, 2012

The point $A$ lies on the sphere, vertically above the centre of the sphere, $O$.

Similarly, point $B$ lies on the sphere, vertically below the centre of the sphere.

The top and bottom faces of the cylinder touch the sphere at points $A$ and $B$ respectively, as shown.

The segment $AB$ passes through the centre of the sphere.

Since $OA$ is a radius of the sphere, it has length $r$.

Similarly, $OB$ has length $r$.

Thus, the length of segment $AB$ is $2r$.

However, segment $AB$ also represents the perpendicular distance between the top and bottom faces of the cylinder and thus has length equal to the height of the cylinder, $h$.

Therefore, the equation relating the height of the cylinder to the radius of the sphere is $h=2r$.

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The volume of the sphere is given by the formula $\frac{4}{3}\pi r^3$.

Since the volume of the sphere is $288\pi$, then $\frac{4}{3}\pi r^3=288\pi$ or $4\pi r^3=3\times 288\pi$, and so $r^3=\frac{3\times 288\pi}{4\pi}=216$.

The volume of the cylinder is given by the formula $\pi r^2h$.

From part (a), $h=2r$.

Thus, the volume of the cylinder is $\pi r^2h=\pi r^2(2r)=2\pi r^3$.

Since $r^3=216$, then the volume of the cylinder is $2\pi(216)=432\pi$.
The shape of the space that Darla is able to travel within is determined by the set of points that are exactly 1 km from the nearest point on the cube.

The surface of the cube is comprised of three types of points.

These are points on a face of the cube, points on an edge of the cube, and points at a vertex of the cube.

To determine the volume of space that Darla is able to occupy, we will consider each of these three types of surface points as separate cases.

Case 1 - Points on a face of the cube.

The question we must answer is, “What set of points is exactly 1 km away from the nearest point that is on a face of the cube?"

Consider that if Darla begins from any point on a face of the cube, the maximum distance that she can travel is 1 km.

To travel 1 km away from that point, but not be nearer to any other point on the cube, Darla must travel in a direction perpendicular to the face of the cube.

If Darla travels 1 km in a direction perpendicular to the face of a cube, beginning from each of the points on a face of the cube, then the shape of this space that she can occupy is another cube of side length 1 km.

This new cube extends directly outward from the original cube.

Since this can be repeated from each of the 6 faces of the original cube, then in this case Darla can occupy a volume of space equal to $6\times1\times1\times1$ or 6 km$^3$, as shown in Figure 1.

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Case 2 - Points on an edge of the cube.

The question we must answer is, “What set of points is exactly 1 km away from the nearest point that is on an edge of the cube?"

Consider point $A$, the midpoint of the edge on which it lies.

Let points $B$ and $C$ be the midpoints of their respective edges also, as shown in Figure 2.

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Darla can travel to both points $B$ and $C$ since point $A$ is on the original cube, 1 km away from each of these points.

However, Darla can also travel from $A$ to any point on the arc $BC$.

This arc is one quarter of the circumference of the circle with centre $A$, radius 1 km, and passing through points $B$ and $C$ (since $\angle BAC=90^{\circ}$).

Darla can repeat this movement, beginning from any point on this edge. Thus, this shape of space that can be occupied is one quarter of a cylinder of radius 1 km and height 1 km, and has a volume of $\frac{1}{4}\pi(1)^2(1)=\frac{1}{4}\pi$ km$^3$ (see Figure 3).

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Since this can be repeated from any point on each of the 12 edges of the original cube, then in this case Darla can occupy a volume of space equal to $12\times \frac{1}{4}\pi$ or $3\pi$ km$^3$, as shown in Figure 4.

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Case 3 - Points at a vertex of the cube.

The question we must answer is, “What set of points is exactly 1 km away from the nearest point that is at a vertex of the cube?" Consider point $P$, a vertex of the original cube.

Let points $Q$, $R$ and $S$ be vertices of external cubes, as shown in Figure 5.

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Darla can travel to points $Q$, $R$ and $S$ since point $P$ is on the original cube, 1 km away from each of these points.

From Case 2, we also know that Darla can travel anywhere along the arcs $QR$, $RS$ and $SQ$.

However, Darla can also travel up to 1 km outward from $P$ to any point on the 3-dimensional surface contained within these 3 arcs.

Since $\angle SPQ=\angle SPR=\angle QPR=90^{\circ}$, this surface is one eighth of the surface of the sphere with centre $P$ and radius 1 km (see Figure 5).

Thus, the volume of the space that can be occupied is $\frac{1}{8}\times\frac{4}{3}\pi(1)^3=\frac{1}{6}\pi$ km$^3$.

Since this can be repeated from each of the 8 vertices of the original cube, then in this case Darla can occupy a volume of space equal to $8\times \frac{1}{6}\pi$ or $\frac{4}{3}\pi$ km$^3$, as shown in Figure 6.

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The total volume of space that Darla can occupy is the sum of the volume of space given by the 3 cases above.

That is, Darla can occupy a volume of space equal to $6+3\pi+\frac{4}{3}\pi$ or $(6+\frac{13\pi}{3})$ km$^3$.