1978 IMO Problems/Problem 2

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:

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

$\left| PD \right|^{2}=\left| PA \right|^{2}+\left| PB \right|^{2}$ [Equation 1]

Using law of cosines:

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

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

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

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

1978 IMO (Problems) • Resources
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3
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1978 IMO Problems/Problem 2

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