Mock AIME 1 Pre 2005 Problems/Problem 8


$ABCD$, a rectangle with $AB = 12$ and $BC = 16$, is the base of pyramid $P$, which has a height of $24$. A plane parallel to $ABCD$ is passed through $P$, dividing $P$ into a frustum $F$ and a smaller pyramid $P'$. Let $X$ denote the center of the circumsphere of $F$, and let $T$ denote the apex of $P$. If the volume of $P$ is eight times that of $P'$, then the value of $XT$ can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Compute the value of $m + n$.


As we are dealing with volumes, the ratio of the volume of $P'$ to $P$ is the cube of the ratio of the height of $P'$ to $P$.

Thus, the height of $P$ is $\sqrt [3]{8} = 2$ times the height of $P'$, and thus the height of each is $12$.

Thus, the top of the frustum is a rectangle $A'B'C'D'$ with $A'B' = 6$ and $B'C' = 8$.

Now, consider the plane that contains diagonal $AC$ as well as the altitude of $P$. Taking the cross section of the frustum along this plane gives the trapezoid $ACC'A'$, inscribed in an equatorial circular section of the sphere. It suffices to consider this circle.

First, we want the length of $AC$. This is given by the Pythagorean Theorem over triangle $ABC$ to be $20$. Thus, $A'C' = 10$. Since the height of this trapezoid is $12$, and $AC$ extends a distance of $5$ on either direction of $A'C'$, we can use a 5-12-13 triangle to determine that $AA' = CC' = 13$.

Now, we wish to find a point equidistant from $A$, $A'$, and $C$. By symmetry, this point, namely $X$, must lie on the perpendicular bisector of $AC$. Let $X$ be $h$ units from $A'C'$ in $ACC'A'$. By the Pythagorean Theorem twice, \begin{align*} 5^2 + h^2 & = r^2 \\ 10^2 + (12 - h)^2 & = r^2 \end{align*} Subtracting gives $75 + 144 - 24h = 0 \Longrightarrow h = \frac {73}{8}$. Thus $XT = h + 12 = \frac {169}{8}$ and $m + n = \boxed{177}$.

See also

Mock AIME 1 Pre 2005 (Problems, Source)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15