A steel cube has edges of length $3 \mathrm{~cm}$, and a cone has a diameter of $8 \mathrm{~cm}$ and a height of $24 \mathrm{~cm}$. The cube is placed in the cone so that one of its interior diagonals coincides with the axis of the cone. What is the distance, in $\mathrm{cm}$, between the vertex of the cone and the closest vertex of the cube?
(A) $6 \sqrt{6}-\sqrt{3}$
(B) $\frac{12 \sqrt{6}-3 \sqrt{3}}{4}$
(C) $6 \sqrt{6}-2 \sqrt{3}$
(D) $5 \sqrt{3}$
(E) $6 \sqrt{6}$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-119.jpg?height=371&width=260&top_left_y=1517&top_left_x=1472)
The first difficult thing about this problem is visualizing the position of the cube inside the cone. From there, we must determine how to calculate the required distance. Since the cube is balanced in an upright position with the axis of the cone coincides with one of the internal diagonals of the cube, then the cube is oriented with one vertex, $A$, pointing downwards, three vertices, $B, C$ and $D$, touching the walls of the cone, and one vertex, $Q$, pointing upwards.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-234.jpg?height=431&width=309&top_left_y=1606&top_left_x=1621)

(We do not need to consider the remaining three vertices.) By symmetry, $B, C$ and $D$ lie in a plane parallel to the base of the cone and form an equilateral triangle.

We draw a diagram to illustrate the configuration of $B$, $C, D$, and $Q$. In the diagram, $Q B$ is the length of the diagonal of the face of a cube (this requires careful visualization of where the faces of the cube are positioned) and $G$ is the intersection of the three medians. By symmetry, vertex $A$ will lie directly beneath $G$ in a line from $Q$. Thus, the line $Q G A$ lies

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-235.jpg?height=321&width=442&top_left_y=284&top_left_x=1278)
along the axis of the cone.

We can now determine the required distance by determining

- the vertical distance from the tip of the cone, $T$, at which $B, C$ and $D$ touch the cone, and
- the distance between $A$ and the plane formed by $B, C$ and $D$, and then subtracting these two distances.

Now for some calculations.

Since each edge of the tetrahedron is a diagonal of a face of the cube, then each has length $\sqrt{3^{2}+3^{2}}=3 \sqrt{2}$. In particular, $B Q, B C, B D$ and $C D$ all have length $3 \sqrt{2}$.

Next, if we extract triangle $B C D$, then and draw the medians $B X$ and $C Y$ which intersect at $G$, then we see that triangle $B G Y$ is a $30-60-90$ triangle, so $B G=\frac{2}{\sqrt{3}}(B Y)=\frac{2}{\sqrt{3}}\left(\frac{1}{2}(B A)\right)=\frac{1}{\sqrt{3}}(3 \sqrt{2})=\sqrt{6}$. This tells us that the distance from the axis of the cone to the points where the cube touches the cone is $\sqrt{6}$.

We can now calculate the distance from $T$ to the plane containing $B, C$ and $D$. To do, this we look at a partial cross-section of the cone labelling $B, T, G$, as well as $O$, the centre of the base of the cone, and $S$, the remaining vertex of the cross-section.

We can see that triangle $T G B$ is similar to triangle $T O S$, so $\frac{T G}{T O}=\frac{G B}{O S}$ or $T G=24\left(\frac{\sqrt{6}}{4}\right)=6 \sqrt{6}$,

(since the radius of the base of the cone is 4).
![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-235.jpg?height=934&width=604&top_left_y=1190&top_left_x=1278)

Lastly, we need to calculate the distance $A G$. To do this, we can calculate $A Q$ and $Q G$ and subtract these two distances.

$A Q$ is the main diagonal of the cube, so $A Q=\sqrt{3^{2}+3^{2}+3^{2}}=3 \sqrt{3}$.

Looking at triangle $B G Q$, which is right-angled, we see that

$Q G=\sqrt{B Q^{2}-B G^{2}}=\sqrt{(3 \sqrt{2})^{2}-(\sqrt{6})^{2}}=2 \sqrt{3}$.

Therefore, $A G=A Q-Q G=\sqrt{3}$.

Therefore, the required distance is $T A=T G-A G=6 \sqrt{6}-\sqrt{3}$.

Answer: (A)