### Always Two

Find all the triples of numbers a, b, c such that each one of them plus the product of the other two is always 2.

### Not Continued Fractions

Which rational numbers cannot be written in the form x + 1/(y + 1/z) where x, y and z are integers?

### Surds

Find the exact values of x, y and a satisfying the following system of equations: 1/(a+1) = a - 1 x + y = 2a x = ay

# Relative Angles

##### Age 14 to 16 Short Challenge Level:

Answer: $60^\circ$

Finding angle B in terms of angle A and angle C
$B=\frac{3}{4}C$ and $B =\frac{3}{2}A$, so \begin{align} \tfrac{3}{4}C&=\tfrac{3}{2}A \\ \Rightarrow \tfrac{1}{4}C&=\tfrac{1}{2}A \\ \Rightarrow C&=2A\end{align}.

Since $A$, $B$ and $C$ are the angles of a triangle, $A+B+C=180$.

So, substituting in $B$ and $C$ in terms of $A$, \begin{align}A+\tfrac32A+2A&=180\\ \Rightarrow \tfrac{9}{2}A&=180\\ \Rightarrow \tfrac{1}{2}A&=20\\ \Rightarrow A&=40\end{align} So $B=\frac{3}{2}\times40=60$.

Finding angles A and C in terms of angle B
$B=\frac{3}{4}C$ so $C=\frac{4}{3}B$.

And $B=\frac{3}{2}A$ so $A=\frac{2}{3}B$.

Since A, B and C are the angles of a triangle, $A+B+C=180$, so \begin{align}\tfrac{2}{3}B+B+\tfrac{4}{3}B&=180\\ \Rightarrow 3B&=180\\ \Rightarrow B&=60\end{align}
You can find more short problems, arranged by curriculum topic, in our short problems collection.