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.

Try more proof problems from this set Olympiad Proof Problems

## Comments

Sort by:

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 \(\frac{\sqrt2}{3}\) . – Milind Prabhu · 10 months, 2 weeks ago

Log in to reply

– Adarsh Kumar · 10 months, 2 weeks ago

Awesome solution!Log in to reply

– Surya Prakash · 10 months, 2 weeks ago

Nice solution!! (+1)Log in to reply

– Sharky Kesa · 10 months, 2 weeks ago

It's just PHP followed by Pythagoras Theorem. Great solution BTW. +1Log in to reply

ni shuo sha? – Jingyang Tan · 10 months, 2 weeks ago

Log in to reply

– Surya Prakash · 10 months, 2 weeks ago

Nándào nǐ bù zhīdào yīngyǔLog in to reply

– Sharky Kesa · 10 months, 2 weeks ago

What? English, please.Log in to reply

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 · 10 months, 2 weeks agoLog in to reply

– Surya Prakash · 10 months, 2 weeks ago

Extremal Principle!!Log in to reply

– Sualeh Asif · 10 months, 2 weeks ago

Actually I solved it in the paper with PHP..Log in to reply

– Sharky Kesa · 10 months, 2 weeks ago

I solved it without paper using Penguin-hole Principle.Log in to reply

– Sualeh Asif · 10 months, 2 weeks ago

WOW Sharky(-the genius) you win!Log in to reply

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 · 10 months, 2 weeks ago

Log in to reply

– Svatejas Shivakumar · 10 months, 2 weeks ago

I like to call it Pig-on-hole Principle.Log in to reply

– Sualeh Asif · 10 months, 2 weeks ago

Easy PHP!Log in to reply