2022 AIME I 第 1 题

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1.

二次多项式 P(x)P(x)Q(x)Q(x) 的首项系数分别为 222-2,这两个多项式的图像都经过 两点 (16,54)(16, 54)(20,53)(20, 53)。求 P(0)+Q(0)P(0) + Q(0)

Quadratic polynomials P(x)P(x) and Q(x)Q(x) have leading coefficients of 22 and 2,-2, respectively. The graphs of both polynomials pass through the two points (16,54)(16, 54) and (20,53).(20, 53). Find P(0)+Q(0).P(0) + Q(0).

答案:116
知识点:多项式一次方程
难度评级:1890
解答:

R(x)=P(x)+Q(x)R(x) = P(x) + Q(x)。首项系数 222-2 抵消,所以 RR 是一次函数。因为两个图像都经过 (16,54)(16, 54)(20,53)(20, 53),所以 R(16)=108R(16) = 108R(20)=106R(20) = 106

RR 的斜率为 1061082016=12\frac{106 - 108}{20 - 16} = -\frac{1}{2},所以 P(0)+Q(0)=R(0)=R(16)+1612=108+8=116. \begin{aligned} P(0) + Q(0) &= R(0) \\ &= R(16) + 16 \cdot \frac{1}{2} \\ &= 108 + 8 = 116. \end{aligned}

Let R(x)=P(x)+Q(x).R(x) = P(x) + Q(x). The leading coefficients 22 and 2-2 cancel, so RR is a linear function. Since both graphs pass through (16,54)(16, 54) and (20,53),(20, 53), we get R(16)=108R(16) = 108 and R(20)=106.R(20) = 106.

The slope of RR is 1061082016=12,\frac{106 - 108}{20 - 16} = -\frac{1}{2}, so P(0)+Q(0)=R(0)=R(16)+1612=108+8=116. \begin{aligned} P(0) + Q(0) &= R(0) \\ &= R(16) + 16 \cdot \frac{1}{2} \\ &= 108 + 8 = 116. \end{aligned}

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