# SAT1000 - P333

Geometry Level 4

Given that $f(x)=\cos 2x$ and $g(x)=\sin x$, find all possible real number $a$ and positive integer $n$, such that $\\ f(x)+a g(x)=0$ has exactly $2013$ roots on the interval $(0,n \pi)$.

How to submit:

• First, find the number of all possible solutions $(a,n)$. Let $N$ denote the number of solutions.
• Then sort the solutions by $a$ from smallest to largest, if $a$ is the same, then sort by $n$ from smallest to largest.
• Let the sorted solutions be: $(a_1,n_1), (a_2,n_2), (a_3,n_3), \cdots ,(a_N,n_N)$, then $M=\displaystyle \sum_{k=1}^N k(a_k+n_k)$.

For instance, if the solution is $(-1,2), (-1,1), (1,3), (0,4)$, the sorted solution will be: $(-1,1), (-1,2), (0,4), (1,3)$, then $N=4$ and $\\ M=\displaystyle \sum_{k=1}^4 k(a_k+n_k)= 1 \times (-1+1) + 2 \times (-1+2) + 3 \times (0+4) + 4 \times (1+3) =30$.

For this problem, submit $\lfloor M+N \rfloor$.

Have a look at my problem set: SAT 1000 problems

×