Problem source

The triangle $ABC$ has side lengths $\left| BC \right| = a$, $\left| CA \right| = b$ and $\left| AB \right| = c$. Equilateral triangles $BXC$, $CYA$  and  $AZB$ are erected on the sides of the  triangle $ABC$, with $X$ on the other side of $BC$ from $A$, and similarly for $Y$ and $Z$. 
Points $L$, $M$ and $N$ are the centres  of rotational symmetry of triangles $BXC$, $CY\!A$ and $AZB$ respectively. 

\begin{questionparts}
\item Show that $| CM| =  \dfrac {\ b} {\sqrt3} \,$ and write down the corresponding expression for $| CL|$. 
\item Use the cosine rule to show that
\[ 6 \left| LM \right|^2 = a^2+b^2+c^2 + 4\sqrt3 \, \Delta \,, \]
where  $\Delta$ is the area of triangle $ABC$. 
Deduce that $LMN$ is an equilateral triangle.
Show further that the areas of triangles $LMN$ and $ABC$ are equal if and only if
\[ a^2+b^2 +c^2 = 4\sqrt3 \, \Delta \,. \]
\item Show that  the conditions
\[ (a -b)^2 = -2ab \big( 1 -\cos(C-60^\circ)\big) \,\]
and 
\[ a^2+b^2 +c^2 = 4\sqrt3 \, \Delta \]
are equivalent.
Deduce that the areas of triangles $LMN$ and $ABC$ are equal if and only if $ABC$ is equilateral.
\end{questionparts}

Solution source

\begin{center}
    \begin{tikzpicture}

        \coordinate (A) at (0,0);

        \coordinate (B) at (3,0);
        \coordinate (C) at (2, 2);

        \coordinate (X) at ($(B)!1!-60:(C)$);
        \coordinate (Y) at ($(C)!1!-60:(A)$);
        \coordinate (Z) at ($(A)!1!-60:(B)$);

        \coordinate (L) at ($ 1/3*(B) + 1/3*(X) + 1/3*(C) $);
        \coordinate (M) at ($ 1/3*(C) + 1/3*(Y) + 1/3*(A) $);
        \coordinate (N) at ($ 1/3*(A) + 1/3*(Z) + 1/3*(B) $);

        \filldraw (A) circle (1.5pt) node[below left] {$A$};
        \filldraw (B) circle (1.5pt) node[below right] {$B$};
        \filldraw (C) circle (1.5pt) node[above] {$C$};
        \filldraw (X) circle (1.5pt) node[right] {$X$};
        \filldraw (Y) circle (1.5pt) node[left] {$Y$};
        \filldraw (Z) circle (1.5pt) node[below] {$Z$};
        
        \draw[thick] (A) -- (B) node[pos=0.5, below] {$c$};
        \draw[thick] (B) -- (C) node[pos=0.5, right] {$a$};
        \draw[thick] (A) -- (C) node[pos=0.5, left] {$b$};

        \draw[dashed] (A) -- (Z) -- (B) -- (X) -- (C) -- (Y) -- (A);

        \filldraw (L) circle (1pt) node[right] {$L$};
        \filldraw (M) circle (1pt) node[left] {$M$};
        \filldraw (N) circle (1pt) node[below] {$N$};
    
    \end{tikzpicture}
\end{center}

\begin{questionparts}
\item Consider the equilateral triangle $CYA$, notice that $YM$ is a vertical line of symmetry, and $\angle ACM = 30^\circ$ therefore $\frac{AC/2}{CM} = \cos 30^\circ \Rightarrow |CM| = \frac{b}{2} \cdot \frac{2}{\sqrt{3}} = \frac{b}{\sqrt{3}}$. Similarly $|CL| = \frac{a}{\sqrt{3}}$

\item $\,$ \begin{align*}
&& |LM|^2 &= |CM|^2 + |CL|^2 - 2 \cdot |CM| \cdot |CL| \cdot \cos \angle MCL \\
&&&= \frac{b^2}{3} + \frac{a^2}{3} - 2 \frac{ab}{3} \cos \left (\angle CMA + \angle CAB + \angle BCL \right) \\
&&&= \frac13 \left (b^2 + a^2 - 2ab \cos \left ( \frac{\pi}{3} + \angle CAB \right) \right) \\
&&&= \frac13 \left (b^2 + a^2 - ab \cos \left ( \angle CAB \right) + \sqrt{3}ab \sin \angle CAB \right) \\ 
&&&= \frac13 \left (b^2 + a^2 - ab \cos  \angle CAB  + 2\sqrt{3} \Delta\right) \\ 
&&&= \frac13 \left (b^2 + a^2 - ab \left (\frac{a^2+b^2-c^2}{2ab} \right)  + 2\sqrt{3} \Delta\right) \\ 
&&&= \frac13 \left ( \frac12(a^2+b^2+c^2) + 2\sqrt{3}\Delta \right) \\
\Rightarrow && 6|LM|^2 &= a^2 + b^2 + c^2 + 4\sqrt{3} \Delta
\end{align*}

However, nothing in our reasoning here was special about $LM$, therefore $LN$ and $MN$ also equal this value, and we find that the triangle is equilateral.

The area of equilateral triangle [LMN] is $\frac{\sqrt{3}}4 |LM|^2$, ie

\begin{align*}
&&& \text{areas are equal} \\
\Leftrightarrow && \Delta &= \frac{\sqrt{3}}4 |LM|^2 \\
&&&= \frac{\sqrt{3}}4 \frac{a^2+b^2+c^2+4\sqrt{3}\Delta}{6} \\
&&&= \frac{\sqrt{3}}{24} (a^2+b^2+c^2) + \frac12 \Delta \\
\Leftrightarrow && \Delta &= \frac{\sqrt{3}}{12}(a^2+b^2+c^2)\\
\Leftrightarrow && 4\sqrt{3}\Delta &=a^2+b^2+c^2\\
\end{align*}

\item $\,$ \begin{align*}
&& (a-b)^2 &= -2ab(1 - \cos(C - 60^{\circ})) \\
\Leftrightarrow && a^2+b^2 - 2ab &=-2ab + 2ab \cos(C - 60^{\circ}) \\
\Leftrightarrow && a^2+b^2 &= ab \cos C+\sqrt{3}ab\sin C \\
\Leftrightarrow && a^2+b^2 &= ab \frac{a^2+b^2-c^2}{2ab} + 2\sqrt{3} \Delta \\
\Leftrightarrow && a^2+b^2+c^2 &= 4\sqrt{3}\Delta
\end{align*}

Since the LHS is non-positive, and the RHS is positive, the only way they can be equal is if they are both $0$, ie $a=b$ and $C = 60^{\circ}$ ie $ABC$ is equilateral.

\end{questionparts}