Wikipedia:Reference desk/Archives/Mathematics/2016 April 10

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April 10

geometry problem

This diagram has 8 sections, indicated by the 7 black diagonal lines. I know the total area of the overall rectangle because I am able to measure it with a ruler in the horizontal and in the vertical. (It measures 66 cm vertically and 49 cm horizontally.) I want the area of each of the 8 sections to be equal. I want the black lines to be parallel to each other. How do I calculate what the spacing between the parallel lines should be? Bus stop (talk) 18:49, 10 April 2016 (UTC)

A 66cm x 49cm rectangle split into 8 regions of equal area by 7 diagonal lines

@Bus stop: Does it matter what angle the lines are at? —  crh ₂₃  (Talk) 19:20, 10 April 2016 (UTC)

In the interests of keeping it simple for myself I'd like to start by just trying to do this with the lines intersecting the horizontal and vertical at 45 degree angles. Bus stop (talk) 19:33, 10 April 2016 (UTC)

It may be simpler not to: an easy way is to make it such that the first line comes from the corner to 66/4 cm up the side (giving the triangle the correct area), do the same in the opposite corner and then use the remaining lines spaced apart by 66/8 cm (vertical spacing, not perpendicular) to form parallelograms of the correct size. I'll add a picture if I can draw one. —  crh ₂₃  (Talk) 19:37, 10 April 2016 (UTC)

@Bus stop: As promised, I have drawn a picture and inserted it here. This is a solution to the problem if the lines can be at any angle: this solution produces round lengths, rather than angles. —  crh ₂₃  (Talk) 20:00, 10 April 2016 (UTC)

Crh23—that is very interesting. You have come up with a simple solution to my problem. It is something I had not thought of. I like a lot of things about it from a design perspective. This is for a painting. That is a nice diagram. Thank you.

Note that you could also put the triangles along the long sides of the rectangle, although the angle would be different there. StuRat (talk) 21:41, 10 April 2016 (UTC)

I think I know what you mean, StuRat. Then the parallelograms would all occur along the short side (the horizontal side) of the rectangle. Interesting, and it may make a good design. But I had in mind a 45 degree angle of intersection between diagonal lines and both the horizontal and the vertical. That would also involve varying distances between the parallel lines. Bus stop (talk) 22:41, 10 April 2016 (UTC)

You can also make all the lines parallel to a diagonal. If the rectangle is (0, 0), (a, 0), (0, b), (a, b) then make lines though (a/2, 0) and (0, b/2), (√2a/2, 0) and (0, √2b/2), (√3a/2, 0) and (0, √3b/2), (a, 0) and (0, b), (a, (2-√3)b/2) and ((2-√3)a/2, b), (a, (2-√2)b/2) and ((2-√2)a/2, b), (a, b/2) and (a/2, b). If the angle is given then the problem is messy but doable. You have to integrate a trapezoidal shaped function, but it's linear on each piece if you divide up the domain into three pieces. The integrals on each piece are easy; quadratic functions. You then need to find where it crosses each of the lines kab/8, which means solving some quadratic equations. --RDBury (talk) 22:36, 10 April 2016 (UTC)

Hi RDBury—thanks for the effort you've expended. After researching the word diagonal I realize I was using it incorrectly. I don't mean diagonal. What I mean are lines that intersect with the vertical and the horizontal at 45 degree angles. I'm sorry for the confusion I introduced. Bus stop (talk) 01:28, 11 April 2016 (UTC)

Each of six parallel stripes covers 1/8 of the rectangle area, so the vertical distance v between the lines should be equal 1/8 of the rectangle's height H. Consequently the height of each triangle is 2/8 H. The angle between the slant lines and the rectangle's base is $\arctan \left(\left({\tfrac {2}{8}}H\right):W\right)=\arctan(H:(4W))$. Its cosine is ${\frac {4W}{H^{2}+(4W)^{2}}}$ and the distance (slant) between lines is v times the cosine:
      ${\frac {2}{8}}H\cdot {\frac {4W}{H^{2}+(4W)^{2}}}={\frac {WH}{H^{2}+(4W)^{2}}}$
--CiaPan (talk) 06:27, 11 April 2016 (UTC)

