AIME 2000 II · 第 12 题

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 $$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 =$\boxed{118}$.

Note: $OA$ is actually $\left< 2, -h, 12-l \right>$. ~fermat_sLastAMC

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