# 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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