Cone of Light

As shown above, a hyperbola can be generated by having a perpendicular plane cutting through the sections of the twin cones. In this case, the wall acts as such a cutting plane while the lamp's light source is giving out cones of light due to the blocking of the lampshades as shown below.

Now suppose the ratio of the cone's radius $R$ and height $h$ = $k = \dfrac{r}{h}$ . The y-axis will be variable for the height, and for variable $r$ , we can set up two more variables: $r^2 = x^2 + z^2$ .

Thus, $k^2 = \dfrac{r^2}{y^2} = \dfrac{x^2 + z^2}{y^2}$ .

Then, $y^2 = \dfrac{x^2 + z^2}{k^2}$ .

$\dfrac{y^2}{h^2}= \dfrac{x^2}{R^2} + \dfrac{z^2}{R^2}$ .

At $z = R$ , $\dfrac{y^2}{h^2} - \dfrac{x^2}{R^2} = 1$

So according to the given equation, it is obvious that such ration $k = \dfrac{3}{4}$ .

Now we can visualize the lampshade as an isosceles trapezoid as shown:

From the question, $AC = 24$ , and $2AB = CD$ .

Suppose $AB = 3n$ for some number $n$ . Then $k = \dfrac{AB}{AO} = \dfrac{3}{4}$ , so $AO = 4n$ .

Similarly, $CD = 6n$ , and $CO = 8n$ .

Thus, $AC = 4n+8n = 12n = 24$ . $n = 2$ .

As a result, the radius of the top opening = $3n = \boxed{6}$ .