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Partition a line segment so that the difference between the square on the greater part and the square on the lesser part is constant.

The point K splits AD into AK and KD, such that $AK^2 - KD^2 = AC^2$, AC is of fixed length

In the figure, point K splits AD into AK and KD, such that $AK^2 - KD^2 = AC^2$, AC is of fixed length.

Can this be achieve by compass and straight edge?

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    Googling tells me that a square on a part refers to a square positioned in the plane with a given line segment as one of its sides. I'm still not sure what you mean by the difference between two of these squares being constant. (Constant with respect to what?)2011-09-19
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    If the length of the segment is $a$, and you are given $b$, are you asking to find $c$ so that $(a-c)^2-c^2=b?$2011-09-19
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    @RossMillikan, exactly.2011-09-19

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If the length of the segment is a, and you are given b, we are asked to find c so that $(a−c)^2−c^2=b=a^2-2ac$ This gives $c=\frac{a^2-b}{2a},$ which is constructible.

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    That should be just $a^2 - 2ac$ without the $-2c^2$.2011-09-19
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    @RahulNarain: Dropped a sign. Thanks. Fixed2011-09-19
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    Could any one demonstrate the construction process?2011-09-19
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    To construct $a^2$ draw a triangle with two sides $1$ and $a$, then construct a similar triangle with $a$ corresponding to $a$. Then you can just subtract $b$ and divide by $a$ again using similar triangles.2011-09-19