Let $\rm ABCD$ be a square, $\rm P$ a point inside it, and $\rm E, F, G, H$ the midpoints of the sides. If we know the areas $\rm [HPED] = 24$,
$\rm [EPFC] = 30$, and $\rm [FPGB] = 48$, you are asked for:
- The area of the shaded region $\rm [GPHA]$.
- The coordinates of $\rm P$ in the affine frame of reference $\{\rm A, \overrightarrow{\rm AB}, \overrightarrow{\rm AD}\}$.
Proposed by F. Damián Aranda Ballesteros on behalf of the RSME, as the juvenile problem of the month (nov. 2022).
It is useful to consider the coordinates discussed in the second point to solve the problem, so let $x$ be the distance from $\rm P$ to the side
$\overline{\rm AD}$, and $y$ that to $\overline{\rm AB}$; let also $\ell$ be the side of the square.
Hence, we can subdivide the figure into the following pieces:
From which we obtain these equations:
$$
\begin{cases}
\frac{1}{4}\ell x + \frac{1}{4}\ell (\ell - y) &= [\rm HPED],\\
\frac{1}{4}\ell (\ell - x) + \frac{1}{4}\ell (\ell - y) &= [\rm EPFC],\\
\frac{1}{4}\ell (\ell - x) + \frac{1}{4}\ell y &= [\rm FPGB];\\
\end{cases} \implies
\begin{cases}
\ell (x + \ell - y) &= 96,\\
\ell (2\ell - x - y) &= 120,\\
\ell (\ell - x + y) &= 192.\\
\end{cases}
$$
Adding the first and the third gives $$ 2\ell^2 = 288 \implies \color{green} \boxed{\ell = 12,} $$ and notice that ${[\rm GPHA]} = \ell x / 4 + \ell y / 4$, so subtracting the second from the above yields $$ 4{[\rm GPHA]} = \ell (x + y) = 168 \implies \color{green} \boxed{{[\rm GPHA]} = 42.} $$
With this results, we can now solve for $x$ and $y$: $$ \begin{cases} x - y &= -4,\\ x + y &= 14, \end{cases} \implies \color{green} \boxed{x = 5,\ y = 9,} $$ so the requested coordinates are finally $(5/12, 9/12)$.
Adding the first and the third gives $$ 2\ell^2 = 288 \implies \color{green} \boxed{\ell = 12,} $$ and notice that ${[\rm GPHA]} = \ell x / 4 + \ell y / 4$, so subtracting the second from the above yields $$ 4{[\rm GPHA]} = \ell (x + y) = 168 \implies \color{green} \boxed{{[\rm GPHA]} = 42.} $$
With this results, we can now solve for $x$ and $y$: $$ \begin{cases} x - y &= -4,\\ x + y &= 14, \end{cases} \implies \color{green} \boxed{x = 5,\ y = 9,} $$ so the requested coordinates are finally $(5/12, 9/12)$.