Olympiad Proof Problem - Day 2

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 P1P_{1}, P2P_{2}, P3P_{3} \ldots P10P_{10} in the square S\mathcal{S} of unit side length. Prove that one can find a pair of points among those ten such that their distance is atmost 23\dfrac{\sqrt{2}}{3} units.


Try more proof problems from this set Olympiad Proof Problems

Note by Surya Prakash
4 years ago

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Divide the unit square into 99 small squares of side length 13\frac {1}{3} . 1010 points are to be chosen within the unit square which means that there will be at least one small square with more than 11 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 23\frac{\sqrt2}{3} .

milind prabhu - 4 years ago

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Nice solution!! (+1)

Surya Prakash - 4 years ago

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

Sharky Kesa - 4 years ago

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Awesome solution!

Adarsh Kumar - 4 years ago

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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

Sualeh Asif - 4 years ago

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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')

Sharky Kesa - 4 years ago

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

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Easy PHP!

Sualeh Asif - 4 years ago

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Extremal Principle!!

Surya Prakash - 4 years ago

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Actually I solved it in the paper with PHP..

Sualeh Asif - 4 years ago

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@Sualeh Asif I solved it without paper using Penguin-hole Principle.

Sharky Kesa - 4 years ago

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@Sharky Kesa WOW Sharky(-the genius) you win!

Sualeh Asif - 4 years ago

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ni shuo sha?

Jingyang Tan - 4 years ago

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What? English, please.

Sharky Kesa - 4 years ago

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Nándào nǐ bù zhīdào yīngyǔ

Surya Prakash - 4 years ago

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