Difference between revisions of "1978 IMO Problems/Problem 2"
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==Solution== | ==Solution== | ||
− | Let <math>R</math> be the radius of the sphere. | + | |
+ | [[File:IMO_1978_P2a.png|400px]] | ||
+ | |||
+ | Let <math>R</math> be the radius of the given fixed sphere. | ||
Let point <math>O</math> be the center of the sphere. | Let point <math>O</math> be the center of the sphere. | ||
Line 11: | Line 14: | ||
Let point <math>E</math> be the point where the line that passes through <math>OP</math> intersects the circle on the side nearest to point <math>A</math> | Let point <math>E</math> be the point where the line that passes through <math>OP</math> intersects the circle on the side nearest to point <math>A</math> | ||
− | Let <math>\alpha=\ | + | Let <math>\alpha=\angle AOP,\;\beta=\angle BPD,\;\theta=\angle APE</math> |
We start the calculations as follows: | We start the calculations as follows: | ||
Line 17: | Line 20: | ||
<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> [Equation 1] | + | <math>\left| AB \right|^{2}=\left| PA \right|^{2}+\left| PB \right|^{2}</math> |
+ | |||
+ | Therefore, <math>\left| PD \right|^{2}=\left| PA \right|^{2}+\left| PB \right|^{2}</math> [Equation 1] | ||
Using law of cosines: | Using law of cosines: | ||
− | <math>R^{2}=\left| OP \right|^{2} + \left| PB \right|^{2} - 2 \left| OP \right| \left| PB \right| cos (\ | + | <math>R^{2}=\left| OP \right|^{2} + \left| PB \right|^{2} - 2 \left| OP \right| \left| PB \right| cos (\angle OPB)</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| cos \left( \frac{\pi}{2}-\theta \right)</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> | ||
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<math>\left| PA \right|^{2} =R^{2}+\left| OP \right|^{2} - 2 \left| OP \right| R cos(\alpha)</math> | <math>\left| PA \right|^{2} =R^{2}+\left| OP \right|^{2} - 2 \left| OP \right| R cos(\alpha)</math> | ||
+ | |||
+ | Since <math>R cos(\alpha) = \left| PA \right| cos(\theta) + \left| OP\right|</math>, then | ||
<math>\left| PA \right|^{2} =R^{2}+\left| OP \right|^{2} - 2 \left| OP \right| \left[ \left| PA \right| cos(\theta) + \left| OP\right| \right]</math> | <math>\left| PA \right|^{2} =R^{2}+\left| OP \right|^{2} - 2 \left| OP \right| \left[ \left| PA \right| cos(\theta) + \left| OP\right| \right]</math> | ||
Line 43: | Line 50: | ||
Now we apply the law of cosines again: | Now we apply the law of cosines again: | ||
− | <math>\left| OD \right|^{2}=\left| OP \right|^{2}+\left| PD \right|^{2}-2\left| OP \right| \left| PD \right|cos(\ | + | <math>\left| OD \right|^{2}=\left| OP \right|^{2}+\left| PD \right|^{2}-2\left| OP \right| \left| PD \right|cos(\angle OPD)</math> |
− | <math>\left| OD \right|^{2}=\left| OP \right|^{2}+\left| PD \right|^{2}-2\left| OP \right| \left| PD \right| cos(\ | + | <math>\left| OD \right|^{2}=\left| OP \right|^{2}+\left| PD \right|^{2}-2\left| OP \right| \left| PD \right| cos(\angle OPB+\angle BPD)</math> |
− | <math>\left| OD \right|^{2}=\left| OP \right|^{2}+\left| PD \right|^{2}-2\left| OP \right| \left| PD \right| cos(\frac{\pi}{2}-\theta+\beta)</math> | + | <math>\left| OD \right|^{2}=\left| OP \right|^{2}+\left| PD \right|^{2}-2\left| OP \right| \left| PD \right| cos \left(\frac{\pi}{2}-\theta+\beta \right)</math> |
