Find the area of the parallelogram $ABCD$
The problem is from Igor's Twitter post. He also gave a solution. In a later comment he said that the original Square could only be a Rectangle, refraining from the same answer. The author of Peter Gallin gave a much appreciated computational solution, which probably remains valid for rectangles as well. I analyzed Igor's solution, which is not explained in detail enough. He writes some equations that lead to the result. Even for this he deserves our thanks.
Solution CiP
I appreciate that Igor's solution is based on the geometric results which also appear in my Post A LEMMA and a COROLLARY ... It will be seen that in this way the Problem can be formulated even in a Parallelogram. The calculations made by Peter Gallin helped me draw a corresponding figure to scale. We thank him too.
We introduce additional notations in the figure:
points $G,\;H,\;K$ and segments $[BG],\;[DG].$ We also call the areas $\textbf{S}_{AFG}=U,\;\textbf{S}_{CEK}=W.$
Applying the Lemma from the previously mentioned post to the quadrilaterals $ADGF,\;BCEG$ we have that the following areas have the values
$$\textbf{S}_{DGH}=\frac{180}{U},\quad \quad \textbf{S}_{BGK}=\frac{9}{W} \;.\tag{1}$$
It remains to express the areas marked with $?,\;??$ in the figure. For this we will use another result, valid in parallelograms, namely that, in the adjacent figure
we have $\textbf{S}_{ABG}=\textbf{S}_{ADC}$ (and of course $\textbf{S}_{BCG}=\textbf{S}_{CDG}$). Apologies for overlapping some notations !! Indeed, for $BH,\;DJ \perp AC$ we have, from congruent triangles $\Delta ADJ \equiv \Delta CBH$ that $BH=DJ$ and sos
$$2 \cdot \textbf{S}_{ABG}=AG \cdot BH=AG \cdot DJ=2 \cdot \textbf{S}_{ADG}$$
Returning to the figure from the problem, we will apply this situation exactly to point $G$, and so we can write
$$\textbf{S}_{ABG}=\textbf{S}_{ADG}\quad \quad \textbf{S}_{CDG}=\textbf{S}_{BCG}\;\;\; \Leftrightarrow$$
$$\Leftrightarrow \;\;\; U+10+?=18+\frac{180}{U} \quad \quad W+1+??=9+\frac{9}{W}\;\;\; \Leftrightarrow$$
$$\Leftrightarrow \;\;\;\;?=\frac{180}{U}-U+8 \quad \quad ??=\frac{9}{W}-W+8 \tag{2}$$
Thus, I found all the information presented in the figure from Igor's solution. (I changed the letters A and B to W and U). All that remains is to write the equations that solve the problem.
We will apply the Corollary we mentioned at the beginning for trapezoids $ABCE$ and $ADCF$ (Note that we don't even have to draw them in full because we would unnecessarily clutter up the figure!) Namely
$\textbf{S}_{BCK}^2=\textbf{S}_{CEK} \cdot \textbf{S}_{ABK}\quad \quad \textbf{S}_{ADH}^2=\textbf{S}_{AFH} \cdot \textbf{S}_{DCH} \tag{3}$
But from the figure with the areas marked on it we also find
$\textbf{S}_{ABK}=U+10+?+\frac{9}{W}\overset{(2)}{=}U+10+\frac{180}{U}-U+8+\frac{9}{W}=\frac{9}{W}+\frac{180}{U}+18$
$\textbf{S}_{CDH}=W+1+??+\frac{180}{U}\underset{(2)}{=}W+1+\frac{9}{W}-W+8+\frac{180}{U}=\frac{9}{W}+\frac{180}{U}+9$
and then the relations (3) are written
$$9^2=W \cdot \left ( \frac{9}{W}+\frac{180}{U}+18 \right )\quad \quad 18^2=U \cdot \left ( \frac{9}{W}+\frac{180}{U}+9\right ) \tag{4}$$
From here we will find $U$ and $W$. Let us also note that the area of the entire parallelogram is
$2\cdot \textbf{S}_{ABC}=U+10+?+\frac{9}{W}+9=\underset{(2)}{=}U+10+\frac{180}{U}-U+8+\frac{9}{W}+9$ so
$$\textbf{S}_{ABCD}=2 \cdot \left (\frac{9}{W}+\frac{180}{U}+27 \right ) \tag{5}$$
We will leave the geometry aside and move on to solving the system of equations (4). After performing the calculations and simplifications, it is written
\begin{cases}10 \frac{W}{U}+W=4\\\frac{U}{W}+U=16\end{cases}
Eliminating the denominators will result
\begin{cases}UW=4U-10W\\UW=16W-U \end{cases}
From here $4U-10W=16W-U\;\Leftrightarrow\;5U=26W$ and substituting $\frac{U}{W}=\frac{26}{5}$ into one of the equations we have $\frac{26}{5}+U=16\;\;\Rightarrow U=\frac{54}{5}$. Then $W=\frac{5}{26}\cdot U=$ $=\frac{5}{26}\cdot \frac{54}{5}=\frac{27}{13}.$
Oh, finally, from (5) we calculate the area of the parallelogram
$$\textbf{S}_{ABCD}=2 \cdot \left (\frac{9}{\frac{27}{13}}+\frac{180}{\frac{54}{5}}+27 \right )=96$$
$\blacksquare$
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