A $10 \mathrm{~m}$ long, rectangular classroom ceiling has two lamps placed on it, which emit conical beams of light with a $90^{\circ}$ opening angle. One of the lamps is located in the center of the ceiling, and it illuminates a circle with a $6 \mathrm{~m}$ diameter on the floor. The other lamp has been rotated so that the illuminated area can accommodate a $10 \mathrm{~m}$ segment in the longitudinal direction of the room, but no light falls on the two opposite walls. How far apart are the two lamps?
I. solution. The circle with a diameter of $6 \mathrm{~m}$ illuminated by the lamp in the center of the room has two endpoints of one of its diameters (points $A$ and $B$), and the lamp (point $C$) form a right-angled triangle (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_dd1f84df673c8a4ae697g-1.jpg?height=309&width=410&top_left_y=266&top_left_x=844)

This triangle is also isosceles because the lamp is not tilted. Thus, $A M = B M = C M = 3 \mathrm{~m}$, so the height corresponding to the hypotenuse of this triangle is $3 \mathrm{~m}$. Based on this, the height of the room is 3 meters.

For the other lamp (Figure 2), a right-angled triangle also forms, but there the hypotenuse is $10 \mathrm{~m}$: $q + p = 10$. From the height theorem: $p q = 9$. Therefore, $p + q = p q + 1$, which can also be written as $(p-1)(1-q)=0$. This is only zero if $p-1=0$ or $1-q=0$.

![](https://cdn.mathpix.com/cropped/2024_05_02_dd1f84df673c8a4ae697g-1.jpg?height=293&width=486&top_left_y=878&top_left_x=814)

One of $p$ and $q$ is 1 meter, then the other is 9 meters. The center of the room is 5 meters away, and this lamp is 1 meter away. Therefore, the lamps are 4 meters apart.

II. solution. In the triangle $A G B$, the angle opposite the side $A B$, $\angle A G B = 90^{\circ}$, so by Thales' theorem, $A E = E B = E G = \frac{10}{2} = 5$.

![](https://cdn.mathpix.com/cropped/2024_05_02_dd1f84df673c8a4ae697g-1.jpg?height=209&width=412&top_left_y=1495&top_left_x=846)

Figure 3

We know that $C E = E D = 3$. Since in the triangle $C F D$ the angle opposite $C D$ is $90^{\circ}$, by Thales' theorem, $E F = 3$.

Finally, applying the Pythagorean theorem to the triangle $E F G$, we get $F G^{2} = E G^{2} - F E^{2} = 5^{2} - 3^{2} = 16$, from which $F G = 4$.