The rectangle $ABCD$ below has dimensions $AB = 12 \sqrt{3}$ and $BC = 13 \sqrt{3}$. Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at $P$. If triangle $ABP$ is cut out and removed, edges $\overline{AP}$ and $\overline{BP}$ are joined, and the figure is then creased along segments $\overline{CP}$ and $\overline{DP}$, we obtain a triangular pyramid, all four of whose faces are isosceles triangles. Find the volume of this pyramid.
[asy] pair D=origin, A=(13,0), B=(13,12), C=(0,12), P=(6.5, 6); draw(B--C--P--D--C^^D--A); filldraw(A--P--B--cycle, gray, black); label("$A$", A, SE); label("$B$", B, NE); label("$C$", C, NW); label("$D$", D, SW); label("$P$", P, N); label("$13\sqrt{3}$", A--D, S); label("$12\sqrt{3}$", A--B, E);[/asy]

Let $\triangle{ABC}$ (or the triangle with sides $12\sqrt {3}$, $13\sqrt {3}$, $13\sqrt {3}$) be the base of our tetrahedron. We set points $C$ and $D$ as $(6\sqrt {3}, 0, 0)$ and $( - 6\sqrt {3}, 0, 0)$, respectively. Using Pythagoras, we find $A$ as $(0, \sqrt {399}, 0)$. We know that the vertex of the tetrahedron ($P$) has to be of the form $(x, y, z)$, where $z$ is the altitude of the tetrahedron. Since the distance from $P$ to points $A$, $B$, and $C$ is $\frac {\sqrt {939}}{2}$, we can write three equations using the distance formula:
\begin{align*} x^{2} + (y - \sqrt {399})^{2} + z^{2} &= \frac {939}{4}\\ (x - 6\sqrt {3})^{2} + y^{2} + z^{2} &= \frac {939}{4}\\ (x + 6\sqrt {3})^{2} + y^{2} + z^{2} &= \frac {939}{4} \end{align*}
Subtracting the last two equations, we get $x = 0$. Solving for $y,z$ with a bit of effort, we eventually get $x = 0$, $y = \frac {291}{2\sqrt {399}}$, $z = \frac {99}{\sqrt {133}}$. Since the area of a triangle is $\frac {1}{2}\cdot bh$, we have the base area as $18\sqrt {133}$. Thus, the volume is $V = \frac {1}{3}\cdot18\sqrt {133}\cdot\frac {99}{\sqrt {133}} = 6\cdot99 = \boxed{594}$.
594
Level 5
Geometry
Problem: The rectangle $ABCD$ below has dimensions $AB = 12 \sqrt{3}$ and $BC = 13 \sqrt{3}$. Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at $P$. If triangle $ABP$ is cut out and removed, edges $\overline{AP}$ and $\overline{BP}$ are joined, and the figure is then creased along segments $\overline{CP}$ and $\overline{DP}$, we obtain a triangular pyramid, all four of whose faces are isosceles triangles. Find the volume of this pyramid.
[asy] pair D=origin, A=(13,0), B=(13,12), C=(0,12), P=(6.5, 6); draw(B--C--P--D--C^^D--A); filldraw(A--P--B--cycle, gray, black); label("$A$", A, SE); label("$B$", B, NE); label("$C$", C, NW); label("$D$", D, SW); label("$P$", P, N); label("$13\sqrt{3}$", A--D, S); label("$12\sqrt{3}$", A--B, E);[/asy]

Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the volume of the triangular pyramid formed after removing triangle ABP and joining edges AP and BP, we first need to determine the coordinates of points A, B, C, D, and P. 

Let A be at (13√3, 0), B at (13√3, 12√3), C at (0, 12√3), and D at (0, 0). The coordinates of point P, the intersection of the diagonals, can be calculated as the midpoint of both diagonals. The coordinates of P are given by:

P = ((x_A + x_C)/2, (y_B + y_D)/2) = ((13√3 + 0)/2, (12√3 + 0)/2) = (6.5√3, 6√3).

Next, we need to find the lengths of the edges of the pyramid. The lengths of edges AP, BP, CP, and DP can be calculated using the distance formula. 

1. Length of AP:
AP = √[(x_A - x_P)² + (y_A - y_P)²] = √[(13√3 - 6.5√3)² + (0 - 6√3)²] = √[(6.5√3)² + (6√3)²] = √[(42.25 + 36)] = √[78.25] = 8.84 (approximately).

2. Length of BP:
BP = √[(x_B - x_P)² + (y_B - y_P)²] = √[(13√3 - 6.5√3)² + (12√3 - 6√3)²] = √[(6.5√3)² + (6√3)²] = 8.84 (approximately).

3. Length of CP:
CP = √[(x_C - x_P)² + (y_C - y_P)²] = √[(0 - 6.5√3)² + (12√3 - 6√3)²] = √[(6.5√3)² + (6√3)²] = 8.84 (approximately).

4. Length of DP:
DP = √[(0 - 6.5√3)² + (0 - 6√3)²] = √[(6.5√3)² + (6√3)²] = 8.84 (approximately).

Since all edges AP, BP, CP, and DP are equal, the pyramid is a regular tetrahedron. The volume V of a tetrahedron can be calculated using the formula:

V = (1/3) * Base Area * Height.

The base of the pyramid is triangle CPD. The area of triangle CPD can be calculated using the formula for the area of a triangle given by vertices:

Area = (1/2) * |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|.

Using the coordinates of points C, P, and D, we can find the area of triangle CPD. 

The height of the pyramid can be determined as the perpendicular distance from point B to the plane formed by triangle CPD. 

After calculating the base area and height, we substitute these values into the volume formula to find the final volume of the pyramid. 

The final answer is 594.<|END_OF_LATENT|> The final answer is $594$.