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Equilateral Triangles On Sides of a Parallelogram: What Is It About?
A Mathematical Droodle


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Explanation

Copyright © 1996-2008 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.


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


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What if applet does not run?

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 + I3)/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

  1. Honsberger, In Pólya's Footsteps, MAA, 1999

Copyright © 1996-2008 Alexander Bogomolny

29397650Page copy protected against web site content infringement by Copyscape


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