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Show that for any triangle it is always possible to construct 3 touching circles with centres at the vertices. Is it possible to construct touching circles centred at the vertices of any polygon?

Overturning Fracsum

Stage: 4 Challenge Level: Challenge Level:2 Challenge Level:2

This can be done in a very similar way to Symmetricality in the May 1999 Six.

You first turn the expressions upside down, then solve for ${1\over x}$, ${1\over y}$ and ${1\over z}$.

Sue Liu from Madras College, St Andrew's sent this solution. Well done Sue. Solve the following system of equations to find the values of $x$, $y$ and $z$.

  1. $${xy\over (x + y)} = {1\over 2}$$
  2. $${yz\over (y + z)} = {1\over 3}$$
  3. $${xz\over (x + z)} = {1\over 7}$$

From (1) we can rearrange the equation to give $y$ in terms of $x$. $$x + y = 2xy$$

  1. $$y = {x\over (2x-1)}$$

where $x \neq {1\over 2}$.

Similarly, from (3) we can rearrange the equation to get $z$ in terms of $x$. $$x + z = 7xz$$

  1. $$z = {x\over (7x - 1)}$$

where $x \neq {1\over 7}$.

From (2) $$y + z = 3yz$$ Substituting from (4) and (5) into this equation: $${x\over 2x-1} + \frac{x}{7x-1} = 3\left(\frac{x}{2x-1}\right)\left(\frac{x}{7x-1}\right)$$ $$x(7x-1) + x(2x-1) = 3x^2$$ $$6x^2 - 2x = 0$$ $$x(3x - 1) = 0$$ Hence $x = 0$ or $x = {1\over 3}$ and $x$ is clearly not zero.

When $x = {1\over 3}$ by substituting in equations (4) and (5) we get $y = -1$ and $z = {1\over 4}$.