Vieta's Solution of a Cubic Equation

This Demonstration shows Vieta's solution of the depressed cubic equation , where . To solve it, draw an isosceles triangle with base and unit legs. Let be the angle at the base and . Draw a second isosceles triangle with base angle and unit legs. The base of the second triangle is a root of the equation.
By construction, , . Let . Since and are similar, . The perpendicular projections of and on also give similar right-angled triangles. So . So and . Eliminating gives .
Since , and applying the law of cosines to , we find .


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[1] G. E. Martin, Geometric Constructions, New York: Springer, 1998 pp. 126–140.
[2] M. Hladnik, "Some Historical Constructions of the Regular Heptagon" (in Slovenian), Obzornik za matematiko in fiziko, 61(4), 2014 pp. 132–145. www.obzornik.si/61.
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