Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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.

share|cite|improve this answer
That should be just $a^2 - 2ac$ without the $-2c^2$. – Rahul Sep 19 '11 at 11:32
@RahulNarain: Dropped a sign. Thanks. Fixed – Ross Millikan Sep 19 '11 at 11:37
Could any one demonstrate the construction process? – qed Sep 19 '11 at 12:00
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.