Equilateral Triangles On Sides of a Parallelogram
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A Mathematical Droodle
Explanation
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Copyright © 1996-2012 Alexander Bogomolny
Equilateral Triangles On Sides of a Parallelogram
The applet suggests the following theorem [Honsberger, pp. 278-279]:
On the sides of a parallelogram ABCD, construct similarly oriented equilateral triangles ABX, BCY, CDZ, and DAW. Not accidentally, the quadrilateral XYZW happens to be a parallelogram. On its sides, in turn, construct equilateral triangles XYP, YZQ, ZWR, WXS with the orientation opposite to that of ABX, BCY, CDZ, and DAW. Not surprisingly PQRS is again a parallelogram. Surprisingly, PQRS and ABCD coincide.
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, W are linear combinations of A, B, C, D with complex coefficients:
| (1) |
X = cA + (1 - c)B,
Y = cB + (1 - c)C,
Z = cC + (1 - c)D,
W = cD + (1 - c)A,
|
where
c = (1 + I 3)/2,
1 - c = (1 - I3)/2.
On the other hand, by construction,
P = cY + (1 - c)X,
Q = cZ + (1 - c)Y,
R = cW + (1 - c)Z,
S = cX + (1 - c)W.
In terms of A, B, C, D these can be written as
| (2) |
| P | = ((1 - c)2 + c2)B + c(1 - c)A + c(1 - c)C |
| | = A - B + C, and similarly |
| Q | = B - C + D, |
| R | = C - D + A, |
| S | = D - A + B. |
|
However, since ABCD is a parallelogram, A - B = D - C, so that, from the first identity in (2),
| P | = A - B + C |
| | = D - C + C |
| | = D, |
and similarly for the other three vertices.
References
- Honsberger, In Pólya's Footsteps, MAA, 1999
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Copyright © 1996-2012 Alexander Bogomolny
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