Subfield: Geometry


Question: In the diagram, $A B C D$ and $P N C D$ are squares of side length 2, and $P N C D$ is perpendicular to $A B C D$. Point $M$ is chosen on the same side of $P N C D$ as $A B$ so that $\triangle P M N$ is parallel to $A B C D$, so that $\angle P M N=90^{\circ}$, and so that $P M=M N$. Determine the volume of the convex solid $A B C D P M N$.

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Solution:
Draw a line segment through $M$ in the plane of $\triangle P M N$ parallel to $P N$ and extend this line until it reaches the plane through $P, A$ and $D$ at $Q$ on one side and the plane through $N, B$ and $C$ at $R$ on the other side.

Join $Q$ to $P$ and $A$. Join $R$ to $N$ and $B$.

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So the volume of solid $A B C D P M N$ equals the volume of solid $A B C D P Q R N$ minus the volumes of solids $P M Q A$ and $N M R B$.

Solid $A B C D P Q R N$ is a trapezoidal prism. This is because $N R$ and $B C$ are parallel (since they lie in parallel planes), which makes $N R B C$ a trapezoid. Similarly, $P Q A D$ is a trapezoid. Also, $P N, Q R, D C$, and $A B$ are all perpendicular to the planes of these trapezoids and equal in length, since they equal the side lengths of the squares.

Solids $P M Q A$ and $N M R B$ are triangular-based pyramids. We can think of their bases as being $\triangle P M Q$ and $\triangle N M R$. Their heights are each equal to 2 , the height of the original solid. (The volume of a triangular-based pyramid equals $\frac{1}{3}$ times the area of its base times its height.)

The volume of $A B C D P Q R N$ equals the area of trapezoid $N R B C$ times the width of the prism, which is 2.

That is, this volume equals $\frac{1}{2}(N R+B C)(N C)(N P)=\frac{1}{2}(N R+2)(2)(2)=2 \cdot N R+4$.

So we need to find the length of $N R$.

Consider quadrilateral $P N R Q$. This quadrilateral is a rectangle since $P N$ and $Q R$ are perpendicular to the two side planes of the original solid.

Thus, $N R$ equals the height of $\triangle P M N$.

Join $M$ to the midpoint $T$ of $P N$.

Since $\triangle P M N$ is isosceles, then $M T$ is perpendicular to $P N$.

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Since $N T=\frac{1}{2} P N=1$ and $\angle P M N=90^{\circ}$ and $\angle T N M=45^{\circ}$, then $\triangle M T N$ is also right-angled and isosceles with $M T=T N=1$.

Therefore, $N R=M T=1$ and so the volume of $A B C D P Q R N$ is $2 \cdot 1+4=6$.

The volumes of solids $P M Q A$ and $N M R B$ are equal. Each has height 2 and their bases $\triangle P M Q$ and $\triangle N M R$ are congruent, because each is right-angled (at $Q$ and at $R$ ) with $P Q=N R=1$ and $Q M=M R=1$.

Thus, using the formula above, the volume of each is $\frac{1}{3}\left(\frac{1}{2}(1)(1)\right) 2=\frac{1}{3}$.

Finally, the volume of the original solid equals $6-2 \cdot \frac{1}{3}=\frac{16}{3}$.
We determine the volume of $A B C D P M N$ by splitting it into two solids: $A B C D P N$ and $A B N P M$ by slicing along the plane of $A B N P$.

Solid $A B C D P N$ is a triangular prism, since $\triangle B C N$ and $\triangle A D P$ are each right-angled (at $C$ and $D$ ), $B C=C N=A D=D P=2$, and segments $P N, D C$ and $A B$ are perpendicular to each of the triangular faces and equal in length.

Thus, the volume of $A B C D P N$ equals the area of $\triangle B C N$ times the length of $D C$, or $\frac{1}{2}(B C)(C N)(D C)=\frac{1}{2}(2)(2)(2)=4$. (This solid can also be viewed as "half" of a cube.)

Solid $A B N P M$ is a pyramid with rectangular base $A B N P$. (Note that $P N$ and $A B$ are perpendicular to the planes of both of the side triangular faces of the original solid, that $P N=A B=2$ and $B N=A P=\sqrt{2^{2}+2^{2}}=2 \sqrt{2}$, by the Pythagorean Theorem.)

Therefore, the volume of $A B N P M$ equals $\frac{1}{3}(A B)(B N) h=\frac{4 \sqrt{2}}{3} h$, where $h$ is the height of the pyramid (that is, the distance that $M$ is above plane $A B N P$ ).

So we need to calculate $h$.

Join $M$ to the midpoint, $T$, of $P N$ and to the midpoint, $S$, of $A B$. Join $S$ and $T$. By symmetry, $M$ lies directly above $S T$. Since $A B N P$ is a rectangle and $S$ and $T$ are the midpoints of opposite sides, then $S T=A P=2 \sqrt{2}$.

Since $\triangle P M N$ is right-angled and isosceles, then $M T$ is perpendicular to $P N$. Since $N T=\frac{1}{2} P N=1$ and $\angle T N M=45^{\circ}$, then $\triangle M T N$ is also right-angled and isosceles with $M T=T N=1$.

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Also, $M S$ is the hypotenuse of the triangle formed by dropping a perpendicular from $M$ to $U$ in the plane of $A B C D$ (a distance of 2) and joining $U$ to $S$. Since $M$ is 1 unit horizontally from $P N$, then $U S=1$.

Thus, $M S=\sqrt{2^{2}+1^{2}}=\sqrt{5}$ by the Pythagorean Theorem.

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We can now consider $\triangle S M T . h$ is the height of this triangle, from $M$ to base $S T$.

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Now $h=M T \sin (\angle M T S)=\sin (\angle M T S)$.

By the cosine law in $\triangle S M T$, we have

$$
M S^{2}=S T^{2}+M T^{2}-2(S T)(M T) \cos (\angle M T S)
$$

Therefore, $5=8+1-4 \sqrt{2} \cos (\angle M T S)$ or $4 \sqrt{2} \cos (\angle M T S)=4$.

Thus, $\cos (\angle M T S)=\frac{1}{\sqrt{2}}$ and so $\angle M T S=45^{\circ}$ which gives $h=\sin (\angle M T S)=\frac{1}{\sqrt{2}}$.

(Alternatively, we note that the plane of $A B C D$ is parallel to the plane of $P M N$, and so since the angle between plane $A B C D$ and plane $P N B A$ is $45^{\circ}$, then the angle between plane $P N B A$ and plane $P M N$ is also $45^{\circ}$, and so $\angle M T S=45^{\circ}$.)

Finally, this means that the volume of $A B N P M$ is $\frac{4 \sqrt{2}}{3} \cdot \frac{1}{\sqrt{2}}=\frac{4}{3}$, and so the volume of solid $A B C D P M N$ is $4+\frac{4}{3}=\frac{16}{3}$.

Final answer: $\frac{16}{3}$