Calculate the dimensions that are in progressive progression.

Question

Give the surface and diagonal of a parallelepiped rectangle. Calculate the known dimensions that are in progressive progression.

(Answer: $\dfrac{2d^2+S-\sqrt{(2d^2+3S)(2d^2-S)}}{4\sqrt{d^2+S}}; \dfrac{S}{2\sqrt{S+d^2}};\dfrac{2d^2+S+\sqrt{(2d^2+3S)(2d^2-S)}}{4\sqrt{d^2+S}}$)

I try

Let $a, b,c$ the dimensions of the parallelepiped rectangle.

$ S = 2ab+2ac+2bc \implies ab+bc+ac=\dfrac{S}{2} $

$ d^2=a^2+b^2+c^2 \implies d=\sqrt{a+b+c}$

$b=\sqrt{ac} \implies b^2=ac$

Can anyone finish?

Answer 1

Let a < b < c

P.G.$(a = \dfrac{b}{r}, b, c=rb)$

$d=\sqrt{a^2+b^2+c^2}=b\sqrt{1+r^2+\dfrac{1}{r^2}} \Longrightarrow b=\dfrac{d}{\sqrt{1+r^2+\dfrac{1}{r^2}}}(I)$

$S=2(ab+ac+bc)=2b^2\left(1+r+\dfrac{1}{r}\right)(II)$

I in II: $\dfrac{r^3+r^2+r}{r^4+r^2+1}=\dfrac{S}{2d^2} \Longrightarrow \dfrac{r(r^2+r+1)}{(r^2+r+1)(r^2-r+1)}=\dfrac{S}{2d^2} \Longrightarrow Sr^2-(S+2d^2)r+S=0.$

$ \therefore r_{1,2}=\dfrac{2d^2+S\pm \sqrt{(2d^2-S)(2d^2+3S)}}{2S}$

By T.Girard: $r_1r_2=1\implies r_2=\dfrac{1}{r_1}$

$\therefore r=r_1=\dfrac{2d^2+S +\sqrt{(2d^2-S)(2d^2+3S)}}{2S}$

$r+\dfrac{1}{r}=r_1+r_2=1+\dfrac{2d^2}{S}(III)$

III in II: $S=2b^2\left(1+\dfrac{2d^2}{S}+1\right) \Longrightarrow \boxed{b=\dfrac{S}{2\sqrt{S+d^2}}}$

$a=\dfrac{b}{r}=br_2, c= br$

$\boxed{a=\dfrac{2d^2+S-\sqrt{(2d^2-S)(2d^2+3S)}}{4\sqrt{S+d^2}}}$

$\boxed{c=\dfrac{2d^2+S+\sqrt{(2d^2-S)(2d^2+3S)}}{4\sqrt{S+d^2}}}$

(Solution by παθμ)