AMC10 2022 A

AMC10 2022 A · Q10

AMC10 2022 A · Q10. It mainly tests Linear equations, Pythagorean theorem.

Daniel finds a rectangular index card and measures its diagonal to be $8$ centimeters. Daniel then cuts out equal squares of side $1$ cm at two opposite corners of the index card and measures the distance between the two closest vertices of these squares to be $4\sqrt{2}$ centimeters, as shown below. What is the area of the original index card?

Daniel 找到一张矩形索引卡，测量其对角线为 $8$ 厘米。然后他在索引卡的对角两个角各剪下边长 $1$ cm 的正方形，并测量这两个正方形最近的两个顶点间的距离为 $4\sqrt{2}$ 厘米，如下图所示。原索引卡的面积是多少？

(A) 14 14

(B) 10\sqrt{2} 10\sqrt{2}

(D) 12\sqrt{2} 12\sqrt{2}

(E) 18 18

Answer

Correct choice: (E)

正确答案：（E）

Solution

Label the bottom left corner of the larger rectangle (without the square cut out) as $A$ and the top right as $D$. $w$ is the width of the rectangle and $\ell$ is the length. Now we have vertices $E, F, G, H$ as vertices of the irregular octagon created by cutting out the squares. Let $I, J$ be the two closest vertices formed by the squares. The distance between the two closest vertices of the squares is thus $IJ=\left(4\sqrt{2}\right).$ Substituting, we get: \[(IJ)^2 = (w-2)^2 + (\ell-2)^2 = \left(4\sqrt{2}\right)^2 = 32 \implies w^2+\ell^2-4w-4\ell = 24.\] Using the fact that the diagonal of the rectangle is $8,$ we get: \[w^2+\ell^2 = 64.\] Subtracting the first equation from the second equation, we get: \[4w+4\ell=40 \implies w+\ell = 10.\] Squaring yields: \[w^2 + 2w\ell + \ell^2 = 100.\] Subtracting the second equation from this, we get: $2w\ell = 36,$ and thus the area of the original rectangle is $w\ell = \boxed{\textbf{(E) } 18}.$

将较大矩形（未剪方形）的左下角标记为 $A$，右上角为 $D$。$w$ 是矩形的宽度，$\ell$ 是长度。现在有顶点 $E, F, G, H$ 为剪下方形后形成的非规则八边形的顶点。设 $I, J$ 为两个正方形形成的最近顶点。两个正方形最近顶点间的距离为 $IJ=4\sqrt{2}$。代入，得： \[(IJ)^2 = (w-2)^2 + (\ell-2)^2 = \left(4\sqrt{2}\right)^2 = 32 \implies w^2+\ell^2-4w-4\ell = 24.\] 利用矩形对角线为 $8$，得： \[w^2+\ell^2 = 64.\] 将第二个方程减去第一个方程，得： \[4w+4\ell=40 \implies w+\ell = 10.\] 平方得： \[w^2 + 2w\ell + \ell^2 = 100.\] 将此减去第二个方程，得 $2w\ell = 36$，因此原矩形面积为 $w\ell = \boxed{\textbf{(E) } 18}$。

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