Consider the square $ABCD$ of side $x,$ and the equilateral triangle $BCE$ as in the figure shown. The square rotates clockwise around $B$ until $A$ overlaps $E,$ then rotates around $E$ until $D$ overlaps $C,$ and so on, until $A$ retakes its initial position. Sketch the path traced by A and find its length. Give the length of the longest horizontal segment with end points on this path. 1. Find the perimeter of a circle sector with central angle $\frac{\pi}{4}$ and a radius of $2.$
2. Calculate the length of the chord between the two ends of the arc of that sector.
3. Compute the sides ratio of a triangle with angles of $30^\circ, 60^\circ$ and $90^\circ.$

Have you tried sketching the movement?

Do you notice any repeating patterns?

How far has point $A$ travelled, in terms of its distance to get back to its initial position, after 4 rotations?

Why not compute the length of each of the 4 arcs traced by $A$?

Have you identified the longest horizontal segment?

To find its length, could you find a triangle with the segment as one of its side?

Why not calculate one angle in the triangle to solve for its side length?

We can construct a sketch of the path by first drawing the triangle and drawing 3 squares that overlap each side of the triangle as shown in the diagram. All the vertices of the squares and triangle in this diagram are visited in a clockwise fashion.

Notice that after rotating four times, we arrive at a situation in which point $A$ has completed one-third of its journey around the triangle, and is taking the position of point $F.$ In the first rotation, the length of its path is the length of arc $AE,$ which is $\frac{\pi x}{6}.$ In the second rotation, point $A$ does not move. In the third rotation, its path length is $\frac{\pi x}{6}.$ In the fourth rotation, its path length is $\frac{\sqrt{2} \pi x}{6}.$ Adding these up and multiplying by three gives us the total path length of $\frac{\pi x(2+\sqrt{2})}{2}.$

From the diagram it is apparent that the longest horizontal chord is $\overline{IF}.$ We calculate this by first noting $\angle{BEC}$ is $\frac{\pi}{3}$ as it is an internal angle of an equilateral triangle. $\angle{IEB}$ and $\angle{CEF}$ are both $\frac{\pi}{6}$ as each of the that created the arcs is $\frac{\pi}{6}.$ Therefore, $\angle{IEF}$ is $\frac{2\pi}{3}.$ Using the fact that the length of $\overline{IE}$ and $\overline{EF}$ is $x,$ we can solve the triangle $\Delta IEF$ to find the length of $\overline{IF}$ which gives the length of the longest horizontal segment is $\sqrt{3}x.$ 