Geometry problem: square and golden ratio

Comment: You can use this property of pentagon which is:

$\phi=\frac{diagonl}{side}$

For prove see https://www.youtube.com/watch?v=lL9VN8ELH1M&pp=ygUMI3dhemlyeGdyYXBo

The construction below is based on this property.

Where $BF=m= \sqrt 3$

Since $AB=1$ then $BE=\phi$

The easy solution. If the vertical side is cut in the golden ratio proportion, then the two pieces are of length $1/\varphi+1/\varphi^2=1$. Then by the Pythagorean theorem, $x=\sqrt{3-1/\varphi^2}=\varphi$. (Hint: we've already seen that $1/\varphi^2=1-1/\varphi$, so substitute this under the square root).

We play with Fibonacci numbers.

$F_{n+4}+2F_{n+3}-F_{n+2}-6F_{n+1}-3F_n$

$=F_{n+3}+F_{n+2}+2F_{n+3}-F_{n+2}-6F_{n+1}-3F_n$

$=3(F_{n+3}-2F_{n+1}-F_n)$

$=3(F_{n+2}+F_{n+1}-2F_{n+1}-F_n)$

$=3(F_{n+2}-F_{n+1}-F_n)=0.$

As $n\to\infty,F_{n+k}/F_n\to\phi^k$, so the above Fibonacci number result implies

$\phi^4+2\phi^3-\phi^2-6\phi-3=0.$

Instead of showing that $\sqrt 3$ entails $\phi$ in the posted figure, the following proceeds conversely, that $\phi$ entails $\sqrt 3$.

I. In Elements, VI, 30 (Dover, tr. Heath), Euclid shows how to cut a given segment $AB$ in extreme and mean ratio, i.e. so that$$\frac{AB}{AE}=\frac{AE}{EB}$$with $AE$ the greater segment: he constructs square $BC$ and applies on side $AC$ rectangle $DC=BC$ (VI, 29), and the proportion follows from VI, 14.

II. Extend Euclid's figure now: construct square $AN=AD$, square $JL=BC$, and join $LB$. Since $N$ also cuts $JK$ in extreme and mean ratio, making $\triangle KLN\cong\triangle EFB$, then $LB$ passes through $N$.

Since $AJ=AE$ has been added to $AB$, $A$ now cuts $JB$ in extreme and mean ratio (Elements, XIII, 5), and supposing $AB=1$, then$$JB=\phi$$Since $MJ=AB=1$, it remains to show that $NB=\sqrt 3$ as in the posted figure.

III. Let $NP\perp NB$ meet $HC$ extended at $P$, let $PQ\perp NP$ meet $BF$ extended at $Q$, extend $KJ$ to $S$, and draw $QR\perp PH$.

Since $NP$, $NB$ are each hypotenuse of right triangles $NPS\cong BNJ$, then rectangle $NBQP$ is a square, in which$$\triangle NPS+\triangle PQR=CFDG=ABHC=1$$Further, since $\triangle ABO\cong\triangle HBF$, then quadrilateral$$CFBO=ABHC=1$$Lastly, since $\triangle QRF\cong \triangle NGO$, then trapezoid$$SNOC+\triangle NGO=ABHC=1$$

Therefore, square$$NBQP=3ABHC$$and$$NB=\sqrt 3$$

It is possible to avoid the complicated quartic that you came up with, by solving a quadratic equation for $x^2$ instead:

Consider the right triangle with hypotenuse $\sqrt3$ and base $x$. Let's call its height $y$. Then $$\frac{x}{y} = \frac{x+1}{1}$$ Or $$\frac{1}{y} - \frac{1}{x} = 1$$ So $$x-y = xy$$ Pythagoras': $$x^2 + y^2 = 3$$ $$x^2 + y^2 - 2xy = 3 - 2(x-y)$$ $$(x-y)^2 + 2(x-y) - 3 = 0$$ Therefore $$x-y = xy = 1$$ The other answer ($x-y = -3$) is not acceptable (why not?)

Pythagoras' again: $$x^2 + \frac{1}{x^2} = 3$$ $$x^4 - 3x^2 + 1 = 0$$ This is the quartic we have to solve, but luckily it is really a quadratic to solve for $x^2$ $$x^2 = \frac{3 \pm \sqrt5}{2}$$ Smells like golden ratio! Let's try to calculate further from $x^2$ to $x$: $$x^2 = \frac{6 \pm 2\sqrt5}{4} = \frac{1+5 \pm 2\sqrt5}{4} = (\frac{1 \pm \sqrt5}{2})^2$$ Finally, noting the constraint that $y = \frac{1}{x} < 1$, we are left with only one answer: $x = \frac{1 + \sqrt5}{2}$

We can calculate the side lengths of the square exactly. For example, if the short leg is $y$, then the long leg is $y\phi$. Then also $y+y\phi=1$. The long leg is now given by $$y\phi=\frac{\phi}{1+\phi}=\frac{1}{\phi}=\phi-1.$$ Here I used $1+\phi=\phi^2$ and $\frac{1}{\phi}=\phi-1$. We can now use the ratio of opposite over adjacent: $$\frac{\phi-1}{x}=\frac{1}{1+x}.$$ I'll leave the final steps for the reader.