Difference between revisions of "1990 AIME Problems/Problem 14"
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== Solution == | == Solution == | ||
| − | {{ | + | [[Image:1990_AIME-14b.png]] |
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| + | Our triangular pyramid has base <math>12\sqrt{3} - 13\sqrt{3} - 13\sqrt{3} \triangle</math>. The area of this [[isosceles triangle]] is easy to find by <math>\frac{1}{2}bh</math>, where we can find <math>h</math> to be <math>\sqrt{399}</math> by the [[Pythagorean Theorem]]. Thus <math>A = \frac 12(12\sqrt{3})\sqrt{399} = 18\sqrt{133}</math>. | ||
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| + | [[Image:1990_AIME-14c.png]] | ||
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| + | To find the volume, we want to use the equation <math>\frac 13Bh = 6\sqrt{133}h</math>, so we need to find the height of the [[tetrahedron]]. By the Pythagorean Theorem, <math>AP = CP = DP = \frac{\sqrt{939}}{2}</math>. If we let <math>P</math> be the center of a [[sphere]] with radius <math>\frac{\sqrt{939}}{2}</math>, then <math>A,C,D</math> lie on the sphere. The cross section of the sphere 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 <math>\triangle ACD</math>. | ||
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| + | From here we just need to perform some brutish calculations. Using the formula <math>A = 18\sqrt{133} = \frac{abc}{4R}</math> (<math>R</math> being the [[circumradius]]), we find <math>R = \frac{12\sqrt{3} \cdot (13\sqrt{3})^2}{4\cdot 18\sqrt{133}} = \frac{13^2\sqrt{3}}{2\sqrt{133}}</math>. By the Pythagorean Theorem, | ||
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| + | <cmath>\begin{eqnarray*}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} \\ | ||
| + | h^2 &=& \frac{13068 \cdot 3}{4 \cdot 133} = \frac{99^2}{133}\\ | ||
| + | h &=& \frac{99}{\sqrt{133}} | ||
| + | \end{eqnarray*}</cmath> | ||
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| + | Finally, we substitute <math>h</math> into the volume equation to find <math>V = 6\sqrt{133}\left(\frac{99}{\sqrt{133}}\right) = \boxed{594}</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=1990|num-b=13|num-a=15}} | {{AIME box|year=1990|num-b=13|num-a=15}} | ||
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| + | [[Category:Intermediate Geometry Problems]] | ||
Revision as of 11:43, 25 November 2007
Problem
The rectangle 
below has dimensions 
and 
. Diagonals 
and 
intersect at 
. If triangle 
is cut out and removed, edges 
and 
are joined, and the figure is then creased along segments 
and 
, we obtain a triangular pyramid, all four of whose faces are isosceles triangles. Find the volume of this pyramid.
Solution
Our triangular pyramid has base 
. The area of this isosceles triangle is easy to find by 
, where we can find 
to be 
by the Pythagorean Theorem. Thus 
.
To find the volume, we want to use the equation 
, so we need to find the height of the tetrahedron. By the Pythagorean Theorem, 
. If we let 
be the center of a sphere with radius 
, then 
lie on the sphere. The cross section of the sphere 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 
.
From here we just need to perform some brutish calculations. Using the formula 
(
being the circumradius), we find 
. By the Pythagorean Theorem,
Finally, we substitute into the volume equation to find 
.
See also
| 1990 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
