Source: 2018 Canadian Open Math Challenge Part B Problem 2
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Let ABCD be a square with side length 1. Points $X$ and $Y$ are on sides $BC$ and $CD$ respectively such that the areas of triangels $ABX$, $XCY$, and $YDA$ are equal. Find the ratio of the area of $\triangle AXY$ to the area of $\triangle XCY$.
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1. **Define Variables and Set Up Equations:**
   Let \( ABCD \) be a square with side length 1. Points \( X \) and \( Y \) are on sides \( BC \) and \( CD \) respectively. Let \( CX = CY = x \). Consequently, \( BX = DY = 1 - x \).

2. **Equate Areas of Triangles:**
   The areas of triangles \( ABX \), \( XCY \), and \( YDA \) are equal. We can express these areas in terms of \( x \):
   - The area of \( \triangle ABX \) is \( \frac{1}{2} \times AB \times BX = \frac{1}{2} \times 1 \times (1 - x) = \frac{1 - x}{2} \).
   - The area of \( \triangle XCY \) is \( \frac{1}{2} \times CX \times CY = \frac{1}{2} \times x \times x = \frac{x^2}{2} \).
   - The area of \( \triangle YDA \) is \( \frac{1}{2} \times DY \times DA = \frac{1}{2} \times (1 - x) \times 1 = \frac{1 - x}{2} \).

   Since the areas are equal, we have:
   \[
   \frac{1 - x}{2} = \frac{x^2}{2}
   \]

3. **Solve for \( x \):**
   \[
   \frac{1 - x}{2} = \frac{x^2}{2}
   \]
   \[
   1 - x = x^2
   \]
   \[
   x^2 + x - 1 = 0
   \]
   Solving this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1}
   \]
   \[
   x = \frac{-1 \pm \sqrt{1 + 4}}{2}
   \]
   \[
   x = \frac{-1 \pm \sqrt{5}}{2}
   \]
   Since \( x \) must be positive and less than 1, we take the positive root:
   \[
   x = \frac{\sqrt{5} - 1}{2}
   \]

4. **Find the Ratio of Areas \( \frac{[AXY]}{[XCY]} \):**
   Let \( AC \cap XY = Z \). We need to find the ratio \( \frac{[AXY]}{[XCY]} \). This ratio is equal to the ratio of the lengths \( \frac{AZ}{CZ} \).

   Since \( AC \) is the diagonal of the square, its length is \( \sqrt{2} \). The coordinates of \( X \) and \( Y \) are \( (1, x) \) and \( (x, 0) \) respectively. The line \( XY \) has the equation:
   \[
   y = -\frac{x}{1-x} (x - 1)
   \]
   The line \( AC \) has the equation \( y = x \). Solving for the intersection \( Z \):
   \[
   x = -\frac{x}{1-x} (x - 1)
   \]
   Simplifying, we find:
   \[
   AZ = \sqrt{2} - \frac{x}{\sqrt{2}}
   \]
   \[
   CZ = \frac{x}{\sqrt{2}}
   \]
   Therefore, the ratio is:
   \[
   \frac{AZ}{CZ} = \frac{\sqrt{2} - \frac{x}{\sqrt{2}}}{\frac{x}{\sqrt{2}}} = \frac{2}{x} - 1
   \]
   Substituting \( x = \frac{\sqrt{5} - 1}{2} \):
   \[
   \frac{2}{\frac{\sqrt{5} - 1}{2}} - 1 = \frac{4}{\sqrt{5} - 1} - 1
   \]
   Rationalizing the denominator:
   \[
   \frac{4(\sqrt{5} + 1)}{(\sqrt{5} - 1)(\sqrt{5} + 1)} - 1 = \frac{4(\sqrt{5} + 1)}{4} - 1 = \sqrt{5} + 1 - 1 = \sqrt{5}
   \]

The final answer is \(\boxed{\sqrt{5}}\).