Problem:
The graph of y=f(x), where f(x) is a polynomial of degree 3 , contains points A(2,4),B(3,9), and C(4,16). Lines AB,AC, and BC intersect the graph again at points D,E, and F, respectively, and the sum of the x-coordinates of D,E, and F is 24 . What is f(0) ?
Answer Choices:
A. β2
B. 0
C. 2
D. 524β
E. 8
Solution:
Let g(x)=f(x)βx2. Then g(2)=g(3)=g(4)=0, so for some constant aξ =0,g(x)=a(xβ2)(xβ3)(xβ4). Thus the coefficients of x3 and x2 in f(x) are a and 1β9a, respectively, so the sum of the roots of f(x) is 9βa1β. If L(x) is any linear function, then the roots of f(x)βL(x) have the same sum. The given information implies that the sets of roots for three such functions are {2,3,x1β}, {2,4,x2β}, and {3,4,x3β}, where
24=x1β+x2β+x3β=3(9βa1β)β2(2+3+4)=9βa3β,
so a=β51β. Therefore f(x)=x2β51β(xβ2)(xβ3)(xβ4), and f(0)=524ββ. (In fact, D=(9,39),E=(8,40),F=(7,37), and the roots of f are 12,1+i, and 1βi.)
The problems on this page are the property of the MAA's American Mathematics Competitions