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2015 Paper 2 Q5
D: 1600.0 B: 1484.9
trigonometry arctan proof by induction telescoping summation tan addition formula right triangle infinite series

In this question, the \(\mathrm{arctan}\) function satisfies \(0\le \arctan x <\frac12 \pi\) for \(x\ge0\,\).

  1. Let \[ S_n= \sum_{m=1}^n \arctan \left(\frac1 {2m^2}\right) \,, \] for \(n=1, 2, 3, \ldots\) . Prove by induction that \[ \tan S_n = \frac n {n+1} \,. \] Prove also that \[ S_n = \arctan \frac n {n+1} \,. \]
  2. In a triangle \(ABC\), the lengths of the sides \(AB\) and \(BC\) are \(4n^2\) and \(4n^4-1\), respectively, and the angle at \(B\) is a right angle. Let \(\angle BCA = 2\alpha_n\). Show that \[ \sum_{n=1}^\infty \alpha_n = \tfrac14\pi \,. \]

Solution:

  1. Claim: \(\tan S_n = \frac n {n+1}\) Proof: (By Induction) Base case: (\(n=1\)): \begin{align*} && \tan \left ( \sum_{m=1}^1 \arctan \left ( \frac{1}{2m^2} \right) \right) &= \tan \left ( \arctan \left ( \frac{1}{2} \right) \right) \\ &&&= \frac12 = \frac{1}{1+1} \end{align*} Therefore the base case is true. Inductive step: Suppose our statement is true for some \(n = k\), ie \begin{align*} && \frac{k}{k+1} &= \tan \left ( \sum_{m=1}^k \arctan \left ( \frac{1}{2m^2} \right) \right) \\ \Rightarrow && \tan S_{k+1} &= \tan \left ( \sum_{m=1}^k \arctan \left ( \frac{1}{2m^2} \right) + \arctan \left ( \frac{1}{2 (k+1)^2} \right) \right) \\ &&&= \frac{\tan S_k + \tan \left ( \arctan \left ( \frac{1}{2 (k+1)^2} \right) \right)}{1-\tan S_k \tan \left ( \arctan \left ( \frac{1}{2 (k+1)^2} \right) \right)} \\ &&&= \frac{\frac{k}{k+1} + \frac{1}{2(k+1)^2}}{1-\frac{k}{k+1} \frac{1}{2(k+1)^2}} \\ &&&= \frac{2k(k+1)^2+(k+1)}{2(k+1)^3-k} \\ &&&= \frac{k+1}{(k+1)+1} \end{align*} Therefore it is true for \(n=k+1\). Conclusion: Therefore by the principle of mathematical induction since our statement is true for \(n=1\) and if it is true for \(n=k\) it is true for \(n=k+1\) it is true for all \(n\geq1\) Since \(S_n < \frac12 \pi\) for all \(n\), we must have \(\arctan \frac{n}{n+1} = S_n\)
  2. \(\tan (2\alpha_n) = \frac{4n^2}{4n^4-1} = \frac{2n^2+2n^2}{(2n^2)(2n^2)-1} = \frac{\frac{1}{2n^2}+\frac{1}{2n^2}}{1-\frac{1}{2n^2}\frac{1}{2n^2}} \Rightarrow \tan (\alpha_n) = \arctan \frac{1}{2n^2}\). In particular \(\displaystyle \sum_{n=1}^{N} \alpha_n = \arctan \frac{n}{n+1} \Rightarrow \sum_{n=1}^{\infty} \alpha_n \to \arctan 1 = \frac{\pi}{4} \)

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2005 Paper 2 Q5
D: 1600.0 B: 1470.1
coordinate geometry right triangle incircle circumcircle tangent area optimisation inequality quadratic

The angle \(A\) of triangle \(ABC\) is a right angle and the sides \(BC\), \(CA\) and \(AB\) are of lengths \(a\), \(b\) and \(c\), respectively. Each side of the triangle is tangent to the circle \(S_1\) which is of radius \(r\). Show that \(2r = b+c-a\). Each vertex of the triangle lies on the circle~\(S_2\). The ratio of the area of the region between~\(S_1\) and the triangle to the area of \(S_2\) is denoted by \(R\,\). Show that $$ \pi R = -(\pi-1)q^2 + 2\pi q -(\pi+1) \;, $$ where \(q=\dfrac{b+c}a\,\). Deduce that $$ R\le \frac1 {\pi( \pi - 1)} \;. $$

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