
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
Let \( A(x; y) \) be a vertex of a square with its center at the origin, given that \(\text{GCD}(x, y) = 2\). Calculate the length of the diagonal of the square \( d = 2 \sqrt{x^2 + y^2} \) and find the area of the square \( S = \frac{d^2}{2} = 2(x^2 + y^2) \). According to the problem, \( S = 10 \text{LCM}(x, y) \). Thus, we have the equation:
\[ 
2(x^2 + y^2) = 10 \text{LCM}(x, y).
\]

Using the property \(\text{GCD}(x, y) \cdot \text{LCM}(x, y) = xy\), we get
\[
\text{LCM}(x, y) = \frac{xy}{\text{GCD}(x, y)} = \frac{xy}{2}.
\]

Substituting \(\text{LCM}(x, y)\) into the above equation, we have
\[
2(x^2 + y^2) = 5xy.
\]

Dividing the last equation by \( y^2 \), we obtain a quadratic equation in terms of \(\frac{x}{y}\):
\[
2 \left(\frac{x}{y}\right)^2 - 5 \frac{x}{y} + 2 = 0.
\]

This equation has solutions \(\frac{x}{y} = 2\) and \(\frac{y}{x} = 2\).

Case 1: \( x = 2y \). Then \(\text{GCD}(x, y) = \text{GCD}(2y, y) = y = 2\), so \( x = 4 \).

Case 2: \( y = 2x \). Then \(\text{GCD}(x, y) = \text{GCD}(x, 2x) = x = 2\), so \( y = 4 \).

Present the answer in LaTex format: \boxed{Your answer}
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