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Origamic (Posted on 2003-09-23) Difficulty: 3 of 5
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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A Slight Improvement on My Last Posting | Comment 9 of 11 |
This is not an easy problem to solve or to discuss an answer lucidly. I am making minor text modifications to my last posting to hopefully improve clarity. With rectangle ABCD where AC and BD are diagonals bring corner A into coincidence with C, creating a crease from X on line AB to Y on line DC. AX must equal CX since the length being folded is constant. It can also be proved that AX = AY = CY by virtue of being similar triangles. Then, diagonal AC must bisect and be perpendicular to XY at a point we can lable Z. Since triangles AZX and ABC are similar we know XZ/BC = AZ/AB. Since XZ = 1/2 x (XY), we know that XY/2 = (AZ)(BC)/(AB) so XY = 2(AZ)(BC)/(AB). But 2AZ is equal to the diagonal, BC equal to the rectangle width and AB equal to the rectangle length. Therefore the length of the line is the square root of (Width squared + Length squared)x Width over Length. For a standard 8 by 11 inch paper this crease would be approx 9.2 inches. I feel better now. Gordon S.
  Posted by Gordon Steel on 2003-10-14 21:35:24
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