Problem source

A piece of uniform wire is bent into three sides of a square $ABCD$ so that the side $AD$ is missing. Show that if it is first hung up by the point $A$ and then by the point $B$ then the angle between the two directions of $BC$ is $\tan^{-1}18.$

Solution source

\begin{center}
\begin{tikzpicture}[scale=1.5]

    \coordinate (A) at (0,0);
    \coordinate (B) at (0,-1);
    \coordinate (C) at (1,-1);
    \coordinate (D) at (1,0);

    \coordinate (G) at (0.5, {-2/3});

    \draw (A) -- (B) -- (C) -- (D);

    \node at (A) [anchor=north east] {$A$};
    \node at (B) [anchor=south east] {$B$};
    \node at (C) [anchor=south west] {$C$};
    \node at (D) [anchor=north west] {$D$};
    \fill (G) circle (0.025); 
    \node at (G) [anchor=north]  {$G$};

    \draw[dashed] (A) -- (G);
    \draw[dashed] (B) -- (G);

    \pic [draw, angle radius=0.4cm, "$\alpha$"] {angle = B--A--G};
    \pic [draw, angle radius=0.4cm, "$\beta$"] {angle = C--B--G};
    
\end{tikzpicture}
\end{center}

In the coordinate system where $A$ is $(0,0)$ and $AD$ is the $x$-axis and $AB$ the $y$-axis and all side lengths are $2$, we find the centre of mass of each of the sides are:

\begin{align*}
AB :& (0,1) \\
BC :& (1,2) \\
CD :& (2,1) \\ 
\\
ABCD:& \l 1, \frac{4}{3} \r
\end{align*}


When hung from $A$, the angle $AB$ makes to the vertical is $\alpha$ and the angle $BC$ makes to the vertical will be $90^{\circ} + \alpha$.

When hung from $B$ the angle $BC$ makes to the vertical will be $\beta$. The value we are interested in therefore is $\beta + 90^{\circ} + \alpha$

\begin{align*}
&& \tan \alpha &= \frac{1}{\frac{4}{3}} \\
&&& = \frac{4}{3} \\
\\
&& \tan \beta &= \frac{\frac{2}{3}}{1} \\
&&&= \frac{2}{3} \\
\\
&& \tan \l \beta + (90^{\circ} + \alpha) \r &= \frac{\tan \beta + \tan\l 90^{\circ} + \alpha  \r}{1 - \tan \beta \tan\l 90^{\circ} + \alpha  \r} \\
&&&= \frac{\frac23 + \frac43}{1- \frac23 \frac43} \\
&&&= \frac{2}{1 - \frac89} \\
&&&= 18
\end{align*}