The positive numbers \(a\), \(b\) and \(c\) satisfy \(bc=a^2+1\). Prove that \[\arctan\left(\frac{1}{a+b}\right)+\arctan\left(\frac{1}{a+c}\right)=\arctan\left(\frac{1}{a}\right).\]
Let
\[\begin{align*} P &= \arctan\left(\frac{1}{a+b}\right) \quad &&\Longrightarrow \quad& \tan P&=\frac{1}{a+b}\\ Q &= \arctan\left(\frac{1}{a+c}\right) \quad &&\Longrightarrow \quad& \tan Q&=\frac{1}{a+c} \\ R &= \arctan\left(\frac{1}{a}\right) \quad &&\Longrightarrow \quad& \tan R&=\frac{1}{a} \end{align*}\]Note that since \(a, b\) and \(c\) are all positive numbers, \(P\), \(Q\) and \(R\) all lie between \(0\) and \(\dfrac{\pi}{2}\).
Then we wish to show that \(P+Q=R\), which is equivalent to showing that \(\tan(P+Q)=\tan R\) (since \(0 < P+Q < \pi\)).
Using the compound angle formula, we also know that \[\tan(P+Q)=\frac{\tan P + \tan Q}{1-\tan P\tan Q}.\] So we find that \[\begin{align*} \tan(P+Q) &= \dfrac{\dfrac{1}{a+b}+\dfrac{1}{a+c}}{1-\dfrac{1}{(a+b)(a+c)}} \\ &= \dfrac{\Bigl(\dfrac{2a+b+c}{(a+b)(a+c)}\Bigr)}{\Bigl(\dfrac{a^2+ab+ac+bc-1}{(a+b)(a+c)}\Bigr)} \\ &= \dfrac{2a+b+c}{a^2+ab+ac+bc-1} \\ &= \dfrac{2a+b+c}{a^2+ab+ac+(a^2+1)-1}\quad\text{as $bc=a^2+1$}\\ &= \dfrac{2a+b+c}{2a^2+ab+ac}\\ &= \dfrac{2a+b+c}{a(2a+b+c)}\\ &= \dfrac{1}{a}\\ &= \tan R, \end{align*}\]so we are done.
The positive numbers \(p\), \(q\), \(r\), \(s\), \(t\), \(u\) and \(v\) satisfy \[st=(p+q)^2+1, \qquad uv=(p+r)^2+1, \qquad qr=p^2+1.\]
Prove that
\[\begin{align*} \arctan\left(\frac{1}{p+q+s}\right)&+\arctan\left(\frac{1}{p+q+t}\right)+\arctan\left(\frac{1}{p+r+u}\right)\\ &+\arctan\left(\frac{1}{p+r+v}\right)=\arctan\left(\frac{1}{p}\right). \end{align*}\]Since \(qr=p^2+1\), we can say immediately from the first part that
\[\arctan\left(\frac{1}{p+q}\right)+\arctan\left(\frac{1}{p+r}\right)=\arctan\left(\frac{1}{p}\right).\]
Now using the first part on the first term on the LHS (since \(st=(p+q)^2+1\)), and on the second term also (since \(uv=(p+r)^2+1\)), we have
\[\arctan\left(\frac{1}{p+q+s}\right)+\arctan\left(\frac{1}{p+q+t}\right)+\arctan\left(\frac{1}{p+r+u}\right) +\arctan\left(\frac{1}{p+r+v}\right)=\arctan\left(\frac{1}{p}\right).\]
Hence show that \[\arctan\left(\frac{1}{13}\right)+\arctan\left(\frac{1}{21}\right)+\arctan\left(\frac{1}{82}\right)+\arctan\left(\frac{1}{187}\right)=\arctan\left(\frac{1}{7}\right).\]
We are looking for values of \(p\), \(q\), \(r\), \(s\), \(t\), \(u\) and \(v\) so that the above result takes this form.
We’ll begin by taking \(p=7\) (to make the right hand side work) and then look for solutions consistent with this. We have the equations \[\begin{align*} 7+q+s &= 13 \qquad& \Longrightarrow&\quad& q+s &= 6\\ 7+q+t &= 21 \qquad& \Longrightarrow&\quad& q+t &= 14\\ 7+r+u &= 82 \qquad& \Longrightarrow&\quad& r+u &= 75\\ 7+r+v &= 187 \qquad& \Longrightarrow&\quad& r+v &= 180 \end{align*}\] and \[\begin{align*} st &= (7+q)^2+1\\ uv &= (7+r)^2+1\\ qr &= 7^2+1=50. \end{align*}\] Using the first two of these we find \(s=6-q\) and \(t=14-q\), and we can then substitute this into \(st=(7+q)^2+1\) to eliminate \(s\) and \(t\), deducing \[\begin{align*} &&(6-q)(14-q)&=(q+7)^2+1&&\quad\\ &\Longrightarrow \quad& 84-20q+q^2&=q^2+14q+50\\ &\Longrightarrow \quad& 34q&=34\\ &\Longrightarrow \quad& q&=1. \end{align*}\]It then follows that \(s=5\), \(t=13\) and \(r=50\) since \(qr=50\). Using this value for \(r\) we also find that \(u=25\) and \(v=130\).
We must also double-check that all seven equations are consistent by subsituting into the equation we didn’t use: \(uv = (p+r)^2+1\). This equation is satisfied, as \[uv=25\times130 = 3250\quad\text{and}\quad (p+r)^2+1 = 57^2+1 = 3249+1=3250.\] So \[\begin{align*} p&=7,& q&=1,& r&=50,& s&=5,\\ t&=13,& u&=25,& v&=130 \end{align*}\]gives us \[\arctan\left(\frac{1}{13}\right)+\arctan\left(\frac{1}{21}\right)+\arctan\left(\frac{1}{82}\right)+\arctan\left(\frac{1}{187}\right)=\arctan\left(\frac{1}{7}\right).\]
