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This question is pretty hard to word without a picture, so I have attached one.

I am wondering if there is a general way to find the area of the green or blue areas given the ordered pair of some interior point (a,b). These lines are going from each vertex of the square and intersecting the interior point. For simplicity, I am assuming the shape is a square with side lengths 1. Of course, finding the area of one of the areas allows you to automatically have the other, so is one of the areas easier to find than the other?

Also, sorry the picture is not the best. I tried to draw it in paint.

Some attempts at a solution have included using just the principles of similar triangles and have yielded some pretty nasty expressions. I've considered attempting the use the shoelace formula as well but I don't know if it will work here. Any help is appreciated. Thanks!

enter image description here

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  • $\begingroup$ How about coordinate geometry? $\endgroup$
    – Arpan
    Apr 5, 2015 at 4:22

2 Answers 2

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Let the figure be a square with each side = s and the point be (a, b).

enter image description here

More than half of the co-ordinates are then known (in terms of s, a, b) except h, k, p and m but they can all be expressed in terms of s, a, and b.

For example, (m, 0) is a point on the line formed by (0, s) and (a, b). The corresponding equation is $\frac {y – b}{x – a} = \frac {s – b}{0 – a}$. (m, 0) lies on this line means $\frac {[0] – b}{[m] – a} = \frac {s – b}{0 – a}$.

Note that it is not necessary to calculate all the triangles area. This is because the sum of the areas all the light green triangles (originally was dark green) is equal to half of the area of the square. Thus, we only need to find the areas of triangles XHK, XAM and XBP.

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  • $\begingroup$ Thank you. How did you know that the area of the light green triangles will be exactly half? It looks realistic from the picture but is this a theorem? $\endgroup$
    – Rellek
    Apr 5, 2015 at 12:17
  • $\begingroup$ I have come up with an expression, although I am having issues with it because it only seems to give correct values within a certain range. I know for sure that if you have the point (.5,.25) that the value for the area of the green is 2/3, but when I put it in my expression it returns 5/6. My expression is: $$\frac{1}{2}+\frac{a}{2}(\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{b}-4)$$ $\endgroup$
    – Rellek
    Apr 5, 2015 at 12:46
  • $\begingroup$ Yes to your first question because the diagonal of a parallelogram (and hence rectangle, square) divides the figure into two equal halves. Need some time to digest your formula. $\endgroup$
    – Mick
    Apr 5, 2015 at 15:56
  • $\begingroup$ Alright. For the formula I really looked at the bases of the triangles XHK, XAM, XBP. For the base of XHK I got a base of $$\frac{a}{1-a}$$ For XAM I got $$\frac{ab}{1-b}$$ And finally for XBP the base came out to be $$\frac{a-ab}{b}$$ and I then summed them up, added 1/2 since the sides are assumed to be 1, and then did a lot of simplifying. $\endgroup$
    – Rellek
    Apr 5, 2015 at 16:29
  • $\begingroup$ Clarification: I multiplied the bases by their heights and 1/2 (for the area) which were much easier to find. I am going to re examine the base I found for XHK. $\endgroup$
    – Rellek
    Apr 5, 2015 at 16:38
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This is a continuation of my answer and it serves as a suggestion in replying the OP’ further query. [I don’t think I can include a picture in the comment.]

Judging from your finding, I suspect the attached can explain why it works only at certain occasions.

enter image description here

Depending on the relative values of a and b, the point X(a, b) will be located at different region. A 2 cases discussion may be necessary.

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  • $\begingroup$ So, by the symmetry of the square, I could just use a matrix to transform any (a,b) so that it has an equivalent pair (a', b') in the bottom half? I think that this might be a good idea, to just rotate the ordered pair by however many radians so that it will lie within the bottom. I am curious as to why (1/2,1/4) yields an incorrect answer though. I imagine that it has something to do with the fact that our $a$ coordinate lies on a centroidal axis? $\endgroup$
    – Rellek
    Apr 5, 2015 at 18:15
  • $\begingroup$ @Rellek I don't think there is anything to do with rotation. Will share more finding several hours later. $\endgroup$
    – Mick
    Apr 6, 2015 at 1:51
  • $\begingroup$ Alright. I randomly thought that perhaps, since the square is symmetric, do you think this area could be related to the distance from the center of the square? I noticed that the points (.25,.5),(.5,.25),(.75,.5), and (.5,.75) are all equidistant from the centroid and all map out the same area (1/3). I thought about drawing a circle of radius 1/2 about the centroid. I haven't checked yet, but maybe any point lying on that circle would map out the same area? Just an idea. $\endgroup$
    – Rellek
    Apr 6, 2015 at 2:14
  • $\begingroup$ @Rellek Data for the bases are all correct and therefore the total area $= \frac{1}{2} + \frac{a}{2} [b (1-b)^{2} + \frac{b^{2}}{1-b} + \frac{a}{1- a}]$. Note that this formula is true only when b>a (i. e. X(a, b) is on the y > x side of the diagonal y = x). If otherwise, you must draw another diagram for those points on the x<y side of the diagonal. $\endgroup$
    – Mick
    Apr 6, 2015 at 11:24

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