B3. You start with a regular pentagon $A B C D E$. Then you draw two circles: one with centre $A$ and radius $A B$, and the other with centre $B$ and radius $B A$. Let the point inside the pentagon at which these two circles intersect be $X$.
What is the size of $\angle D E X$ ?
SOLUTION
As the pentagon $A B C D E$ is regular we know that $A E=A B=$ $B C$. Hence $B$ and $E$ lie on the circle with centre $A$ and radius $A B$, and $A$ and $C$ lie on the circle with centre $B$ and radius $B A$.
$A X$ is a radius of the circle centre $A, B X$ is a radius of the circle centre $B, A B$ is a side length of the regular pentagon and so these three lengths are equal. Hence triangle $A X B$ is equilateral and $\angle X A B=60^{\circ}$.

Given that $\angle E A B$ is an interior angle of a regular pentagon, it follows that $\angle E A X=\angle E A B-\angle X A B=\left(\frac{540}{5}\right)^{\circ}-60^{\circ}=$ $108^{\circ}-60^{\circ}=48^{\circ}$.
$A E=A X$ as they are both radii of the circle centre $A$. Hence triangle $A E X$ is isosceles. It follows that $\angle X E A=\angle A X E=$ $\left(\frac{180-48}{2}\right)^{\circ}=66^{\circ}$.
As $\angle D E A$ is an interior angle of a regular pentagon, it follows that $\angle D E X=\angle D E A-\angle X E A=108^{\circ}-66^{\circ}=42^{\circ}$.