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.

 Find the perimeter of a circle sector with central angle
$\frac{\pi}{4}$
and a radius of$2.$
 Calculate the length of the chord between the two ends of the arc of that sector.
 Compute the sides ratio of a triangle with angles of
$30^\circ, 60^\circ$
and$90^\circ.$
 Find the perimeter of a circle sector with central angle

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 onethird 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$\sqrt{3}x.$