This is the second proof problem I am posting. Try these problems and post your working. I will eventually post my solution to all of these notes, named "Olympiad Proof Problems", three days after I post them. So here is the problem.

Consider any ten distinct points \(P_{1}\), \(P_{2}\), \(P_{3}\) \(\ldots\) \(P_{10}\) in the square \(\mathcal{S}\) of unit side length. Prove that one can find a pair of points among those ten such that their distance is atmost \(\dfrac{\sqrt{2}}{3}\) units.

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TopNewestDivide the unit square into \(9\) small squares of side length \(\frac {1}{3}\) . \(10\) points are to be chosen within the unit square which means that there will be at least one small square with more than \(1\) point. The maximum distance between any two points in such a square is when they are at diagonally opposite points of that square. By the Pythagorean theorem this distance is \(\frac{\sqrt2}{3}\) .

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Awesome solution!

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Nice solution!! (+1)

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It's just PHP followed by Pythagoras Theorem. Great solution BTW. +1

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ni shuo sha?

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Nándào nǐ bù zhīdào yīngyǔ

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What? English, please.

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Here is a variant of the above question:

Consider a equilateral triangular dartboard of side 2ft. If you throw five darts with no misses, then show that at least two will be within one feet of each other.

## Pakistan First Round 2015

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Extremal Principle!!

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Actually I solved it in the paper with PHP..

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Once again, PHP followed by some simple maths.

Connect the medians of the triangle to create for equilateral triangles of side length 1ft. Now, at least 2 darts go into on of these triangles by PHP. The furthest distance 2 points can be separated in an equilateral triangle is the 2 vertices (pick any other 2 points and they are closer). This length is the side length, which is 1ft. Therefore, at least 2 darts are within 1ft of each other.

(Note: PHP stands for Pigeonhole Principle (or as I like to call it 'Penguin-hole Principle')

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I like to call it Pig-on-hole Principle.

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Easy PHP!

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