2 problems found
The positive numbers \(a\), \(b\) and \(c\) satisfy \(bc=a^2+1\). Prove that $$ \arctan\left(\frac1 {a+b}\right)+ \arctan\left(\frac1 {a+c}\right)= \arctan\left(\frac1 a \right). $$ The positive numbers \(p\), \(q\), \(r\), \(s\), \(t\), \(u\) and \(v\) satisfy $$ st = (p+q)^2 + 1 \;, \ \ \ \ \ \ uv=(p+r)^2 + 1 \;, \ \ \ \ \ \ qr = p^2+1\;. $$ Prove that $$ \arctan \! \!\left(\!\frac1 {p+q+s}\!\right) + \arctan \! \!\left(\!\frac 1{p+q+t}\!\right) + \arctan \! \!\left(\!\frac 1 {p+r+u}\!\right) + \arctan \! \!\left(\!\frac1 {p+r+v}\!\right) =\arctan \! \!\left( \! \frac1 p \! \right) . $$ Hence show that $$ \arctan\left(\frac1 {13}\right) +\arctan\left(\frac1 {21}\right) +\arctan\left(\frac1 {82}\right) +\arctan\left(\frac1 {187}\right) =\arctan\left(\frac1 {7}\right). $$ [Note that \(\arctan x\) is another notation for \( \tan^{-1}x \,.\,\)]
Solution: \begin{align*} && \tan \left (\arctan\left(\frac1 {a+b}\right)+ \arctan\left(\frac1 {a+c}\right) \right) &= \frac{\frac1{a+b}+\frac1{a+c}}{1-\frac{1}{(a+b)(a+c)}} \\ &&&= \frac{a+c+a+b}{(a+b)(a+c)-1} \\ &&&= \frac{2a+b+c}{a^2+ab+ac+bc-1} \\ &&&= \frac{2a+b+c}{2a^2+ab+ac} \\ &&&= \frac{1}{a} \\ &&&= \tan \arctan \frac1a\\ \Rightarrow && \arctan\left(\frac1 {a+b}\right)+ \arctan\left(\frac1 {a+c}\right) &= \arctan \frac{1}{a} + n \pi \end{align*} Since \(\arctan x \in (-\frac{\pi}{2}, \frac{\pi}{2})\) the LHS \(\in (0, \pi)\) so \(n = 0\). \begin{align*} a=p+q, b = s, c = t:&& \arctan \! \!\left(\!\frac1 {p+q+s}\!\right) + \arctan \! \!\left(\!\frac 1{p+q+t}\!\right) &= \arctan \left ( \frac{1}{p+q} \right) \\ a=p+r, b= u, c = v && \arctan \! \!\left(\!\frac 1 {p+r+u}\!\right) + \arctan \! \!\left(\!\frac1 {p+r+v}\!\right) &= \arctan \! \!\left(\!\frac1 {p+r}\!\right) \\ a = p, b = q, c = r:&& \arctan \left ( \frac{1}{p+q} \right) +\arctan \! \!\left(\!\frac1 {p+r}\!\right) &= \arctan \left ( \frac1p \right) \end{align*} and the result follows. Taking \(p = 7\) we need to solve \[ \begin{cases} q+s &= 6 \\ q+t &= 14 \\ r+u &= 75 \\ r+v &= 180 \end{cases} \] also satisfying \(qr = 50\) etc, so say \(q = 1, r = 50, s = 5, v=25\)
Prove that \(\displaystyle \arctan a + \arctan b = \arctan \l {a + b \over 1-ab} \r\,\) when \(0 < a < 1\) and \(0 < b < 1\,\). Prove by induction that, for \(n \ge 1\,\), \[ \sum_{r = 1}^n \arctan \l {1 \over r^2 + r + 1} \r = \arctan \l {n \over n+2} \r \] and hence find \[ \sum_{r = 1}^\infty \arctan \l {1 \over r^2 + r + 1} \r\,. \] Hence prove that \[ \sum_{r = 1}^\infty \arctan \l {1 \over r^2 - r + 1} \r = {\pi \over 2}\,. \]
