Here I want to link the words 'Round 1 Problem 1', 'Round 1 Problem 2' to the words 'A 1', 'A 2' of the second page (The picture just below). So that whenever I click on Round 1 Problem 1 it takes me to A 1 Now again in the second page I want to link the word 'A 1' to the words of 'A 1' the third page (The picture just below) so that whenever I click on A 1 on second page it takes me to the Solution of the A 1
I am using hyperlink
and hypertarget
for this. Here is the code i am using.
colframe=black,colback=Chartreuse1,
boxrule=0.5pt,arc=4pt,boxsep=0pt,left=6pt,right=6pt,top=6pt,bottom=6pt}
\begin{center}
\section*{\huge{\underline{Round 1 Problems}}}
\mybox{\hyperlink{R1P1}{Round 1 P1}} \\ \vspace{4mm}
\mybox{Round 1 P2}\\ \vspace{4mm}
\mybox{Round 1 P3} \\ \vspace{4mm}
\mybox{Round 1 P4} \\ \vspace{4mm}
\mybox{Round 1 P5}} \\ \vspace{4mm}
\mybox{Round 1 P6} \\ \vspace{4mm}
\section*{\huge{\underline{Round 2 Problems}}}
\mybox{Round 2 P1} \\ \vspace{4mm}
\mybox{Round 2 P2} \\ \vspace{4mm}
\mybox{Round 2 P3} \\ \vspace{4mm}
\mybox{Round 2 P4} \\ \vspace{4mm}
\mybox{Round 2 P5} \\ \vspace{4mm}
\mybox{Round 2 P6} \\ \vspace{4mm}
\end{center}
\mybox{{Problem 1 : }}
Prove the inequality which states that if you let $x_{1}, x_{2}, \ldots, x_{n}$ be positive real numbers, with $n \geq 2,$ then you have the inequality
$$
\frac{x_{1}}{x_{2}+x_{3}+\cdots+x_{n}}+\frac{x_{2}}{x_{1}+x_{3}+x_{4}+\cdots+x_{n}}+\cdots+\frac{x_{n}}{x_{1}+x_{2}+\cdots+x_{n-1}} \geq \frac{n}{n-1}
$