### Areas and Ratios

Do you have enough information to work out the area of the shaded quadrilateral?

### Napoleon's Hat

Three equilateral triangles ABC, AYX and XZB are drawn with the point X a moveable point on AB. The points P, Q and R are the centres of the three triangles. What can you say about triangle PQR?

### Exhaustion

Find the positive integer solutions of the equation (1+1/a)(1+1/b)(1+1/c) = 2

# Shape and Territory

##### Age 16 to 18 Challenge Level:

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.