Should you need noncongruent triangles of equal area, here’s how to get them. Segment BX is their altitude (height). Segments AC, CD, DE, EF, FG, GH, and HJ (triangle bases) are constructed to be congruent, so each of these non-overlapping, differently-colored triangles has the same area because they have the same base and height lengths. Extend AJ as a line, add more points, using the same spacing, and you can have as many such triangles as you desire.
Much work has been done on the medians of triangles: segments that connect vertices with the midpoints of opposite sides. This morning, I decided to explore what happens if the trideans are examined, rather than the medians.
The reason you don’t know the word “tridean” is simple: I just made it up. It’s related to a median, though. To find trideans, you must first trisect, rather than bisect, each side of a triangle. This gives you two equally-spaced points on each side. To form a tridean, simply connect one of those points to the opposite vertex. Every triangle has six trideans, and they split the triangle into a set of non-overlapping polygons, as you can see in the diagram.
When I examined the area of these polygons, I started finding unexpected things right away. As you can see, polygons of the same color in this diagram have the same area, even though they are non-congruent. Moreover, these areas are interesting fractions of the area of the entire triangle. The six yellow triangles, for example, each have 1/21st of the area of the entire triangle. Each blue triangle is 1/70th the area of the large triangle. Each green quadrilateral is 1/14th the area of the large triangle. 21, 70, and 14 have one factor in common: the number seven. Seven? I officially have NO idea why sevens are popping up all over the place in this investigation, but there they are.
The red hexagon in the middle has 1/10th the area of the entire triangle, and this number surprised me as well.
The three orange pentagons took a little more work. As you can see, dividing the area of the large triangle by the area of one of these orange pentagons yields 9.5454, with the “54” repeating. This decimal is 105/11. (Eleven?) At least 105 has three as a factor (as does 21, from earlier); three is the number I expected to pop out all over the place, but it shows up little in this investigation. However, what are the other factors of 105? There’s five — and, yet again, seven.
These sevens are everywhere in this thing, and I have no idea why.
Now, of course, I have proven none of this. This is merely a demonstration and explanation of something I think is a new discovery. I did change the shape of the triangle many ways, as a test, and each of these area ratios remained constant.
If anyone can shed some light on any of this — especially all these sevens — please comment.
This was created by augmenting a great dodecahedron with more great dodecahedra, and then augmenting the result with even more of them.
The software I used was Stella 4d, which you can find right here.
…with help from many of their hexagonal friends.
A regular polygon has some number of sides (n), and its sides and diagonals form a certain number of triangles (t).
For a triangle, n=3 and t=1.
For a square, n=4. There are four triangles congruent to the one shown in orange, and four more like the one shown in light blue, so adding these gives t=8.
For a pentagon, with n=5, there are five of the purple triangles, five of the green triangles, five of the red triangles, and five of the yellow triangles, for a total of t=20.
For a hexagon, there are two of the orange triangles, six of the yellow ones, twelve of the red ones, six of the light blue ones, twelve of the purple ones, twenty-four of the green ones, twelve of the dark blue ones, six of the grey ones, six of the black ones, twelve of the pink ones, and six of the brown ones. That’s 2+6+12+6+12+24+12+6+6+12+6 = t = 104.
Three yields one, four yields eight, five yields 20, and six yields 104. At the moment, I don’t have the patience to count the triangles in a heptagon, but it would clearly be, well, quite a few.
There may or may not be a formula for this; any pattern eludes me, so far. I am reminded of the alkane series in chemistry: one isomer each of methane, ethane, and propane, two of butane, three of pentane, and so on to 75 for decane, and beyond. All efforts to find a formula for the number of isomers, in terms of the number of carbon atoms, have failed (to date). For now, these are both filed under “unsolved problems.”
UPDATE: A friend of mine has shown me that this polygon problem has, in fact, been solved, and he provided this link: http://oeis.org/A006600 — apparently I missed some of the triangles in the pentagon (the red and yellow combined, for example), as well as the hexagon. The correct numbers for those two polygons are 35 and 110, respectively. Aside from this update, I’m not changing the rest, for I need reminders of my own fallibility. This will do nicely.
I am confident that this “beat frequency” effect will continue as this process of combining trigonometric and exponential functions continues.
http://www.software3d.com/stella.php — That’s where you can find software to make things like this.
The Platonic Icosahedron has twenty faces which are equilateral triangles. In the Golden Icosahedron, twelve of those triangles (the yellow ones) have been replaced by acute, isosceles triangles with a leg:base ratio which is the Golden Ratio.
To try the software I used to make this, just visit http://www.software3d.com/stella.php.
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