Aaron takes a square sheet of paper, with one corner labeled $A$. Point $P$ is chosen at random inside of the square and Aaron folds the paper so that points $A$ and $P$ coincide. He cuts the sheet along the crease and discards the piece containing $A$. Let $p$ be the probability that the remaining piece is a pentagon. Find the integer nearest to $100p$.

[i]Proposed by Aaron Lin[/i]
1. **Understanding the Problem:**
   - We start with a square sheet of paper with one corner labeled \( A \).
   - A point \( P \) is chosen randomly inside the square.
   - The paper is folded such that \( A \) coincides with \( P \).
   - The paper is cut along the crease and the piece containing \( A \) is discarded.
   - We need to find the probability \( p \) that the remaining piece is a pentagon.

2. **Analyzing the Fold and Cut:**
   - When the paper is folded such that \( A \) coincides with \( P \), the crease line is the perpendicular bisector of the segment \( AP \).
   - The crease line will divide the square into two regions.
   - We need to determine the conditions under which the remaining piece (after discarding the piece containing \( A \)) is a pentagon.

3. **Geometric Considerations:**
   - For the remaining piece to be a pentagon, the crease line must intersect the square in such a way that it creates five sides.
   - This happens when the crease line intersects two adjacent sides of the square and one of the sides is cut into two segments.

4. **Determining the Region for \( P \):**
   - The crease line will create a pentagon if it intersects two adjacent sides of the square.
   - This occurs when \( P \) is within a certain region of the square.
   - Specifically, \( P \) must be within the region where the perpendicular bisector of \( AP \) intersects two adjacent sides of the square.

5. **Calculating the Area of the Region:**
   - The region where \( P \) must lie can be determined by symmetry and geometric properties.
   - The region is a quarter of the square, as the perpendicular bisector of \( AP \) must intersect two adjacent sides of the square.
   - Therefore, the area of the region where \( P \) must lie is \( \frac{1}{4} \) of the area of the square.

6. **Probability Calculation:**
   - The probability \( p \) that the remaining piece is a pentagon is the ratio of the area of the region where \( P \) must lie to the total area of the square.
   - Since the area of the region is \( \frac{1}{4} \) of the square, the probability \( p \) is \( \frac{1}{4} \).

7. **Finding the Integer Nearest to \( 100p \):**
   - We need to find the integer nearest to \( 100p \).
   - Since \( p = \frac{1}{4} \), we have \( 100p = 100 \times \frac{1}{4} = 25 \).

The final answer is \( \boxed{25} \).