AMC12 2005 B

AMC12 2005 B · Q24

AMC12 2005 B · Q24. It mainly tests Coordinate geometry, Trigonometry (basic).

All three vertices of an equilateral triangle are on the parabola $y = x^2$, and one of its sides has a slope of $2$. The $x$-coordinates of the three vertices have a sum of $m/n$, where $m$ and $n$ are relatively prime positive integers. What is the value of $m + n$?

一个正三角形的三个顶点都在抛物线 $y = x^2$ 上，其中一边斜率为 $2$。三顶点的 $x$ 坐标之和为 $m/n$，其中 $m$ 和 $n$ 互质正整数。求 $m + n$ 的值。

(A) 14 14

(B) 15 15

(D) 17 17

(E) 18 18

Answer

Correct choice: (A)

正确答案：（A）

Solution

Let the three points be at $A = (x_1, x_1^2)$, $B = (x_2, x_2^2)$, and $C = (x_3, x_3^2)$, such that the slope between the first two is $2$, and $A$ is the point with the least $y$-coordinate. Then, we have $\textrm{Slope of }AC = \frac{x_1^2 - x_3^2}{x_1 - x_3} = x_1 + x_3$. Similarly, the slope of $BC$ is $x_2 + x_3$, and the slope of $AB$ is $x_1 + x_2 = 2$. The desired sum is $x_1 + x_2 + x_3$, which is equal to the sum of the slopes divided by $2$. To find the slope of $AC$, we note that it comes at a $60^{\circ}$ angle with $AB$. Thus, we can find the slope of $AC$ by multiplying the two complex numbers $1 + 2i$ and $1 + \sqrt{3}i$. What this does is generate the complex number that is at a $60^{\circ}$ angle with the complex number $1 + 2i$. Then, we can find the slope of the line between this new complex number and the origin: \[(1+2i)(1+\sqrt{3}i)\] \[= 1 - 2\sqrt{3} + 2i + \sqrt{3}i\] \[\textrm{Slope } = \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}}\] \[= \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}} \cdot \frac{1 + 2\sqrt{3}}{1 + 2\sqrt{3}}\] \[= \frac{8 + 5\sqrt{3}}{-11}\] \[= \frac{-8 - 5\sqrt{3}}{11}.\] The slope $BC$ can also be solved similarly, noting that it makes a $120^{\circ}$ angle with $AB$: \[(1+2i)(-1+\sqrt{3}i)\] \[= -1 - 2\sqrt{3} - 2i + \sqrt{3}i\] \[\textrm{Slope } = \frac{\sqrt{3} - 2}{-2\sqrt{3} - 1}\] \[= \frac{2 - \sqrt{3}}{1 + 2\sqrt{3}} \cdot \frac{1 - 2\sqrt{3}}{1 - 2\sqrt{3}}\] At this point, we start to notice a pattern: This expression is equal to $\frac{2 + \sqrt{3}}{1 - 2\sqrt{3}} \cdot \frac{1 + 2\sqrt{3}}{1 + 2\sqrt{3}}$, except the numerators of the first fractions are conjugates! Notice that this means that when we multiply out, the rational term will stay the same, but the coefficient of $\sqrt{3}$ will have its sign switched. This means that the two complex numbers are conjugates, so their irrational terms cancel out. Our sum is simply $2 - 2\cdot\frac{8}{11} = \frac{6}{11}$, and thus we can divide by $2$ to obtain $\frac{3}{11}$, which gives the answer $\boxed{\mathrm{(A)}\ 14}$.

设三点为 $A=(x_1,x_1^2)$、$B=(x_2,x_2^2)$、$C=(x_3,x_3^2)$，使得前两点连线的斜率为 $2$，且 $A$ 是 $y$ 坐标最小的点。则 $\textrm{Slope of }AC=\frac{x_1^2-x_3^2}{x_1-x_3}=x_1+x_3$。同理，$BC$ 的斜率为 $x_2+x_3$，而 $AB$ 的斜率为 $x_1+x_2=2$。所求和为 $x_1+x_2+x_3$，它等于三条边斜率之和除以 $2$。为求 $AC$ 的斜率，注意它与 $AB$ 成 $60^{\circ}$ 角。因此可将复数 $1+2i$ 与 $1+\sqrt{3}i$ 相乘。这样得到与复数 $1+2i$ 成 $60^{\circ}$ 角的复数，然后求该新复数与原点连线的斜率： \[(1+2i)(1+\sqrt{3}i)\] \[= 1 - 2\sqrt{3} + 2i + \sqrt{3}i\] \[\textrm{Slope } = \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}}\] \[= \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}} \cdot \frac{1 + 2\sqrt{3}}{1 + 2\sqrt{3}}\] \[= \frac{8 + 5\sqrt{3}}{-11}\] \[= \frac{-8 - 5\sqrt{3}}{11}.\] 同理可求 $BC$ 的斜率，注意它与 $AB$ 成 $120^{\circ}$ 角： \[(1+2i)(-1+\sqrt{3}i)\] \[= -1 - 2\sqrt{3} - 2i + \sqrt{3}i\] \[\textrm{Slope } = \frac{\sqrt{3} - 2}{-2\sqrt{3} - 1}\] \[= \frac{2 - \sqrt{3}}{1 + 2\sqrt{3}} \cdot \frac{1 - 2\sqrt{3}}{1 - 2\sqrt{3}}\] 此时可观察到一个规律：该表达式等于 $\frac{2 + \sqrt{3}}{1 - 2\sqrt{3}} \cdot \frac{1 + 2\sqrt{3}}{1 + 2\sqrt{3}}$，只是第一分式的分子互为共轭！这意味着展开相乘时，有理项保持不变，而 $\sqrt{3}$ 的系数符号相反。因此这两个复数互为共轭，其无理项相互抵消。于是斜率之和为 $2-2\cdot\frac{8}{11}=\frac{6}{11}$，再除以 $2$ 得 $\frac{3}{11}$，答案为 $\boxed{\mathrm{(A)}\ 14}$。

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