AMC12 2020 A

AMC12 2020 A · Q17

AMC12 2020 A · Q17. It mainly tests Rational expressions, Area & perimeter.

The vertices of a quadrilateral lie on the graph of $y = \ln x$, and the $x$-coordinates of these vertices are consecutive positive integers. The area of the quadrilateral is $\ln \frac{91}{90}$. What is the $x$-coordinate of the leftmost vertex?

一个四边形的顶点位于$y = \ln x$的图像上，其$x$坐标为四个连续的正整数。该四边形的面积为$\ln \frac{91}{90}$。最左顶点的$x$坐标是多少？

(A) 6 6

(B) 7 7

(D) 12 12

(E) 13 13

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Let the $x$-coordinates of the four vertices be $a$, $a+1$, $a+2$, and $a+3$. The area of the quadrilateral can be computed as the sum of the areas of three trapezoids with horizontal height $1$, minus the area of a trapezoid with horizontal height $3$. Thus the area is $$ \frac12(\ln a+\ln(a+1))+\frac12(\ln(a+1)+\ln(a+2))+\frac12(\ln(a+2)+\ln(a+3))-\frac32(\ln a+\ln(a+3)), $$ which simplifies to $$ \ln(a+1)+\ln(a+2)-\ln a-\ln(a+3)=\ln\frac{(a+1)(a+2)}{a(a+3)}. $$ The requested value of $a$, the $x$-coordinate of the leftmost vertex of the quadrilateral, can be found by solving the quadratic equation $$ \frac{(a+1)(a+2)}{a(a+3)}=\frac{91}{90} $$ or by noting that the area is a decreasing function of $a$ and $$ \frac{91}{90}=\frac{13\cdot7}{6\cdot15}=\frac{13\cdot14}{12\cdot15}. $$ Hence $a=12$. Note: The area of the quadrilateral can also be calculated using the Shoelace Formula.

答案（D）：设四个顶点的 $x$ 坐标分别为 $a$、$a+1$、$a+2$ 和 $a+3$。四边形的面积可以表示为三个水平高为 $1$ 的梯形面积之和，减去一个水平高为 $3$ 的梯形面积。因此面积为 $$ \frac12(\ln a+\ln(a+1))+\frac12(\ln(a+1)+\ln(a+2))+\frac12(\ln(a+2)+\ln(a+3))-\frac32(\ln a+\ln(a+3)), $$ 化简得 $$ \ln(a+1)+\ln(a+2)-\ln a-\ln(a+3)=\ln\frac{(a+1)(a+2)}{a(a+3)}. $$ 所求的 $a$（四边形最左端顶点的 $x$ 坐标）可通过解二次方程 $$ \frac{(a+1)(a+2)}{a(a+3)}=\frac{91}{90} $$ 得到；或者注意到面积关于 $a$ 是递减函数，并且 $$ \frac{91}{90}=\frac{13\cdot7}{6\cdot15}=\frac{13\cdot14}{12\cdot15}. $$ 因此 $a=12$。注：四边形的面积也可以用鞋带公式（Shoelace Formula）计算。

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