A Balkan Mathematical Olympiad problem

Reference: Complex Numbers from A to …Z by Titu Andreescu and Dorin Andrica

Balkan Mathematical Olympiad, 1985.

Problem:

Let O be the circumcenter of the triangle ABC, let D be the mid-point of the segment AB, and let E be the centroid of the triangle ACD. Prove that lines CD and OE are perpendicular if and only if AB=AC.

Solution:

Let O be the origin of the complex plane and let a, b, c, d, e be the coordinates of points A, B, C, D, E respectively. Then,

d=\frac{a+b}{2} and e=\frac{a+c+d}{3}=\frac{3a+b+2c}{6}

Using the real product of complex numbers, if R is the circumradius of triangle ABC, then

a.a=b.b=c.c=R^{2}

Lines CD and DE are perpendicular if and only if (d-c).e=0. That is,

(a+b-2c).(3a+b+2c)=0.

The last relation is equivalent to

3a.a+a.b+2a.c+3a.b+b.b+2b.c-6a.c-2b.c-4c.c=0, that is, a.b=a.c — call this equation I.

On the other hand, AB=AC is equivalent to |b-a|^{2}=|c-a|^{2}. That is,

(b-a).(b-a)=(c-a).(c-a)

or, b.b-2a.b+a.a=c.c-2a.c+a.a, hence, a.b=a.c —- equation II

The relations (1) and (2) show that CD is perpendicular to OE, if and only if AB=AC.

Ref: Complex Numbers from A to Z by Titu Andreescu and Dorin Andrica

Thanks Prof Andreescu!

from,

Nalin Pithwa

 

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