SAT1000 - P801

Geometry Level pending

As shown above, the ellipse EE has equation: x216+y212=1\dfrac{x^2}{16}+\dfrac{y^2}{12}=1, and point CC is the center of the circle: x2+y24x+2=0x^2+y^2-4x+2=0 If PP is a point on ellipse EE, l1,l2l_1, l_2 both pass through PP and they are both tangent to circle CC, and the product of the slope of l1,l2l_1, l_2 is equal to 12\dfrac{1}{2}, then find the all possible coordinate(s) of point PP.

How to submit:

  • First, find the number of all possible solutions. Let NN denote the number of solutions.

  • Then Sort the solutions by x-coordinate from smallest to largest, if the x-coordinate is the same, then sort by y-coordinate from smallest to largest.

  • Let the sorted solutions be: (x1,y1),(x2,y2),(x3,y3),,(xN,yN)(x_1,y_1), (x_2,y_2), (x_3,y_3), \cdots ,(x_N,y_N), then M=k=1Nk(xk+yk)M=\displaystyle \sum_{k=1}^N k(x_k+y_k) .

For instance, if the solution is: (1,2),(1,1),(1,3),(0,4)(-1,2), (-1,1), (1,3), (0,4)

Then 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, M=k=14k(xk+yk)=1×(1+1)+2×(1+2)+3×(0+4)+4×(1+3)=30M=\displaystyle \sum_{k=1}^4 k(x_k+y_k)= 1 \times (-1+1) + 2 \times (-1+2) + 3 \times (0+4) + 4 \times (1+3) =30 .

For this problem, submit 1000(M+N)\lfloor 1000(M+N) \rfloor.

Have a look at my problem set: SAT 1000 problems


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