AMC12 2017 B

AMC12 2017 B · Q23

AMC12 2017 B · Q23. It mainly tests Polynomials, Coordinate geometry.

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 $f(0)$?

$y = f(x)$的图像，其中$f(x)$是三次多项式，包含点$A(2, 4)$、$B(3, 9)$和$C(4, 16)$。直线$AB$、$AC$和$BC$分别再次与图像相交于点$D$、$E$和$F$，且$D$、$E$、$F$的$x$坐标之和为24。求$f(0)$？

(A) -2 -2

(B) 0 0

(D) \frac{24}{5} \frac{24}{5}

(E) 8 8

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Let $g(x)=f(x)-x^2$. Then $g(2)=g(3)=g(4)=0$, so for some constant $a\ne 0$, $g(x)=a(x-2)(x-3)(x-4)$. Thus the coefficients of $x^3$ and $x^2$ in $f(x)$ are $a$ and $1-9a$, respectively, so the sum of the roots of $f(x)$ is $9-\frac{1}{a}$. If $L(x)$ is any linear function, then the roots of $f(x)-L(x)$ have the same sum. The given information implies that the sets of roots for three such functions are $\{2,3,x_1\}$, $\{2,4,x_2\}$, and $\{3,4,x_3\}$, where $$ 24=x_1+x_2+x_3=3\left(9-\frac{1}{a}\right)-2(2+3+4)=9-\frac{3}{a}, $$ so $a=-\frac{1}{5}\cdot\frac{1}{24}$. Therefore $f(x)=x^2-\frac{1}{5}(x-2)(x-3)(x-4)$, and $f(0)=\frac{24}{5}$. (In fact, $D=(9,39)$, $E=(8,40)$, $F=(7,37)$, and the roots of $f$ are $12$, $1+i$, and $1-i$.)

答案（D）：令 $g(x)=f(x)-x^2$。则 $g(2)=g(3)=g(4)=0$，所以对某个常数 $a\ne 0$，有 $g(x)=a(x-2)(x-3)(x-4)$。因此，$f(x)$ 中 $x^3$ 与 $x^2$ 的系数分别为 $a$ 与 $1-9a$，所以 $f(x)$ 的根之和为 $9-\frac{1}{a}$。若 $L(x)$ 是任意一次函数，则 $f(x)-L(x)$ 的根之和不变。题给信息表明，这三种函数对应的根集分别为 $\{2,3,x_1\}$、$\{2,4,x_2\}$、$\{3,4,x_3\}$，其中 $$ 24=x_1+x_2+x_3=3\left(9-\frac{1}{a}\right)-2(2+3+4)=9-\frac{3}{a}, $$ 因此 $a=-\frac{1}{5}\cdot\frac{1}{24}$。所以 $f(x)=x^2-\frac{1}{5}(x-2)(x-3)(x-4)$，并且 $f(0)=\frac{24}{5}$。（事实上，$D=(9,39)$，$E=(8,40)$，$F=(7,37)$，且 $f$ 的根为 $12$、$1+i$、$1-i$。）

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