1978 IMO Problems/Problem 2
balls
Solution
Let be the radius of the given fixed sphere.
Let point be the center of the sphere.
Let point be the 4th vertex of the face of the parallelepiped that contains points , , and .
Let point be the point where the line that passes through intersects the circle on the side nearest to point
Let
We start the calculations as follows:
Therefore, [Equation 1]
Using law of cosines:
[Equation 2]
Using law of cosines again we also get:
Since , then
[Equation 3]
Substituting [Equation 2] and [Equation 3] into [Equation 1] we get:
[Equation 4]
Now we apply the law of cosines again:
Since, and then,
[Equation 5]
Substituting [Equation 4] into [Equation 5] we get:
Notice that all of the terms with cancel and thus we're left with:
regardless of . [Equation 6]
Now we need to find
Since points , , and are on the plane perpendicular to the plane with points , , and , then these points lie on the big circle of the sphere. Therefore the distance can be found using the formula:
Solving for we get:
[Equation 7]
Now we need to get which will be using the formula:
[Equation 8]
Substituting [Equation 6] and [Equation 7] into [Equation 8] we get:
This results in:
which is constant regardless of and constant regardless of where points , , and are located as long as they're still perpendicular to each other.
In space, this is a sphere with radius which is equal to
Therefore, the locus of vertex is a sphere of radius with center at , where is the radius of the given sphere and the distance from the center of the given sphere to point
~ Tomas Diaz. orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1978 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |