AMC12 2007 B

AMC12 2007 B · Q20

AMC12 2007 B · Q20. It mainly tests Systems of equations, Coordinate geometry.

The parallelogram bounded by the lines $y=ax+c$, $y=ax+d$, $y=bx+c$, and $y=bx+d$ has area $18$. The parallelogram bounded by the lines $y=ax+c$, $y=ax-d$, $y=bx+c$, and $y=bx-d$ has area $72$. Given that $a$, $b$, $c$, and $d$ are positive integers, what is the smallest possible value of $a+b+c+d$?

由直线 $y=ax+c$, $y=ax+d$, $y=bx+c$, $y=bx+d$ 围成的平行四边形面积为$18$。由直线 $y=ax+c$, $y=ax-d$, $y=bx+c$, $y=bx-d$ 围成的平行四边形面积为$72$。已知$a$、$b$、$c$、$d$为正整数，求$a+b+c+d$的最小可能值。

(A) 13 13

(B) 14 14

(D) 16 16

(E) 17 17

Answer

Correct choice: (D)

正确答案：（D）

Solution

Plotting the parallelogram on the coordinate plane, the 4 corners are at $(0,c),(0,d),\left(\frac{d-c}{a-b},\frac{ad-bc}{a-b}\right),\left(\frac{c-d}{a-b},\frac{bc-ad}{a-b}\right)$. Because $72= 4\cdot 18$, we have that $4(c-d)\left(\frac{c-d}{a-b}\right) = (c+d)\left(\frac{c+d}{a-b}\right)$ or that $2(c-d)=c+d$, which gives $c=3d$ (consider a homothety, or dilation, that carries the first parallelogram to the second parallelogram; because the area increases by $4\times$, it follows that the stretch along the diagonal, or the ratio of side lengths, is $2\times$). The area of the triangular half of the parallelogram on the right side of the y-axis is given by $9 = \frac{1}{2} (c-d)\left(\frac{d-c}{a-b}\right)$, so substituting $c = 3d$: \[\frac{1}{2} (c-d)\left(\frac{c-d}{a-b}\right) = 9 \quad \Longrightarrow \quad 2d^2 = 9(a-b)\] Thus $3|d$, and we verify that $d = 3$, $a-b = 2 \Longrightarrow a = 3, b = 1$ will give us a minimum value for $a+b+c+d$. Then $a+b+c+d = 3 + 1 + 9 + 3 = \boxed{\mathbf{(D)} 16}$.

将平行四边形画在坐标平面上，四个顶点为$(0,c),(0,d),\left(\frac{d-c}{a-b},\frac{ad-bc}{a-b}\right),\left(\frac{c-d}{a-b},\frac{bc-ad}{a-b}\right)$。因为$72= 4\cdot 18$，所以有$4(c-d)\left(\frac{c-d}{a-b}\right) = (c+d)\left(\frac{c+d}{a-b}\right)$，即$2(c-d)=c+d$，从而$c=3d$（可考虑一个相似变换或伸缩，将第一个平行四边形变为第二个；由于面积增大$4$倍，可知沿对角线的伸缩、或边长比为$2$倍）。位于$y$轴右侧的平行四边形一半（三角形）的面积为$9 = \frac{1}{2} (c-d)\left(\frac{d-c}{a-b}\right)$，代入$c = 3d$： \[\frac{1}{2} (c-d)\left(\frac{c-d}{a-b}\right) = 9 \quad \Longrightarrow \quad 2d^2 = 9(a-b)\] 因此$3|d$，并验证$d = 3$, $a-b = 2 \Longrightarrow a = 3, b = 1$会使$a+b+c+d$取得最小值。于是$a+b+c+d = 3 + 1 + 9 + 3 = \boxed{\mathbf{(D)} 16}$。

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