Maximizing and Minimizing the Diameter - II
This post continues on the questions posed in the last one. Thanks to Prof. Roman Karasev, who has provided the answers given below.
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Question: In 2D, if both area and perimeter are specified, which convex shape has maximum/minimum diameter?
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Answer 1 - Minimum diameter: The mean width w of a convex figure is always proportional to its perimeter, p. Indeed, p = pi * w. (http://en.wikipedia.org/wiki/Mean_width)
This gives the partial answer: Keeping perimeter fixed, as the specified area is varied (within a range as shown below), the convex figure that minimizes the diameter is always a figure of *constant width*. (http://en.wikipedia.org/wiki/Curve_of_constant_width).
Among all figures of constant width with same perimeter, p, the circle has greatest area and the Reuleaux triangle the least. With the same perimeter specified, if the requried area is reduced below Reuleaux triangle, the diameter of the required convex figure will be *more* than the Reuleaux value - that is because the figure will not be of constant width and its average width (= p/pi) cannot change (since perimeter is not changed) and the diameter is the maximum value of width.
Of course, with perimeter p, there is no convex region possible with area greater than that of the circle with that perimeter.
The question remains: what shape will minimize diameter when perimeter is p and the required area is *less than* that of the Reuleaux triangle with perimeter p?
Lemma: As noted above, among all constant width convex 2D regions of fixed perimeter, Circle has largest and Reuleaux the smallest area. It is also true that for *every* intermediate area (and with the same perimeter), there exist some convex region(s) of constant width.
The proof is remarkably simple: is $K$ and $L$ are convex bodies of constant width $1$, then $(1-t)K+tL$ is a convex body of constant width $1$ for every $t\in [0, 1]$. Since the area of $(1-t)K+tL$ varies continuously, it spans a segment.
Answer 2: Maximum diameter:
Prof. Karasev's answer: "Maximum diameter: the 2D shape is the intersection of two congruent disks with appropriate diameters and distance between centers (how we draw a convex lens). This follows from physical considerations of the constant pressure. In 3D, the convex shape which maximizes diameter with volume and surface area specified could be a spindle - the analog of the lens in 2D"
Explanation: http://math.scichina.com:8081/sciAe/fileup/PDF/08ya1119.pdf proves the following 'final theorem': the convex 2D region fully containing a given line segment and having a specified perimeter and the *largest possible* area is a *symmetric lens* formed by two identical arcs and the given segment as diameter. The above quoted answer follows readily (indeed, if some other figure were the answer, a lens exists with the same diameter segment and containing the same area but with *less perimeter*. And for any lens L, another lens with same area as L and greater perimeter can be formed, with larger diameter than L). I do not yet understand the Physics part of the story.
--------
Question: In 2D, if both area and perimeter are specified, which convex shape has maximum/minimum diameter?
--------
Answer 1 - Minimum diameter: The mean width w of a convex figure is always proportional to its perimeter, p. Indeed, p = pi * w. (http://en.wikipedia.org/wiki/Mean_width)
This gives the partial answer: Keeping perimeter fixed, as the specified area is varied (within a range as shown below), the convex figure that minimizes the diameter is always a figure of *constant width*. (http://en.wikipedia.org/wiki/Curve_of_constant_width).
Among all figures of constant width with same perimeter, p, the circle has greatest area and the Reuleaux triangle the least. With the same perimeter specified, if the requried area is reduced below Reuleaux triangle, the diameter of the required convex figure will be *more* than the Reuleaux value - that is because the figure will not be of constant width and its average width (= p/pi) cannot change (since perimeter is not changed) and the diameter is the maximum value of width.
Of course, with perimeter p, there is no convex region possible with area greater than that of the circle with that perimeter.
The question remains: what shape will minimize diameter when perimeter is p and the required area is *less than* that of the Reuleaux triangle with perimeter p?
Lemma: As noted above, among all constant width convex 2D regions of fixed perimeter, Circle has largest and Reuleaux the smallest area. It is also true that for *every* intermediate area (and with the same perimeter), there exist some convex region(s) of constant width.
The proof is remarkably simple: is $K$ and $L$ are convex bodies of constant width $1$, then $(1-t)K+tL$ is a convex body of constant width $1$ for every $t\in [0, 1]$. Since the area of $(1-t)K+tL$ varies continuously, it spans a segment.
Answer 2: Maximum diameter:
Prof. Karasev's answer: "Maximum diameter: the 2D shape is the intersection of two congruent disks with appropriate diameters and distance between centers (how we draw a convex lens). This follows from physical considerations of the constant pressure. In 3D, the convex shape which maximizes diameter with volume and surface area specified could be a spindle - the analog of the lens in 2D"
Explanation: http://math.scichina.com:8081/sciAe/fileup/PDF/08ya1119.pdf proves the following 'final theorem': the convex 2D region fully containing a given line segment and having a specified perimeter and the *largest possible* area is a *symmetric lens* formed by two identical arcs and the given segment as diameter. The above quoted answer follows readily (indeed, if some other figure were the answer, a lens exists with the same diameter segment and containing the same area but with *less perimeter*. And for any lens L, another lens with same area as L and greater perimeter can be formed, with larger diameter than L). I do not yet understand the Physics part of the story.
posted by R.Nandakumar @ 8:12 PM
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