1) Prove that a (convex or concave) polygon having the perimeter of 2 can be placed inside a circle which has the diameter of 1 pefectly

Hint from teacher:using symmetry of point.

2) With each point X inside convex polygon:A(1)A(2)...A(2009),we set S(X)=XA(1)+XA(2)+...+XA(2009)

Do 2 point P,Q exist inside the polygon that they satisfy:

PQ=1 and |S(P)-S(Q)|=2007

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## Comments

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TopNewestIt is best to separate out your questions into 2 discussion posts, since they are very different questions.

For your second question, you will need to specify extra conditions. For example, if the polygon is too small, then we might not be able to find 2 points such that \(PQ = 1\), or allow for \( S(P) \geq 2007 \). I have a slight suspicion that the vertices of the polygon lie on the unit circle.

Here's another hint for question 1, with an approach that is very different from what you teacher suggests. This is the first approach that I thought of, which is influenced by how I approached similar problems in the past.

Hint:Kelly's Theorem.Log in to reply

Thank Sir.I have just examined Helly Theorem and surprisingly,i prove problem 1 immediately.The most beautiful theorem that I have ever seen :))

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Can you write up your solution so that others can learn from it? You have to be careful with some details.

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