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Difference between revisions of "1978 IMO Problems/Problem 2"
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==Solution==
 
==Solution==
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{{solution}}
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Let <math>R</math> be the radius of the sphere.
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Let point <math>O</math> be the center of the sphere.
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Let point <math>D</math> be the 4th vertex of the face of the parallelepiped that contains points <math>P</math>, <math>A</math>, and <math>B</math>.
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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>
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Let <math>\alpha=\measuredangle AOP</math> ;  <math>\beta=\measuredangle BPD</math> ;  <math>\theta=\measuredangle APE</math>
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We start the calculations as follows:
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<math>\left| AB \right|= \left| PD \right|</math>
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<math>\left| PD \right|^{2}=\left| PA \right|^{2}+\left| PB \right|^{2}</math>
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Using law of cosines:
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<math>R^{2}=\left| OP \right|^{2} + \left| PB \right|^{2} - 2 \left| OP \right| \left| PB \right| cos (\measuredangle OPB)</math>
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<math>R^{2}=\left| OP \right|^{2} + \left| PB \right|^{2} - 2 \left| OP \right| \left| PB \right| cos (\frac{\pi}{2}-\theta)</math>
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<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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{{alternate solutions}}
  
 
== See Also == {{IMO box|year=1978|num-b=1|num-a=3}}
 
== See Also == {{IMO box|year=1978|num-b=1|num-a=3}}

Revision as of 13:22, 11 November 2023

Problem

We consider a fixed point $P$ in the interior of a fixed sphere$.$ We construct three segments $PA, PB,PC$, perpendicular two by two$,$ with the vertexes $A, B, C$ on the sphere$.$ We consider the vertex $Q$ which is opposite to $P$ in the parallelepiped (with right angles) with $PA, PB, PC$ as edges$.$ Find the locus of the point $Q$ when $A, B, C$ take all the positions compatible with our problem.

Solution

Let $R$ be the radius of the sphere.

Let point $O$ be the center of the sphere.

Let point $D$ be the 4th vertex of the face of the parallelepiped that contains points $P$, $A$, and $B$.

Let point $E$ be the point where the line that passes through $OP$ intersects the circle on the side nearest to point $A$

Let $\alpha=\measuredangle AOP$ ; $\beta=\measuredangle BPD$ ; $\theta=\measuredangle APE$

We start the calculations as follows:

$\left| AB \right|= \left| PD \right|$

$\left| PD \right|^{2}=\left| PA \right|^{2}+\left| PB \right|^{2}$

Using law of cosines:

$R^{2}=\left| OP \right|^{2} + \left| PB \right|^{2} - 2 \left| OP \right| \left| PB \right| cos (\measuredangle OPB)$

$R^{2}=\left| OP \right|^{2} + \left| PB \right|^{2} - 2 \left| OP \right| \left| PB \right| cos (\frac{\pi}{2}-\theta)$

$R^{2}=\left| OP \right|^{2} + \left| PB \right|^{2} - 2 \left| OP \right| \left| PB \right| sin (\theta)$




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
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