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CTK Exchange
Owen
Member since Nov-11-05
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Jul-24-10, 11:24 AM (EST) |
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"Fermat Point"
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I don't understand the second solution for finding the Fermat point P of a triangle (P is defined to be the point minimizing the sum AP + BP + CP for a given triangle ABC). https://www.cut-the-knot.org/Generalization/fermat_point.shtml The solution assumes such a point P exists and then argues that one can adjust the triangle to be isosceles, and then, ultimately equilateral, in which case P is simply the centroid of triangle ABC. But, I don't understand how this finds P in the original triangle. (The argument says to slide A to a point along ray PA until AB=BC, but this supposes that we know where P is already.) What am I missing here? |
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alexb
Charter Member
2558 posts |
Jul-24-10, 11:49 AM (EST) |
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2. "RE: Fermat Point"
In response to message #0
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>I don't understand the second solution for finding the >Fermat point P of a triangle (P is defined to be the point >minimizing the sum AP + BP + CP for a given triangle ABC). What the second solution does is proving that, provided the Fermat point exists, the segments joining it to the vertices meet at 120° angles. This is so because it is true for equilateral triangles. How does this work?If Fermat point P exists for ÄABC, then the proof shows that moving a vertex, say A, towards P does not cause P to move. If P remains in place when A moves along AP and the same holds for B and C, then, without changing the lines AP, BP, and CP, we can transform ÄABC into an equilateral triangle. For the latter, due to the symmetry, the lines AP, BP, CP do meet at 120°. Since the lines have not moved under our transformations, the same holds for the original triangle. > >https://www.cut-the-knot.org/Generalization/fermat_point.shtml > >The solution assumes such a point P exists and then argues >that one can adjust the triangle to be isosceles, and then, >ultimately equilateral, in which case P is simply the >centroid of triangle ABC. But, I don't understand how this >finds P in the original triangle. It does not. P is the Fermat point of the original triangle because it has been so chosen. (The argument says to >slide A to a point along ray PA until AB=BC, but this >supposes that we know where P is already.) What am I >missing here? We are not looking for a point, but only to establish that certain angles are all 120°. This is so because
- We may transform a given triangle into an equilateral one without changing the lines AP, BP, CP.
- For an equilateral triangles those lines obviously form 120° angles.
<\ol>
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Owen
Member since Nov-11-05
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Jul-24-10, 05:30 PM (EST) |
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3. "RE: Fermat Point"
In response to message #2
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OK, thanks Alex. I understand now. The definition of the Fermat point at the top said it was the one that minimized the sum of the distances. Nothing is said there (or at the beginning of the solution) that one is just trying to show that each angle is 120 degrees, and I missed the last line of the solution where this connection was made. My real goal is to construct the equivalent point for a tetrahedron. It'seems your proof can be adapted to show: If P is the Fermat point of tetrahedron ABCD, then P will also be the Fermat point of tetrahedron A'B'C'D', where A', B', C', and D' lie on the rays PA, PB, PC, and PD, respectively. Then, we reach the conclusion that each "angle" (the four 3-D trihedral angles from P to the three vertices of each of the four faces) formed should have the same "measure". I suppose this measure will be pi, or 1/4 the surface area of the unit'sphere, just like 2pi/3 is 1/3 the perimeter of the unit circle in the 2-D case. I may be getting ahead of myself, but the hope is that appropriate tetrahedrons can be built on each face of ABCD (analagous to the equilateral triangles built on each edge of triangle ABC in the 2-D case) in order to actually construct P. |
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alexb
Charter Member
2558 posts |
Jul-24-10, 06:14 PM (EST) |
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4. "RE: Fermat Point"
In response to message #3
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>OK, thanks Alex. I understand now. The definition of the >Fermat point at the top said it was the one that minimized >the sum of the distances. Nothing is said there (or at the >beginning of the solution) that one is just trying to show >that each angle is 120 degrees, This is a consequence of the construction and also follows from the relationship with the Napoleon triangles. >and I missed the last line >of the solution where this connection was made. > No, no. The last line simply describe exceptional cases where not all angles in a triangle are less 120°. That the cases are bound to be exceptional follows from the preceding paragraphs where the rays at Fermat point meet at 120°. >My real goal is to construct the equivalent point for a >tetrahedron. It'seems your proof can be adapted to show: >If P is the Fermat point of tetrahedron ABCD, then P will >also be the Fermat point of tetrahedron A'B'C'D', where A', >B', C', and D' lie on the rays PA, PB, PC, and PD, >respectively. Yes, this argument appears to work in 3D as well. The minimum point does not change when a vertex slides along a ray through that point. >Then, we reach the conclusion that each "angle" (the four >3-D trihedral angles from P to the three vertices of each of >the four faces) formed should have the same "measure". I >suppose this measure will be pi, or 1/4 the surface area of >the unit'sphere, just like 2pi/3 is 1/3 the perimeter of the >unit circle in the 2-D case. Probably something along these lines. >I may be getting ahead of myself, but the hope is that >appropriate tetrahedrons can be built on each face of ABCD >(analagous to the equilateral triangles built on each edge >of triangle ABC in the 2-D case) in order to actually >construct P. I am not sure about that. Conditions for concurrency may be more involved in 3D. |
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Owen
Member since Nov-11-05
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Jul-24-10, 06:36 PM (EST) |
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5. "RE: Fermat Point"
In response to message #4
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>No, no. The last line simply describe exceptional cases where not >all angles in a triangle are less 120°. That the cases are bound to >be exceptional follows from the preceding paragraphs where the rays >at Fermat point meet at 120°. All I was saying was that the solution never says that the goal is to prove that the point is the one which makes all angles 120 degrees. The number 120 only appears (explicitly, anyway) in the last sentence, though if one knows where the proof is headed, it is certainly present earlier. Anyway, I understand, now. Thanks. >I am not sure about that. Conditions for concurrency may be more >involved in 3D.
I'll let you know if I find an elegant construction.
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Owen
Member since Nov-11-05
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Aug-07-10, 04:26 PM (EST) |
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7. "RE: Fermat Point"
In response to message #6
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OK, here is the story. It'seems that the last word on the 3-D version of the Fermat-Toricelli point (i.e., the point P that minimizes the sum AP + BP + CP + DP for a tetrahedron ABCD) appears in an article by Folke Erikkson in a 1997 issue of the Mathematical Gazette. Such a point P will be an interior point of the tetrahedron if each of the four solid angles, one at each vertex, have measure less than pi. (If an angle has measure pi or greater, P will coincide with that vertex.) When P is an interior point of ABCD, the four solid angles (P, ABC), (P, ABD), (P, ABC), (P, BCD) will have measure pi. (This is analogous to the 2-D case, where each of the angles APB, APC, and BPC have measure 120 degrees.) There is some additional structure for those angles, too, but evidently the geometric construction of P is an open problem. Several of the 2-D proofs you give on your side for showing 1) P is Fermat point --> congruent angles from P to the vertices, and 2) Congruent angles from P to vertices --> P is Fermat point seem to extend to the 3-D case. However, the obvious analogs of the constructions of P do not work. As you suspected, concurrency of lines is much trickier in space. |
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