17. The equilateral triangle $P Q R$ has side length 1 . The lines $P T$ and $P U$ trisect the angle $R P Q$, the lines $R S$ and $R T$ trisect the angle $Q R P$ and the lines $Q S$ and $Q U$ trisect the angle $P Q R$.
What is the side length of the equilateral triangle $S T U$ ?
A $\frac{\cos 80^{\circ}}{\cos 20^{\circ}}$
B $\frac{1}{3} \cos 20^{\circ}$
C $\cos ^{2} 20^{\circ}$
D $\frac{1}{6}$
$\mathrm{E} \cos 20^{\circ} \cos 80^{\circ}$
SOLUTION
A Since the triangle $P Q R$ is equilateral, all its three angles are $60^{\circ}$. So the trisectors divide these into three $20^{\circ}$ angles. In particular $\angle Q P U=20^{\circ}$ and $\angle U P T=20^{\circ}$. Because of the symmetry of the figure, $Q U=P U=P T$. So the triangles $P Q U$ and $P U T$ are isosceles. In these triangles, we let $X$ be the foot of the perpendicular from $U$ to $P Q$, and $Y$ be the foot of perpendicular from $P$ to $U T$.
In the right-angled triangle $P X U, \angle X P U=20^{\circ}$, and so $\frac{P X}{P U}=\cos 20^{\circ}$. Since the length of $P X$ is $\frac{1}{2}$ it follows that $P U$ has length $\frac{1}{2} \div \cos 20^{\circ}$. In the right-angled triangle $P Y U, \angle U P Y=10^{\circ}$ and therefore $\angle Y U P=80^{\circ}$.

Therefore, from triangle $P Y U, \cos 80^{\circ}=\frac{U Y}{P U}$ and hence $U Y$ has length $P U \cos 80^{\circ}$, that is,
$$
\frac{1}{2} \times \frac{\cos 80^{\circ}}{\cos 20^{\circ}} \text {. }
$$

Since the length of $U Y$ is half the length of $U T$, it follows that the length of $U T$ is $\frac{\cos 80^{\circ}}{\cos 20^{\circ}}$.
ReMARKS
In this problem, it is clear that, because the triangle $P Q R$ is equilateral, so also is the triangle STU. It is a remarkable fact, discovered by Frank Morley (1860-1937), that:

In every triangle $P Q R$, the triangle $S T U$ formed by the points where the trisectors of the angles of $P Q R$ meet is an equilateral triangle.

The triangle $S T U$ is called the Morley triangle of the triangle $P Q R$.
No very easy proofs of this theorem are known.
A proof may be found on pages 186-187 of The Geometry of the Triangle by Gerry Leversha, UKMT, 2013. This book may be ordered from the UKMT website. The virtue of the proof in this book is that it uses purely geometrical ideas. Other proofs use trigonometry and complex numbers.