Two identical cubes are stacked and they are placed inside a cone. The height of the cone is $34$ and the radius is $3$. The cubes have the same height. What is the volume of the space not taken up by the cubes?
I have made a bit of progress at this problem by trying to take cross section, but I haven't been able to reach a solution.
Start by drawing the cross section, an isosceles triangle of height $34$ and base $6$ (twice the radius). Now let's say the base is $AB$ and the vertex (tip of the cone) is $C$. The middle of the base is $O$. So we have $OA=OB=3$ and $OC=34$.
Now say that the side of a cube is $x$, so the height of two cubes is going to be $2x$. Let's draw the line parallel to $AB$ at height $2x$. It will intersect the sides at $A'$ and $B'$ and the $OC$ line at $O'$. Now here is the subtle thing: the length of $A'B'$ is not $x$ but $x\sqrt2$. Why is that? Because if the $A'$ and $B'$ are corners of a cube, and $O'$ is on the center of the face, then $A'B'$ is a diagonal of the face of the cube. Then $O'A'=x\frac{\sqrt 2}2$. Then noticing that $\triangle OCA$ is similar to $\triangle O'CA'$ we can write: $$\frac{O'A'}{OA}=\frac{O'C}{OC}$$ or:$$\frac{x\frac{\sqrt 2}2}{3}=\frac{34-2x}{34}$$ You can now calculate $x$. The volume of the two cubes is $2x^3$, which you need to subtract from the volume of the cone.