Fill Till The Top!

Consider the top view, the two blue Circles represent the top balls and the green Circles represent the bottom two Circles. As seen in the figure, the horizontal distance between the centers of top ball and bottom ball can be calculated by Pythagoras Theorem,

$= \sqrt{2.5^2 + 2.5^2} = 2.5\sqrt{2} cm$

Now, consider a right triangle with the following sides-

$1)$ Hypotenuse connects to the centers of top and bottom balls $= 5cm.$

$2)$ One leg is the horizontal distance between the centers $=2.5\sqrt{2}cm.$

$3)$ Other leg is the vertical distance between the centers.

Using Phythsgoras , Other leg $=2.5\sqrt2cm$

Total height of the balls, $h =( 2.5\sqrt2 + 5)cm$

Volume of water = Volume of Cylinder upto $h -$ Volume of $4$ balls

$= π(5^2)(5 + 2.5\sqrt2)cm^3 - 4\times \dfrac {4}{3} (2.5)^3 cm^3 =\boxed{ \dfrac {125π(2 + 3\sqrt2)}{6}cm^3}$

Therefore, $a + b + c + d = 125 + 2 + 3 +6 =\boxed{136}$

Connecting the centers of the $4$ balls creates a tetrahedron with sides of $a = 5$ , whose midsphere radius is $\frac{a}{\sqrt{8}} = \frac{5}{\sqrt{8}}$ . This means the height of the water is $h = \frac{5}{2} + \frac{5}{\sqrt{8}} + \frac{5}{\sqrt{8}} + \frac{5}{2} = 5(\frac{\sqrt{2}}{2} + 1)$ .

The amount of water needed is the volume of the cylinder (with the height $h$ above) minus the volume of the $4$ balls, which is $V = \pi \cdot 5^2 \cdot 5(\frac{\sqrt{2}}{2} + 1) - 4 \cdot \frac{4}{3} \cdot \pi \cdot (\frac{5}{2})^3 = \frac{125\pi(2 + 3\sqrt{2})}{6}$ .

Therefore, $a = 125$ , $b = 2$ , $c = 3$ , and $d = 6$ , and $a + b + c + d = \boxed{136}$ .