In the diagram, $P A B C D$ is a pyramid with square base $A B C D$ and with $P A=P B=P C=P D$. Suppose that $M$ is the midpoint of $P C$ and that $\angle B M D=90^{\circ}$. Triangular-based pyramid $M B C D$ is removed by cutting along the triangle defined by the points $M, B$ and $D$. The volume of the remaining solid $P A B M D$ is 288 . What is the length of $A B$ ?

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## PART B

For each question in Part B, your solution must be well-organized and contain words of explanation or justification. Marks are awarded for completeness, clarity, and style of presentation. A correct solution, poorly presented, will not earn full marks.
Let the side length of the square base $A B C D$ be $2 a$ and the height of the pyramid (that is, the distance of $P$ above the base) be $2 h$.

Let $F$ be the point of intersection of the diagonals $A C$ and $B D$ of the base. By symmetry, $P$ is directly above $F$; that is, $P F$ is perpendicular to the plane of square $A B C D$.

Note that $A B=B C=C D=D A=2 a$ and $P F=2 h$. We want to determine the value of $2 a$.

Let $G$ be the midpoint of $F C$.

Join $P$ to $F$ and $M$ to $G$.

![](https://cdn.mathpix.com/cropped/2024_04_17_ca28acd3f90ebca4605fg-06.jpg?height=372&width=409&top_left_y=562&top_left_x=912)

Consider $\triangle P C F$ and $\triangle M C G$.

Since $M$ is the midpoint of $P C$, then $M C=\frac{1}{2} P C$.

Since $G$ is the midpoint of $F C$, then $G C=\frac{1}{2} F C$.

Since $\triangle P C F$ and $\triangle M C G$ share an angle at $C$ and the two pairs of corresponding sides adjacent to this angle are in the same ratio, then $\triangle P C F$ is similar to $\triangle M C G$.

Since $P F$ is perpendicular to $F C$, then $M G$ is perpendicular to $G C$.

Also, $M G=\frac{1}{2} P F=h$ since the side lengths of $\triangle M C G$ are half those of $\triangle P C F$.

The volume of the square-based pyramid $P A B C D$ equals $\frac{1}{3}\left(A B^{2}\right)(P F)=\frac{1}{3}(2 a)^{2}(2 h)=\frac{8}{3} a^{2} h$.

Triangular-based pyramid $M B C D$ can be viewed as having right-angled $\triangle B C D$ as its base and $M G$ as its height.

Thus, its volume equals $\frac{1}{3}\left(\frac{1}{2} \cdot B C \cdot C D\right)(M G)=\frac{1}{6}(2 a)^{2} h=\frac{2}{3} a^{2} h$.

Therefore, the volume of solid $P A B M D$, in terms of $a$ and $h$, equals $\frac{8}{3} a^{2} h-\frac{2}{3} a^{2} h=2 a^{2} h$.

Since the volume of $P A B M D$ is 288 , then $2 a^{2} h=288$ or $a^{2} h=144$.

We have not yet used the information that $\angle B M D=90^{\circ}$.

Since $\angle B M D=90^{\circ}$, then $\triangle B M D$ is right-angled at $M$ and so $B D^{2}=B M^{2}+M D^{2}$.

By symmetry, $B M=M D$ and so $B D^{2}=2 B M^{2}$.

Since $\triangle B C D$ is right-angled at $C$, then $B D^{2}=B C^{2}+C D^{2}=2(2 a)^{2}=8 a^{2}$.

Since $\triangle B G M$ is right-angled at $G$, then $B M^{2}=B G^{2}+M G^{2}=B G^{2}+h^{2}$.

Since $\triangle B F G$ is right-angled at $F$ (the diagonals of square $A B C D$ are equal and perpendicular), then

$B G^{2}=B F^{2}+F G^{2}=\left(\frac{1}{2} B D\right)^{2}+\left(\frac{1}{4} A C\right)^{2}=\frac{1}{4} B D^{2}+\frac{1}{16} A C^{2}=\frac{1}{4} B D^{2}+\frac{1}{16} B D^{2}=\frac{5}{16} B D^{2}=\frac{5}{2} a^{2}$

Since $2 B M^{2}=B D^{2}$, then $2\left(B G^{2}+h^{2}\right)=8 a^{2}$ which gives $\frac{5}{2} a^{2}+h^{2}=4 a^{2}$ or $h^{2}=\frac{3}{2} a^{2}$ or $a^{2}=\frac{2}{3} h^{2}$.

Since $a^{2} h=144$, then $\frac{2}{3} h^{2} \cdot h=144$ or $h^{3}=216$ which gives $h=6$.

From $a^{2} h=144$, we obtain $6 a^{2}=144$ or $a^{2}=24$.

Since $a>0$, then $a=2 \sqrt{6}$ and so $A B=2 a=4 \sqrt{6}$.

## Part B