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Geometry problem I generated. Help would be greatly appreciated.
Two lines originate from the same point A and run at a given angle (theta) continuously away from each other. A third line, AG, originates from the same point at half the given angle, and perpendicular to it run lines BB', CC', DD', etc. These lines are bisected by AG. All polygons formed inside the original two lines AF and AF' must have area x.
My goal is to describe this pattern in which the triangle and descending trapezoids share the same area x. I want to do this by finding some equation that scales the original height, h, generating the next polygon height in the sequence.
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Oh woops, I said AG is the bisector, I meant AA'.
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ratio a:b sides similar triangle
area a^2:b^2
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>>8998640
Would you mind elaborating a little bit? I'm not sure what you're saying.
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>>8998673
hes saying if you scale a triangle by some factor x then the area scales by a factor x^2
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>>8998537
You may be interested in problem 53 from the ancient Egyptian Rhind Papyrus, anon. It (seems to) consider an analogous version of your problem, regarding various trapezoids and an upper similar triangle generated by some given isosceles triangle. See the triangle at the bottom of pic related, and notice how it looks like your diagram.
That said, the problem is phrased in an extremely goofy way (and has incomplete information) which makes studying it more of a historical or philological problem than a mathematical problem as-such. So don't expect any big enlightenment toward your own problem if you check it out, just know that there's an ancient historical version of your problem.
Full disclosure: I wrote the wiki describing this problem.
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>>8998852
pic related. the bottom triangle is what is referred to. Notice the "parallels" with information next to them.
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>>8998640
This anon has it. The second triangle (ACC') must be twice the size of the first (ABB') so the difference in areas is the desired x. It is similar to the first triangle, so to get double the area we multiply the lengths by root(2). So the change in height is (root(2)-1)h, where h is the original triangle's height.
For the second trapezoid we do the same thing. The isosceles triangle it is part of, ADD' must be 3 times the size of the first (ie area of 3x). So we increase all the sides by a root(3). The difference in heights is (root(3)-root(2))h.
In general the nth trapezoid has height of (root(n+1)-root(n))h.
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>>8998854
Any other references you'd recommend besides the wiki?
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>>8998673
if h increases by factor k
area of trapez BB'C'C = h^2tantheta(k+1)(k-1)
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>>8998903
The wiki is based upon Chace so there's that, but once you actually read it it amounts to the same thing.
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Op gave up
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