A square piece of paper $A B C D$ is white on one side and grey on the other side. Initially, the paper is flat on a table with the grey side down. Point $E$ is on $B C$ so when the paper is folded along $A E, B$ lands on diagonal $A C$. Similarly, point $F$ is on $D C$ so that when the paper is folded along $A F, D$ lands on $A C$. After these folds, the resulting shape is kite $A E C F$. What fraction of the area of $A E C F$ is grey?

![](https://cdn.mathpix.com/cropped/2024_04_17_b6707dc71c8b7ccd30afg-4.jpg?height=509&width=1000&top_left_y=453&top_left_x=622)
Solution 1

Suppose $A B C D$ has side length $x$ and let $G$ be on $A C$ so that $E G$ is perpendicular to $A C$. Set $G E=y$.

![](https://cdn.mathpix.com/cropped/2024_04_17_54c374478de6843636d1g-14.jpg?height=504&width=501&top_left_y=1664&top_left_x=866)

When the paper is folded, $B$ lands on $A C$ and $E$ does not move.

This means that $\angle B A E=\angle G A E$. As well, $\angle A G E=\angle A B E=90^{\circ}$ by construction.

Since $\triangle A B E$ and $\triangle A G E$ have two angles in common, they must have three angles in common. They also share side $A E$, so they must be congruent.

The area of $A E C F$ is the sum of the areas of $\triangle A E C$ and $\triangle A F C$.

By symmetry, $\triangle A F C$ is congruent to $\triangle A E C$, which means the area of $A E C F$ is

$$
2\left(\frac{1}{2} \times A C \times G E\right)=(A C)(G E)
$$

The length of $A C$ is $\sqrt{2} x$ since it is a diagonal of a square with side length $x$.

Therefore, the area of $A E C F$ is $\sqrt{2} x y$.

The area of the shaded region is twice the area of $\triangle A G E$, and since $\triangle A G E$ is congruent to $\triangle A B E$, we have $A G=A B=x$, so the shaded region has area

$$
2\left(\frac{1}{2} \times A G \times G E\right)=(A G)(G E)=x y
$$

Therefore, the fraction we seek is $\frac{x y}{\sqrt{2} x y}=\frac{1}{\sqrt{2}}$.

## Solution 2

Let the side length of the square be 1 , set $B E=x$, and define $G$ as in Solution 1 .

Using the argument from Solution 1, we get that $G E=x$ as well.

Since $A C$ is the diagonal of a square, $\triangle A B C$ is isosceles with $\angle A B C=90^{\circ}$, which means $\angle A C E=45^{\circ}$.

Since $\angle C G E=90^{\circ}$ by construction, $\triangle C G E$ is an isosceles right triangle, so $C E=\sqrt{2} G E$.

We also have that $C E=B C-B E=1-x$ and $G E=x$, so this means $1-x=\sqrt{2} x$.

Solving for $x$, we get $x=\frac{1}{\sqrt{2}+1}=\frac{1}{\sqrt{2}+1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1}=\sqrt{2}-1$.

Using the argument from Solution 1, $\triangle A B E, \triangle A G E, \triangle A G F$, and $\triangle A D F$ are all congruent. This means the area of the shaded region is equal to twice the area of $\triangle A B E$, which is

$$
2\left(\frac{1}{2} \times A B \times B E\right)=x=\sqrt{2}-1
$$

The area of $A E F C$ is the area of the square minus the combined area of $\triangle A B E$ and $\triangle A D F$. This means the area of $A E F C$ is 1 minus twice the area of $\triangle A B E$, or $1-(\sqrt{2}-1)=2-\sqrt{2}$. Therefore, the ratio is

$$
\frac{\sqrt{2}-1}{2-\sqrt{2}}=\frac{\sqrt{2}-1}{\sqrt{2}(\sqrt{2}-1)}=\frac{1}{\sqrt{2}}
$$

ANSWER: $\frac{1}{\sqrt{2}}$