3. A covered football field of rectangular shape with a length of $90 \mathrm{m}$ and a width of 60 m is being designed, which should be illuminated by four spotlights, each hanging at some point on the ceiling. Each spotlight illuminates a circle, the radius of which is equal to the height at which the spotlight hangs. It is necessary to find the minimum possible ceiling height, under which the following conditions are met: every point on the football field is illuminated by at least one spotlight; the ceiling height must be a multiple of 0.1 m (for example, 19.2 m, 26 m, 31.9 m).
Answer: 27.1

Let in rectangle $ABCD$ the diagonals intersect at point $O$. Place the projectors on the ceiling above the points that are the midpoints of segments $AO, BO, CO$, and $DO$, at a height equal to a quarter of the diagonal of the rectangle. Then the first projector will fully illuminate a circle, inside which is a rectangle with diagonal $AO$, the second projector will illuminate a rectangle with diagonal $BO$, and so on. In the end, the entire field will be illuminated. In this case, the height of the ceiling is $\frac{\sqrt{90^{2}+60^{2}}}{4}=\frac{5}{2} \sqrt{117}=\frac{15}{2} \sqrt{13}$ m. It is necessary to prove that there is no more "efficient" arrangement. For this, we apply the pigeonhole principle. If we take 5 points on the field: points $A, B, C, D$, and $O$ (it is important that the distance between any two of these points is not less than half the diagonal of the rectangle), and assume that the diameter of the circle is less than half the diagonal, then one projector will illuminate no more than one of these four points. Therefore, at least one of these 5 points will remain unlit.

It remains to estimate the value of $\frac{15}{2} \sqrt{13}$ from above. Since $\frac{15}{2} \sqrt{13}=\sqrt{731.25}$, and $27^{2}=729<731.25$ $<734.41=27.1^{2}$, the height of the ceiling should be 27.1 m.