Pentagon in a Square???Sounds Strange!!!

Probability Level 5

Aaron takes a square sheet of paper with one corner labelled as A A . Point P P is chosen at random inside the square and Aaron folds the paper so that A A and P P coincide. He cuts the sheet along the crease and discards the piece containing A A . Let p p be the probability that the remaining piece is a pentagon. Find the integer nearest to 100 p 100p .


The answer is 57.

Yash Singhal by Yash Singhal , 22, India

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2 solutions

EDIT : point A is the bottom right vertex of square \text{EDIT : point A is the bottom right vertex of square}

It is not very hard to think that locus of limiting position of the required point P \text{P} is indeed the boundry of shaded region , with the knowledge of this fact the problem becomes almost trivial .

We want all points that lie inside \color{#20A900}{\text{inside}} of the boundry of the shaded region ,

So required probability \text{required probability} = area of shaded region total area of square \dfrac{\text{area of shaded region}}{\text{total area of square}}

Take the side length 1 1 {as the side length doesn't effect probability} and on simple calculations you will find that the area of shaded region is π 2 − 1 \dfrac{\pi}{2} -1 which would be numerically equal to the probability as area of square equals 1 1 .

Thus required probability is 0.570796 \boxed{0.570796}

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Michael Mendrin
Jun 12, 2015

To buttress Siddarth's solution, check this graphic out to see how those arcs are indeed circular

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Thanks sir, I was wondering if there was any software/app that you used to draw that , would you mind telling me?

Siddharth Bhatnagar - 6 years ago

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An old mathematica software. Still works.

Michael Mendrin - 6 years ago

You can also use geogebra, the software is free and can easily be downloaded.

Will Wombell - 5 years, 9 months ago

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