Problem
If three balls with a diameter of $8 \ cm$ are arranged on a flat surface so that each ball touches the other two as shown in the figure, and a ball with a diameter of $7 \ cm$ is placed on top of them, find the height of the object.
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Solution
If you draw a three-dimensional figure faithfully to the problem statement, it will look like the following:
Let the diameters of the three balls in the bottom row be $x$, and the diameter of the one ball in the top row be $y$. Here, $A$ is the vertex of this object, $B$ and $C$ are the feet of the perpendicular line drawn from $A$ to the equilateral triangle $O_1O_2O_3$ and the plane, respectively, and $D$ is the tangent point of the spheres centered at $O_1$ and $O_2$.
Regarding the right triangle $O_4 O_1 D$, the following can be said from the Pythagorean theorem:
$$O_1 D^2+DO_4^2=O_4 O_1^2,$$
$$(\frac{x}{2})^2+DO_4^2=(\frac{x+y}{2})^2,$$
$$DO_4^2=\frac{xy}{2}+\frac{y^2}{4},$$
$$∴ \quad DO_4=\sqrt{\frac{xy}{2}+\frac{y^2}{4}}.$$
Next, since $B$ is the center of gravity of the equilateral triangle $O_1 O_2 O_3$, we know that
$$BD=\frac{\sqrt{3}}{2} x×\frac{1}{3}=\frac{\sqrt{3}}{6} x.$$
Therefore, for the right triangle $O_4 BD$, the following holds:
$$O_4 B^2+BD^2=DO_4^2,$$
$$O_4B^2=DO_4^2-BD^2=\frac{xy}{2}+\frac{y^2}{4}-\frac{x^2}{12}=\frac{3y^2+6xy-x^2}{12},$$
$$∴ \quad O_4 B=\frac{1}{2} \sqrt{\frac{3y^2+6xy-x^2}{3}}.$$
Because the height of this object is $AC=AO_4+O_4 B+BC$, we have
$$AC=\frac{x+y+\sqrt{\frac{3y^2+6xy-x^2}{3}}}{2}. \qquad [*]$$
Since $x=8$ and $y=7$ from the problem statement, by substituting these into $[*]$, we get the following:
$$AC=\frac{8+7+\sqrt{\frac{3×7^2+6×8×7-8^2}{3}}}{2}=\frac{15+\sqrt{\frac{419}{3}}}{2}≒13.409.$$
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$$Ans.\ 13.409 \ cm.\quad $$
Reference
Yoshikazu Yamakawa, ed. (1997) Okayama ken no Sangaku (Sangaku in Okayama Prefecture), p.51; p.303.