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# Shape and Territory

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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. 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.

A and B are two fixed points on a circle and RS is a variable diamater. What is the locus of the intersection P of AR and BS?

Make a conjecture about the sum of the squares of the odd positive integers. Can you prove it?