 1 year, 4 months ago

That is, more or less, what the question asks. If this condition holds, he must have drawn \(7n\) segments. Prove why.Log in to reply

– Vishnu C · 1 year, 4 months ago

So you're basically saying that the condition will hold when pigs fly.Log in to reply

– Vishnu C · 1 year, 4 months ago

That's an amusing way to think. But wait! Swine flu! So pigs flew! So the condition is true! :)Log in to reply