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Difference between revisions of "Mock AIME 1 Pre 2005 Problems/Problem 8"

 
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#REDIRECT [[Mock AIME 1 Pre 2005 Problems/Problem 9]]
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== Problem ==
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<math>ABCD</math>, a [[rectangle]] with <math>AB = 12</math> and <math>BC = 16</math>, is the base of [[pyramid]] <math>P</math>, which has a height of <math>24</math>. A plane parallel to <math>ABCD</math> is passed through <math>P</math>, dividing <math>P</math> into a [[frustum]] <math>F</math> and a smaller pyramid <math>P'</math>. Let <math>X</math> denote the center of the circumsphere of <math>F</math>, and let <math>T</math> denote the apex of <math>P</math>. If the volume of <math>P</math> is eight times that of <math>P'</math>, then the value of <math>XT</math> can be expressed as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Compute the value of <math>m + n</math>.
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== Solution ==
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As we are dealing with volumes, the ratio of the volume of <math>P'</math> to <math>P</math> is the cube of the ratio of the height of <math>P'</math> to <math>P</math>.
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Thus, the height of <math>P</math> is <math>\sqrt [3]{8} = 2</math> times the height of <math>P'</math>, and thus the height of each is <math>12</math>.
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Thus, the top of the frustum is a rectangle <math>A'B'C'D'</math> with <math>A'B' = 6</math> and <math>B'C' = 8</math>.
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Now, consider the plane that contains diagonal <math>AC</math> as well as the altitude of <math>P</math>. Taking the cross section of the frustum along this plane gives the trapezoid <math>ACC'A'</math>, inscribed in an equatorial circular section of the sphere. It suffices to consider this sphere.
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First, we want the length of <math>AC</math>. This is given by the Pythagorean Theorem over triangle <math>ABC</math> to be <math>20</math>. Thus, <math>A'C' = 10</math>. Since the height of this trapezoid is <math>12</math>, and <math>AC</math> extends a distance of <math>5</math> on either direction of <math>A'C'</math>, we can use a 5-12-13 triangle to determine that <math>AA' = CC' = 13</math>.
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Now, we wish to find a point equidistant from <math>A</math>, <math>A'</math>, and <math>C</math>. By symmetry, this point, namely <math>X</math>, must lie on the perpendicular bisector of <math>AC</math>. Let <math>X</math> be <math>h</math> units from <math>A'C'</math> in <math>ACC'A'</math>. By the Pythagorean Theorem twice,
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<cmath>
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\begin{align*} 5^2 + h^2 & = r^2 \\
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10^2 + (12 - h)^2 & = r^2 \end{align*}
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</cmath>
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Subtracting gives <math>75 + 144 - 24h = 0 \Longrightarrow h = \frac {73}{8}</math>. Thus <math>XT = h + 12 = \frac {169}{8}</math> and <math>m + n = \boxed{177}</math>.
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== See also ==
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{{Mock AIME box|year=Pre 2005|n=1|num-b=7|num-a=9|source=14769}}
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[[Category:Intermediate Geometry Problems]]

Revision as of 19:26, 21 March 2008

Problem

$ABCD$, 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 $m + n$.

Solution

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 sphere.

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
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