Quant, Math & Computer Science Puzzles for Interview Preparation & Brain Teasing
A collection of ~225 Puzzles with Solutions (classified by difficulty and topic)

Dec 13, 2012

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.

13 comments:

  1. the line through their centres of masses

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  2. Actually, ignore my last answer

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  3. 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. Is it a theorem ? A lemma ? :- If one First, finds any line that bisects the area of a triangle....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? I d not see this so easily!!

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  5. 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. 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. Vintage problem ... but is it solvable by basic principles or do we need to know computational geometry etc for solving this?

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  12. 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. "Any line through the incenter of the triangle divides it into 2 parts with equal area (and perimeter too). "
    ---------- The assertion is not necessarily correct.

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