# Fun Equilateral Triangle Problem

$$\Delta ABC$$ 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

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 $$\boxed{AD = \sqrt{5}-1}$$

• thanks. ı was too lazy to use cos theorem and ı learned my lesson so ı wont do it again Apr 25, 2021 at 22:57

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'.

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 $$AD = \frac{2}{\varphi} = \sqrt{5}-1.$$