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 17

The figure shows a non-overlapping trace on a $$4\times4$$ grid which visits all points exactly once. Imagine the same type of trace on an $$n\times n$$ grid, where $$n$$ can be arbitrarily large. Using the fact that $$\sum_{k=1}^\infty {1\over k}$$ diverges to infinity, show that the sum of all acute angles of the trace also diverges to infinity when $$n$$ tends to infinity, despite the angles tending to 0 the closer the path gets to the grid's diagonal.

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## Warm-up Questions

1. Let's prove that $$\sum_{k=1}^{\infty}{\frac{1}{k}}$$ diverges. For each $$n,$$ how many terms in the series are between $$\frac{1}{2^n}$$ (inclusive) and $$\frac{1}{2^{n+1}}$$ (exclusive)? By bounding each term $$\frac{1}{k}$$ in the original series with the largest $$\frac{1}{2^n}$$ such that $$\frac{1}{2^n} \leq \frac{1}{k},$$ compute this new sum. Hence argue why $$\sum_{k=1}^{\infty}{\frac{1}{k}}$$ diverges.

2. (Source: Wikipedia) Using this diagram, express $$\sin \theta, \cos \theta$$ and $$\tan \theta$$ in terms of $$x.$$ Hence show that $$\cos(\arcsin x) = \sqrt{1-x^2}.$$ Find a similar expression for $$\cos(\arctan x).$$

## Hints

• Hint 1
To show the sum of all the angles diverges, it suffices to show the sum of a subset of the angles diverges.
• Hint 2
Express the $$i^{th}$$ largest angle $$\theta_i$$ that touches the top of the square in terms of $$i$$.
• Hint 3
The angle $$\theta_i$$ can be expressed as the difference between two angles.
• Hint 4
The larger angle is $$\frac{\pi}{4}$$, and the tangent of the smaller angle can be expressed as the ratio between the sides of the triangle along the grid.
• Hint 5
Remember that the function $$\tan x$$ is concave in the region $$(0,\frac{\pi}{4}).$$
• Hint 6
For positive integer $$i,$$ $$0 < \frac{i}{i+1} < 1,$$ so $$\arctan \frac{i}{i+1}$$ will lie in the region $$(0,\frac{\pi}{4}).$$
• Hint 7
By drawing the graphs $$y = ({\frac{\pi}{4}})^{-1}x$$ and $$y = \tan x,$$ you can see that $$\tan x$$ lies below the line in the region $$(0,\frac{\pi}{4}).$$
• Hint 8
This means that $$\arctan \frac{i}{i+1} \leq \frac{\pi}{4}\frac{i}{i+1}.$$

## Solution

The full sum $$S$$ is greater than the sum of only the angles that touch the top of the square. Call this subsequence of angles $$\theta_i$$. By considering the right-angled triangles with sides $$i$$ and $$i+1, \theta_i$$ can be derived as: \begin{align} \theta_i &= \frac{\pi}{4} - \arctan{\frac{i}{i+1}} \\ &\geq \frac{\pi}{4} - \frac{\pi}{4}\frac{i}{i+1} \\ &= \frac{\pi}{4}(\frac{1}{i+1}) \end{align} where the inequality is due to the concavity of $$\tan x$$ in the region $$(0, \frac{\pi}{4})$$ and $$\tan \frac{\pi}{4}=1.$$ We then have that: $$S \geq \sum_{i=0}^\infty \theta_i \geq \frac{\pi}{4}\sum_{i=0}^{\infty}\frac{1}{i+1},$$ and the latter diverges. Hence $$S$$ diverges.

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