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## 'Shape and Territory' printed from http://nrich.maths.org/

Sue Liu, S5, Madras College sent in a good solution which shows
that if $A, B$ and $C$ are angles in a triangle and $$\tan (A - B)
+ \tan (B - C) + \tan (C - A) = 0$$ then the triangle is isosceles.
Can you prove a stronger result? We start with the expression
$$\tan (A - B) + \tan (B - C) + \tan (C - A) = 0.$$ Write $X = A -
C$ and $Y = B - C$, then the given expression becomes $$\tan (X -
Y) + \tan Y + \tan -X = 0.$$ This gives $$\tan (X - Y) = \tan X -
\tan Y$$ and we know the identity $$\tan (X - Y) = {{\tan X - \tan
Y}\over {1 - \tan X \tan Y}}.$$ Hence either $$\tan X = \tan Y
\quad (1)$$ or $$\tan X \tan Y = 0 \quad (2)$$ In case (1) we show
that the angles $X$ and $Y$ are equal. $$|X - Y| = |A - B| < A +
B < 180 ^\circ$$ and the tan function is periodic
with period 180 degrees so $X = Y.$ This gives $A - C = B - C$
hence $A = B$, so the triangle is isosceles. In case (2), either
$\tan X = 0$ or $\tan Y = 0$, hence $A = C$ or $B = C$ and in all
the cases the triangle is isosceles.