Conference rusure::math

The curve is an ellipse, with the antennae as foci. 

    I had guessed the 90 deg case was a circle. There is some theorem that
    says if you form an inscribed triangle where one side is a diameter,
    the opposite angle will be ninety degrees. This would place the
    antennas at each end of the diameter, and the circle is the set of
    points you can observe the antennas from.
    
    But I also guessed that for other angles, you still get circles.
    
    I have trouble with your solution. If in fact they are ellipses with
    the antennae at the foci, then the special case of a circle would put
    them in the middle of the circle - not on the circle! Something's wrong
    here. I think ellipses are for constant total length from the observer
    to each antenna, not constant angle.
    
    Is there some theorem that says for a given line through a circle (is
    that called a chord?), there is some constant angle formed in the
    triangle I described above? The diameter would just be a special case.
    
    You can be arbitrarily close to one antenna and always find a point
    near it that forms the desired angle with the other antenna. This tells
    me the antennae must lie on the curve, whatever they may be...

    What if for a nearly zero angle, you had a huge circle where both
    antennae were on the circle? As you change the angle towards ninety
    degrees, the circle shrinks and the antennae "move apart" (normalized
    to the size of the circle) until they are opposite extremes - on a
    diameter. But what happens for 91 degrees? I'm sure there are still
    solutions, you must stand somewhere nearly between them. Whatever this
    curve is, it must disappear when you get beyond 180 degrees. I think at
    180 degrees the curve is a line segment joining the two antennae.
    
    But I'm pretty sure they should all be circles somehow. Any idea if
    this is so, and if so, how it can be proved in a simple to understand
    way?

>The curve is an ellipse, with the antennae as foci. 

Too hasty, Lynn. Tsk, tsk.

Draw a circle and a chord, say horizontal and somewhat below the center of 
the circle. From any point on the circle and above the chord draw the 
angle subtended by the chord, and do the same from any point on the circle
below the chord. The two angles so constructed are independent of which
points on the circle are used, and the two angles sum to 180 deg. (This is
easiest to see if the two points so selected are diagonally opposite - then
the angles at the ends of the chord are both 90 deg.) 

So the locus of points in the problem is the larger part of the circle if
the fixed angle is < 90, and the smaller part if the angle is > 90. 

Sorry about the first cut answer.

Lynn

    I'll agree that the construction you describe makes the sum of the
    angles 180 deg when the two points are diagonally opposite. Since the
    whole shape is a quadrilateral,  360-90-90 = 180.
    
    I guess the part that surprises me is that "the angle subtended by the
    chord" is a constant, regardless of the point chosen to measure the
    angle. It sounds like that is common knowledge, the way you said it.
    I've probably just forgotten my basic geometry... Is there a simple
    proof of this fact?

    So in simple terms, when the angle is less than 90 deg, the curve looks
    like the "view through binoculars" used in the movies. When the angle
    is greater than 90, the curve looks like the cross section of a convex
    lens.
    
    Is that it?

>    I guess the part that surprises me is that "the angle subtended by the
>    chord" is a constant, regardless of the point chosen to measure the
>    angle. ... Is there a simple proof of this fact?

Yes, but it's a little hard to put in a note. Think of a circle with center 
o and radius r. Pick points a and b on the circumference. These correspond 
to the antennae. Pick a point c also on the circumference (it's easiest to 
visualize if it's on the opposite side of the circle somewhere, but it can 
be anywhere on the circumference). Draw the diameter from c thru o to c', 
say between a and b. Let X be the angle subtended by ab at the center o.
The diameter c-c' splits X in two, X' and X", with X'+X"=X. (One of these 
angles can be negative.) Now construct the triangles aoc and boc. In the 
isosoles triangle aoc the central angle is 180-X', so each of the equal 
angles is (180 - (180-X'))/2 = X'/2. Similarly, in boc the angle at c is
X"/2. So the angle acb is X'/2+X"/2 = X/2, and that is determined by the
choice of a and b: the arc ab is the angle X in radians. 

So what we've shown is that the angle acb is half ab, so is the same for 
all points c on the circle. The other way around, the locus of points c 
subtending the same angle is the circle; computing the radius of the circle 
I leave as an exercise for the reader B^).

Yes, the complete loci for both sides of the line ab is two overlapping 
circles.

    Thanks for the explanation. When you got to part that angle acb is half
    of angle aob, it all came back to me. I guess I missed the real
    significance of it! 
    
    This means you could use a carpenter's rule (those L-shaped things?)
    and a couple of nails to draw circles.