Difference between revisions of "2000 AIME II Problems/Problem 8"

Revision as of 18:31, 11 November 2007

Problem

In trapezoid $ABCD$ , leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD}$ , and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001}$ , find $BC^2$ .

Solution

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 == Problem ==
-A container in the shape of a right circular cone is 12 inches tall and its base has a 5-inch radius. The liquid that is sealed inside is 9 inches deep when the cone is held with its point down and its base horizontal. When the liquid is held with its point up and its base horizontal, the liquid is <math>m - n\sqrt [3]{p},</math> where <math>m,</math> <math>n,</math> and <math>p</math> are positive integers and <math>p</math> is not divisible by the cube of any prime number. Find <math>m + n + p</math>.
+In trapezoid <math>ABCD</math>, leg <math>\overline{BC}</math> is perpendicular to bases <math>\overline{AB}</math> and <math>\overline{CD}</math>, and diagonals <math>\overline{AC}</math> and <math>\overline{BD}</math> are perpendicular. Given that <math>AB=\sqrt{11}</math> and <math>AD=\sqrt{1001}</math>, find <math>BC^2</math>.
 == Solution ==

2000 AIME II (Problems • Answer Key • Resources)
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Difference between revisions of "2000 AIME II Problems/Problem 8"

Revision as of 18:31, 11 November 2007

Problem

Solution

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