7. Given a triangle $ABC$. On side $AC$, a point $Q$ is chosen in such a way that the length of the segment $MK$, where $M$ and $K$ are the feet of the perpendiculars dropped from point $Q$ to sides $AB$ and $BC$ respectively, is minimized. In this case, $QM=1$, $QK=\sqrt{2}$, and $\angle B=45^{0}$. Find the area of triangle $ABC$.
Solution. Let the lengths of the perpendiculars dropped from point $Q$ to the base $AC$ be denoted as $d_{1}$ and $d_{2}$; let $\angle B=\beta$. The quadrilateral $MBKQ$ is inscribed in a circle, and $BQ$ is its diameter. Using the formula for the radius of the circumcircle of triangle $MBK$, we have

$$
BQ = \frac{MK}{\sin \beta} \Rightarrow MK = BQ \cdot \sin \beta
$$

Since the angle $\beta$ is fixed, the length of segment $MK$ is smaller the shorter the length of $BQ$. Therefore, point $Q$ is the foot of the perpendicular dropped from point $B$ to $AC$, and $BQ$ is the height (the foot of the perpendicular $Q$ lies on the side $AC$ itself, not on its extension, since angles $A$ and $C$ are acute; if, for example, angle $A$ were obtuse, point $M$ would lie on the extension of side $AB$, not on the side itself); let $BQ = h$. We will find the area of $\triangle ABQ$, assuming for now that $h$ is a known quantity. We have $AQ = \frac{h}{\sin \angle A} = \frac{h}{\sqrt{1 - \frac{d_{1}^{2}}{h^{2}}}}$. Then $S_{ABQ} = \frac{h^{4}}{\sqrt{h^{2} - d_{1}^{2}}}$. Similarly, $S_{BQC} = \frac{h^{4}}{\sqrt{h^{2} - d_{2}^{2}}}$. The desired area is their sum

$$
S_{ABC} = S_{ABQ} + S_{BQC} = \frac{h^{4}}{2} \cdot \left( \frac{1}{\sqrt{h^{2} - d_{1}^{2}}} + \frac{1}{\sqrt{h^{2} - d_{2}^{2}}} \right)
$$

It remains to find $h$. Since $h = BQ$, it follows from (1) that $h = \frac{MK}{\sin \beta}$. We will find $MK$ from $\Delta MQK$ (in which $\angle MQK = 180^{\circ} - \beta$) using the cosine theorem: $MK^{2} = d_{1}^{2} + d_{2}^{2} + 2 d_{1} d_{2} \cos \beta$. Thus,

$$
h = \frac{MK}{\sin \beta} = \frac{\sqrt{d_{1}^{2} + d_{2}^{2} + 2 d_{1} d_{2} \cos \beta}}{\sin \beta}
$$

To use (2), first for convenience, we will calculate $h^{2} - d_{1}^{2} = \frac{d_{1}^{2} + d_{2}^{2} + 2 d_{1} d_{2} \cos \beta - d_{1}^{2} \sin^{2} \beta}{\sin^{2} \beta} = \frac{(d_{1} \cos \beta + d_{2})^{2}}{\sin^{2} \beta}$. Similarly, $h^{2} - d_{2}^{2} = \frac{(d_{2} \cos \beta + d_{1})^{2}}{\sin^{2} \beta}$. Substituting the obtained expressions into (2), we find $S_{ABC} = \frac{(d_{1}^{2} + d_{2}^{2} + 2 d_{1} d_{2} \cos \beta)^{2}}{2 (d_{1} \cos \beta + d_{2})(d_{2} \cos \beta + d_{1}) \sin \beta}$. Using the numerical data from the problem, we get the answer.

Answer: $S_{ABC} = \frac{25}{6}$.

Interregional Olympiad for Schoolchildren Based on Departmental Educational Organizations in Mathematics