Track / Stage 4 / 282 of 340 #542 of 2000

Problem 542

AMC 12 late, AIME early

Geometry Difficulty 4.9 Find the answer

4. As shown in Figure 3, in the square paper piece $A B C D$ , $E$ is the midpoint of $B C$ . Fold the square so that point $A$ coincides with $E$ , and flatten it with the crease being $M N$ . Then the ratio of the area of trapezoid $A D M N$ to the area of trapezoid $B C M N$ is $\qquad$

Your answer

Official solution

4. $\frac{3}{5}$ .

As shown in Figure 5, let the side length of the square be 2. Then $A E=\sqrt{5}$ .

Draw a perpendicular from point $M$ to $A B$ , intersecting at point $P$ .

It is easy to see that $M N$ is the perpendicular bisector of $A E$ .
Thus, $\triangle M P N \cong \triangle$
$A B E$
$\Rightarrow P N=B E=1 \text {. }$

By $\triangle A N T \backsim \triangle A E B$ , we get $A N=\frac{5}{4}$ . Then $D M=A P=\frac{1}{4}$ .
$\text { Therefore, } \frac{S_{\text {trapezoid } A D M N}}{S_{\text {trapezoid } B C M N}}=\frac{A N+D M}{M C+N B}=\frac{3}{5} \text {. }$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.