1 problem 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 \(q = 1, r = 50, s = 5, v=25\)