Five rays in space

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Source: Vietnam NMO 1987 Problem 6

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#1 h Feb 4, 2009, 3:18 AM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

Prove that among any five distinct rays $Ox$ , $Oy$ , $Oz$ , $Ot$ , $Or$ in space there exist two which form an angle less than or equal to $90^{\circ}$ .

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#2 h Feb 9, 2009, 8:23 PM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

We assume for the sake of contradiction that all angles determined by any two of the five rays are greater than $90^\circ$ .

Consider a sphere centered at $O$ , and let $Ox,Oy,Oz,Ot,Or$ intersect the sphere at $x, y, z, t, r$ respectively.
The plane perpendicular to $Ox$ at $O$ cuts the sphere into two hemispheres.
Let $A$ be the hemisphere containing $x$ . $y$ cannot lie on $A$ , because if it did, the angle $\angle xOy$ would be less than or equal to $90^\circ$ .
By the same argument $z, t, r$ cannot lie on $A$ . Define hemispheres $B$ , $C$ , $D$ , and $E$ similarly, to contain $y, z, t, r$ , respectively and have similar properties.

Each of the five hemispheres has a measure of $2 \pi$ steradians. Because of the assumption, any two hemispheres must overlap at a solid angle of less than $\pi$ steradians. Let the intersection of $A$ and $B$ have measure $\theta < \pi$ .
The union of $A$ and $B$ must have measure $2\pi + 2\pi - \theta = 4\pi - \theta$ , and a complement $G$ of measure $4\pi - (4\pi - \theta) = \theta$ , and be bounded by two semi-great-circles that meet at two points, $p$ and $q$ . The complement $G$ is contained within what it would be if its measure were its upper bound of $\pi$ .

The intersection of $G$ and $C$ has a minimum of $\dfrac{\theta}{2}$ as $z$ approaches either $p$ or $q$ , so it is greater than $\dfrac{\theta}{2}$ . Thus the union of $A, B, C$ has measure greater than $(4\pi - \theta) + \dfrac{\theta}{2} = 4\pi - \dfrac{\theta}{2}$ , so the complement $H$ of this union has measure less than $\dfrac{\theta}{2} < \dfrac{\pi}{2}$ and is contained within what it would be if $z$ were at $p$ or $q$ .

$H$ is contained within what it would be if $z$ were at $p$ or $q$ and, since $H$ is a subset of $G$ , within what $H$ would be if $\theta$ equalled $\pi$ . This would be a $90^\circ$ - $90^\circ$ - $90^\circ$ spherical triangle, so $H$ is a subset of a $90^\circ$ - $90^\circ$ - $90^\circ$ spherical triangle. Since $t$ and $r$ must lie on $H$ , and therefore a $90^\circ$ - $90^\circ$ - $90^\circ$ spherical triangle, $\angle tOr \le 90^\circ$ . We have contradicted our assumption, so we are done.

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#3 h May 20, 2010, 6:45 PM • 1 Y

Y by Adventure10

This is my solution.

Let $\vec{x}, \vec{y}, \vec{z}, \vec{t}, \vec{r}$ be the the vector in the rays' directions.
We can set up our coordinate system such that, we may assume that all rays start at the origin $O$ and we may assume that $\vec{x} = \vec{k}$ , that is pointing straight upwards. So that means the other four vectors must have a negative $z$ component, that is, pointing at the bottom half-space.

Let $P$ be the plane $z = -1$ and let $Y,Z,R,T$ be the intersection of $O_y, O_z, O_r, O_t$ with $P$ . These points must exist because all those four rays are pointing "downwards". We now have a pyramid $O.YZRT$ and let $S$ be the point $(0,0,-1)$ , that is, the projection of point $O$ onto $S$ .

Examine the 2D figure on plane $P$ . We have a quadrilateral $YZRT$ , and a point $S$ . Draw the following four rays: $\vec{SY}, \vec{SZ}, \vec{SR}, \vec{ST}$ all originating from $S$ . It's easy to see that two of those rays must form an angle less than or equal to 90 degrees, for the sum of all four angles must equal to 360 degrees.

WLOG, suppose the angle between $\vec{SY}$ and $\vec{SZ}$ is $\theta \leq 90^o$ . But not that $SO$ is perpendicular to plane $P$ , and hence perpendicular to all vectors on plane $P$ .
$\vec{OY} . \vec{OZ} = (\vec {SY} - \vec{SO})(\vec{SZ} - \vec{SO}) = \vec{SY} . \vec{SZ} + SO^2$ because $SO \perp SY, SZ$

But since the angle between $\vec{SY}$ and $\vec{SZ}$ is $\leq 90^o$ then $\vec{SY} . \vec{SZ} \geq 0$ which means $\vec{OY} . \vec{OZ} > 0$ , which means that the rays $OY$ and $OZ$ form an acute angle.

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