Origami Probability

Geometry Level 4

A square sheet of paper has its center marked with point $O$ . A random point is uniformly chosen on the square paper, and this point is labeled $P$ . The paper is then folded so that point $P$ coincides with point $O$ . Let the probability that a pentagon is formed after the fold be $N$ . Find the value of $\lfloor 1000N \rfloor$ .

The answer is 429.

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Michael Mendrin
May 11, 2014

Let the sides of the square be 2. The distance from the center to any corner is √2. If P, located in any quadrant, is further than √2 from all the other corners, then the paper folded is a pentagon. Hence, the probability is 2-(π/2) = 0.429204... That is, probability = [AQCR] /' [ABCD]. If P were to fall anywhere on the circular arcs QC or RC, then the fold will pass through a corner.

Pentagon

Hmm... A little too brief for comprehensive understanding of your solution... I may post one later if nobody else posts a more detailed solution.

EDIT: I guess the above solution can pass as clear enough now.

Daniel Liu - 7 years, 1 month ago

Please post now!

Arya Samanta - 7 years, 1 month ago

Thank you , Liu

Michael Mendrin - 7 years, 1 month ago

why P is to be located further than √2, it seems within √2 can do the same. Please make this point clear Michael.

Amarnath Chowdhury - 7 years ago

Not sure what is your question, but first divide the square into quadrants, so that we only need to check one quadrant, for reasons of symmetry. Then if P falls within ACQR, the fold passes through top and left sides of the square, so that it's still a 5-sided figure. Otherwise, if P falls outside of ACQR, the fold passes through the right and left sides of the square, or through the top and bottom sides of the square, so that it becomes a 4-sided figure.

Michael Mendrin - 7 years ago

shouldn't it be 4/5

libprince libprince - 7 years ago

Origami Probability

The answer is 429.

1 solution

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