AMC12 2008 B

AMC12 2008 B · Q17

AMC12 2008 B · Q17. It mainly tests Quadratic equations, Graphs (coordinate plane).

Let $A$, $B$ and $C$ be three distinct points on the graph of $y=x^2$ such that line $AB$ is parallel to the $x$-axis and $\triangle ABC$ is a right triangle with area $2008$. What is the sum of the digits of the $y$-coordinate of $C$?

设 $A$、$B$、$C$ 是抛物线 $y=x^2$ 上三个不同的点，使得直线 $AB$ 平行于 $x$ 轴，且 $\triangle ABC$ 是面积为 $2008$ 的直角三角形。点 $C$ 的 $y$ 坐标的各位数字之和是多少？

(A) 16 16

(B) 17 17

(D) 19 19

(E) 20 20

Answer

Correct choice: (C)

正确答案：（C）

Solution

Supposing $\angle A=90^\circ$, $AC$ is perpendicular to $AB$ and, it follows, to the $x$-axis, making $AC$ a segment of the line $x=m$. But that would mean that the coordinates of $C$ are $(m, m^2)$, contradicting the given that points $A$ and $C$ are distinct. So $\angle A$ is not $90^\circ$. By a similar logic, neither is $\angle B$. This means that $\angle C=90^\circ$ and $AC$ is perpendicular to $BC$. Let C be the point $(n, n^2)$. So the slope of $BC$ is the negative reciprocal of the slope of $AC$, yielding $m+n=\frac{1}{m-n}$ $\Rightarrow$ $m^2-n^2=1$. Because $m^2-n^2$ is the length of the altitude of triangle $ABC$ from $AB$, and $2m$ is the length of $AB$, the area of $\triangle ABC=m(m^2-n^2)=2008$. Since $m^2-n^2=1$, $m=2008$. Substituting, $2008^2-n^2=1$ $\Rightarrow$ $n^2=2008^2-1=(2000+8)^2-1=4000000+32000+64-1=4032063$, whose digits sum to $18 \Rightarrow \textbf{(C)}$.

假设 $\angle A=90^\circ$，则 $AC$ 垂直于 $AB$，从而也垂直于 $x$ 轴，使得 $AC$ 是直线 $x=m$ 上的一段。但这将意味着 $C$ 的坐标为 $(m, m^2)$，与题设 $A$ 与 $C$ 为不同点矛盾。所以 $\angle A$ 不是 $90^\circ$。同理，$\angle B$ 也不是。因此 $\angle C=90^\circ$，且 $AC$ 垂直于 $BC$。设 $C$ 为点 $(n, n^2)$。于是 $BC$ 的斜率是 $AC$ 斜率的负倒数，得到 $m+n=\frac{1}{m-n}$ $\Rightarrow$ $m^2-n^2=1$。因为 $m^2-n^2$ 是从 $AB$ 到 $C$ 的高的长度，而 $2m$ 是 $AB$ 的长度，所以 $\triangle ABC$ 的面积为 $m(m^2-n^2)=2008$。由于 $m^2-n^2=1$，得 $m=2008$。代入得 $2008^2-n^2=1$ $\Rightarrow$ $n^2=2008^2-1=(2000+8)^2-1=4000000+32000+64-1=4032063$，其各位数字之和为 $18 \Rightarrow \textbf{(C)}$。

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