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ecoist
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Sum of Similars is Similar
« on: Jan 19th, 2008, 10:22am » |
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Let ABC and XYZ be triangles in the plane labelled counterclockwise. Suppose that, for some real r>0, rAB=XY, rBC=YZ, and rAC=XZ (i.e., XYZ is similar, and similarly oriented, to ABC). Show that the triangle whose vertices are A+X, B+Y, and C+Z is a single point or is similar to ABC. (Example: A=(-3,1), B=(-1,2), C=(0,1), X=(3,3), Y=(2,0), and Z=(0,0). Here, r=sqrt(2). Then the triangle (A+X)(B+Y)(C+Z) is similar (congruent, in this example) to ABC.)
Thanks, Hippo, for pointing out my oversight. The sum could be zero, and I should have allowed r=0.
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| « Last Edit: Jan 19th, 2008, 8:30pm by ecoist » |
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Hippo
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Re: Sum of Similars is Similar
« Reply #1 on: Jan 19th, 2008, 3:07pm » |
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If the triangle collapses to one vertex ... (in complex plane A,B,C are roots of x^3=1, XYZ are roots of x^3=-1), otherwise ...
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Icarus
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Re: Sum of Similars is Similar
« Reply #2 on: Jan 19th, 2008, 4:48pm » |
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Of course, this is not just true in the plane. It is true in any euclidean space of dimension > 1.
Even Hippo's counter-example could be considered to work, with the r value for the sum being 0, if one is willing to stretch the definitions.
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ecoist
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Re: Sum of Similars is Similar
« Reply #3 on: Jan 19th, 2008, 9:05pm » |
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Isn't the following a counterexample to extending the result to any Euclidean space of dimension >1, Icarus? A=(0,0,0), B=(1,0,0), C=(1,1,0), X=(0,0,1), Y=(0,0,0), and Z=(1,0,0). Similarity is not enough in the plane; similarly oriented is also required. Don't know what "similarly oriented" means in 3-space. Interchanging X with Z in the counterexample is also a counterexample.
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towr
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Re: Sum of Similars is Similar
« Reply #4 on: Jan 20th, 2008, 7:12am » |
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I would propose "similarly oriented" means that the congruent sides are parallel (AB || XY and BC || YZ and AC || XZ).
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ecoist
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Re: Sum of Similars is Similar
« Reply #5 on: Jan 20th, 2008, 9:47am » |
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Maybe my choice of words is bad, towr, sorry. The triangles (0,0)(3,0)(2,1) and (0,0)(-3,0)(-2,1) are similar but not similarly oriented. One has to flip the former about the y-axis to get the latter; and their sum is not similar to either, equalling (0,0)(0,0)(0,2). The puzzle holds only for those triangles in the plane which are similar without flipping.
Your idea suggests the definition: the angles between AB and XY, between BC and YZ, and between AC and XZ all be equal. Hmm! Wonder now if this definition allows generalizing to dimensions greater than 2!
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| « Last Edit: Jan 20th, 2008, 9:59am by ecoist » |
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Hippo
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Re: Sum of Similars is Similar
« Reply #6 on: Jan 20th, 2008, 1:04pm » |
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on Jan 20th, 2008, 7:12am, towr wrote:| I would propose "similarly oriented" means that the congruent sides are parallel (AB || XY and BC || YZ and AC || XZ). |
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In that case it would be much easier. ... it avoids skew lines (for concurrent lines AX, BY, CZ this is easily visualizable ... when ABC and XYZ are drawn on parallel planes ... (A+X)/2,(B+Y)/2,(C+Z)/2; ((A+X)+0)/2, ((B+Y)+0)/2, ((C+Z)+0)/2))
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| « Last Edit: Jan 20th, 2008, 1:05pm by Hippo » |
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Icarus
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Re: Sum of Similars is Similar
« Reply #7 on: Jan 20th, 2008, 2:16pm » |
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My interpretation of "similarly oriented" is vector-parallelism. It strikes me that the problem is a simple algebraic calculation using vectors.
Side AB of the triangle is represented by the vector B - A. The condition says that (Y-X)=r(B-A), (Z-Y)=r(C-B), (X-Z) = r(A-C). For the sum triangle, the sides are ((B+Y)-(A+X) = (B-A) + (Y-X) = (1+r)(B-A). Similarly, the other two sides are (1+r)(C-B) and (1+r)(A-C). Thus the sum triangle is similar to ABC. This calculation works in any number of dimensions.
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Eigenray
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Re: Sum of Similars is Similar
« Reply #8 on: Jan 20th, 2008, 2:48pm » |
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That's what I thought at first too, but I think here it means the transformation taking one to the other is orientation preserving. That is, it's a product of translation, rotation, and scaling. Mathworld calls this "directly similar", and the result is essentially the fundamental theorem of directly similar figures.
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ecoist
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Re: Sum of Similars is Similar
« Reply #9 on: Jan 20th, 2008, 3:33pm » |
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Thanks, Eigenray. I made up this problem years ago, not knowing that someone else had done it first. I found it interesting that "directly similar" plane figures form a group!
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