Who listens to their architect anyway?

Now suppose the lamp has a radius $r$ and is suspended a height $h$ off the ground in a room with height $2h$. Again, the radius of the shadow on the ceiling is $R.$

For whatever reason, the restaurant’s architect insists that she wants $r$, $h$, and $R$, as measured in feet, to all be whole numbers. What is the smallest value of $R$ for which this is possible?

Let's imagine that this restaurant is named, say Cafe Pythagoras. Abstracting the Classic problem's solution by having a circle of radius $r$ centered at $(0,h),$ then we have the minimal distance from the lamp to the reflected path of the light from the ground to the ceiling as $d^* = 2h \sin \theta.$ Therefore, we see that $$R = \min \{ 3h \tan \theta \mid 2h \sin \theta \geq r \} = 3h \tan \left( \sin^{-1} \left(\frac{r}{2h} \right) \right) = \frac{3hr}{\sqrt{4h^2 - r^2}}.$$

Since the architect at Cafe Pythagoras wants to make sure that $r$, $h$ and $R$ are all integers, we should start searching through Pythagorean triples for possible solutions, since if the denominator $\sqrt{4h^2 -r^2}$ is not an integer, then there is no way for $R$ to be an integer. In particular we need $2h$ to be the hypotenuse of a Pythagorean triple and either $r=2a$ or $r=2b$. Naturally, let's start with the triple $(3,4,5)$. So we set $h=5$ and try $r=6,$ which yields the undesirable $R = \frac{45}{4} \not\in \mathbb{N}.$ If instead, $h=5$ and $r=8$, then we get $R = 20,$ so at least we know that this your architect is not asking for the impossible. By taking scaled multiples, say $h=5k$, $r=8k$ and $R=20k,$ for any $k \in \mathbb{N},$ we see that in fact the architect can have infinitely many solutions, with the smallest one generated by the (3,4,5) primitive Pythagorean triple having a shadow radius of $R=20$.

Let's assume that we have some integers $a, b, c \in \mathbb{N}$ with $a^2 + b^2 = c^2,$ $\gcd (a,b,c) = 1,$ and $c \gt 5$ and see if we can come up with any other solutions for our Pythagorean architect. Without loss of generality, let's see what would happen if we set $h = c$ and $r = 2b.$ In this case we get $$R = \frac{6bc}{\sqrt{4h^2 - (2b)^2}} = \frac{6bc}{2a} = \frac{3bc}{a}.$$ Since $\gcd(a,b,c) = 1,$ if $a \ne 3,$ then $R \not\in \mathbb{N}.$ Since the only primitive Pythagorean theorem that contains $3$ is $(3,4,5),$ we see that if $(a,b,c) \ne (3,4,5),$ then $R \not\in\mathbb{N}.$

Therefore, we have confirmed two very important things: Firstly, that the smallest possible value of $R$ at Cafe Pythagoras is $R=20,$ when $h=5$ and $r=8.$ And, lastly, but perhaps equally important, I don't think that I'll be going to Cafe Pythagoras. With a sizeable 8' radius orb floating with their centers 5' off the ground, that leaves only 1' of clearance remaining between the orbs and the ground. Even if I got somehow situated beneath the giant lamp by sitting on the floor I guess, the enormous 20' circular ceiling shadows would likely throw off the aesthetics a bit. Perhaps I'm just too old to be the target audience for this restaurant. I hear the food is good though!