Geometry Contruction - Bisect Areas of 2 Triangles

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.

Comments

  1. takaki7:45 PM, December 29, 2012

    the line through their centres of masses

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  2. takaki8:02 PM, December 29, 2012

    Actually, ignore my last answer

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  3. Betzalel12:32 AM, March 02, 2013

    I would use the following methodology.

    First, 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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  4. Vivek 6:52 PM, March 06, 2013

    This comment has been removed by the author.

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  5. Vivek 4:34 PM, March 17, 2013

    One 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..

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  6. Vivek 10:41 PM, March 23, 2013

    I 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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  7. Vivek 4:25 PM, April 04, 2013

    This comment has been removed by the author.

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  8. Vivek 7:04 AM, April 06, 2013

    This comment has been removed by the author.

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  9. Vivek 7:08 AM, April 06, 2013

    This comment has been removed by the author.

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  10. Vivek 7:12 AM, April 06, 2013

    This comment has been removed by the author.

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  11. Vivek 2:38 PM, April 06, 2013

    This comment has been removed by the author.

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  12. Sravan3:16 PM, May 25, 2014

    Any 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.

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

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  13. Vivek 1:48 AM, July 24, 2014

    This comment has been removed by the author.

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