An Anticomplementary Triangle Surprise
What Is It About?
A Mathematical Droodle

29 November 2015, Created with GeoGebra

The applet suggests the following theorem [Honsberger, pp. 276-278] from triangle geometry:

On the sides of ΔABC, construct similarly oriented equilateral triangles ABZ, BCX, CAY. In turn, on the sides of ΔXYZ, form triangles XYR, YZP, ZXQ with an orientation opposite to that of ABZ, BCX, CAY. Then ΔPQR is the anticomplementary triangle of ΔABC.

( ΔPQR is anticomplementary to ΔABC if ΔABC is the medial triangle of ΔPQR.)

29 November 2015, Created with GeoGebra

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,

where

  c = (1 + i3)/2,
1 - c = (1 - i3)/2.

On the other hand, by construction,

  P = (1 - c)Z + cY,
Q = (1 - c)X + cZ,
R = (1 - c)Y + cX.

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.

We want to show that, e.g., PQ||AB and that C is the midpoint of PQ. From (2),

 
PQ= Q - P
 = (A - B + C) - (-A + B + C)
 = 2(A - B),
 = -2·AB,

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,

so that indeed C is the midpoint of PQ.

References

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

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