Difference between revisions of "2000 AIME II Problems/Problem 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

@@ Line 28: / Line 28: @@
-== Solution 3 (1434) ==
+== 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.
@@ Line 35: / Line 35: @@
-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>.
+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
 == 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
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