AMC10 2021 B

AMC10 2021 B · Q21

AMC10 2021 B · Q21. It mainly tests Vieta / quadratic relationships (basic), Triangles (properties).

A square piece of paper has side length $1$ and vertices $A,B,C,$ and $D$ in that order. As shown in the figure, the paper is folded so that vertex $C$ meets edge $\overline{AD}$ at point $C'$, and edge $\overline{BC}$ intersects edge $\overline{AB}$ at point $E$. Suppose that $C'D = \frac{1}{3}$. What is the perimeter of triangle $\bigtriangleup AEC' ?$

一张边长为$1$的正方形纸片，顶点依次为$A,B,C,D$。如图所示，将纸片折叠使顶点$C$与边$\overline{AD}$上的点$C'$重合，且边$\overline{BC}$与边$\overline{AB}$相交于点$E$。已知$C'D=\frac{1}{3}$。求三角形$\bigtriangleup AEC'$的周长。

(A) 2 2

(B) 1+\frac{2}{3}\sqrt{3} 1+\frac{2}{3}\sqrt{3}

(D) 1 + \frac{3}{4}\sqrt{3} 1 + \frac{3}{4}\sqrt{3}

(E) \frac{7}{3} \frac{7}{3}

Answer

Correct choice: (A)

正确答案：（A）

Solution

We can set the point on $CD$ where the fold occurs as point $F$. Then, we can set $FD$ as $x$, and $CF$ as $1-x$ because of symmetry due to the fold. It can be recognized that this is a right triangle, and solving for $x$, we get, \[x^2 + \left(\frac{1}{3}\right)^2 = (1-x)^2 \rightarrow x^2 + \frac{1}{9} = x^2 - 2x + 1 \rightarrow x=\frac{4}{9}\] We know this is a 3-4-5 triangle because the side lengths are $\frac{3}{9}, \frac{4}{9}, \frac{5}{9}$. We also know that $EAC'$ is similar to $C'DF$ because angle $EC'F$ is a right angle. Now, we can use similarity to find out that the perimeter is just the perimeter of $C'DF \times \frac{AC'}{DF}$. That's just $\frac{4}{3} \times \frac{\frac{2}{3}}{\frac{4}{9}} = \frac{4}{3} \times \frac{3}{2} = 2$. Therefore, the final answer is $\boxed{\textbf{(A)} ~2}$

我们可以将折叠发生在$CD$上的点设为$F$。则设$FD=x$，$CF=1-x$，因为折叠的对称性。可以看出这是一个直角三角形，解$x$得， \[x^2 + \left(\frac{1}{3}\right)^2 = (1-x)^2 \rightarrow x^2 + \frac{1}{9} = x^2 - 2x + 1 \rightarrow x=\frac{4}{9}\] 我们知道这是一个3-4-5三角形，因为边长为$\frac{3}{9}, \frac{4}{9}, \frac{5}{9}$。我们还知道$EAC'$与$C'DF$相似，因为$\angle EC'F$是直角。现在，我们可以用相似性来发现周长就是$C'DF$的周长乘以$\frac{AC'}{DF}$。那就是$\frac{4}{3} \times \frac{\frac{2}{3}}{\frac{4}{9}} = \frac{4}{3} \times \frac{3}{2} = 2$。因此，最终答案是$\boxed{\textbf{(A)} ~2}$。

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