Summary edit

Task edit

Given is a square of side length ${\displaystyle a_{0}}$. Inserted is a second square of side length ${\displaystyle a_{2}}$. What is the length of ${\displaystyle a_{2}}$ that leads to a golden ratio of the areas of the second square and the remaining area within the larger square?

General solution edit

If there a two areas ${\displaystyle F_{1}}$ and ${\displaystyle F_{2}}$ then they fulfill the golden ratio if:
${\displaystyle \quad {\frac {F_{1}+F_{2}}{F_{1}}}={\frac {F_{1}}{F_{2}}}}$
What is the size of ${\displaystyle F_{1}}$ relative to ${\displaystyle F_{1}+F_{2}=F_{0}}$ ?
Starting with: ${\displaystyle \quad {\frac {F_{0}}{F_{1}}}={\frac {F_{1}}{F_{2}}}}$ we derive
${\displaystyle \quad \Leftrightarrow F_{0}\cdot F_{2}=F_{1}^{2}}$
${\displaystyle \quad \Leftrightarrow F_{0}\cdot (F_{0}-F_{1})=F_{1}^{2}\quad }$, by replacing ${\displaystyle F_{2}=F_{0}-F_{1}}$
${\displaystyle \quad \Leftrightarrow F_{0}^{2}-F_{0}\cdot F_{1}=F_{1}^{2}\quad }$, binomial formula
${\displaystyle \quad \Leftrightarrow F_{0}^{2}-F_{0}\cdot F_{1}+\left({\frac {1}{2}}\cdot F_{1}\right)^{2}=F_{1}^{2}+\left({\frac {1}{2}}\cdot F_{1}\right)^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow \left(F_{0}-{\frac {1}{2}}\cdot F_{1}\right)^{2}=F_{1}^{2}+{\frac {1}{4}}\cdot F_{1}^{2}\quad }$, binomial formula
${\displaystyle \quad \Leftrightarrow \left(F_{0}-{\frac {1}{2}}\cdot F_{1}\right)^{2}={\frac {5}{4}}\cdot F_{1}^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow \left(F_{0}-{\frac {1}{2}}\cdot F_{1}={\frac {\sqrt {5}}{2}}\cdot F_{1}\right)v\left(F_{0}-{\frac {1}{2}}\cdot F_{1}=-{\frac {\sqrt {5}}{2}}\cdot F_{1}\right)\quad }$, drawing the root
${\displaystyle \quad \Leftrightarrow \left(F_{0}={\frac {1}{2}}\cdot F_{1}+{\frac {\sqrt {5}}{2}}\cdot F_{1}\right)v\left(F_{0}={\frac {1}{2}}\cdot F_{1}-{\frac {\sqrt {5}}{2}}\cdot F_{1}\right)\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow F_{0}={\frac {1+{\sqrt {5}}}{2}}\cdot F_{1}\quad }$, excluding the second term since ${\displaystyle F_{0}>0}$
${\displaystyle \quad \Leftrightarrow F_{1}={\frac {2}{1+{\sqrt {5}}}}\cdot F_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow F_{1}={\frac {2\cdot (1-{\sqrt {5}})}{(1+{\sqrt {5}})\cdot (1-{\sqrt {5}})}}\cdot F_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow F_{1}={\frac {2\cdot (1-{\sqrt {5}})}{1-5}}\cdot F_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow F_{1}={\frac {2\cdot ({\sqrt {5}}-1)}{4}}\cdot F_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow F_{1}={\frac {{\sqrt {5}}-1}{2}}\cdot F_{0}\approx 0.618\cdot F_{0}\quad }$, rearranging

Specific solution edit

In the case of two squares ${\displaystyle {\overline {ABCD}}}$ and ${\displaystyle {\overline {EBFG}}}$ we have: ${\displaystyle F_{0}=a_{0}^{2}}$ and ${\displaystyle F_{1}=a_{1}^{2}}$
${\displaystyle \quad \Rightarrow a_{1}^{2}={\frac {{\sqrt {5}}-1}{2}}\cdot a_{0}^{2}\quad }$, filling in the general solution
${\displaystyle \quad \Rightarrow a_{1}={\sqrt {\frac {{\sqrt {5}}-1}{2}}}\cdot a_{0}\quad \approx 0.786\cdot a_{0}}$, using only the non-complex root

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