Grade 11 math

Let's use $(A_1A_2)$ as x-axis and $(A_1B_1)$ as y-axis.

Considering the right triangle $A_1 A_2 B_2$ with hypotenuse $A_2B_2 = L$ and shorter leg $A_1B_2 = L/2$, using the Pythagorean theorem we get $A_1A_2 = \sqrt{L² - (\frac L2)²} = \frac{\sqrt3}2L$.

Consider the midpoint C of $A_1A_2$, which is vertically below the midpoint $B_3$, so that $C B_3$ is perpendicular to $A_1A_2$. (In other words, C is the intersection of the parallel to $(A_1B_1)$ through $B_3$, with $(A_1A_2)$.)
So there's a right angle at C in the triangle $C A_3 B_3$. The hypotenuse is $A_3B_3 = L$ and the smaller leg is $C B_3 = L/4$ (since the midpoint $B_3$ is at half the height of the midpoint $B_2$ which is at half the height of $B_1$ which is L). So, $C A_3 = \sqrt{L² - (L/4)²} = L \sqrt{15}/4 $, and

$A_1A_3 / A_1A_2 = (A_1C + CA_3)/ A_1A_2 = ½ + (\sqrt{15} /4)/ (\sqrt 3/2) = ½ + \sqrt 5/2 ≈ 1.618 \approx1.62$, a.k.a. the Golden Ratio.