[ Area of a Triangle (using two sides and the angle between them)] Area

On the segment $AB$ lie points $C$ and $D$, with point $C$ between points $A$ and $D$. Point $M$ is taken such that lines $AM$ and $MD$ are perpendicular, and lines $CM$ and $MB$ are also perpendicular. Find the area of triangle $AMB$, given that the measure of angle $CMD$ is $\alpha$, and the areas of triangles $AMD$ and $CMB$ are $S_{1}$ and $S_{2}$, respectively.
Denote: $A M=x, C M=y, D M=z, B M=t$ and form a system of equations.

## Solution

Let $A M=x, C M=y, D M=z, B M=t, S_{\Delta \mathrm{AMB}}=S$. Then

$$
\begin{gathered}
S_{1}=\frac{1}{2} x z, S_{2}=\frac{1}{2} y t, S_{\Delta \mathrm{CMD}}=\frac{1}{2} y z \sin \alpha, S_{1}+S_{2}-S_{\Delta \mathrm{CMD}}=S \\
S=\frac{1}{2} x t \sin \angle A M B=\frac{1}{2} x t \sin \left(180^{\circ}-\alpha\right)=\frac{1}{2} x t \sin \alpha
\end{gathered}
$$

We obtain the system of equations

$$
\left\{\begin{array}{l}
x t y z=4 S_{1} S_{2} \\
S_{1}+S_{2}-\frac{1}{2} y z \sin \alpha=S
\end{array}\right.
$$

or

$$
\left\{\begin{array}{l}
x t y z=4 S_{1} S_{2} \\
S_{1}+S_{2}-\frac{1}{2} y z \sin \alpha=\frac{1}{2} x t \sin \alpha
\end{array}\right.
$$

or

$$
\left\{\begin{array}{l}
x t \cdot y z=4 S_{1} S_{2} \\
x t+y z=\frac{2\left(S_{1}+S_{2}\right)}{\sin \alpha}
\end{array}\right.
$$

From this system, we find that

$$
x t=\frac{S_{1}+S_{2} \pm \sqrt{\left(S_{1}+S_{2}\right)^{2}-4 S_{1} S_{2} \sin ^{2} \alpha}}{\sin \alpha}
$$

Therefore,

$$
S=\frac{1}{2} x t \sin \alpha=\frac{1}{2}\left(S_{1}+S_{2} \pm \sqrt{\left(S_{1}+S_{2}\right)^{2}-4 S_{1} S_{2} \sin ^{2} \alpha}\right) .
$$

Since $S>S_{1}$ and $S>S_{2}$, then $S>\frac{S_{1}+S_{2}}{2}$. Therefore, the condition of the problem is satisfied only by

$$
S=\frac{1}{2}\left(S_{1}+S_{2}+\sqrt{\left(S_{1}+S_{2}\right)^{2}-4 S_{1} S_{2} \sin ^{2} \alpha}\right)
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_e3b1ddf947c7dd2094eeg-07.jpg?height=338&width=900&top_left_y=277&top_left_x=585)

## Answer

$\frac{1}{2}\left(S_{1}+S_{2}+\sqrt{\left(S_{1}+S_{2}\right)^{2}-4 S_{1} S_{2} \sin ^{2} \alpha}\right)$