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 Origamic (Posted on 2003-09-23)
A rectangular sheet of paper is folded so that two diagonally opposite corners come together. The crease thus formed is as long as the longer side of the rectangle.

What is the ratio of the longer side of the rectangle to the shorter?

 See The Solution Submitted by DJ Rating: 4.4167 (12 votes)

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 No Subject | Comment 2 of 11 |
Let the rectangle's corners be A, B, C, and D. Assume the fold unites corners A and C. The fold creates a line EF. Let E be on AB and F be on CD. From the folding, AF and FC are congruent. Let the center of the rectangle be G. AC is perpendictular to EF and the lines intersect at G. G also bisects AC and EF.

Since AD is the shorter side, let AD=1 and AB=x (and EF=x). Let DF=a. Then FC=x-a (and AF=x-a), FG=x/2, and AG=(sqrt(1+x^2))/2.

ADF is a right triangle, then AD^2+DF^2=AF^2: 1 + a^2 = (x-a)^2. AFG is also a right triangle, then AF^2=FG^2+AG^2: (x-a)^2 = (1+x^2)/4 + x^2/4

Solving the pair of equations yeilds x=1.27202, hence the ratio of the sides is 1.27202
 Posted by Brian Smith on 2003-09-23 15:13:33

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