10.3. Different lines ${ }^{a}$ and ${ }^{b}$ intersect at point O. Consider all possible segments $\mathrm{AB}$ of length $l$, with endpoints A and B lying on ${ }^{a}$ and $^{b}$ respectively, and denote by $\mathrm{P}$ the point of intersection of the perpendiculars to the lines ${ }^{a}$ and ${ }^{b}$, dropped from A and B respectively. Find the geometric locus of points $\mathrm{P}$.
Answer. The circle with center O and radius $\frac{l}{\sin x}$, where $x$ is the angle between the lines $a$ and $b$. Solution. Angles RAO and RBO are right angles, so points A, B, O, and R lie on a circle with diameter OR. Therefore, R belongs to the circumcircle of triangle AOB, the radius of which, by the sine theorem, is $\frac{l}{2 \sin x}$, where $x$ is the angle between the lines $a$ and $b$. As we have already noted, OR is the diameter of this circle, so OR is always equal to $\frac{l}{\sin x}$, which means that point R always lies on the circle with center O and radius $\frac{l}{\sin x}$.

The only exceptional case of this construction is when the lines are perpendicular and the segment lies entirely on one of them, with one of the points A and B coinciding with O. In this case, point P will coincide with the other of A and B, and the distance to it will still be $l = \frac{l}{\sin x}$.

On the other hand, consider any point P on this circle, and drop perpendiculars PA and PB from it to the lines $a$ and $b$ respectively. Then RO is the diameter of the circumcircle of triangle PAB, so the radius of this circle is half of this diameter, i.e., $\frac{l}{2 \sin x}$. Angle APB is $x$ or $180 - x$, so by the sine theorem, AB is equal to $2 \frac{l}{2 \sin x} \cdot \sin (180 - x) = l$, and point P is the intersection of the perpendiculars PA and PB, drawn from the ends of the segment AB of length $l$, with the ends A and B lying on $a$ and $b$ respectively. The exceptional case of perpendicular lines is verified similarly.

Grading criteria. It is shown that point P lies on the specified circle: 4 points. It is shown that any point on this circle is point P from the problem statement for some A and B: 3 points. No mention of the exceptional case: minus 1 point.