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Difference between revisions of "2000 AIME II Problems/Problem 12"

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== Problem ==
 
== Problem ==
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The points <math>A</math>, <math>B</math> and <math>C</math> lie on the surface of a [[sphere]] with center <math>O</math> and radius <math>20</math>. It is given that <math>AB=13</math>, <math>BC=14</math>, <math>CA=15</math>, and that the distance from <math>O</math> to [[triangle]] <math>ABC</math> is <math>\frac{m\sqrt{n}}k</math>, where <math>m</math>, <math>n</math>, and <math>k</math> are positive integers, <math>m</math> and <math>k</math> are relatively prime, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n+k</math>.
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The points <math>A</math>, <math>B</math> and <math>C</math> lie on the surface of a [[sphere]] with center <math>O</math> and radius <math>20</math>. It is given that <math>AB=13</math>, <math>BC=14</math>, <math>CA=15</math>, and that the distance from <math>O</math> to <math>\triangle ABC</math> is <math>\frac{m\sqrt{n}}k</math>, where <math>m</math>, <math>n</math>, and <math>k</math> are positive integers, <math>m</math> and <math>k</math> are relatively prime, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n+k</math>.
  
 
== Solution ==
 
== Solution ==

Revision as of 23:20, 10 March 2010

Problem

The points $A$, $B$ and $C$ lie on the surface of a sphere with center $O$ and radius $20$. It is given that $AB=13$, $BC=14$, $CA=15$, and that the distance from $O$ to $\triangle ABC$ is $\frac{m\sqrt{n}}k$, where $m$, $n$, and $k$ are positive integers, $m$ and $k$ are relatively prime, and $n$ is not divisible by the square of any prime. Find $m+n+k$.

Solution

Let $D$ be the foot of the perpendicular from $O$ to the plane of $ABC$. By the Pythagorean Theorem on triangles $\triangle OAD$, $\triangle OBD$ and $\triangle OCD$ we get:

\[DA^2=DB^2=DC^2=20^2-OD^2\]

It follows that $DA=DB=DC$, so $D$ is the circumcenter of $\triangle ABC$.

By Heron's Formula the area of $\triangle ABC$ is (alternatively, a $13-14-15$ triangle may be split into $9-12-15$ and $5-12-13$ right triangles):

\[K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-15)(21-14)(21-13)} = 84\]

From $R = \frac{abc}{4K}$, we know that the circumradius of $\triangle ABC$ is:

\[R = \frac{abc}{4K} = \frac{(13)(14)(15)}{4(84)} = \frac{65}{8}\]

Thus by the Pythagorean Theorem again,

\[OD = \sqrt{20^2-R^2} = \sqrt{20^2-\frac{65^2}{8^2}} = \frac{15\sqrt{95}}{8}.\]

So the final answer is $15+95+8=\boxed{118}$.

See also

2000 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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