Four Equilateral Triangles Question

1st solution Let $a=AB=BC=CA$ , $x=AD=AE=AF$ , $y=BF=BG=FG$ , $z=AF=AH=FH$ . Let $\theta=\angle BAD$ , $\varphi=\angle GBC$ .

First we prepare some angle relations:
$\angle EAC=60{}^\circ -\angle CAD=\angle CAB-\angle CAD=\theta,$ $\angle ABF=60{}^\circ -\angle FBD=\angle GBC-\angle FBD=\varphi,$ $\angle ADB=\angle DAC+\angle ACD=\left( 60{}^\circ -\theta \right)+60{}^\circ =120{}^\circ -\theta.$

Now, on $\triangle ECA$ , $\left[ ECA \right]=6\Rightarrow \dfrac{1}{2}ax\sin \theta =6\Rightarrow x=\dfrac{12}{a\sin \theta } \ \ \ \ \ (1)$ On $\triangle CGB$ , $\left[ CGB \right]=5\Rightarrow \dfrac{1}{2}ay\sin \varphi =5\Rightarrow y=\dfrac{10}{a\sin \varphi } \ \ \ \ \ (2)$ On $\triangle ABH$ , $\left[ ABH \right]=2\Rightarrow \dfrac{1}{2}az\sin \left( 60{}^\circ -\theta \right)=2\Rightarrow z=\dfrac{4}{a\sin \left( 60{}^\circ -\theta \right)} \ \ \ \ \ (3)$ By sine rule On $\triangle AFB$ , $\dfrac{z}{\sin \varphi }=\dfrac{y}{\sin \theta }\overset{\left( 2 \right)}{\mathop{\Rightarrow }}\,\dfrac{z}{\sin \varphi }=\dfrac{10}{a\sin \varphi \sin \theta }\Rightarrow z=\dfrac{10}{a\sin \theta } \ \ \ \ \ (4)$

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$\begin{aligned} \left( 3 \right),\left( 4 \right)\Rightarrow \dfrac{10}{a\sin \theta }=\dfrac{4}{a\sin \left( 60{}^\circ -\theta \right)} & \Leftrightarrow 5\sin \left( 60{}^\circ -\theta \right)=2\sin \theta \\ & \Leftrightarrow 5\left( \dfrac{\sqrt{3}}{2}\cos \theta -\dfrac{1}{2}\sin \theta \right)=2\sin \theta \\ & \Leftrightarrow 5\sqrt{3}\cos \theta =9\sin \theta \\ & \Leftrightarrow \cot \theta =\dfrac{9}{5\sqrt{3}} \ \ \ \ \ (5)\\ \end{aligned}$ By sine rule On $\triangle ADB$ , $\dfrac{x}{\sin 60{}^\circ }=\dfrac{a}{\sin \left( 120{}^\circ -\theta \right)}\overset{\left( 1 \right)}{\mathop{\Rightarrow }}\,\dfrac{12}{a\sin \theta \cdot \sin 60{}^\circ }=\dfrac{a}{\sin \left( 120{}^\circ -\theta \right)}\Rightarrow {{\alpha }^{2}}=\dfrac{8\sqrt{3}\sin \left( 120{}^\circ -\theta \right)}{\sin \theta } \ \ \ \ \ (6)$ Finally, for the area of $\triangle ABC$ , $\begin{aligned} & \left[ ABC \right]=\dfrac{{{a}^{2}}\sqrt{3}}{4} \\ & \overset{\left( 6 \right)}{\mathop{=}}\,\dfrac{\sqrt{3}}{4}\cdot \dfrac{8\sqrt{3}\sin \left( 120{}^\circ -\theta \right)}{\sin \theta } \\ & =6\dfrac{\dfrac{\sqrt{3}}{2}\cos \theta +\dfrac{1}{2}\sin \theta }{\sin \theta } \\ & =3\sqrt{3}\cot \theta +3 \\ & \overset{\left( 5 \right)}{\mathop{=}}\,3\sqrt{3}\dfrac{9}{5\sqrt{3}}+3 \\ & =\boxed{8.4} \\ \end{aligned}$

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2nd solution We use the same notation as in the previous solution. It is easy to see that triangles with the same colour fill, or shading are congruent, i.e. $\triangle AEC\cong \triangle ADB \ \ \ \ \ \triangle BGC\cong \triangle BFA \ \ \ \ \ \triangle AHB\cong \triangle AFC$ So, we have $\begin{aligned} \left[ ABC \right] & =\left[ ADB \right]+\left[ AFC \right]+\left[ CFD \right] \\ & =\left[ AEC \right]+\left[ AHB \right]+\left[ CFD \right] \\ & =6+2+\left[ CFD \right] \\ & =8+\left[ CFD \right] \ \ \ \ \ (1)\\ \end{aligned}$ The only thing left to find is $\left[ CFD \right]$ . Let’s work for this.

Triangles $BFA$ and $BDF$ share a common altitude from vertex $B$ , hence, $\begin{aligned} \dfrac{\left[ BFA \right]}{\left[ BDF \right]}=\dfrac{AF}{FD} & \Rightarrow \dfrac{AF}{FD}=\dfrac{\left[ BGC \right]}{\left[ ADB \right]-\left[ ABF \right]}=\dfrac{\left[ BGC \right]}{\left[ AEC \right]-\left[ BGC \right]}=\dfrac{5}{6-5}=5 \\ & \Rightarrow \dfrac{AF}{FD}=5 \\ \end{aligned}$ Furthermore, triangles $AFC$ and $CFD$ share a common altitude from vertex $C$ , hence, $\dfrac{\left[ AFC \right]}{\left[ CFD \right]}=\dfrac{AF}{FD}\Rightarrow \dfrac{2}{\left[ CFD \right]}=5\Rightarrow \left[ CFD \right]=0.4 \ \ \ \ \ (2)$ Finally, $\left( 1 \right),\left( 2 \right)\Rightarrow \left[ ABC \right]=8+0.4=\boxed{8.4}$