1990 AIME Problems/Problem 14

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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.

Solution

Solution 1

Our triangular pyramid has base $12\sqrt{3} - 13\sqrt{3} - 13\sqrt{3} \triangle$ . The area of this isosceles triangle is easy to find by $[ACD] = \frac{1}{2}bh$ , where we can find $h_{ACD}$ to be $\sqrt{399}$ by the Pythagorean Theorem. Thus $A = \frac 12(12\sqrt{3})\sqrt{399} = 18\sqrt{133}$ .

To find the volume, we want to use the equation $\frac 13Bh = 6\sqrt{133}h$ , so we need to find the height of the tetrahedron. By the Pythagorean Theorem, $AP = CP = DP = \frac{\sqrt{939}}{2}$ . If we let $P$ be the center of a sphere with radius $\frac{\sqrt{939}}{2}$ , then $A,C,D$ lie on the sphere. The cross section of the sphere that contains $A,C,D$ is a circle, and the center of that circle is the foot of the perpendicular from the center of the sphere. Hence the foot of the height we want to find occurs at the circumcenter of $\triangle ACD$ .

From here we just need to perform some brutish calculations. Using the formula $A = 18\sqrt{133} = \frac{abc}{4R}$ (where $R$ is the circumradius), we find $R = \frac{12\sqrt{3} \cdot (13\sqrt{3})^2}{4\cdot 18\sqrt{133}} = \frac{13^2\sqrt{3}}{2\sqrt{133}}$ (there are slightly simpler ways to calculate $R$ since we have an isosceles triangle). By the Pythagorean Theorem,

$\begin{align*}h^2 &= PA^2 - R^2 \\&= \left(\frac{\sqrt{939}}{2}\right)^2 - \left(\frac{13^2\sqrt{3}}{2\sqrt{133}}\right)^2\\&= \frac{939 \cdot 133 - 13^4 \cdot 3}{4 \cdot 133} = \frac{13068 \cdot 3}{4 \cdot 133} = \frac{99^2}{133}\\h &= \frac{99}{\sqrt{133}}\end{align*}$

Finally, we substitute $h$ into the volume equation to find $V = 6\sqrt{133}\left(\frac{99}{\sqrt{133}}\right) = \boxed{594}$ .

Solution 2

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:

$\begin{eqnarray*}x^{2} + (y - \sqrt {399})^{2} + z^{2} &=& \frac {939}{4}\\(x - 6\sqrt {3})^{2} + y^{2} + z^{2} &=& \frac {939}{4}\\(x + 6\sqrt {3})^{2} + y^{2} + z^{2} &=& \frac {939}{4}\end{eqnarray*}$

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 $V = \frac {1}{3}\cdot18\sqrt {133}\cdot\frac {99}{\sqrt {133}} = 6\cdot99 = 594$ .

1990 AIME (Problems • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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