AMC12 2009 A

AMC12 2009 A · Q14

AMC12 2009 A · Q14. It mainly tests Quadratic equations, Coordinate geometry.

A triangle has vertices $(0,0)$, $(1,1)$, and $(6m,0)$, and the line $y = mx$ divides the triangle into two triangles of equal area. What is the sum of all possible values of $m$?

一个三角形的顶点为 $(0,0)$、$(1,1)$ 和 $(6m,0)$，直线 $y = mx$ 将该三角形分成两个面积相等的三角形。所有可能的 $m$ 值之和是多少？

(A) $-\frac{1}{3}$ $-\frac{1}{3}$

(B) $-\frac{1}{6}$ $-\frac{1}{6}$

(D) $\frac{1}{3}$ $\frac{1}{3}$

(E) $\frac{1}{2}$ $\frac{1}{2}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Let's label the three points as $A=(0,0)$, $B=(1,1)$, and $C=(6m,0)$. Clearly, whenever the line $y=mx$ intersects the inside of the triangle, it will intersect the side $BC$. Let $D$ be the point of intersection. The triangles $ABD$ and $ACD$ have the same height, which is the distance between the point $A$ and the line $BC$. Hence they have equal areas if and only if $D$ is the midpoint of $BC$. The midpoint of the segment $BC$ has coordinates $\left( \frac{6m+1}2, \frac 12 \right)$. This point lies on the line $y=mx$ if and only if $\frac 12 = m \cdot \frac{6m+1}2$. This simplifies to $6m^2 + m - 1 = 0$. Using Vieta's formulas, we find that the sum of the roots is $\boxed{\textbf{(B)} - \!\frac {1}{6}}$. For illustration, below are pictures of the situation for $m=1.5$, $m=0.5$, $m=1/3$, and $m=-1/2$. Note : In contests when you get problems like these, your AOPS knowledge will not be enough. AOPs books are just the basic, always prep for more.

将三点标为 $A=(0,0)$，$B=(1,1)$，$C=(6m,0)$。显然，只要直线 $y=mx$ 与三角形内部相交，它就会与边 $BC$ 相交。设交点为 $D$。三角形 $ABD$ 与 $ACD$ 具有相同的高，即点 $A$ 到直线 $BC$ 的距离。因此它们面积相等当且仅当 $D$ 是 $BC$ 的中点。线段 $BC$ 的中点坐标为 $\left( \frac{6m+1}2, \frac 12 \right)$。该点在直线 $y=mx$ 上当且仅当 $\frac 12 = m \cdot \frac{6m+1}2$，化简得 $6m^2 + m - 1 = 0$。由韦达定理，根的和为 $\boxed{\textbf{(B)} - \!\frac {1}{6}}$。为说明起见，下图展示了 $m=1.5$、$m=0.5$、$m=1/3$ 与 $m=-1/2$ 时的情况。注：在竞赛中遇到这类题，仅靠 AOPS 的知识是不够的。AOPS 的书只是基础，务必准备更多。

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