Problem source

$ABCD$ is a uniform rectangular lamina and $X$ is a point on $BC\,$. The lengths of $AD$, $AB$ and $BX$ are  $p\,$, $q$ and $r$ respectively. 
The triangle $ABX$ is cut off the lamina.
Let $(a,b)$ be the position of the centre of gravity of the lamina, where the axes are such that the coordinates of $A\,$, $D$ and $C$  are $(0,0)\,$, $(p,0)$ and $(p,q)$ respectively. Derive equations for $a$ and $b$ in terms of $p\,$, $q$ and $r\,$.
 
When the resulting trapezium is freely suspended from the point $A\,$, the side $AD$ is inclined at $45^\circ$ below the horizontal. 
Show that $\displaystyle r = q - \sqrt{q^2 - 3pq + 3p^2}\,$. You should justify carefully the choice of sign in front of the square root.

Solution source

\begin{center}
    \begin{tikzpicture}[scale=2]
        \def\p{2};
        \def\q{1};
        \coordinate (A) at (0,0);
        \coordinate (B) at (0,\q);
        \coordinate (C) at (\p,\q);
        \coordinate (D) at (\p,0);

        \coordinate (X) at ($(B)!0.3!(C)$);

        \draw (A) -- (X) -- (C) -- (D) -- cycle;
        \draw[dashed] (A) -- (B) -- (X) -- cycle;

        \node[left] at (A) {$A$};
        \node[left] at (B) {$B$};
        \node[right] at (C) {$C$};
        \node[left] at ($(A)!0.5!(B)$) {$q$};
        \node[below] at ($(A)!0.5!(D)$) {$p$};
        \node[above] at ($(B)!0.5!(X)$) {$r$};
        \node[right] at (D) {$D$};
        \node[above] at (X) {$X$};
        
    \end{tikzpicture}
\end{center}

\begin{array}{c|c|c|c}
& ABX & ABCD & AXCD \\ \hline 
\text{area} & \frac12 q r & pq & q(p - \frac12 r) \\
\text{com} & \binom{\frac{r}{3}}{\frac{2q}{3}} & \binom{p/2}{q/2} & \binom{a}{b}
\end{array}

\begin{align*}
&& q(p-\frac12 r) \binom{a}{b} &= pq\binom{p/2}{q/2} - \frac12 q r  \binom{\frac{r}{3}}{\frac{2q}{3}} \\
\Rightarrow && \binom{a}{b} &= \frac{2}{2p-r}\binom{p^2/2-\frac16r^2}{pq/2-\frac13qr} \\
&&&= \binom{\frac{p^2-\frac13 r^2}{2p-r}}{\frac{pq-\frac23qr}{2p-r}}
\end{align*}

\begin{center}
    \begin{tikzpicture}[scale=2]
        \def\p{1};
        \def\q{2};
        \def\r{1};
        \begin{scope}[rotate=225]
            \coordinate (A) at (0,0);
            \coordinate (B) at (0,\q);
            \coordinate (C) at (\p,\q);
            \coordinate (D) at (\p,0);
    
            \coordinate (X) at ($(B)!0.3!(C)$);
            \coordinate (G) at ({(\p*\p-1/3*\r*\r)/(2*\p-\r)},{(\p*\q-2/3*\q*\r)/(2*\p-\r)});
        \end{scope}

        \draw (A) -- (X) -- (C) -- (D) -- cycle;
        \draw[dashed] (A) -- (B) -- (X) -- cycle;

        \filldraw (G) circle (1pt);

        \node[above] at (A) {$A$};
        \node[right] at (B) {$B$};
        \node[below] at (C) {$C$};
        \node[right] at ($(A)!0.5!(B)$) {$q$};
        \node[left] at ($(A)!0.5!(D)$) {$p$};
        \node[right] at ($(B)!0.5!(X)$) {$r$};
        \node[left] at (D) {$D$};
        \node[right] at (X) {$X$};
        \node[right] at (G) {$G$};

        \coordinate (t) at (-2,0);
        \coordinate (s) at ($(A)!1.2!(G)$);

        \draw[dashed] (-1,0) -- (1.5,0);
        \draw[dashed] (A) -- (s);

        \pic [draw, angle radius=0.5cm, angle eccentricity=1.25, "$45^\circ$"] {angle = t--A--D};
        \pic [draw, angle radius=0.5cm, angle eccentricity=1.25, "$45^\circ$"] {angle = D--A--G};
        
    \end{tikzpicture}
\end{center}

We must have:

\begin{align*}
&& 1 &= \frac{p^2-\frac13r^2}{pq-\frac23qr} \\
\Rightarrow && pq-\frac23qr &= p^2 - \frac13 r^2 \\
\Rightarrow && 0 &=r^2-2q r + 3p(q-p) \\
\Rightarrow && 0 &= (r-q)^2 -q^2+3pq-3p^2 \\
\Rightarrow && r&= q \pm \sqrt{q^2-3pq+3p^2}
\end{align*}

Suppose $r > q$, then $p > q > r$ and we have a shape which looks like this


\begin{center}
    \begin{tikzpicture}[scale=2]
        \def\p{3};
        \def\q{1};
        \def\r{2};
        \begin{scope}[rotate=225]
            \coordinate (A) at (0,0);
            \coordinate (B) at (0,\q);
            \coordinate (C) at (\p,\q);
            \coordinate (D) at (\p,0);
    
            \coordinate (X) at ($(B)!0.5!(C)$);
            \coordinate (G) at ({(\p*\p-1/3*\r*\r)/(2*\p-\r)},{(\p*\q-2/3*\q*\r)/(2*\p-\r)});
        \end{scope}

        \draw (A) -- (X) -- (C) -- (D) -- cycle;
        \draw[dashed] (A) -- (B) -- (X) -- cycle;

        \filldraw (G) circle (1pt);

        \node[above] at (A) {$A$};
        \node[right] at (B) {$B$};
        \node[below] at (C) {$C$};
        \node[right] at ($(A)!0.5!(B)$) {$q$};
        \node[left] at ($(A)!0.5!(D)$) {$p$};
        \node[right] at ($(B)!0.5!(X)$) {$r$};
        \node[left] at (D) {$D$};
        \node[right] at (X) {$X$};
        \node[right] at (G) {$G$};

        \coordinate (t) at (-2,0);
        \coordinate (s) at ($(A)!1.2!(G)$);

        \draw[dashed] (-1,0) -- (1.5,0);
        \draw[dashed] (A) -- (s);

        \pic [draw, angle radius=0.5cm, angle eccentricity=1.25, "$45^\circ$"] {angle = t--A--D};
        \pic [draw, angle radius=0.5cm, angle eccentricity=1.25, "$45^\circ$"] {angle = D--A--G};
        
    \end{tikzpicture}
\end{center}


which definitely wouldn't have $G$ hanging below $A$.