Maximum Angle between Perpendicular BisectorsDate: 06/11/99 at 06:09:03 From: Brecht Hommez Subject: The maximum angle between two perpendicular bisectors Hello Dr. Math, First of all, congratulations on your excellent website. It has already been a great help to me (a Ph.D. student in high energy physics). Right now, I have a very annoying problem and I wonder if there is a solution. Suppose you have two circles with the same radius and they do not intersect. Step 1: Choose a point on each circle and consider the line segment formed by these two points. Then construct the perpendicular bisector of this line segment. Step 2: Repeat step 1, but with different choices of points on each of the two circles. Thus then we have two perpendicular bisectors. Now, my question is: which four points do I have to choose (two points in step 1 and two points in step 2), to obtain the maximum angle between the two perpendicular bisectors? I really hope that you can help me - after days of thinking about this, I couldn't find the answer. Thanks, Brecht Hommez Date: 06/11/99 at 16:49:22 From: Doctor Peterson Subject: Re: The maximum angle between two perpendicular bisectors Hello, Brecht! I'm glad we've been helpful. We can simplify the problem considerably by ignoring the perpendicular bisectors, because the angle between the perpendiculars is the same as the angle between the segments themselves. Then we can consider only the slope of each segment, ignoring details like where they intersect. This makes it fairly obvious that the maximum angle will occur when the two segments are tangent to the two circles, one sloped as far as possible to the left and the other as far as possible to the right: *********** *** *** *** *** * * * * * * * A * * + * * / | \ r * * / | \ * P3+ Q+------------+P1 * | x * *** | *** \ *** d/2| *** / \ *********** / \ | / \ |y /sqrt((d/2)^2 - r^2) \ | / \ | / \ | / \ | / \ | / \|/ *O /|\ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / *********** \ / *** | *** \ *** | *** * | * P2+ | +P4 * \ | / * * \ | / * * + * * B * * * * * * * *** *** *** *** *********** I don't know in what form you need to find the pairs of points, but we can see that the segments intersect at the midpoint of the segment connecting the centers of the circles, and the maximum angle will be 2 arcsin(2r/d) where r is the radius of the circles and d is the distance between their centers. The four points will be symmetrical about this point and about the connecting line. If you need x and y coordinates of the points, and the circles are in something like the position I've shown, you can get them from these similar triangles: Using APQ and AOP: PQ OP x sqrt((d/2)^2 - r^2) -- = -- --- = ------------------- PA OA r d/2 Using POQ and AOP: OQ OP y sqrt((d/2)^2 - r^2) -- = -- ------------------- = ------------------- OP OA sqrt((d/2)^2 - r^2) d/2 Let me know if I've misunderstood the problem or if there is more you need. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 06/14/99 at 05:45:38 From: Brecht Hommez Subject: Re: The maximum angle between two perpendicular bisectors Hello Dr Peterson, Thanks for the quick answer to my problem, but ... I 'm still having trouble. In fact, the problem I sent to you was the derivative problem of the original problem. The solution of the derivative problem does not help me further with the original problem and therefore I think it may be wise to send the original problem to you: Suppose you have three circles (call them circle 1, circle 2, and circle 3) with the same radius and they do not intersect. Step 1: Choose a point on each of two different circles (e.g. circle 1 and circle 2) and consider the line segment formed by these two points. Then construct the perpendicular bisector of this line segment. Step 2: Repeat step 1, but with two circles not the same as the circles used in step 1 (e.g. for step 2 you choose circle 2 and circle 3) Step 3: Thus we have two perpendicular bisectors. We calculate the point where they intersect. Let's call this point P. We repeat these 3 steps for a lot of points on each circle and each of the 3 combinations of different circles in step 1 and step 2: Combination 1: Step 1: circle 1 and circle 2 Step 2: circle 2 and circle 3 Combination 2: Step 1: circle 1 and circle 2 Step 2: circle 1 and circle 3 Combination 3: Step 1: circle 1 and circle 3 Step 2: circle 2 and circle 3 I have written a computer program that calculates all these points P and they seem to lie on a polygon with 6 vertices. Now my question is: Is there a way to calculate analytically the (x,y)-coordinates of the 6 vertices of this polygon? Thanks for considering my question, Brecht Hommez Date: 06/15/99 at 09:04:32 From: Doctor Peterson Subject: Re: The maximum angle between two perpendicular bisectors Hello again, Brecht. Your problem is an interesting one, and I have spent some time thinking about it. I do not have all the details worked out, but I can at least tell you the general idea, which may be all you need. If we consider first only one pair of circles, I have convinced myself that the locus of the perpendicular bisectors of segments joining pairs of points on the circles is the interior of a hyperbola whose semiaxes are r (the radius of the circles) and sqrt(d^2 - r^2) (half the length of the tangents from my earlier answer; the angle I gave is the angle between the asymptotes). If the circles of radius r have their centers at (-d, 0) and (d, 0), the equation of this hyperbola is x^2 y^2 --- - --------- = 1 r^2 d^2 - r^2 The circles A and B are symmetrical about the midpoint O, through which I have drawn a secant CDEF; for reasons of symmetry I can say that the perpendicular bisectors of CE and DF are the extrema of the set of all perpendicular bisectors of segments connecting the two circles with the same slope. I have drawn the locus of these two bisectors in blue and green respectively; the locus of point P, in red, is one branch of the hyperbola which forms the envelope of these lines. It can be seen that this is a hyperbola, because the difference BP - AP is (BD + DP) - (FP - AF) = (DP - FP) + (BD + AF) = 0 + 2r which is constant, because P is on the bisector and DP = FP. Now if we draw these hyperbolas for all three pairs of circles, we see that the locus of points of intersection of pairs of lines will be the union of the pairwise intersections of the hyperbolic regions. That is, the bisectors for one pair of circles may meet those for another pair of circles where their two hyperbolic regions intersect, which will look like a curved quadrilateral. If I understand your explanation correctly, I would expect the result to look like a curved hexagram (star) rather than a hexagon. Here is the intersection of two regions, one between two horizontal lines (approximating the branches of one hyperbola), and the other between slanted lines (representing another hyperbola): / / / ----+---------+ /........./ /........./ /........./ +---------+---- / / / and here is the union of all three such quadrilaterals: + /.\ /...\ +---+-----+---+ \./.......\./ +.........+ /.\......./.\ +---+-----+---+ \.../ \./ + It appears that you are getting instead the intersection of the three hyperbolic regions, which may mean I am slightly misunderstanding your intention; perhaps you are choosing one point on each of the three circles and finding the intersection of the three bisectors (that is, finding the locus of circumcenters of triangles with one vertex on each circle). Then your locus will be either the "curved hexagon," or at least a subset of it; apparently it's the whole thing: + / \ / \ +---+-----+---+ \ /.......\ / +.........+ / \......./ \ +---+-----+---+ \ / \ / + As you can see, since you have not an actual polygon but a slightly curved region, it will take more than the coordinates of the vertices to define it fully (though they would define a hexagon that would contain your region, and appears to be a good approximation). But what I have given you should be enough to find those points and to define the curves if needed. Now that I have worked this out, I am very curious as to the possible application of it! What are the circles, and why is this region important? - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/