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Japan Junior Mathematical Olympiad

Geometry Difficulty 4.5 AIME Prove it Japan

Let $ABCDEFG$ be a regular heptagon. Suppose $P$ , $Q$ , $R$ , $S$ are points on the line segments $AB$ , $BC$ , $CD$ , $EF$ , respectively, for which $BP = CQ = DR = FS = \frac{1}{3}$ are satisfied. Let $T$ be the point of intersection of the line segments $PR$ and $QS$ . Find the value of $\angle PTS$ .

Here by $XY$ we mean the length of the line segment $XY$ .

Solution

$\boxed{\frac{720^{\circ}}{7}}$

Let $U$ , $V$ , $W$ be points on the sides $DE$ , $FG$ , $GA$ , respectively, dividing respective side in $2 : 1$ ratio. Then, it is easy to see that $PQRUSVW$ becomes also a regular heptagon. Hence these $7$ points lie on the circumference of a same circle, and divides the circumference into $7$ arcs of equal length. From this it follows that $\angle PRS$ is $\frac{1}{2}$ of the angle subtended by the arc $PWVS$ of the circum-circle of this heptagon at its center, and thus $\angle PRS = \frac{1}{2} \cdot \frac{3}{7} \cdot 360^{\circ}$ . Similarly, we have $\angle QSR = \frac{1}{2} \cdot \frac{1}{7} \cdot 360^{\circ}$ . Consequently, we get $\angle PTS = \angle PRS + \angle QSR = \frac{720^{\circ}}{7}$ .

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