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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 Solution (I think!) | Comment 5 of 13 |
Designate a rectangle as ABCD where AC and BD are the diagonals. Bring corner A into coincidence with corner C, creating a crease of XY where X lies on line AB and Y on line CD. AX must equal CX (since the length is a constant). We can also prove that XC = AX = CY = AY by virtue of being similar triangles. Thus diagonal AC is perpendicular to and bisects XY at a point we may label Z. Since triangles AZX and ABC are similar, we know that XZ/BC = AZ/AB. And, since XY = 2XZ, we can rewrite this as 2 x (AZ) x (BC)/AB. But this is the same as (W/L) x Square root of (L squared + Y squared). Gordon S.
  Posted by Gordon Steel on 2003-09-30 12:06:49
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