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.
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- 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
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Have you tried sketching the movement?
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Do you notice any repeating patterns?
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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?
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To find its length, could you find a triangle with the segment as one of its side?
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Why not calculate one angle in the triangle to solve for its side length?
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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$\sqrt{3}x.$