# Practice Paper 4 Question 12

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.

## Related topics

## Warm-up Questions

- 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 \(90^\circ.\)

## Hints

- Hint 1Have you tried sketching the movement?
- Hint 2Do you notice any repeating patterns?
- Hint 3How far has point \(A\) travelled, in terms of its distance to get back to its initial position, after 4 rotations?
- Hint 4Why not compute the length of each of the 4 arcs traced by \(A\)?
- Hint 5Have you identified the longest horizontal segment?
- Hint 6To find its length, could you find a triangle with the segment as one of its side?
- Hint 7Why not calculate one angle in the triangle to solve for its side length?

## Solution

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 \(\sqrt{3}x.\)

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