# Constant square difference partition point

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

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?) – anon Sep 19 '11 at 7:53
If the length of the segment is $a$, and you are given $b$, are you asking to find $c$ so that $(a-c)^2-c^2=b?$ – Ross Millikan Sep 19 '11 at 9:08
@RossMillikan, exactly. – qed Sep 19 '11 at 11:20

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.
That should be just $a^2 - 2ac$ without the $-2c^2$. – Rahul Sep 19 '11 at 11:32
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. – Ross Millikan Sep 19 '11 at 12:12