The CSAT and Practice[+] are designed by the Climb Foundation to help candidates. We are advocates for more opportunity to shine and less opportunity to fail, and we strive to level the playing field. 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. The above links are provided as is. They are not affiliated with the Climb Foundation unless otherwise specified.

Warm-up Questions

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 $$90^\circ.$$

Hints

• Hint 1
Have you tried sketching the movement?
• Hint 2
Do you notice any repeating patterns?
• Hint 3
How far has point $$A$$ travelled, in terms of its distance to get back to its initial position, after 4 rotations?
• Hint 4
Why not compute the length of each of the 4 arcs traced by $$A$$?
• Hint 5
Have you identified the longest horizontal segment?
• Hint 6
To find its length, could you find a triangle with the segment as one of its side?
• Hint 7
Why 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 $$\sqrt{3}x.$$ If you have queries or suggestions about the content on this page or the CSAT Practice Platform then you can write to us at . Please do not write to this address regarding general admissions or course queries.