300 on top of 3 100s

First we have to observe the tetrahedron. IMAGE

The base is an equilateral triangle. The distance of the centre of the base from one of the vertices of the base is $200sin60^o \times \frac{2}{3}= \frac{200}{\sqrt 3}$ . Now apply Pythagorean theorem on the orange triangle to get $H$ (heihht). IMAGE

Thus, the maximum distance of a point on the larger sphere from the plane= $H+100+300=200 \sqrt\frac{11}{3} +400$ . GIF of the result is $782$ .

Note: I have used $H$ to denote height. Please don't confuse it with the plane named H whose notation I have not used anywhere.

Centers of all 4 spheres form a pyramid with equilateral triangular base of side $200$ and say height $h$

The distance from the vertex of the equilateral triangle to its centroid can be found by 30-60-90 triangles to be $\frac{200}{\sqrt{3}}.$

By the Pythagorean Theorem, we have

$\left(\frac{200}{\sqrt{3}}\right)^2 + h^2 = 400^2 \implies h = 100\cdot\sqrt{\frac{44}{3}} \approx 382.97$

$d = 100 + 300 + 100\cdot\sqrt{\frac{44}{3}} \approx 782.97$

Answer: $\boxed{782}$

![alt text] (http://s16.postimg.org/x4rmpmgxx/geom2.png)

So, $\lfloor AB \rfloor=\lfloor AO+TB \rfloor=\lfloor 400+382.79 \rfloor=\fbox{782}$

I would argue that you can balance a sphere on another sphere in a mathematical situation. Thus making it $100+100+300+300=\boxed{800}$

the centres of three base spheres form an equilateral triangle of side =200 which inturn a tetrahedron with centre of large sphere with slant height 400 whose projection on the base is 200/sqrt(3) so the height of the tetrahedron is given by sqrt((400 400)-(200 200/3)) and this must be added to base height and radius of the large sphere which is approx..782.97

let O1, O2, O3, O4 be the centers of the circles(O4 is radius of larger circle).
O1, O2, O3, O4 form a pyramid with equilateral triangle(O1,O2,O3) of side 200 as base and and edges of length 400 connected to O4.
the median of the equilateral triangle(M) , one of O1,02,O3, and O4 forms a right triangle with MO4 as height of pyramid.
height of pyramid = 100 $\sqrt{16-4/3}$ ~ 382.97.
height of highest point on sphere = 382.97+100+300= 782.97

This is exactly the same as 2004 AMC 10 #25 except with different numbers:

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=149&t=131337&