Triangle in a sphere

Geometry Level pending

The points $A$ , $B$ , and $C$ lie on the surface of a sphere with center $O$ and radius $20$ . It is given that straight lines $AB=13$ , $BC=14$ , $CA=15$ .

And that the shortest distance from $O$ to any part of the 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 $m+n+k$ .

Clarification: The triangle $ABC$ is a Euclidean geometry triangle, not a spherical triangle.

The answer is 118.

1 solution

Thanos Petropoulos
Nov 21, 2020

Since the center $O$ of the sphere is equidistant from points $A$ , $B$ and $C$ , it lies on the line which is perpendicular to the plane of $\triangle ABC$ passing through the circumcenter $D$ of the triangle.

Let $r=AD$ be the circumradius of $\triangle ABC$ . Then, $r=\frac{abc}{4A}$ where $A$ is the area of $\triangle ABC$ .

By Heron ‘s formula we find $A=84$ , so $r=\frac{abc}{4A}=\frac{13\cdot 14\cdot 15}{4\cdot 84}=\frac{65}{8}$ Using Pythagorean theorem on $\triangle OAD$ we can evaluate the shortest distance $d$ from $O$ to $\triangle ABC$

$d=OD=\sqrt{O{{A}^{2}}-A{{D}^{2}}}=\sqrt{{{R}^{2}}-{{r}^{2}}}=\sqrt{{{20}^{2}}-{{\left( \frac{65}{8} \right)}^{2}}}=\frac{15\sqrt{95}}{8}$ For the answer, $m+n+k=15+95+8=\boxed{118}$ .

Triangle in a sphere

The answer is 118.

1 solution

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