Napoleon's Theorem,
A second proof with complex numbers
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On each side of a triangle, erect an equilateral triangle, lying exterior to the original triangle. Then the segments connecting the centroids of the three equilateral triangles themselves form an equilateral triangle.
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Let the original triangle be ABC with equilateral triangle ABC1, CAB1, and BCA1 built on its sides. Think of all the vertices involved as complex numbers. We shall apply a classical criterion to the three equilateral triangles. Let j be a suitable rotation through 120°. Then the fact that triangles ABC1, CAB1, and BCA1 are equilateral may be expressed as
| (1) | A + jB + j2C1 = 0 |
| (2) | C + jA + j2B1 = 0 |
| (3) | B + jC + j2A1 = 0 |
The center of ΔABC1 is given by P = (A + B + C1)/3, and similarly for centers Q and R of triangles CAB1 and BCA1: Q = (C + A + B1)/3 and R = (B + C + A1)/3. We want to show that P + jQ + j2R = 0. Indeed,
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| 3(P + jQ + j2R) | = A + B + C1 + j(C + A + B1) + j2(B + C + A1) |
| | = (B + jA1 + j2C) + j(A + jB + j2C1) + j2(C + jA + j2B1) |
| | = 0. |
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Napoleon's Theorem
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Copyright © 1996-2012 Alexander Bogomolny
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