Rectangles and Cubics

Geometry Level 3

The problem below is a variation of a brillant problem of the week.

The graph of the cubic function above has has real roots at $x = p, x = -p$ and $x = 0$ and the lines $BC$ and $AD$ are tangent to the curve at $B$ and $D$ respectively.

If the two tangent points and two non-zero -intercepts are joined together to form a rectangle, find the ratio of the longer side to the shorter side to five decimal places.

The answer is 1.73205.

1 solution

Rocco Dalto
Nov 28, 2018

$m(x) = x(x - p)(x + p) = x^3 - p^2x \implies \dfrac{dm}{dx}|_{x = x_{0}} = 3x_{0}^2 - p^2 = \dfrac{x_{0}(x_{0} - p)(x_{0} + p)}{x_{0} - p} \implies$
$3x_{0}^2 - p^2 = x_{0} + px_{0} \implies 2x_{0}^2 - px_{0} - p^2 = 0 \implies x_{0} = p, -\dfrac{p}{2}$

$x_{0} \neq p \implies x_{0} = -\dfrac{p}{2} \implies B:(-\dfrac{p}{2}, \dfrac{3p^3}{8})$ and $D:(\dfrac{p}{2}, -\dfrac{3p^3}{8})$ .

$AC = BD = 2p = \dfrac{p}{4}\sqrt{16 + 9p^4} \implies 64 = 16 + 9p^4 \implies p^4 = \dfrac{16}{3} \implies p = \dfrac{2}{\sqrt[4]{3}}.$

$AB = \dfrac{p}{8}\sqrt{16 + 9p^4} = \dfrac{2}{\sqrt[4]{3}}$

and

$BC = \dfrac{3p}{8}\sqrt{16 + p^4} = 2\sqrt[4]{3}$

$\implies \dfrac{BC}{AB} = \sqrt{3} \approx \boxed{1.73205}$ .

Rectangles and Cubics

The answer is 1.73205.

1 solution

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