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 Origamic III (Posted on 2012-10-07)
This is in continuation of Origamic II.

A sheet of paper has the exact shape of a rectangle (denoted by ABCD) where the length of AB is greater than or equal to the length of AD. The vertex A is folded onto the vertex C, resulting in the crease EF (E on AB and F on CD).

The paper is thereafter unfolded and, the vertex A is folded onto F so that, the length of the resulting crease is equal to AB.

Is ABCD always a square? If so, prove it - otherwise, give a counter example.

 No Solution Yet Submitted by K Sengupta No Rating

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 Solution Comment 1 of 1
`Clearly if ABCD is a square, then the lengthof the second fold will equal the length of AB.`
`But, there exists another rectangle ABCD suchthat the length of the second fold equals thelength of AB.`
`The fold EF is perpendicular to AC and passesthrough its midpoint X. The second fold GH( G on AB and H on AD ) is perpendicular to AFand passes through its midpoint Y.`
`Let |AB| = x and |AD| = 1.`
`Right triangles ABC and AXE are similar, therefore,`
`          |EX|          |EX|     |CB|     1     --------------- = ------ = ------ = ---      sqrt(x^2+1)/2     |AX|     |AB|     x`
`                  or`
`     |EX| = sqrt(x^2+1)/(2x).`
`          |AE|          |AE|     |AC|          --------------- = ------ = ------       sqrt(x^2+1)/2     |AX|     |AB|         `
`                        sqrt(x^2+1)                     = -------------                             x`
`                  or`
`     |AE| = (x^2+1)/(2x).`
`Right triangles AXE, AXF, and CXF are concruent, therefore,`
`     |CF| = |AF| = |AE| = (x^2+1)/(2x)  and`
`     |FD| = |CD|-|CF| = x-(x^2+1)/(2x) `
`          = (x^2-1)/(2x).`
`Right tringles ADF, AYH, and GAH are similar,therefore,`
`          |YH|         |YH|     |DF|        -------------- = ------ = ------      (x^2+1)/(4x)     |AY|     |AD|     `
`                       (x^2-1)/(2x)                    = --------------                            1`
`                  or`
`     |YH| = (x^4-1)/(8x^2).`
`          |AH|         |AH|     |AF|        -------------- = ------ = ------      (x^2+1)/(4x)     |AY|     |AD|     `
`                       (x^2+1)/(2x)                    = --------------                            1`
`                  or`
`     |AH| = (x^2+1)^2/(8x^2).   `
`      (x^2+1)^2/(8x^2)     |AH|     |YH|        ------------------ = ------ = ------             x             |GH|     |AH|     `
`                            (x^4-1)/(8x^2)                        = ------------------                           (x^2+1)^2/(8x^2) `
`                  or`
`     (x^2+1)^3 = 8x^3(x^2-1).            (*)`
`Equation (*) has four complex roots, onereal root which does not have H on AD, and      x ~= 1.510577.`
`This agrees with Geometer's Sketchpad.       `

 Posted by Bractals on 2012-10-08 20:09:52

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