Waste less time on Facebook — follow Brilliant.
×

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 \(P_{1}\), \(P_{2}\), \(P_{3}\) \(\ldots\) \(P_{10}\) 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

Note by Surya Prakash
2 years ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

Divide 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 - 2 years ago

Log in to reply

Awesome solution!

Adarsh Kumar - 2 years ago

Log in to reply

Nice solution!! (+1)

Surya Prakash - 2 years ago

Log in to reply

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

Sharky Kesa - 2 years ago

Log in to reply

ni shuo sha?

Jingyang Tan - 2 years ago

Log in to reply

Nándào nǐ bù zhīdào yīngyǔ

Surya Prakash - 2 years ago

Log in to reply

What? English, please.

Sharky Kesa - 2 years ago

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 - 2 years ago

Log in to reply

Extremal Principle!!

Surya Prakash - 2 years ago

Log in to reply

Actually I solved it in the paper with PHP..

Sualeh Asif - 2 years ago

Log in to reply

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

Sharky Kesa - 2 years ago

Log in to reply

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

Sualeh Asif - 2 years ago

Log in to reply

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 - 2 years ago

Log in to reply

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

Brilliant Member - 2 years ago

Log in to reply

Easy PHP!

Sualeh Asif - 2 years ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...