Maxwell Theorem via the Center of Gravity
Michel Cabart
30 April, 2008
Below we offer a proof of Maxwell's theorem that is based on the notion of barycenter. Maxwell's theorem states the following fact:
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Given ΔABC and a point G, the sides of ΔMNP are parallel to the cevians in ΔABC through G. Then the cevians in ΔMNP parallel to the sides of ΔABC are concurrent.
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Below, the vector joining point A to B will be written in bold, so that, for example, AB = - BA.
In ΔABC with A', B', C' on sides opposite the vertices A, B, and C, the fact that the cevians AA', BB', CC' concur in point G is equivalent to either of the two conditions:
| | There is a triple of real numbers (a, b, c), unique up to a non-zero factor, such that |
| (1) | aGA + bGB + cGC = 0 |
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| | There is a triple of real numbers (a, b, c), unique up to a non-zero factor, such that |
| (2) |
A' = Z(B, b; C, c)
B' = Z(A, a; C, c)
C' = Z(A, a; B, b),
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where Z(X, x; Y, y) denotes the barycenter of two material points X and Y with masses x at X and y at Y.
Let's suppose lines AA', BB' and CC' intersect.
Step 1: NP, PM, MN are parallel to GA, GB, GC means there exists x, y, z such that NP = xGA, PM = yGB, MN = zGC. As NP + PM + MN = 0, xGA + yGB + zGC = 0. A comparison with (1) shows that (x, y, z) is a multiple of (a, b, c). We can assume (x, y, z) = (a, b, c). Thus NP = aGA, PM = bGB, MN = cGC.
Step 2: MM', NN', PP' are parallel to BC, AC, AB meaning there is (m, n, p) such that MM' = mBC, NN' = nAC, PP' = pAB. The first equality yields
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MM' = m(GC - GB) = (m/c)MN + (m/b)MP
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so that
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M' = Z(N, 1/c; P, 1/b) = Z(N, b; C, c)
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by multiplying by bc. Similarly,
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N' = Z(M, a; P, c) and
P' = Z(M, a; N, b).
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This proves the theorem thanks to (2).
Barycenter and Barycentric Coordinates
- 3D Quadrilateral - a Coffin Problem
- Barycentric Coordinates
- Barycentric Coordinates: a Tool
- Barycentric Coordinates and Geometric Probability
- Ceva's Theorem
- Determinants, Area, and Barycentric Coordinates
- Maxwell Theorem via the Center of Gravity
- Medians in a Quadrilateral
- Three glasses puzzle
- Van Obel Theorem and Barycentric Coordinates
Copyright © 1996-2009 Alexander Bogomolny
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