**Source:**Asked to me by Sankeerth Rao (EE IITB 4th year Student)

**Problem:**

Given any two triangles in a plane construct a line which bisects both their areas.

**Background:**

In fact the existence of such a line is true in a very general setting - for any two polygons in a plane there exists a line which bisects both their areas. In fact its true for any two Jordan measurable sets in a plane. Further generalized version is called the Ham Sandwich Theorem and is proved using Borsuk Ulam Theorem.

the line through their centres of masses

ReplyDeleteActually, ignore my last answer

ReplyDeleteI would use the following methodology.

ReplyDeleteFirst, find any line that bisects the area of the first triangle. (Easy to do.) Then there is a point on this line such that if you rotate the line about this point, the area in the triangle on both sides of the line will always be equal.

Next, if you rotate the line about this point, you'll at some point in time bisect the area of the second triangle.

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ReplyDeleteOne approach to solve this problem is to find a solution by co-ordinate geometry and then try to back engineer the solution into a geometry solution..

ReplyDeleteI think it should be possible to solve this problem by a set of rotational and translational transformations superimposed on one another. How?? ... will have to be thought. But just a hunch.

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ReplyDeleteAny line through the incenter of the triangle divides it into 2 parts with equal area (and perimeter too). This is not too difficult to prove.

ReplyDeleteSo, consider the line passing through the incenters of the given triangles.

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