SAT1000 - P333

Geometry Level 4

Given that f(x)=cos2xf(x)=\cos 2x and g(x)=sinxg(x)=\sin x, find all possible real number aa and positive integer nn, such that f(x)+ag(x)=0\\ f(x)+a g(x)=0 has exactly 20132013 roots on the interval (0,nπ)(0,n \pi).

How to submit:

  • First, find the number of all possible solutions (a,n)(a,n). Let NN denote the number of solutions.
  • Then sort the solutions by aa from smallest to largest, if aa is the same, then sort by nn from smallest to largest.
  • Let the sorted solutions be: (a1,n1),(a2,n2),(a3,n3),,(aN,nN)(a_1,n_1), (a_2,n_2), (a_3,n_3), \cdots ,(a_N,n_N), then M=k=1Nk(ak+nk)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)(-1,2), (-1,1), (1,3), (0,4), the sorted solution will be: (1,1),(1,2),(0,4),(1,3)(-1,1), (-1,2), (0,4), (1,3), then N=4N=4 and M=k=14k(ak+nk)=1×(1+1)+2×(1+2)+3×(0+4)+4×(1+3)=30\\ 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 M+N\lfloor M+N \rfloor.

Have a look at my problem set: SAT 1000 problems


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