AMC10 2022 A

AMC10 2022 A · Q5

AMC10 2022 A · Q5. It mainly tests Quadratic equations, Coordinate geometry.

Square $ABCD$ has side length $1$. Points $P$, $Q$, $R$, and $S$ each lie on a side of $ABCD$ such that $APQCRS$ is an equilateral convex hexagon with side length $s$. What is $s$?

正方形 $ABCD$ 边长为 $1$。点 $P$、$Q$、$R$ 和 $S$ 各位于 $ABCD$ 的一条边上，使得 $APQCRS$ 是一个边长为 $s$ 的等边凸六边形。$s$ 的值为？

(A) \frac{\sqrt{2}}{3} \frac{\sqrt{2}}{3}

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

(D) 1 - \frac{\sqrt{2}}{4} 1 - \frac{\sqrt{2}}{4}

(E) \frac{2}{3} \frac{2}{3}

Answer

Correct choice: (C)

正确答案：（C）

Solution

Note that $BP=BQ=DR=DS=1-s.$ It follows that $\triangle BPQ$ and $\triangle DRS$ are congruent isosceles right triangles. In $\triangle BPQ,$ we have $PQ=BP\sqrt2,$ or \begin{align*} s &= (1-s)\sqrt2 \\ s &= \sqrt2 - s\sqrt2 \\ \left(\sqrt2+1\right)s &= \sqrt2 \\ s &= \frac{\sqrt2}{\sqrt2 + 1}. \end{align*} Therefore, the answer is \[s = \frac{\sqrt2}{\sqrt2 + 1}\cdot\frac{\sqrt2 - 1}{\sqrt2 - 1} = \boxed{\textbf{(C) } 2 - \sqrt{2}}.\]

注意到 $BP=BQ=DR=DS=1-s$。由此，$\triangle BPQ$ 和 $\triangle DRS$ 是全等的等腰直角三角形。在 $\triangle BPQ$ 中，有 $PQ=BP\sqrt2$，即 \begin{align*} s &= (1-s)\sqrt2 \\ s &= \sqrt2 - s\sqrt2 \\ \left(\sqrt2+1\right)s &= \sqrt2 \\ s &= \frac{\sqrt2}{\sqrt2 + 1}. \end{align*} 因此，答案是 \[s = \frac{\sqrt2}{\sqrt2 + 1}\cdot\frac{\sqrt2 - 1}{\sqrt2 - 1} = \boxed{\textbf{(C) } 2 - \sqrt{2}}.\]

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