Fun Equilateral Triangle Problem

Question

$\Delta ABC$ is an equilateral triangle with a side length of $4$ units.$\: $$\: $ $\angle CAF = \angle EBC =\angle FAB$ .$\: $$\: $ $D \in \left | AF \right |\: ,\: E \in \left | CD \right |\: ,\: F \in \left | BE \right | $ $\: $ Find the length of $\left | AD \right |$

By given angles, its easy to see that $\Delta DEF$ is an equilateral triangle and by similarity, $\left | EF \right |$ is $2$. Area of $\Delta ABC = 4S = 4\sqrt3 \Rightarrow S = \sqrt{3} $. I couldn't get any further from this point

Answer 1

Let $AD = x$. Then, $CD = 2 + x$ and $AC = 4$. Because we know $\angle ADC = \frac{2\pi}{3}$, we can apply the Law of Cosines on $\triangle ACD$:

$$AC^{2} = AD^{2} + CD^{2} - 2(AD)(CD)\cos\bigg(\frac{2\pi}{3}\bigg)$$

$$4^{2} = x^{2} + (2 + x)^{2} + x(2 + x)$$

$$3x^{2}+6x - 12 = 0$$

$$x^{2} + 2x - 4 = 0$$

The solutions are $x = -1\pm\sqrt{5}$. Taking the positive solution, we find $\boxed{AD = \sqrt{5}-1}$

Answer 2

Clearly the inner equilateral triangle with one-fourth the total area has sides of length 2. So each outer triangle has sides measuring $x$ and $x+2$ where $x$ is the length asked for in the problem, and an included angle measuring 120°.

Now apply the side-angle-side formula for the area of a triangle:

$(1/2)(x)(x+2)\sin(120°)=\sqrt{3}$

So with $\sin(120°)=\sqrt3/2$ we get $x(x+2)=4$ from which (taking the positive root) $\color{blue}{x=\sqrt5-1}$.

The three interior segments divide one another in what was originally called "extreme and mean ratio". The term "golden ratio" for $(1+\sqrt5)/2$ seems to have been a Renaissance invention.

Incidentally, the angle $\theta$ is close to but not exactly 45°. More exactly it's about 44°28'.

Answer 3

No trigonometry:

Draw $FC$. Then $$\frac{|\triangle DEF|}{|\triangle CEF|} = \frac{DE}{CE}$$ and $$\frac{|\triangle CEF|}{|\triangle BCF|} = \frac{EF}{BF} = \frac{DE}{CE}.$$ So if $|\triangle CEF| = T$, we have $$\frac{S}{T} = \frac{T}{S-T} = \frac{1}{(S/T)-1},$$ hence $$\frac{FD}{AD} = \frac{DE}{CE} = \frac{S}{T} = \frac{1 + \sqrt{5}}{2} = \varphi,$$ the golden ratio. But since $|\triangle ABC| = 4|\triangle DEF| = 4S$, it follows that $FD = \frac{AB}{2} = 2$, hence $$AD = \frac{2}{\varphi} = \sqrt{5}-1.$$