A geometry problem by Mr X

Let's consider the rectangle in the $xy$ plane such that the vertices are labelled as in the picture. $h$ represent the altitude and $b$ the base. The number $\{1,...,4\}$ represent the four lines, which equation follows*:

$1.\quad$ $\displaystyle y=\frac{v}{b}x$

$2.\quad$ $\displaystyle y=\frac{h}{u}x$

$3.\quad$ $\displaystyle y=\frac{(v-h)}{b}x+h$

$4.\quad$ $\displaystyle y=\frac{h(x-b)}{u-b}x$

Now, we can find the coordinates of $A$ (intersecting $2$ and $3$ ), $B$ (intersecting $3$ and $4$ ) and $C$ (intersecting $1$ and $4$ ). We get:

$\displaystyle A\left(\frac{bhu}{bh+hu-uv},\frac{bh^2}{bh+hu-uv}\right)$

$\displaystyle B\left(\frac{bhu}{bv+hu-uv},\frac{bhv}{bv+hu-uv}\right)$

$\displaystyle C\left(\frac{bh^2}{b(h+v)-uv},\frac{bhu}{b(h+v)-uv}\right)$

Let's consider the triangle with area $7$ . The area can be written as $\frac{1}{2}\cdot u\cdot (h-A_y)=7$ . Similary, for the bigger triangle we have $\frac{1}{2}\cdot v \cdot (b-C_x)=22$ . The remaning polygon can be see as the union of a triangle and a trapezium, so its area is $\frac{1}{2}(B_x-u)(h-B_y)+\frac{1}{2}([h-B_y]+[h-v])(b-B_x)=27$ . In our rectangle this three condition must be satisfied at the same time, so:

$\displaystyle \begin{cases} \displaystyle \frac{1}{2}\cdot u\cdot (h-A_y)=7 \\[3pt] \displaystyle \frac{1}{2}\cdot v \cdot (b-C_x)=22 \\[3pt] \displaystyle \frac{1}{2}(B_x-u)(h-B_y)+\frac{1}{2}([h-B_y]+[h-v])(b-B_x)=27 \end{cases}$

Expliciting* the equations:

$\begin{cases} \displaystyle \frac{hu^2(h-v)}{2(bh+hu-uv)}=7 \\[5pt] \displaystyle \frac{(b-u)(h-v)(bv+hu)}{2(bv+hu-uv)}=27 \\[5pt] \displaystyle \frac{b^2v}{2(bh+bv-uv)}=22 \end{cases}$

The system has different solution, but the only one with the geometrical requirements ( $0 \leq u \leq b$ , $0\leq v \leq h$ , $b>0$ and $h>0$ ) is $h=\frac{236.8}{b}, b\neq 0$ . So, the rectangle area $\displaystyle A=bh=\boxed{236.8}$ . It's important to notice that there are uncountably infinite combination of $(b,h)$ (so infinite rectangles) that solves the problem. Infact, our system contains three non-linear equation, while the unknowns are four. Nevertheless, the area of the infinite rectangles that solve the problem is always the same.

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*: I omitted the calculation because are quite laborious. Even the system has been solved with the calculator, choosing at the end the more congenial solution.