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

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== Solution 3 (1434) ==
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== Solution 3 ==
 
The distance from <math>O</math> to <math>ABC</math> forms a  right triangle with the circumradius of the triangle and the radius of the sphere.  
 
The distance from <math>O</math> to <math>ABC</math> forms a  right triangle with the circumradius of the triangle and the radius of the sphere.  
  
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The circumradius of a <math>13</math>, <math>14</math>, <math>15% triangle is </math>\frac{65}{8}<math>, so the distance from </math>O<math> to </math>ABC<math> is given by </math>\sqrt{20^2-(\frac{65}{8})^2} = frac{15\sqrt{95}}{8}<math>, and </math>15+95+8 = \boxed{118}$.
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The circumradius of a <math>13</math>, <math>14</math>, <math>15</math> triangle is <math>\frac{65}{8}</math>, so the distance from <math>O</math> to <math>ABC</math> is given by <math>\sqrt{20^2-(\frac{65}{8})^2} = \frac{15\sqrt{95}}{8}</math>, and <math>15+95+8 = \boxed{118}</math>.
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-skibbysiggy
  
 
== See also ==
 
== See also ==

Latest revision as of 17:12, 17 November 2024

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 1

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 $15+95+8=\boxed{118}$.

Solution 2 (Vectors)

We know the radii to $A$,$B$, and $C$ form a triangular pyramid $OABC$. We know the lengths of the edges $OA = OB = OC = 20$. First we can break up $ABC$ into its two component right triangles $5-12-13$ and $9-12-15$. Let the $y$ axis be perpendicular to the base and $x$ axis run along $BC$, and $z$ occupy the other dimension, with the origin as $C$. We look at vectors $OA$ and $OC$. Since $OAC$ is isoceles we know the vertex is equidistant from $A$ and $C$, hence it is $7$ units along the $x$ axis. Hence for vector $OC$, in form $<x,y,z>$ it is $<7, h, l>$ where $h$ is the height (answer) and $l$ is the component of the vertex along the $z$ axis. Now on vector $OA$, since $A$ is $9$ along $x$, and it is $12$ along $z$ axis, it is $<-2, h, 12- l>$. We know both vector magnitudes are $20$. Solving for $h$ yields $\frac{15\sqrt{95} }{8}$, so Answer = $\boxed{118}$.


Solution 3

The distance from $O$ to $ABC$ forms a right triangle with the circumradius of the triangle and the radius of the sphere.


The hypotenuse has length $20$, since it is the radius of the sphere.


The circumradius of a $13$, $14$, $15$ triangle is $\frac{65}{8}$, so the distance from $O$ to $ABC$ is given by $\sqrt{20^2-(\frac{65}{8})^2} = \frac{15\sqrt{95}}{8}$, and $15+95+8 = \boxed{118}$.

-skibbysiggy

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

this is highly trivial for an AIME #12

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