Solution: \begin{align*} && \arctan a &\in (0, \tfrac{\pi}{4}) \\ && \arctan b &\in (0, \tfrac{\pi}{4}) \\ \Rightarrow && \arctan a+\arctan b &\in (0, \tfrac{\pi}{2}) \\ && \tan \left ( \arctan a+\arctan b \right) &= \frac{\tan \arctan a + \tan \arctan b}{1 - \tan \arctan a \tan \arctan b} \\ &&&= \frac{a+b}{1-ab} \in (0, \infty) \\ \Rightarrow && \arctan \left ( \frac{a+b}{1-ab} \right) &\in (0, \tfrac{\pi}{2}) \\ \Rightarrow && \arctan a + \arctan b &= \arctan \left ( \frac{a+b}{1-ab} \right) \end{align*} Claim: \(\displaystyle \sum_{r = 1}^n \arctan \l {1 \over r^2 + r + 1} \r = \arctan \l {n \over n+2} \r\) Proof: (By Induction): Base case (\(n=1\)): \begin{align*} && LHS &= \sum_{r=1}^1 \arctan \left ( \frac{1}{r^2+r+1} \right) \\ &&&= \arctan \left ( \frac{1}{3} \right) \\ && RHS &= \arctan \left ( \frac{1}{1+2} \right)\\ &&&= \arctan \left ( \frac{1}{3} \right) = LHS \end{align*} Inductive step, suppose true for \(n = k\), ie \begin{align*} && \sum_{r = 1}^k \arctan \l {1 \over r^2 + r + 1} \r &= \arctan \l {k \over k+2} \r \\ \Rightarrow && \sum_{r = 1}^{k+1} \arctan \l {1 \over r^2 + r + 1} \r &= \sum_{r = 1}^k \arctan \l {1 \over r^2 + r + 1} \r+ \arctan \left ( \frac{1}{(k+1)^2+(k+1)+1} \right) \\ &&&= \arctan \l {k \over k+2} \r+\arctan \left ( \frac{1}{(k+1)^2+(k+1)+1} \right) \\ &&&= \arctan \left ( \frac{{k \over k+2}+\frac{1}{(k+1)^2+(k+1)+1} }{1-\frac{k}{k+2}\frac{1}{(k+1)^2+(k+1)+1} } \right) \\ &&&= \arctan \left ( \frac{k((k+1)^2+k+1+k)+(k+2) }{(k+2)((k+1)^2+(k+1)+1)-k} \right) \\ &&&= \arctan \left ( \frac{k^3+3k^2+4k+2 }{k^3+5k^2+8k+6} \right) \\ &&&= \arctan \left ( \frac{(k+1)(k^2+2k+2) }{(k+3)(k^2+2k+2)} \right) \\ &&&= \arctan \left ( \frac{k+1 }{(k+1)+2} \right) \\ \end{align*} Therefore it is true for \(n = k+1\), therefore it is true for all \(n \geq 1\) by the principle of mathematical induction. \begin{align*} && S &= \lim_{n \to \infty} \sum_{r = 1}^n \arctan \l {1 \over r^2 + r + 1} \r \\ &&&= \lim_{n \to \infty} \arctan \l \frac{n}{n+2} \r \\ &&&= \lim_{n \to \infty} \arctan \l \frac{1}{1+2/n} \r \\ &&&=\arctan\l \lim_{n \to \infty} \frac{1}{1+2/n} \r \\ &&&= \frac{\pi}{4} \end{align*} \begin{align*} && \sum_{r = 1}^\infty \arctan \l {1 \over r^2 - r + 1} \r &= \sum_{r = 0}^\infty \arctan \left( \frac{1}{ (r+1)^2 - (r+1) + 1} \right) \\ &&&= \sum_{r = 0}^\infty \arctan \left( \frac{1}{ r^2+r+1} \right) \\ &&&= \arctan \l \frac{1}{0^2+0+1} \r + \frac{\pi}{4} \\ &&&= \frac{\pi}{2} \end{align*}