### Kissing

Two perpendicular lines are tangential to two identical circles that touch. What is the largest circle that can be placed in between the two lines and the two circles and how would you construct it?

### Good Approximations

Solve quadratic equations and use continued fractions to find rational approximations to irrational numbers.

### Target Six

Show that x = 1 is a solution of the equation x^(3/2) - 8x^(-3/2) = 7 and find all other solutions.

# Symmetrically So

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

In this case make the substitution $y=x+4$, which transforms the equation into
$$(y-1)^4+(y+1)^4=20.$$
This is perhaps the simplest form of the equation under a linear transformation as the two factors are now at least 'similar'.

If I expand these brackets then the coefficients will match but with some opposing signs, so much cancellation will occur:
$$(y^4-4y^3+6y^2-y+1)+(y^4+4y^3+6y^2+y+1)=20.$$
So, the odd factors cancel to give
$$2y^4+12y^2-18=0$$
This then becomes the simple equation
$$y^4+6y^2-9=0\,,$$
which is a quadratic equation in $y^2$ with solutions
$$y^2 = \frac{-6\pm\sqrt{36+36}}{2}=-3(1\pm\sqrt 2).$$
One of these solutions is positive; taking the square root gives two real values for $y$ as
$$y = \pm \sqrt{3\left(\sqrt{2}-1)\right)}$$
Therefore two real solutions to the equation are
$$x= -4 \pm \sqrt{3\left(\sqrt{2}-1\right)}$$