<math>\left| OD \right|^{2}=\left| OP \right|^{2}+\left| PD \right|^{2}-2\left| OP \right| \left| PD \right| sin(\theta-\beta)</math> | <math>\left| OD \right|^{2}=\left| OP \right|^{2}+\left| PD \right|^{2}-2\left| OP \right| \left| PD \right| sin(\theta-\beta)</math> | ||
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<math>\left| OD \right|^{2}=\left| OP \right|^{2}+\left| PD \right|^{2}-2\left| OP \right| \left[\left| PB \right|sin(\theta)-\left| PA \right|cos(\theta) \right]</math> [Equation 5] | <math>\left| OD \right|^{2}=\left| OP \right|^{2}+\left| PD \right|^{2}-2\left| OP \right| \left[\left| PB \right|sin(\theta)-\left| PA \right|cos(\theta) \right]</math> [Equation 5] | ||
− | Substituting [Equation | + | Substituting [Equation 4] into [Equation 5] we get: |
<math>\left| OD \right|^{2}=\left| OP \right|^{2}+2R^{2}-2\left| OP \right|^{2}+2\left| OP \right| \left[ \left| PB \right| sin(\theta) - \left| PA \right| cos(\theta) \right]-2\left| OP \right| \left[\left| PB \right|sin(\theta)-\left| PA \right|cos(\theta) \right]</math> | <math>\left| OD \right|^{2}=\left| OP \right|^{2}+2R^{2}-2\left| OP \right|^{2}+2\left| OP \right| \left[ \left| PB \right| sin(\theta) - \left| PA \right| cos(\theta) \right]-2\left| OP \right| \left[\left| PB \right|sin(\theta)-\left| PA \right|cos(\theta) \right]</math> | ||
+ | |||
+ | Notice that all of the terms with <math>\theta</math> cancel and thus we're left with: | ||
<math>\left| OD \right|^{2}=2R^{2}-\left| OP \right|^{2}</math> regardless of <math>\theta</math>. [Equation 6] | <math>\left| OD \right|^{2}=2R^{2}-\left| OP \right|^{2}</math> regardless of <math>\theta</math>. [Equation 6] | ||
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Substituting [Equation 6] and [Equation 7] into [Equation 8] we get: | Substituting [Equation 6] and [Equation 7] into [Equation 8] we get: | ||
− | <math>\left| OQ \right|^{2}=2R^{2}-\left| OP \right|^{2}+R^{2}-\left| OP \right|^{2}</math> | + | <math>\left| OQ \right|^{2}=2R^{2}-\left| OP \right|^{2}+R^{2}-\left| OP \right|^{2}</math> |
− | |||
− | |||
+ | This results in: | ||
+ | <math>\left| OQ \right|^{2}=3R^{2}-2\left| OP \right|^{2}</math> | ||
+ | which is constant regardless of <math>\theta</math> and constant regardless of where points <math>A</math>, <math>B</math>, and <math>C</math> are located as long as they're still perpendicular to each other. | ||
+ | In space, this is a sphere with radius <math>\left| OQ \right|</math> which is equal to <math>\sqrt{3R^{2}-2\left| OP \right|^{2}}</math> | ||
+ | Therefore, the locus of vertex <math>Q</math> is a sphere of radius <math>\sqrt{3R^{2}-2\left| OP \right|^{2}}</math> with center at <math>O</math>, where <math>R</math> is the radius of the given sphere and <math>\left| OP \right|</math> the distance from the center of the given sphere to point <math>P</math> | ||
+ | ~ Tomas Diaz. orders@tomasdiaz.com | ||
+ | {{alternate solutions}} | ||
− | + | == See Also == | |
− | + | {{IMO box|year=1978|num-b=1|num-a=3}} | |
− | [[Category: | + | [[Category:Geometry Problems]] |
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Latest revision as of 12:00, 13 March 2024
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 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 |