Difference between revisions of "1978 IMO Problems/Problem 2"
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<math>\left| AB \right|= \left| PD \right|</math> | <math>\left| AB \right|= \left| PD \right|</math> | ||
− | <math>\left| PD \right|^{2}=\left| PA \right|^{2}+\left| PB \right|^{2}</math> | + | <math>\left| PD \right|^{2}=\left| PA \right|^{2}+\left| PB \right|^{2}</math> [Equation 1] |
Using law of cosines: | Using law of cosines: | ||
Line 25: | Line 25: | ||
<math>R^{2}=\left| OP \right|^{2} + \left| PB \right|^{2} - 2 \left| OP \right| \left| PB \right| cos (\frac{\pi}{2}-\theta)</math> | <math>R^{2}=\left| OP \right|^{2} + \left| PB \right|^{2} - 2 \left| OP \right| \left| PB \right| cos (\frac{\pi}{2}-\theta)</math> | ||
− | <math>R^{2}=\left| OP \right|^{2} + \left| PB \right|^{2} - 2 \left| OP \right| \left| PB \right| sin (\theta)</math> | + | <math>R^{2}=\left| OP \right|^{2} + \left| PB \right|^{2} - 2 \left| OP \right| \left| PB \right| sin (\theta)</math> [Eqiatopn 2] |
Revision as of 14:41, 11 November 2023
Problem
We consider a fixed point in the interior of a fixed sphere
We construct three segments
, perpendicular two by two
with the vertexes
on the sphere
We consider the vertex
which is opposite to
in the parallelepiped (with right angles) with
as edges
Find the locus of the point
when
take all the positions compatible with our problem.
Solution
Let be the radius of the 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:
[Equation 1]
Using law of cosines:
[Eqiatopn 2]
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 |