Archimedes' Riddle

The volume of the cone = $\dfrac{\pi}{3}R^{2}h$ .

The volume of the sphere = $\dfrac{4\pi}{3}r^{3}$ .

With half of the original height, the cone's radius will be halved as well because within a conic 2-D plane, they are two similar triangles. Hence, the volume of the remaining water = $\dfrac{\pi}{3}(\dfrac{R}{2})^{2}(\dfrac{h}{2}) = \dfrac{1}{8}(\dfrac{\pi}{3}R^{2}h)$ .

The remaining water is $\dfrac{1}{8}$ of the full cone's volume $V$ , and when added this to the sphere's volume, it will equal to half of the full cone's volume as half of the water would be poured out to result in these volumes.

Therefore, sphere's volume = $(\dfrac{1}{2} -\dfrac{1}{8})V = (\dfrac{3}{8})V$ .

Now substituting in terms of $R$ , $h$ , $r$ variables:

$\dfrac{4\pi}{3}r^{3} = (\dfrac{3}{8})\dfrac{\pi}{3}R^{2}h$

$r^{3} = (\dfrac{3}{32})R^{2}h$

Now we have to find the radius of the inscribed sphere within the cone as a circle inscribed in a triangle:

From the image above, the radius $r$ is perpendicular to all three tangents, and so the area of the whole triangle equals the sum of three triangles with height $r$ = $(\dfrac{1}{2})r(2R + 2\sqrt{R^2 + h^2})$ .

But we know that this triangle has base $2R$ and height $h$ .

Hence, $Rh = r(R + \sqrt{R^2 + h^2})$ . $r = \dfrac{Rh}{R + \sqrt{R^2 + h^2}} = \dfrac{h}{1 + \sqrt{1 + (\frac{h}{R})^2}}$ .

Substituting the term into the previous equation:

$r^{3} = (\dfrac{h}{1 + \sqrt{1 + (\frac{h}{R})^2}})^3 = (\dfrac{3}{32})R^{2}h$ .

Suppose $\dfrac{h}{R} = k$ ; $h = Rk$ .

$(\dfrac{Rk}{1 + \sqrt{1 + k^2}})^3 = (\dfrac{3}{32})R^{2}(Rk)$ .

$(\dfrac{k}{1 + \sqrt{1 + k^2}})^3 = (\dfrac{3}{32})k$ .

$\dfrac{k^2}{(1 + \sqrt{1 + k^2})^3} = \dfrac{3}{32}$ .

Then let $x = \sqrt{1 + k^2}$ :

$\dfrac{x^{2}-1}{(1 + x)^3} = \dfrac{3}{32}$ .

$x^{2} - 1 = (\dfrac{3}{32})((1 + x)^3)$

$32(x + 1)(x - 1) = 3(x + 1)^3$

$0 = (x+1)(3(x^2 + 2x +1) - 32(x-1))$

$0 = (x+1)(3x^2 - 26x +35)$

$0 = (x + 1)(3x-5)(x-7)$

$x$ can't be $-1$ because the square root value is positive. Therefore, $x = \dfrac{5}{3}$ or $x=7$ .

If $x=7$ , $\sqrt{1 + k^2} = 7$ . $k = \sqrt{7^2 - 1} = \sqrt{48} = 4\sqrt{3}$ .

If $x=\dfrac{5}{3}$ ), $\sqrt{1 + k^2} = \dfrac{5}{3}$ ). $k = \sqrt{(\dfrac{5}{3}) ^2 - 1} = \dfrac{4}{3}$ .

Since both $R$ and $h$ are integers, its ratio must be rational number. Thus, $k$ can't be $4\sqrt{3}$ .

Hence, $h = \dfrac{4R}{3}$ .

Plugging in this term into the original equation for $r$ :

$r = \dfrac{h}{1 + \sqrt{1 + k^2}} = \dfrac{h}{1 + \sqrt{1 + (\frac{4}{3})^2}} = \dfrac{h}{\frac{8}{3}}$ .

Thus, $8r = 3h$ , and since they are co-prime, $r = 3$ ; $h = 8$ . That makes $R = 2r = 6$ .

As a result, $R + h + r = 6 + 8 + 3 = \boxed{17}$ .

$V_{cone}=\frac{πR^2h}{3}$ $V_{ball}=\frac{4πr^3}{3}$ The volume of the ball, and another half cone are spilled, and a 1/8 cone's volume is left: $V_{cone}- V_{ball}-\frac{1}{2}V_{cone}=\frac{1}{8}V_{cone}$ $3V_{cone}=8V_{ball}$ $πR^2h=\frac{32}{3}πr^3$ Multiply both sides by $\frac{3}{πR^3}$ : $3\frac{h}{R}=32(\frac{r}{R})^3$ If the ball is touching the cone and the water surface, let β be the angle between the plane of the water surface and a line from the edge to the ball's centre. The water surface and a line from edge to apex make an angle 2β. So we can show that $\frac{r}{R}=\tan β$ then $\frac{h}{R}=\tan 2β$ . $3\tan 2β=32\tan^3 β$

Using the double angle formula $\tan{2β} = \frac{2 \tan{β}}{1- \tan^2{β}}$ , and setting $x=\tanβ$ we get $\frac{6x}{1-x^2}=32x^3$ which simplifies to $16x^4 - 16 x^2 + 3 = 0$ so that $x^2= \frac{16 \pm \sqrt{16^2-4×3×16} }{2×16} = \frac{2 \pm 1 }{4}$ so, knowing that $x>0$ , either $x=\frac{1}{2}$ or $x= \frac{1}{2}\sqrt{3}$ .

For $x= \frac{1}{2}\sqrt{3}$ , $\frac{ r}{R}$ is not rational, so this will not lead to a solution of the problem.

For $x=\frac{1}{2}$ we get $\frac{r}{R}=\frac{1}{2}$ and $\tan {2β}=\frac{h}{R}=\frac{4}{3}$ . So here we get the ratio $r:R:h=3÷6÷8$ which are coprime, hence the answer is $R+h+r=\boxed{17}$