A container in the shape of a triangular prism stands on one of its triangular faces. Three spheres of radius 1 are placed inside the container, each touching the triangular bottom. Each sphere touches two of the rectangular faces of the container and each sphere touches the other two spheres. A fourth sphere of radius 1 is placed on top of the three spheres, touching each of the three spheres and the top of the prism. The volume of the prism is closest to
(A) 48.00
(B) 47.75
(C) 47.50
(D) 47.25
(E) 47.00
To determine the volume of the prism, we calculate the area of its base and the height of the prism.

First, we calculate the area of its base.

Any cross-section of the prism parallel to its base has the same shape, so we take a cross-section 1 unit above the base.

Since each of the spheres has radius 1, this triangular cross-section will pass through the centre of each of the spheres and the points of tangency between the spheres and the rectangular faces. Let the vertices of the triangular cross-section be $A, B$ and $C$, the centres of the spheres be $X$, $Y$ and $Z$, and the points of tangency of the spheres (circles) to the faces of the prism (sides of the triangle) be $M, N, P, Q, R$, and $S$, as shown:

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-072.jpg?height=369&width=393&top_left_y=702&top_left_x=926)

Join $X$ to $M, S, Y, Z, A$, join $Y$ to $N, P, Z, B$, and $Z$ to $Q, R, C$.

We determine the length of $B C$. A similar calculation will determine the lengths of $A B$ and $A C$. By symmetry, $A B=A C=B C$.

Note that $Y P$ and $Z Q$ are perpendicular to $B C$ since radii are perpendicular to tangents.

Also, $Y P=Z Q=1$ since the radius of each sphere (and hence of each circle) is 1 .

Since $Z Y P Q$ has right angles at $P$ and $Q$ and has $Y P=Z Q=1$, then $Z Y P Q$ is a rectangle. Therefore, $Y Z=P Q$.

Since $Y Z$ passes through the point at which the circles touch, then $Y Z$ equals the sum of the radii of the circles, or 2 .

Thus, $P Q=Y Z=2$.

Since $A B=B C=C A$, then $\triangle A B C$ is equilateral and so $\angle A B C=60^{\circ}$.

Now $Y B$ bisects $\angle A B C$, by symmetry.

This means that $\angle Y B P=30^{\circ}$, which means that $\triangle B Y P$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.

Since $Y P=1$, then $B P=\sqrt{3}$, using the ratios of side lengths in a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.

Similarly, $Q C=\sqrt{3}$.

Therefore, $B C=B P+P Q+Q C=\sqrt{3}+2+\sqrt{3}=2+2 \sqrt{3}$.

This means that $A B=B C=C A=2+2 \sqrt{3}$.

To calculate the area of $\triangle A B C$, we drop a perpendicular from $A$ to $T$ on $B C$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-072.jpg?height=363&width=398&top_left_y=2044&top_left_x=923)

Since $\angle A B C=60^{\circ}$, then $\triangle A B T$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.

Since $A B=2+2 \sqrt{3}$, then $A T=\frac{\sqrt{3}}{2} A B=\frac{\sqrt{3}}{2}(2+2 \sqrt{3})$.

Therefore, the area of $\triangle A B C$ is $\frac{1}{2}(B C)(A T)=\frac{1}{2}(2+2 \sqrt{3})\left(\frac{\sqrt{3}}{2}(2+2 \sqrt{3})\right)=\frac{\sqrt{3}}{4}(2+2 \sqrt{3})^{2}$.

This means that the area of the base of the prism (which is $\triangle A B C$ ) is $\frac{\sqrt{3}}{4}(2+2 \sqrt{3})^{2}$.

Now we calculate the height of the prism.

Let the centre of the top sphere be $W$.

The vertical distance from $W$ to the top face of the prism equals the radius of the sphere, which is 1 .

Similarly, the vertical distance from the bottom face to the plane through $X, Y$ and $Z$ is 1 .

To finish calculating the height of the prism, we need to determine the vertical distance between the cross-section through $X, Y, Z$ and the point $W$.

The total height of the prism equals 2 plus this height.

Since the four spheres touch, then the distance between any pair of centres is the sum of the radii, which is 2 .

Therefore, $W X=X Y=Y Z=W Z=W Y=X Z=2$. (This means that $W X Y Z$ is a tetrahedron with equal edge lengths.)

We need to calculate the height of this tetrahedron.

Join $W$ to the centre, $V$, of $\triangle X Y Z$.

By symmetry, $W$ is directly above $V$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-073.jpg?height=303&width=371&top_left_y=1014&top_left_x=931)

Let $G$ be the midpoint of $Y Z$. This means that $Y G=G Z=1$.

Join $V$ to $Y$ and $G$.

Since $V$ is the centre of $\triangle X Y Z$, then $V G$ is perpendicular to $Y Z$ at $G$.

Since $\triangle X Y Z$ is equilateral, then $\angle V Y G=\frac{1}{2} \angle X Y Z=30^{\circ}$. This is because $V$, the centre of equilateral $\triangle X Y Z$, lies on the angle bisectors of each of the angles.

$\triangle Y V G$ is another $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, and so $Y V=\frac{2}{\sqrt{3}} Y G=\frac{2}{\sqrt{3}}$.

Finally, $\triangle W Y V$ is right-angled at $V$.

Thus, $W V=\sqrt{W Y^{2}-Y V^{2}}=\sqrt{2^{2}-\frac{4}{3}}=\sqrt{\frac{8}{3}}$.

This means that the height of the prism is $1+1+\sqrt{\frac{8}{3}}$.

The volume of the prism is equal to the area of its base times its height, which is equal to $\frac{\sqrt{3}}{4}(2+2 \sqrt{3})^{2} \cdot\left(2+\sqrt{\frac{8}{3}}\right)$ which is approximately 46.97.

Of the given answers, this is closest to 47.00 .

ANsWER: (E)