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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>.
  
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== Solution ==
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== Solution 1==
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Let <math>D</math> be the foot of the perpendicular from <math>O</math> to the plane of <math>ABC</math>. By the [[Pythagorean Theorem]] on triangles <math>\triangle OAD</math>, <math>\triangle OBD</math> and <math>\triangle OCD</math> we get:
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Let <math>D</math> be the foot of the [[perpendicular]] from <math>O</math> to the plane of <math>ABC</math>. By the [[Pythagorean Theorem]] on triangles <math>\triangle OAD</math>, <math>\triangle OBD</math> and <math>\triangle OCD</math> we get:
  
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<math>DA^2=DB^2=DC^2=20^2-OD^2</math>
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<cmath>DA^2=DB^2=DC^2=20^2-OD^2</cmath>
  
 
It follows that <math>DA=DB=DC</math>, so <math>D</math> is the [[circumcenter]] of <math>\triangle ABC</math>.
 
It follows that <math>DA=DB=DC</math>, so <math>D</math> is the [[circumcenter]] of <math>\triangle ABC</math>.
  
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By [[Heron's Formula]] the area of <math>\triangle ABC</math> is:
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By [[Heron's Formula]] the area of <math>\triangle ABC</math> is (alternatively, a <math>13-14-15</math> triangle may be split into <math>9-12-15</math> and <math>5-12-13</math> [[right triangle]]s):
  
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<math>K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-15)(21-14)(21-13)} = \sqrt{(21)(8)(7)(6)} = \sqrt{(3\cdot 7)(2^3)(7)(2\cdot 3)} =\sqrt{2^4\cdot 3^2\cdot 7^2} = 2^2\cdot 3\cdot 7 = 84</math>
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<cmath>K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-15)(21-14)(21-13)} = 84</cmath>
  
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Now the [[circumradius]] of <math>\triangle ABC</math> is:
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From <math>R = \frac{abc}{4K}</math>, we know that the [[circumradius]] of <math>\triangle ABC</math> is:
  
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<math>R = \frac{abc}{4K} = \frac{(13)(14)(15)}{4(84)} = \frac{65}{8}</math>
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<cmath>R = \frac{abc}{4K} = \frac{(13)(14)(15)}{4(84)} = \frac{65}{8}</cmath>
  
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Thus by the Pythagorean Theorem again,
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Thus by the [[Pythagorean Theorem]] again,
  
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<math>OD = \sqrt{20^2-R^2} = \sqrt{20^2-\frac{65^2}{8^2}} = \sqrt{\frac{160^2-65^2}{8^2}} = \frac{\sqrt{(160-65)(160+65)}}{8} = \frac{\sqrt{(95)(225)}}{8} = \frac{15\sqrt{95}}{8}</math>.
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<cmath>OD = \sqrt{20^2-R^2} = \sqrt{20^2-\frac{65^2}{8^2}} = \frac{15\sqrt{95}}{8}.</cmath>
  
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So the final answer is <math>15+95+8=\boxed{118}</math>
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So the final answer is <math>15+95+8=\boxed{118}</math>.
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== Solution 2 (Vectors) ==
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 +
We know the radii to <math>A</math>,<math>B</math>, and <math>C</math> form a triangular pyramid <math>OABC</math>. We know the lengths of the edges <math>OA = OB = OC = 20</math>. First we can break up <math>ABC</math> into its two component right triangles <math>5-12-13</math> and <math>9-12-15</math>. Let the <math>y</math> axis be perpendicular to the base and <math>x</math> axis run along <math>BC</math>, and <math>z</math> occupy the other dimension, with the origin as <math>C</math>. We look at vectors <math>OA</math> and <math>OC</math>. Since <math>OAC</math> is isoceles we know the vertex is equidistant from <math>A</math> and <math>C</math>, hence it is <math>7</math> units along the <math>x</math> axis. Hence for vector <math>OC</math>, in form <math><x,y,z></math> it is <math><7, h, l></math> where <math>h</math> is the height (answer) and <math>l</math> is the component of the vertex along the <math>z</math> axis. Now on vector <math>OA</math>, since <math>A</math> is <math>9</math> along <math>x</math>, and it is <math>12</math> along <math>z</math> axis, it is <math><-2, h, 12- l></math>. We know both vector magnitudes are <math>20</math>. Solving for <math>h</math> yields <math>\frac{15\sqrt{95} }{8}</math>, so Answer = <math>\boxed{118}</math>.
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== Solution 3 ==
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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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 +
 
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The hypotenuse has length <math>20</math>, since it is the radius of the sphere.
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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>.
 +
 
 +
-skibbysiggy
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 +
== See also ==
 
{{AIME box|year=2000|n=II|num-b=11|num-a=13}}
 
{{AIME box|year=2000|n=II|num-b=11|num-a=13}}
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 +
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}
 +
this is highly trivial for an AIME #12

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

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this is highly trivial for an AIME #12

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