*x*

^{2}-

*sx*+

*p*= 0, , using the standard construction tools of compass and straightedge. The method has been attributed to critic Thomas Carlyle.

Construct the circle in the Cartesian plane with center and passing through

*A*(0,1). By symmetry, the circle also passes through and .

Because the circle has center and passes through

*A*(0,1), the equation of the circle is

This reduces to

*x*

^{2}-

*sx*+

*y*

^{2}- (

*p*+ 1)

*y*+

*p*= 0

So the

*x*-intercepts of the circle are the solutions to

*x*

^{2}-

*sx*+

*p*= 0.

## Alternate justification:

The segment joining the*x*-intercepts has a length , hence

*x*

_{1}+

*x*

_{2}=

*s*.

intercepts the arc

*AX*

_{1}

*X*

_{2}, and intercepts the opposite arc, hence the two angles are supplementary. But is also supplementary with , so is congruent to , which in turn impies that . Hence

,

which implies that , so

*x*_{1 }*x*_{2}=*p*.Thus (

*x*-

*x*

_{1})(

*x*-

*x*

_{2}) =

*x*

^{2}-

*sx*+

*p*, and the solutions to

*x*

^{2}-

*sx*+

*p*= 0 are

*x*

_{1 }and

*x*

_{2}.

## 3 comments:

you might want to change the y coordinate of your circle center -- shows as \frac{p-1}{2} when you want \frac{p+1}{2}

you may want to change the y-coordinate of your circle center -- it reads \frac{p-1}{2} when you want \frac{p+1}{2}

Thanks.

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