AMC8 2015

AMC8 2015 · Q25

AMC8 2015 · Q25. It mainly tests Quadratic equations, Similarity.

One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?

从这个5英寸正方形的四个角上切下1英寸正方形。能放入剩余空间的最大正方形的面积（平方英寸）是多少？

(A) 9 9

(B) $12\frac{1}{2}$ $12\frac{1}{2}$

(D) $15\frac{1}{2}$ $15\frac{1}{2}$

(E) 17 17

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Let $EQ = c$ and $TQ = s$ as indicated in the figure. Triangles $QUV$ and $FEQ$ are similar since $\angle FQE$ and $\angle QVU$ are congruent because both are complementary to $\angle VQU$. So $\frac{QU}{UV}=\frac{FE}{EQ}$ and thus $QU=\frac{1}{c}$. Then $AB=1+c+\frac{1}{c}+1=5$ and so $c+\frac{1}{c}=3$. Since the area of square $ABCD$ equals the sum of areas of square $QRST$, four unit squares, four $1\times c$ triangles, and four $\frac{1}{c}\times 1$ triangles, it follows that $25=s^2+4\left(1+\frac{c}{2}+\frac{1}{2c}\right)$ $=s^2+4+2\left(c+\frac{1}{c}\right)$ $=s^2+4+2\cdot 3$ Therefore, the area of square $QRST=s^2=15$.

答案（C）：如图所示，令 $EQ=c$，$TQ=s$。因为 $\angle FQE$ 与 $\angle QVU$ 全等（它们都与 $\angle VQU$ 互余），所以三角形 $QUV$ 与 $FEQ$ 相似。因此 $\frac{QU}{UV}=\frac{FE}{EQ}$ 从而 $QU=\frac{1}{c}$。接着 $AB=1+c+\frac{1}{c}+1=5$，所以 $c+\frac{1}{c}=3$。由于正方形 $ABCD$ 的面积等于正方形 $QRST$、4 个单位正方形、4 个 $1\times c$ 的三角形以及 4 个 $\frac{1}{c}\times 1$ 的三角形面积之和，可得 $25=s^2+4\left(1+\frac{c}{2}+\frac{1}{2c}\right)$ $=s^2+4+2\left(c+\frac{1}{c}\right)$ $=s^2+4+2\cdot 3$ 因此，正方形 $QRST$ 的面积为 $s^2=15$。

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