An Anticomplementary Triangle Surprise: What Is It About?
A Mathematical Droodle
The applet suggests the following theorem [Honsberger, pp. 276-278] from triangle geometry:
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On the sides of ABC, construct similarly oriented equilateral triangles ABZ, BCX, CAY. In turn, on the sides of XYZ, form triangles XYR, YZP, ZXQ with an orientation opposite to that of ABZ, BCX, CAY. Then PQR is the anticomplementary triangle of ABC.
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( PQR is anticomplementary to ABC if ABC is the medial triangle of PQR.)
As on several other occasions (e.g., Three Isosceles Triangles, When a Triangle is Equilateral?, and others), we can make a good use of complex numbers. Points X, Y, Z are linear combinations of A, B, C with complex coefficients:
| (1) |
X = (1 - c)B + cC,
Y = (1 - c)C + cA,
Z = (1 - c)A + cB,
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where
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c = (1 + i 3)/2,
1 - c = (1 - i 3)/2.
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On the other hand, by construction,
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P = (1 - c)Z + cY,
Q = (1 - c)X + cZ,
R = (1 - c)Y + cX.
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In terms of A, B, C these can be written as
| (2) |
| P | = ((1 - c)2 + c2)A + c(1 - c)B + c(1 - c)C |
| | = -A + B + C, and similarly |
| Q | = A - B + C, |
| R | = A + B - C. |
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We want to show that, e.g., PQ||AB and that C is the midpoint of PQ. From (2),
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| PQ | = Q - P |
| | = (A - B + C) - (-A + B + C) |
| | = 2(A - B), |
| | = -2·AB, |
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which tells us that PQ is parallel to AB and is twice as long. Further,
| (4) |
| (P + Q)/2 | = ((A - B + C) + (-A + B + C))/2 |
| | = (2C)/2 |
| | = C, |
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so that indeed C is the midpoint of PQ.
References
- Honsberger, In Pólya's Footsteps, MAA, 1999
Copyright © 1996-2008 Alexander Bogomolny
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