For constant $k>1$, 2 points $X,\ Y$ move on the part of the first quadrant of the line, which passes through $A(1,\ 0)$ and is perpendicular to the $x$ axis, satisfying $AY=kAX$. Let a circle with radius 1 centered on the origin $O(0,\ 0)$ intersect with line segments $OX,\ OY$ at $P,\ Q$ respectively. Express the maximum area of $\triangle{OPQ}$ in terms of $k$.

[i]2011 Tokyo Institute of Technology entrance exam, Problem 3[/i]
1. **Define the problem and set up the coordinates:**
   - Given a line passing through \( A(1, 0) \) and perpendicular to the \( x \)-axis, the line equation is \( x = 1 \).
   - Points \( X \) and \( Y \) move on this line in the first quadrant.
   - The condition \( AY = k \cdot AX \) implies that the distance from \( A \) to \( Y \) is \( k \) times the distance from \( A \) to \( X \).

2. **Determine the coordinates of \( X \) and \( Y \):**
   - Since \( X \) and \( Y \) are on the line \( x = 1 \), their coordinates can be written as \( X(1, y_1) \) and \( Y(1, y_2) \).
   - The distance \( AX = y_1 \) and \( AY = y_2 \).
   - Given \( AY = k \cdot AX \), we have \( y_2 = k \cdot y_1 \).

3. **Find the intersection points \( P \) and \( Q \) with the unit circle:**
   - The line segments \( OX \) and \( OY \) intersect the unit circle \( x^2 + y^2 = 1 \) at points \( P \) and \( Q \).
   - The coordinates of \( P \) and \( Q \) can be parameterized as \( P(\cos \alpha, \sin \alpha) \) and \( Q(\cos \beta, \sin \beta) \), where \( \alpha \) and \( \beta \) are the angles \( \angle AOX \) and \( \angle AOY \) respectively.

4. **Relate the angles \( \alpha \) and \( \beta \) to the coordinates:**
   - Since \( \tan \alpha = y_1 \) and \( \tan \beta = y_2 \), we have \( \tan \beta = k \cdot \tan \alpha \).

5. **Express the area of \( \triangle OPQ \):**
   - The area of \( \triangle OPQ \) can be found using the formula for the area of a triangle with vertices at the origin and two points on the unit circle:
     \[
     \text{Area} = \frac{1}{2} \left| \cos \alpha \sin \beta - \sin \alpha \cos \beta \right|
     \]
   - Using the sine difference identity, this becomes:
     \[
     \text{Area} = \frac{1}{2} \left| \sin(\beta - \alpha) \right|
     \]

6. **Maximize the area:**
   - To maximize \( \left| \sin(\beta - \alpha) \right| \), we need to maximize \( \beta - \alpha \).
   - Given \( \tan \beta = k \cdot \tan \alpha \), we use the tangent addition formula:
     \[
     \tan(\beta - \alpha) = \frac{\tan \beta - \tan \alpha}{1 + \tan \beta \tan \alpha} = \frac{k \tan \alpha - \tan \alpha}{1 + k \tan \alpha \tan \alpha} = \frac{(k-1) \tan \alpha}{1 + k \tan^2 \alpha}
     \]
   - The maximum value of \( \left| \sin(\beta - \alpha) \right| \) occurs when \( \beta - \alpha = \frac{\pi}{2} \), but we need to consider the constraints of the problem.

7. **Use AM-GM inequality to find the maximum value:**
   - By the AM-GM inequality, we have:
     \[
     1 + k \tan^2 \alpha \geq 2 \sqrt{k} \tan \alpha
     \]
   - Thus:
     \[
     \left| \frac{(k-1) \tan \alpha}{1 + k \tan^2 \alpha} \right| \leq \left| \frac{(k-1) \tan \alpha}{2 \sqrt{k} \tan \alpha} \right| = \left| \frac{k-1}{2 \sqrt{k}} \right|
     \]

8. **Conclusion:**
   - The maximum area of \( \triangle OPQ \) is:
     \[
     \text{Max Area} = \frac{1}{2} \left| \frac{k-1}{k+1} \right|
     \]

The final answer is \(\boxed{\frac{1}{2} \left( \frac{k-1}{k+1} \right)}\)