Actually the problem isn't as messy as I thought when the angle is given. Restate and generalize the problem as this: Given a rectangle height H and width W and a number x between 0 and 1, divide the rectangle into two sections of area xHW and (1-x)HX by a line with slope m≠0, so that the xHW area is below and the (1-x)HW area is above. We can take |m|=1 (i.e. 45°) by rescaling the vertical axis. The we can take m=-1 (upper left to lower right) by reflecting through a vertical line. Also assume H≥W, otherwise reflect through a diagonal line. There three possibilities; either the lower left region is a triangle, or the upper right region is a triangle, or both regions are trapezoids (=trapezia outside North America, see a thread from a few weeks ago). In the first case you want an isosceles right triangle of area xHW, and this would have sides √(2xHW) and would work as long as √(2xHW)≤W, which is equivalent to x≤W/2H. For the second case you want an isosceles right triangle of are (1-x)HW and this would have sides √(2(1-x)HW) and would work for x≥1-W/2H. In the third case, the height of the lower trapezoid, meaning the distance between the parallels, is W. Also the difference between the lengths of the parallel sides is W. If u and v are these lengths then you get v-u=W and W(u+v)/2=xHW, which solves to u=xH-W/2, v=xH+W/2 and this works for W/2H < x < 1-W/2H.

For this particular problem, W = 49, H = 66, and x has values 1/8, 2/8, ... 7/8. W/2H is between 2/8 and 3/8 so the first two cuts go through the left and bottom sides of the rectangle. The first section is a right triangle with sides √808.5 ≈ 28.43. The second cut would go through the left and bottom sides at distance √1617 ≈ 40.21 from the bottom left corner of the rectangle. The third cut would go through the vertical sides at distances .25 and 49.25 from the bottom. The fourth cut would go through the vertical sides at distances 8.5 and 57.5 from the bottom. The remaining cuts are symmetrical with the first three. The result is backwards from the picture but I'm assuming that doesn't matter. --RDBury (talk) 10:41, 11 April 2016 (UTC)

RDBury—I think you calculated correctly. I'm not good with math. I lightly indicated by pencil line the results of your calculations on the canvas and it is as I had imagined. Some areas are "fatter" and some areas are more attenuated. Thank you for your assistance. Bus stop (talk) 18:23, 11 April 2016 (UTC)

I want to calculate this for a square canvas. I don't happen to have a square canvas but I want to see what it would look like. I've taken an average sheet of copy paper and trimmed off one edge so that it now measures 21.5 cm by 21.5 cm. My problem is the same. I want to draw 7 straight lines at a 45 degree angle sloping from lower left to upper right so that the resulting 8 shapes are equal in area. Multiplying 21.5 times 21.5 I get a total area of 462.25 sq. cm. Dividing by 8, I get 57.78125 sq. cm. The square root of 57.78125 is 7.6013978. This is the length of the legs of my triangles found at the lower right and the upper left of my "canvas". But this is where I get stumped. How do I determine where I draw my 5 other lines? Thank you. Bus stop (talk) 21:28, 12 April 2016 (UTC)

I realize that one of those five lines will be a diagonal of the square "canvas". It will connect the lower left corner with the upper right corner. Bus stop (talk) 21:34, 12 April 2016 (UTC)

In this case your cuts are parallel to the diagonal, so RDBury's calcuations in that case are what you need. (They come from observing that your successive cuts, starting from the diagonal, cut off a series of similar triangles of prescribed area.) --JBL (talk) 22:03, 12 April 2016 (UTC)

Hi Joel B. Lewis—I can't grasp this. I know that the square can be divided by a diagonal. Doing so results in two isosceles right triangles. In such a triangle one leg squared plus the other leg squared equals the hypotenuse squared. But I think that the area of the various isosceles right triangles found within the triangles which resulted from drawing the diagonal are used to find the points from which the lines are drawn to divide the square into the sections that I have specified. The area of the various triangles would be 1/8 of the total square, 2/8 of the total square, and 3/8 of the total square. I am not able to find out the areas of these triangles using just the leg and the hypotenuse. I am thus stumped as to where to place my lines. Thanks for your assistance. Bus stop (talk) 05:21, 14 April 2016 (UTC)

