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Let $R$ be a $3\times 5$ rectangle divided as shown in the figure below, where the division points are arbitrary. The the sum of the areas of the subrectangles is always $15$. Prove that this is similarly true for a general rectangle and an arbitrary subdivision.
I found this https://math.stackexchange.com/a/771043/533661, but I haven't learnt the Green's Theorem yet. Are there other ways to prove this statement? Thanks.
enter image description here

2 Answers2

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For any such subdivision, extend all the horizontal and vertical segments to the boundary of the rectangle. Then you have a sum of areas of rectangles arranged in a grid.

The dimensions of the small rectangles will be differences of the $x$ and $y$ coordinates of the given points. If you write out the sum carefully you can make it telescope to give you the area of the containing rectangle.

The algebra and notation will be tedious but straightforward.

Ethan Bolker
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  • The OP asks explicitly about "arbitrary" division points. How can you fulfill the request by writing out given coordinates, as you suggest? – David G. Stork Jun 12 '20 at 22:00
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Here, as the OP requests, are arbitrary division points (not some given set):

$$A = \color{red}{x_2 y_1}\color{black} + \color{purple}{x_1 (h - y_1)}\color{black} + \color{blue}{(w - x_2) y_2}\color{black} + \color{green}{(w - x_1)(h - y_2)}\color{black} + \color{yellow}{(x_2 - x_1)(y_2 - y_1)}\color{black} = w h .$$

enter image description here

The same principle holds within any component rectangle as well, of course, yielding the same total area.

David G. Stork
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  • Gosh... I was expecting a veritable swarm of downvotes on @EthanBolker's answer, which merely states in words what my earlier (and downvoted) solution provides with detailed (and I trust useful) calculations. Bring them on!! – David G. Stork Jun 12 '20 at 21:52