A cube with edge length 8 is balanced on one of its vertices on a horizontal table such that the diagonal from this vertex through the interior of the cube to the farthest vertex is vertical. When the sun is directly above the top vertex, the shadow of the cube on the table is a regular hexagon. The area of this shadow can be written in the form $a \sqrt{b}$, where $a$ and $b$ are positive integers and $b$ is not divisible by any perfect square larger than 1 . What is the value of $a+b$ ?
Let $O$ be the vertex of the cube farthest away from the table.

Let $A, B$ and $C$ be the vertices of the cube connected to $O$ with edges.

Since the cube has edge length 8 , then $O A=O B=O C=8$.

Note that $\angle A O B=\angle A O C=\angle B O C=90^{\circ}$, which means that each of $\triangle A O B, \triangle A O C$ and $\triangle B O C$ is a right-angled isosceles triangle, which means that $A B=\sqrt{2} A O=8 \sqrt{2}$, and so $A B=A C=B C=8 \sqrt{2}$.

Let $P, Q$ and $R$ be the vertices that complete the square faces $P A O B, Q A O C$ and $R B O C$. From directly above, the cube looks like this:

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-021.jpg?height=399&width=371&top_left_y=2069&top_left_x=931)

When the sun is directly overhead, the shadow of the cube will look exactly like the area of the "flat" hexagon $A P B R C Q$. In mathematical terms, we are determining the area of what is called a projection.

To find the area of this figure, we need to know some lengths, but we have to be careful because not all of the edges in the edges in the diagram above are "flat".

We do know that points $A, B$ and $C$ are in the same horizontal plane, and we also know that $A B=A C=B C=8 \sqrt{2}$. This means that these are true lengths in that the points $A, B$ and $C$ are this distance apart.

Note that flat quadrilateral $P A O B$ is divided into two regions of equal area by $A B$. Similarly, $Q A O C$ is divided into two regions of equal area by $A C$, and $R B O C$ is divided into two regions of equal area by $B C$.

In other words, the area of $\triangle A B C$ is half of the area of hexagon $A P B R C Q$, so to find the area of $A P B R C Q$, we double the area of $\triangle A B C$.

So we need to calculate the area of equilateral $\triangle A B C$ which has side length $8 \sqrt{2}$.

Let $M$ be the midpoint of $B C$. Thus, $B M=C M=4 \sqrt{2}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-022.jpg?height=320&width=341&top_left_y=827&top_left_x=946)

Since $\triangle A B C$ is equilateral, then $A M$ is perpendicular to $B C$.

By the Pythagorean Theorem in $\triangle A M C$,

$$
A M=\sqrt{A C^{2}-M C^{2}}=\sqrt{(8 \sqrt{2})^{2}-(4 \sqrt{2})^{2}}=\sqrt{128-32}=\sqrt{96}
$$

Therefore, the area of $\triangle A B C$ is

$$
\frac{1}{2} \cdot C B \cdot A M=\frac{1}{2} \cdot 8 \sqrt{2} \cdot \sqrt{96}=4 \sqrt{192}=4 \sqrt{64 \cdot 3}=4 \cdot 8 \sqrt{3}=32 \sqrt{3}
$$

This means that the area of the hexagonal shadow is $64 \sqrt{3}$.

Since this is in the form $a \sqrt{b}$ where $a$ and $b$ are positive integers and $b$ is not divisible by the square of an integer larger than 1 , then $a=64$ and $b=3$ and so $a+b=67$.

ANSWER: 67