AMC10 2021 A

AMC10 2021 A · Q24

AMC10 2021 A · Q24. It mainly tests Area & perimeter, Coordinate geometry.

The interior of a quadrilateral is bounded by the graphs of $(x+ay)^2 = 4a^2$ and $(ax-y)^2 = a^2$, where $a$ is a positive real number. What is the area of this region in terms of $a$, valid for all $a > 0$?

一个四边形的内部由图象 $(x+ay)^2 = 4a^2$ 和 $(ax-y)^2 = a^2$ 围成，其中 $a$ 是正实数。这个区域的面积用 $a$ 表示，对所有 $a > 0$ 有效是多少？

(A) \frac{8a^2}{(a+1)^2} \frac{8a^2}{(a+1)^2}

(B) \frac{4a}{a+1} \frac{4a}{a+1}

(D) \frac{8a^2}{a^2+1} \frac{8a^2}{a^2+1}

(E) \frac{8a}{a^2+1} \frac{8a}{a^2+1}

Answer

Correct choice: (D)

正确答案：（D）

Solution

The cases for $(x+ay)^2 = 4a^2$ are $x+ay = \pm2a,$ or two parallel lines. We rearrange each case and construct the table below: The cases for $(ax-y)^2 = a^2$ are $ax-y=\pm a,$ or two parallel lines. We rearrange each case and construct the table below: Since the slopes of intersecting lines $(1)\cap(1*), (1)\cap(2*), (2)\cap(1*),$ and $(2)\cap(2*)$ are negative reciprocals, we get four right angles, from which the quadrilateral is a rectangle. Two solutions follow from here: Recall that for constants $A,B,C_1$ and $C_2,$ the distance $d$ between parallel lines $\begin{cases} Ax+By+C_1=0 \\ Ax+By+C_2=0 \end{cases}$ is \[d=\frac{\left|C_2-C_1\right|}{\sqrt{A^2+B^2}}.\] From this formula: - The distance between lines $(1)$ and $(2)$ is $\frac{4a}{\sqrt{1+a^2}},$ the length of this rectangle. - The distance between lines $(1*)$ and $(2*)$ is $\frac{2a}{\sqrt{a^2+1}},$ the width of this rectangle. The area we seek is \[\frac{4a}{\sqrt{1+a^2}}\cdot\frac{2a}{\sqrt{a^2+1}}=\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.\]

$(x+ay)^2 = 4a^2$ 的情况为 $x+ay = \pm2a$，即两条平行线。我们重新排列每个情况并构建下表： $(ax-y)^2 = a^2$ 的情况为 $ax-y=\pm a$，即两条平行线。我们重新排列每个情况并构建下表：由于相交线 $(1)\cap(1*)$、$(1)\cap(2*)$、$(2)\cap(1*)$ 和 $(2)\cap(2*)$ 的斜率为负互为倒数，我们得到四个直角，由此四边形是一个矩形。由此有两种解法：回想对于常数 $A,B,C_1$ 和 $C_2$，平行线 $\begin{cases} Ax+By+C_1=0 \\ Ax+By+C_2=0 \end{cases}$ 之间的距离 $d$ 为 \[d=\frac{\left|C_2-C_1\right|}{\sqrt{A^2+B^2}}.\] 由此公式： - 线 $(1)$ 和 $(2)$ 之间的距离为 $\frac{4a}{\sqrt{1+a^2}}$，即此矩形的长度。 - 线 $(1*)$ 和 $(2*)$ 之间的距离为 $\frac{2a}{\sqrt{a^2+1}}$，即此矩形的宽度。所求面积为 \[\frac{4a}{\sqrt{1+a^2}}\cdot\frac{2a}{\sqrt{a^2+1}}=\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.\]

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