Cube $A B C D E F G H$ has edge length 6 , and has $P$ on $B G$ so that $B P=P G$. What is the volume of the three-dimensional region that lies inside both square-based pyramid $E F G H P$ and square-based pyramid $A B C D G$ ?

![](https://cdn.mathpix.com/cropped/2024_04_17_1deb74b62155eda4bdbfg-2.jpg?height=393&width=439&top_left_y=2156&top_left_x=1358)

## 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.
We start by determining the heights above the bottom of the cube of the points of intersection of the edges of the pyramids.

For example, consider square $A F G B$ and edges $A G$ and $F P$. We call their point of intersection $X$.

We assign coordinates to the various points using the fact that the edge length of the cube is 6 : $F(0,0), G(6,0), B(6,6), A(0,6), P(6,3)$ ( $P$ is the midpoint of $B G)$.

![](https://cdn.mathpix.com/cropped/2024_04_17_238abf67dc2e378be7cdg-05.jpg?height=504&width=550&top_left_y=518&top_left_x=842)

Line segment $A G$ has slope -1 , and so has equation $y=-x+6$.

Line segment $F P$ has slope $\frac{3}{6}=\frac{1}{2}$ and so has equation $y=\frac{1}{2} x$.

To find the coordinates of $X$, we equate expressions in $y$ to obtain $-x+6=\frac{1}{2} x$ which gives $\frac{3}{2} x=6$ or $x=4$, and so $y=-4+6=2$.

Therefore, point $X$ is a height of 2 above square $E F G H$.

Using a similar argument, the point of intersection between $P H$ and $G C$ is 2 units above square EFGH.

To see why the point of intersection of $G D$ and $P E$ is also 2 units above $E F G H$, we note that rectangle $D E G B$ has a height of 6 (like square $A F G B$ ) and a width of $6 \sqrt{2}$. As a result, we can think of obtaining rectangle $D E G B$ by stretching square $A F G B$ horizontally by a factor of $\sqrt{2}$. This horizontal stretch will not raise or lower the point of intersection between $G D$ and $P E$ and so this point is also two units above $E F G H$.

Now, imagine drawing a plane through the three points of intersection of the edges of the pyramids.

Since each of these points is 2 units above $E F G H$, this plane must be horizontal and will also intersect $B G 2$ units above $G$, forming a square. (The points of intersection form a square because every horizontal cross-section of both pyramids is a square.) This square has side length 2 because the $x$-coordinate of $X$ was 4 , which is 2 units from $B G$ in that coordinate system.

This square divides the common three-dimensional region into two square-based pyramids.

One of these pyramids points upwards and has fifth vertex $P$. This pyramid has a square base with edge length 2 and a height of $3-2=1$, since $P$ is 3 units above $G$ and the base of the pyramid is 2 units above $G$.

The other pyramid points downwards and has fifth vertex $G$. This pyramid has a square base with edge length 2 and a height of 2.

Thus, the volume of the region is $\frac{1}{3} \cdot 2^{2} \cdot 1+\frac{1}{3} \cdot 2^{2} \cdot 2=\frac{4}{3}+\frac{8}{3}=4$.

## Part B