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Fitting circles under a curve

  1. May 29, 2012 #1
    I thought of a problem a few days ago and I have no idea as to its solution. I posted this on Reddit and xkcd forums earlier but not much has been solved apart from the area of one circle. Suppose you have a boundary formed by the curve y=e^(-x), and the lines x=0 and y=0. In this boundary you place the largest possible circle you can, which touches the y-axis, the x-axis and y=e^(-x). You then place the next largest possible circle to the right of this one, which touches the first circle, y=e^(-x) and the x-axis. This process is repeated indefinitely. If there are an infinite number of circles formed in this way, what is the sum of their areas?

    I suspect that the answer will be pretty complicated. Just working out the x-value of the point at which the first circle touches the curve involves working out an x such that x-sqrt((1+e^(2x))x^(2))+sqrt(2xe^(3x))-e^(x)=0, which is a transcendental equation. Someone pointed out that this equation is equivalent to the more compact but still transcendental 4xcosh x = (1+x)^2. As for the remaining infinite circles, I have to idea.

  2. jcsd
  3. May 30, 2012 #2


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    Hey Randommelon and welcome to the forums.

    The way I see you coming up with an analytic solution is to do the following: for each circle you will need to maximize a diagonal that goes from the point of the left-most point of the circle (i.e. the left-most point which has an infinite-tangent line) where the tangent of the diagonal is 1. You do this for every single circle after this.

    You also need to check if you actually can do this (I have a feeling you can't). Basically if you can show that for this circle and portion of the function that the function cuts the circle in more than one spot, then you can't form a circle and you have proven that it's impossible.

    Remember that the height of your starting point for your diagonal vector is going to have a height that is the radius of your circle that you are solving for and the end-point of that vector will be at the function itself, if the circle really exists.

    But like I inferred above, it would be a lot easier for you to actually show using simultaneous equations and the definition of the function e^(-x) whether you get the function crossing more than point for the circle to show whether a circle can be constructed.
  4. May 30, 2012 #3
    Cool problem. Even though the equation for the first radius does not look to have an analytic solution, the problem is still tantalizing. If you assume that the first radius is R, then the curve itself has a sort of self-similarity with respect to translations and vertical scaling that almost makes you think you could get a reasonably simple recursion for all subsequent circle radii. Unfortunately if you translate and vertically scale a circle, you get an ellipse, so it is not so simple to transform one circle into the next one in the sequence. But perhaps there is a way....
  5. May 30, 2012 #4
    I've been thinking about this one and have a few related problems. First of all

    1. Let X be the set of circles that are tangent to both the x axis and the curve y=e^{-x}. Given any element C_0 of X, we can find a sequence of elements of X such that any two consecutive elements are tangent to each other. (We can extend this in both directions, not just toward the tail end of the curve). The original problem takes the element of X that is tangent to the y axis as the first element of the sequence. The proof of this shouldnt' be too hard. Start with a circle that does not touch C_0 and "slide" it toward C_0 (adjusting the radius as you go to stay inside X) until you just touch C_0.

    2. Consider the set of all points where two elements of X touch. These points form a curve. Would it be too good to be true for that curve to have the form ke^(-x), where k is between 0 and 1?

    3. Does there exist a smooth 1 parameter family of transformations of the plane that acts transitively on X? In other words, is there a vector field on the plane that induces a flow that transforms elements of X into elements of X. A naive first guess here would be x' = 1 and y'=-y because that preserves the x axis and the curve y=e^{-x}. But the corresponding family of transformations is (x+t, e^{-t}y), which transforms circles into ellipses. If such a flow existed, the divergence of the vector field would tell us the rate of change of the areas of the elements of X as we slid them along.
  6. May 30, 2012 #5
    Thanks for your responses. Vargo, I was wondering the same thing about idea 2; it would be very interesting if they did lie on such a curve. Would that imply their centers also lie on the curve?
  7. May 30, 2012 #6
    yes, I think their centers do lie on a smooth curve, which is probably asymptotically like (1/2)e^(-t), but I managed to check that the centers do not lie exactly on a line ke^{-t}
  8. May 30, 2012 #7
    I don't know if this helps, but I found that the radius of a circle r can be expressed explicitly in terms of where the circle touches e^(-x), at x. r = 2cosh(x)-sqrt(e^(2x)+1).
  9. May 30, 2012 #8
    Well, using that you can fnid the center of the circle in terms of x.

    Suppose you have two circles that belong to X and are tangent to each other. If their centers are (x_1,y_1) and (x_2,y_2) respectively, then the fact that the distance to their centers equals the sum of the radii equals y_1+y_2 yields:

    Now the center of the first circle depends on a single variable t (the x coordinate of the point where it touches the curve y=e^-x. The center of the second circle also depends on a single coordinate s. So by plugging into the above equation, we get an equation between the variables t and s that must be satisfied if the two circles are tangent to each other. Unfortunately that equation is too complicated for me to draw any conclusions from. I think I'm stumped.
  10. May 30, 2012 #9
    Might I ask what the equation you got that related s and t was? If a circle intersects at t, then its center lies at x=t+1-sqrt(1+e^(-2t)) (and from that I derived the equation r = 2cosh(t)-sqrt(e^(2t)+1)). Is that the equation you used?
  11. May 30, 2012 #10
    Given t (the x-coordinate where the circle touches the curve), we have
    x(t), the x-coordinate of the center of the circle which is given by the formula you wrote there. And you have y(t), the y-coordinate of the center of the circle. The y coordinate is equal to the radius r(t) which you also derived.

    Ok, now suppose we have another circle which is tangent to the curve at s. Then its center is at x(s),y(s)=r(s).

    The squared distance between the centers of the circle is given by the pythagorean theorem:
    (x(t) - x(s))^2 + (y(t)-y(s))^2.
    If these two circles are tangent to each other, then that distance must also equal the sum of the radii. So we get
    (x(t) - x(s))^2 + (y(t)-y(s))^2 = (y(t)+y(s))^2,
    using the fact that y(t)=r(t).
    Without any reference to the dependence on t, those equations algebraically reduce to
    (x(t)-x(s))^2 = 4y(t)y(s).

    This is the equation I was referring to. We have formulas for x(t) and y(t), so we can substitute them to get an explicit equation between s and t that controls whether the two circles are tangent to each other.
  12. May 31, 2012 #11
    Now I'm stumped too. You're right, if we use x(t)=t+1-sqrt(1+e^(-2t)) and y(t)=2cosh(t)-sqrt(e^(2t)+1)) then (x(t)-x(s))^2 = 4y(t)y(s) produces a really horrible equation. Wolfram Alpha can't even find the next point. Maybe this isn't the best method?
  13. May 31, 2012 #12
    For now I am well and truly stumped on that one. Here is an alternate problem that might be of interest.... Replace y=e^(-x) with y=1/x. Otherwise make the same construction of circles. Will the sum of their areas be a convergent series?
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