AMC 10B, 202!, Problem 21, Question and solution

[Deepan Dutta]

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 the edge \(\overline{AD}\) at point \(C^{\prime}\) and the edge \(\overline{BC}\) intersects the edge \(\overline{AB}\) at the point \(E\). Suppose that \(C^{\prime}D=\frac{1}{3}\). What is the perimeter of triangle \(\triangle AEC^{\prime}\).

• This topic was modified 1 year, 3 months ago by Deepan Dutta.
• This topic was modified 1 year, 3 months ago by Deepan Dutta.

[Deepan Dutta]

All the edges are \(1\) unit.

Let \(\overline{DF}=x\) unit.

\(\therefore\) \(\overline{C^{\prime}F}=1-x\)

As, \(\angle C^{\prime}DF=90^o\) we can apply the \(\textbf{Pythagorean Theorem}\) on that triangle,

\({\frac{1}{3}}^2+x^2=(1-x)^2\)

\(\Rightarrow x=\frac{4}{9}\)

\(\therefore\) \(FC^{\prime}=\frac{5}{9}\)

Now, if the \(\angle FC^{\prime}D\) be \(\theta\) then \(\angle EC^{\prime}A=90^o-\theta\).

So, \(\triangle AEC^{\prime}\sim\triangle DFC^{\prime}\)

\(\therefore\) \(\frac{DF}{AC^{\prime}}=\frac{C^{\prime}D}{AE}=\frac{C^{\prime}F}{C^{\prime}E}\)

From, \(\frac{DF}{AC^{\prime}}=\frac{\frac{4}{9}}{\frac{2}{3}}=\frac{2}{3}\)

So, \(\frac{C^{\prime}D}{AE}=\frac{2}{3}\Rightarrow AE=\frac{1}{3}\times\frac{3}{2}=\frac{1}{2}\)

\(\frac{C^{\prime}F}{C^{\prime}E}=\frac{2}{3}\Rightarrow C^{\prime}E =\frac{5}{9}\times\frac{3}{2}=\frac{5}{6}\).

So the perimeter of the \(\triangle AEC^{\prime}=\frac{2}{3}+\frac{1}{2}+\frac{5}{6}=\boxed{2}\).

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