### Baby Circle

A small circle fits between two touching circles so that all three circles touch each other and have a common tangent? What is the exact radius of the smallest circle?

### Absurdity Again

What is the value of the integers a and b where sqrt(8-4sqrt3) = sqrt a - sqrt b?

### Strange Rectangle 2

Find the exact values of some trig. ratios from this rectangle in which a cyclic quadrilateral cuts off four right angled triangles.

# Ab Surd Ity

##### Stage: 5 Challenge Level:

Congratulations to Hyeyoun from St Paul's Girls School, London, to Sue Liu of Madras College, St Andrew's, Scotland, Sanjay from The Perse School, Cambridge and Bill from Alcester Grammar School for your solutions.

We take the square root symbol in the question to signify the positive square root. The tactic here is to square both sides and then find the correct square root. If $\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}=X$, then $$X^2= \left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right) \left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right).$$ The right hand side equals $$2 + \sqrt{3}-2\left(\sqrt{2+\sqrt{3}}\times \sqrt{2-\sqrt{3}}\right) +2-\sqrt{3},$$ and $$\sqrt{2+\sqrt{3}}\times\sqrt{2-\sqrt{3}}=1.$$ Therefore $$X^2 = 2+\sqrt{3}-2+2-\sqrt{3}=2.$$ Does $X=-\sqrt{2}$ or $+\sqrt{2}$?

Well $2+\sqrt{3}> 2-\sqrt{3}$, so $\sqrt{2+\sqrt{3}}> \sqrt{2-\sqrt{3}}$, so $X$ is positive and we have $X=\sqrt{2}$.

Note that we could take each square root to be positive or negative. If so, then the question is much harder and there are more solutions for $X$. For example, we could take $\sqrt{3} = 1{\cdot}732\cdots$; then $\sqrt{2+\sqrt{3}}$ has two values (approximately $\pm 1{\cdot}93$), and $\sqrt{2-\sqrt{3}}$ has two values (approximately $\pm 0{\cdot}52$). It follows that $X$ has four values (approximately $\pm 2{\cdot}45$ and $\pm 1{\cdot}41$). Alternatively, we could take $\sqrt{3} = -1{\cdot}732\cdots$, and then we would get even more solutions.

In the second part there are again many solutions (because square roots have two values and cube roots have three values). To simplify the solution we restrict ourselves to real cube roots. We want to find $$X = {\root 3\of {2+\sqrt{5}}} + {\root 3\of {2-\sqrt{5}}}.$$ One way to do this is to write $a = {\root 3\of {2+\sqrt{5}}}$ and $b= {\root 3\of {2-\sqrt{5}}}$, and use the equation $$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 = a^3 + 3ab(a + b) + b^3.$$ As $X = a + b$, we have $$X^3 = (2 + \sqrt{5}) + 3X{\root 3\of {(2 + \sqrt{5})(2 - \sqrt{5})}} +(2 - \sqrt{5}).$$ As $\root 3\of {(2 + \sqrt{5})(2 - \sqrt {5}} = \root 3\of {-1} = -1$ this gives $X^3 + 3X - 4 = 0$ and hence $$(X - 1)(X^2 + X + 4) = 0.$$ As $X^2 + X + 4 = 0$ has only complex solutions, and we are looking for the real values of $X$, so we have $X = 1$, that is $${\root 3\of {2 + \sqrt{5}}} + {\root 3\of {2 - \sqrt{5}}} = 1.$$ You can check this on your calculator!