The return of me.

I will be using radians and will use many obscure and specific trigonometric identities, noted by $\stackrel{*}{=}$ . Also, I will define $q=\frac{\sqrt{3}}{2}$ and $2s=t=\frac{\tau}{6}=\frac{2\pi}{6}$ for convenience. Also, I am scaling the dimensions so that the length on the bottom is 1 instead of twelve; after everything is done I'm just gonna multiply $a$ by 12.

I will Cartesian coordinates, with the bottom left corner of the orange triangle at the origin, and the bottom line of that triangle on the x-axis. Let all the points of intersection be labelled $A$ - $H$ , starting with the origin $A$ , and moving clockwise around the figure. Some notes:

• The blue equilateral triangles are $\triangle ABD$ and $\triangle DEG$

• The orange triangle is $\triangle ADG$ .

• Because $\angle BDA=\angle GDE=t$ and $D$ is on $\overline{BE}$ , then $\angle ADG=t$

• The angle $\angle CHF$ has angle $\frac{\tau}{4}$

• $C$ is the midpoint of $\overline{BD}$

• $F$ is the midpoint of $\overline{EG}$

• $H$ is on $\overline{AG}$

• $\overline{GE}$ is parallel to $\overline{AD}$ . This is because the angles $\angle ADG$ and $\angle DGE$ are equal.

Let $h=|AD|$ , and $\beta=\angle DAG$ . Then, the equation for $\overline{AD}$ is: $y=\tan(\beta)x$ , and likewise $\overline{AB}: y=\tan(\beta+t)x$ . Thus, $D=(h\cos(\beta),h\sin(\beta))$ and $B=(h\cos(\beta+t),h\sin(\beta+t))$ , from which we can gather that $C=(h\frac{\cos(\beta+t)+\cos(\beta)}{2},h\frac{\sin(\beta+t)+\sin(\beta)}{2})\stackrel{*}{=}(qh\cos(\beta+s),qh\sin(\beta+s))$ . Finally, we gather that $\overline{BD}:y=\tan(\beta-t)(x-h\cos(\beta))+h\sin(\beta)$ .

Now, we get the equation for $\overline{DG}$ . We know that because $\angle BDA=\angle GDE=t$ , then the upper angle of the triangle is also $\angle ADG=t$ . Thus, $\overline{DG}:y=\tan(\beta-2t)(x-h\cos(\beta))+h\sin(\beta)$ . We then calculate the y-intercept to find $G$ , and constrain $G=(1,0)$ by defining $h$ . $0=\tan(\beta-2t)(x-h\cos(\beta))+h\sin(\beta)\implies x=h(\cos(\beta)-\frac{\sin(\beta)}{\tan(\beta-2t)})\stackrel{*}{=}qh\sec(\beta-s)=1\implies h=\frac{1}{q\sec(\beta-s)}$ .

From here, we find the equation for $\overline{GE}:y=\tan(\beta)(x-1)$ (remember it is parallel to $\overline{AD}$ ). We then calculate the intersection of this line with $\overline{BD}:y=\tan(\beta-t)(x-h\cos(\beta))+h\sin(\beta)$ to get $\tan(\beta)(x-1)=\tan(\beta-t)(x-h\cos(\beta))+h\sin(\beta)=\tan(\beta-t)\left(x-\frac{\cos(\beta)}{q\sec(\beta-s)}\right)+\frac{\sin(\beta)}{q\sec(\beta-s)}$

$\implies (\tan(\beta)-\tan(\beta-t))x=\tan(\beta)-\tan(\beta-t)\frac{\cos(\beta)}{q\sec(\beta-s)}+\frac{\sin(\beta)}{q\sec(\beta-s)}\implies x=\frac{\tan(\beta)-\tan(\beta-t)\frac{\cos(\beta)}{q\sec(\beta-s)}+\frac{\sin(\beta)}{q\sec(\beta-s)}}{\tan(\beta)-\tan(\beta-t)}\stackrel{*}{=} 1+\frac{1}{q}\sin(\beta)\cos(\beta)$ . We then solve for $y$ , and use this to find the equation for $E$ . $y=\tan(\beta)(x-1)=\tan(\beta)(\frac{1}{q}\sin(\beta)\cos(\beta))=\frac{1}{q}\sin^2(\beta) \implies E=(1+\frac{1}{q}\sin(\beta)\cos(\beta),\frac{1}{q}\sin^2(\beta)) \implies F=(1+\frac{1}{2q}\sin(\beta)\cos(\beta),\frac{1}{2q}\sin^2(\beta))$ .

Finally, we will define $a=|AH|$ , find $\overline{HD}$ , and write an equation for $a$ that checks if $F$ is on the perpendicular. $\overline{HD}:y=\frac{h\sin(\beta+t)}{h\cos(\beta+t)-a}(x-a)$ and the perpendicular takes the negative inverse of the slope to get $\bot\overline{HD}:y=\frac{a-h\sin(\beta+t)}{h\cos(\beta+t)}(x-a)$ . Plugging the values of $F$ into the equation, we get $\frac{1}{2q}\sin^2(\beta)=\frac{a-h\sin(\beta+t)}{h\cos(\beta+t)}(1+\frac{1}{2q}\sin(\beta)\cos(\beta)-a)$ . This is parabolic in $a$ , so for any given angle $\beta$ there are almost always two solutions. However, only one of them is true for all $\beta$ ; when $a=0.75$ . Thus, this is the value we are looking for.

Multiplying by 12 to get the answer to the problem, we find that the answer is $12*0.75=9$ .