Square with 5 diagonals split into strips of equal area

@Bus stop: You do not need to use the hypotenuse to do any of the necessary calculations. Since the triangles are right and isosceles, their area is ${\frac {1}{2}}l^{2}$, where $l$ is the length of one leg. Lets say the square to be divided has dimensions 1 unit by 1 unit (so you can just multiply any lengths found by the side length of your square). We'll call the lengths of the legs of the successively larger triangles $l_{1},l_{2},l_{3},$ etc. As you said, the first triangle has area of ${\frac {1}{8}}$ of the total, and solving ${\frac {1}{2}}l_{1}^{2}={\frac {1}{8}}$ gives $l_{1}=0.5$. The next triangle will have double the area, so the lengths will increase by a factor of ${\sqrt {2}}$, giving $l_{2}={\frac {\sqrt {2}}{2}}$. The next triangle will have triple the area of the first, so by the same process we deduce $l_{3}={\frac {\sqrt {3}}{2}}$. Since the next triangle has area of ${\frac {4}{8}}={\frac {1}{2}}$, $l_{4}$ is the center diagonal, and the rest of the triangles are the same reflected around that centre line. Picture is on the right.—  crh ₂₃  (Talk) 07:35, 14 April 2016 (UTC)

Hi Crh23—You say "Lets say the square to be divided has dimensions 1 unit by 1 unit (so you can just multiply any lengths found by the side length of your square)." I'm sure there is wisdom to assuming that the square is "1 unit by 1 unit" but I think this may be contributing to my inability to understand this. (I'm also just dense.) I have a square piece of paper in front of me. It measures 21.5 cm by 21.5 cm. You say "Since the triangles are right and isosceles, their area is ${\frac {1}{2}}l^{2}$, where $l$ is the length of one leg." Therefore the largest triangle has an area of 231.125 sq. cm. I know that there are 3 smaller triangles within it. Their areas would be 173.34375 sq. cm., 115.5625 sq. cm., and 57.78125 sq. cm. You also say that "the rest of the triangles are the same reflected around that centre line." I understand that. I'm just not sure what calculation I use to find the lengths of the legs of those smaller triangles. I'm also surprised that the smallest triangle has a leg that is exactly half the length of the side of the square. Thanks. Bus stop (talk) 18:44, 14 April 2016 (UTC)

A square split into eight equal triangles by way of six line segments

Alright Bus stop, I'll rephrase slightly. Firstly, to convert the lengths from my diagram of 1 unit x 1 unit to your square paper of 21.5 cm x 21.5 cm you multiply the lengths I had by 21.5 cm (this is based on Similarity). Secondly, I have inserted a picture on the right to show you why the smallest triangle has a side length of half that of the square (it seems intuitively strange but is true, humans are bad at judging area) The area of an isosceles right triangle is half of that of the square on one of its sides (as the triangle is half of a square split along its diagonal): the square made up of triangles 1 and 2 in the picture on the right has twice the area of each triangle. Thus to make a triangle of area, for example, 173.34375 cm², you would find the side length of a square that has twice that area (i.e. 346.6875 cm²). To do this we take the square root of the area of the square:√(346.6875 cm²)=18.62 cm. Has that cleared things up? —  crh ₂₃  (Talk) 21:11, 14 April 2016 (UTC)

Thanks. I think I got it. I realize this isn't rocket science. But I can only presume I have a mental block. I may be back when I have a canvas of known dimensions, if I can't figure it out, or if I want my conclusions checked. Sincerely, thanks for trying. I actually think a lot about mathematics vis-a-vis designs for paintings. I can do some calculations on my own. I have a painting that I like a lot consisting entirely of rectangles of various configurations. In that painting I also wanted the areas of all the rectangles to be equal. The math for that was easy to do. Bus stop (talk) 04:46, 16 April 2016 (UTC)