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Homework Help: An analysis/geometry problem

  1. Jan 29, 2005 #1
    Hi everyone. Thanks in advance for your help. I've been working on this problem that is in my analysis book and I'm not sure how to go about it.

    I'm required to find the supremum of the epsilon neighborhood around the point (3, 5/2) such that the entire neighborhood is contained is the set S = {the closed triangle with vertices (2,0), (2,2), (4,4)}.

    Just so that non-analysis acquainted people can still help me: the epsilon neighborhood that they are asking for is basically a circle with some radius epsilon. They want me to find this radius, I think.

    So I thought about taking the equation of the sphere around this point (3, 5/2) and finding where it intersects each of the sides (using the equation of a line), but I ended up going nowhere with this. I was going to take the derivative and find the max or something... I just got lost.

    I was also thinking about the idea that the circle would be tangent to one of these lines at the point where the radius is perpendicular to that side of the triangle. So I found the equation of a couple of the lines and took the negative reciprocal, and set those equal to the sphere equation, but that didn't work either! Please help me. Thanks :)
  2. jcsd
  3. Jan 29, 2005 #2


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    Well, of course the circle will have to be tangent to the edge of the triangle it touches. If it weren't, then the edge would go through the circle. In other words, any line that touches a circle at only one point is tangent to it, and you can easily verify this. You can show it by considering a circle centered about the origin of radius r, and a line going through the point (0, r). Show that it touches the circle at only one point iff it's slope is zero, i.e. if y = mx + r is the equation of the line, and y² + x² = r² is the equation of the circle, then this system of two equations has some solution for x not equal to zero iff m is not equal to zero.

    To solve this problem, find the point (a, b) found on one of the arms of the triangle such that (a - 3, b - 2.5) is perpendicular to the arm and it's length is shortest. You have three sides of the triangle, so you will have three values for (a, b). You want to choose the (a, b) such that the length of (a - 3, b - 2.5) is minimized.

    I will find one of the (a,b) for you, and it's length.

    One the side connecting (2,2) and (2,0), the point (a,b) would satisfy:

    (a - 3, b - 2.5) . (2 - 2, 2 - 0) = 0 (the dot product of two perpendicular vectors is zero)
    (a - 3, b - 2.5) . (0, 2) = 0
    2(b - 2.5) = 0
    b = 2.5

    Now (a, b) has to be on the line segment joining (2,2) and (2, 0), so (2,2)-(a,b) has to be collinear with (a,b)-(2,0), i.e., for some real k:

    (2,2)-(a,b) = k[(a,b) - (2,0)]
    (2-a,2-b) = (ka - 2k, kb)
    (2-a,-0.5) = (ka - 2k, 2.5k)

    So k = -0.2, so:

    2 - a = (-0.2)a - 2(-0.2)
    2 - a = 0.4 - 0.2a
    1.6 = 0.8a
    a = 2

    Now, your intuition should already have told you that 'a' would be 2, and that since the edge connecting (2,0) and (2,2) is vertical, that the perpendicular segment from (3, 2.5) and that edge would be horizontal, so b would be 2.5. Also, note how (a,b) = (2, 2.5) is not even on the edge connecting (2,0) and (2,2) (it's above it). If you drew a picture, your intuition would also have told you that this is the wrong edge to be looking at. The length of (a - 3,b - 2.5) = (-1, 0) is:

    [tex]\sqrt{(a-3)^2 + (b-2.5)^2} = \sqrt{(-1)^2 + 0^2} = \sqrt{1} = 1[/tex]

    That calculation was entirely pointless except to show you how it's done just in case you didn't know (but I bet you did). So do the same for the other two sides, and the one with the smallest length will be your [itex]\epsilon[/itex].
  4. Jan 30, 2005 #3
    Thank you very much for your help I ended up making some silly errors when figuring out the equation of the line. In particular, I said that the equation of the line connecting (2,0) and (4,4) was 2x, not 2x-4. This through my calculations off quite a bit.

    Once again though, thank you very much for your help on this problem. I pretty much gave up on working in it until I saw your post. Just for curiosity the answer is 1/(2*sqrt(5))
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