Bisection

Starting with any point P on the closed curve, which has an "antipodal point" P', which is halfway around the curve, so that line PP' bisects the perimeter. Let A and B be the areas created by PP' dividing the enclosed area. Let's say A > B, where A is on the right side of the line PP', going from P to P'. Move the point P along the perimeter until it reaches where P' was. The new area A' is the same as the old area B, so that A' < B', which is the opposite of what it was at the beginning. Therefore, there has to be a point between P and P' where the areas had to be the same.

We have to assume a certain amount about the curve to make things work. I shall suppose that the curve can be given by a polar equation $r \; = \; f(\theta)$ where $f$ is continuously differentiable and $f(\theta) >0$ for all $\theta$ . Then the arc-length function $s(\theta) \; = \; \int_0^\theta \sqrt{f(\phi)^2 + f'(\phi)^2}\,d\phi$ is a continuous strictly increasing function on $[0,2\pi]$ from $0$ to $S$ , where $S$ is the perimeter of the curve.Indeed it is a continuous strictly increasing function on $[0,4\pi]$ from $0$ to $2S$ . Thus the function $s \,:\, [0,4\pi] \to [0,2S]$ has a continuous inverse. Thus the function $\beta(\theta) \; = \; s^{-1}\big(s(\theta) + \tfrac12S\big) \qquad 0 \le \theta \le 2\pi$ is a continuous function. If a point $Q$ on the curve has angle $\theta$ then the point $Q'$ with angle $\beta(\theta)$ is such that the arc length from $Q$ to $Q'$ is equal to $\tfrac12S$ . The area cut off on one side of the chord $QQ'$ is equal to $a(\theta) \; = \; \tfrac12\int_\theta^{\beta(\theta)} f(\phi)^2\,d\phi - \tfrac12f(\theta)f(\beta(\theta))\sin\big(\beta(\theta)-\theta\big)$ and this is a continuous function of $\theta$ .

Since $a(0) + a(\beta(0)) = A$ , the total area inside the curve, we deduce that either $a(0) \ge \tfrac12A \ge a(\beta(0))$ or else $a(0) \le \tfrac12A \le a(\beta(0))$ . By the Intermediate Value Theorem, we deduce that there exists some $\theta$ between $0$ and $\beta(0)$ such that $a(\theta) = \tfrac12A$ , which is what we want.

You could calculate the "center of mass" in a matter analagous to how one would do so in a physics calculation. By the definition of the center of mass, any line passing through that point will divide the area evenly. Imagine then extending a line randomly from that point. If the perimeters on each side are not equal, you can simply rotate the line by the necessary amount to make them equal.

First, let me restate the last sentence in the problem with one of the two possible interpretations I would give it.

There is at least one straight line that splits the enclosed region R into two regions, or two collections of regions, of equal area; and, splits the curve into two curves, or two collections of curves, where the sum of the lengths of the curves in each collection is the same.

Proof: (without rigorous detail)

Construct a circle about the region R. Pick a point A on the circle then find another point B on the circle where line segment AB divides the circle into two segments and the region R into regions of equal area. Line segment AB will also divide the curve into two curves or sets of curves. The length totals of each set may or may not be equal.

If one set of curves has a greater lengths total than the other, then take points X starting at A, and Y starting at B, and move them clockwise about the circle. Do so in continuous way so that XY maintains the equal area split of region R.

At some point in the process, before point X reaches B and Y reaches A, the collection of curves (separated by XY) with smaller length total will become the collection of curves with the greater length total. Somewhere in the process, while maintaining an equal area split, the length total in each collection will be equal.

This probably isn't a solid solution, but it's enough to convince me that I'm right. It seems reasonable to assume there exists a "smooth" way of rotating and and translating the line so that we change its orientation (i.e. a continuous funtion, of sorts), but maintain equal areas on either side of the line.

Draw the horizontal line that bisects the area of the curve (this exists by sliding the horizontal line downward from the top until equal areas are achieved). Imagine our line has a "top" and a "bottom", and I have a meter that reads the perimeter on the "top" side and the "bottom" side. Let the top perimeter be T and the bottom perimeter be B. If T=B, then we are done. Otherwise, use the assumed "smooth" rotation/translation to spin the line 180 degrees. Then the bottom side is now on the top, and the top side is now on the bottom. The old bottom side reads T, and the top side reads B. Before the rotation, the difference of the two readings is T-B. After, it is B-T. Since one must be negative and the other positive, and it is reasonable to assume that both T and B are continuous functions of the line orientation, the IVT says the difference must have been 0 at some point, so at some point we had equal perimeters.

Consider the center of mass of closed figure, C, any line through C will divide it into two equal areas. Consider a line through C intersecting the curves at A and B. Let 'x' be the measure of length along the curve from A to B, say clockwise and 'y' be from B to A in the same direction. Consider the measure ' d = x - y '. Now move the point A along the curve (say in clockwise direction) and get new line A B , such that it passes through C, till the points A & B are interchanged. Now the new measure, d is negative of previous measure with same magnitude. This implies that there exist a line A'B' passing through C, somewhere when moving the line from AB to BA such that measure d =0, in other words A'B' divides the perimeter.

NOTE : In case of non convex curve where the line through Center of Mass can intersect the curve at more than two points take points A and B on either side of C

This problem is better understood with fractal geometry. This curve (map of India) actually has an infinite perimeter but encloses a finite area. Hence we can divide the area in two equal halves and not worry about the perimeter. Also look up the coastline paradox for this (very interesting). :D :D

We need not think on such an intricate solution. Just take the example of a circle in which we have at least one diameter. Now a diameter bisects the circumference and also the area of the circle. True is the case with an ellipse like an oval or geoid shape. So we get at least two figures that satisfy the given conditions. Now on a Cartesian plane, a parabola or hyperbola subtend some area that can be bisected by a line which also bisects the length of the curve-bolas. So by PCI i.e. Principle Of Geometric Induction, the statement is true.

I like this question very much.

There are an infinite number of lines that bisect the area. Choose a line with an arbitrary slope. As we "slide" a line from one side to the other, the area on one side of the line changes from 0 to a. Since an infinitely small change in the position of the line results in an infinitely small change in the area, the area is a continuous function of the position of the line. By the intermediate value theorem, there exists a line where the area on one side is a/2, which is half the area.

Now consider an arbitrary line that bisects the area. As the direction of the line changes, the portion of the perimeter does as well. Let the portion be x. Now rotate the line 180 degrees, and now the portion is p-x. Again, by the intermediate value theorem, the perimeter is p/2.

Considering a curve AB, there will be one straight line passing at point C where this will bisect the perimeter and the area of curve and that is TRUE statement.

K.K.GARG,India