3. Three lines intersect at one point 0. Outside these lines, a point M is taken and perpendiculars are dropped from it to them. The points $\mathrm{H}_{1}, \mathrm{H}_{2}$ and $\mathrm{H}_{3}$ are the bases of these perpendiculars. Find the ratio of the length of the segment OM to the radius of the circle circumscribed around the triangle $\mathrm{H}_{1} \mathrm{H}_{2} \mathrm{H}_{3}$.

Answer: 2.
Solution

![](https://cdn.mathpix.com/cropped/2024_05_06_fa440bccba1e99f6def8g-1.jpg?height=376&width=478&top_left_y=159&top_left_x=1611)

$2, \prime \prime$

![](https://cdn.mathpix.com/cropped/2024_05_06_fa440bccba1e99f6def8g-1.jpg?height=358&width=463&top_left_y=194&top_left_x=2204)

First, consider two intersecting lines and a point $M$ lying outside these lines. $\mathrm{H}_{1}$ and $\mathrm{H}_{2}$ are the feet of the perpendiculars dropped from point $\mathrm{M}$ to these lines. The possible cases are:

1) $\mathrm{H}_{1}$ and $\mathrm{H}_{2}$ lie on opposite sides of the line $0 M$
2) $\mathrm{H}_{1}$ and $\mathrm{H}_{2}$ lie on the same side of the line $0 M$;
3) $\mathrm{H}_{1}$ or $\mathrm{H}_{2}$ coincides with point 0

In the first case, $\angle \mathrm{MH} \mathrm{H}_{1} \mathrm{O} + \angle \mathrm{MH} \mathrm{H}_{2} \mathrm{O} = 90^{\circ} + 90^{\circ} = 180^{\circ}$. The quadrilateral $\mathrm{MH}_{1} \mathrm{OH}_{2}$ is cyclic, i.e., points $\mathrm{M}, \mathrm{H}_{1}, \mathrm{O}$, and $\mathrm{H}_{2}$ lie on the same circle, with the diameter of this circle being segment $OM$. In the second case, the quadrilateral $\mathrm{MH}_{1} \mathrm{H}_{2} \mathrm{O}$ is also cyclic. Points $M, \mathrm{H}_{1}, \mathrm{O}$, and $\mathrm{H}_{2}$ again lie on the same circle with diameter $OM$. In the third case, it is obvious that for points $M, H_{1}, 0$, and $H_{2}$, we get the same result.

Draw a third line and mark point $\mathrm{H}_{3}$. Then, obviously, points $\mathrm{H}_{1}$, $\mathrm{H}_{2}$, and $\mathrm{H}_{3}$ lie on the circle with diameter $OM$, and the desired ratio is 2.

## Comments on Grading

The problem is solved, but in proving that $\mathrm{M}, \mathrm{H}_{1}, \mathrm{O}$, and $\mathrm{H}_{2}$ lie on the same circle, only case 1) or case 2) is considered -5 points.