Each of two different lines parallel to the the axis $Ox$ have exactly two common points on the graph of the function $f(x)=x^3+ax^2+bx+c$. Let $\ell_1$ and $\ell_2$ be two lines parallel to $Ox$ axis which meet the graph of $f$ in points $K_1, K_2$ and $K_3, K_4$, respectively. Prove that the quadrilateral formed by $K_1, K_2, K_3$ and $ K_4$ is a rhombus if and only if its area is equal to $6$ units.
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Tags: geometry, function, rhombus, Vieta, geometry proposed
thesamyaandoctor
18.01.2011 22:27
pleasee help. I need soln teach my students Samyaan.
peregrinefalcon88
14.06.2013 23:42
I will prove that the 4 points always form a rhombus. The area condition can instantly be seen as absurd because scaling all the roots of a cubic by a factor c, will scale the image of its graph by a factor of c in both dimensions, hence any area is attainable. p(x) = x^3+bx+c, where we have made a change of variables to remove the ax^2 term. p'(x) = 0 iff 3x^2+b = 0, so if $a_1$ and $a_2$ are the 0s p'(x), then we have $a_1+a_2 = 0$. We also have that $p(x) = p(a_1)+(x-a_1)^2(x-d_1) = p(a_2)+(x-a_2)^2(x-d_2)$, so we have from vieta's that $0 = 2a_1+d_1 = 2a_2+d_2$ and $a_1^2+2a_1d_1 = a_2^2+2a_2d_2$ from which it is not hard to see that $a_1+a_2 = d_1+d_2$ which attains the desired result.
MrezaAttaran
27.12.2023 17:35
peregrinefalcon88 wrote: which attains the desired result. I think you proved that quadrilateral is always parallelogram ,not rhombus.
torch
27.12.2023 22:50
ten year bump goes crazy