[b]p13.[/b] Let $\vartriangle ABC$ be an equilateral triangle with side length $1$. Let the unit circles centered at $A$, $B$, and $C$ be $\Omega_A$, $\Omega_B$, and $\Omega_C$, respectively. Then, let $\Omega_A$ and $\Omega_C$ intersect again at point $D$, and $\Omega_B$ and $\Omega_C$ intersect again at point $E$. Line $BD$ intersects $\Omega_B$ at point $F$ where $F$ lies between $B$ and $D$, and line $AE$ intersects $\Omega_A$ at $G$ where $G$ lies between $A$ and $E$. $BD$ and $AE$ intersect at $H$. Finally, let $CH$ and $FG$ intersect at $I$. Compute $IH$.


[b]p14.[/b] Suppose Bob randomly fills in a $45 \times 45$ grid with the numbers from $1$ to $2025$, using each number exactly once. For each of the $45$ rows, he writes down the largest number in the row. Of these $45$ numbers, he writes down the second largest number. The probability that this final number is equal to $2023$ can be expressed as $\frac{p}{q}$ where $p$ and $q$ are relatively prime positive integers. Compute the value of $p$.


[b]p15.[/b] $f$ is a bijective function from the set $\{0, 1, 2, ..., 11\}$ to $\{0, 1, 2, ... , 11\}$, with the property that whenever $a$ divides $b$, $f(a)$ divides $f(b)$. How many such $f$ are there? 
[i]A bijective function maps each element in its domain to a distinct element in its range.
[/i]

PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
1. To find the probability that the second largest number in the list of the largest numbers from each row is 2023, we need to consider the placement of the numbers 2023, 2024, and 2025 in the grid.
2. The desired result will occur when 2024 and 2025 are in the same row, but 2023 is in a different row from those two. The placement of the other 2022 numbers is irrelevant.
3. First, note that there are a total of 
   \[
   \binom{2025}{3}
   \]
   ways to place the numbers 2023, 2024, and 2025 in the grid.
4. We place 2024 and 2025 first. There are 
   \[
   45 \cdot \binom{45}{2}
   \]
   ways to choose the row and then choose the exact placement of 2024 and 2025 in that row.
5. Next, we place 2023. There are 
   \[
   44 \cdot 45
   \]
   ways to choose the placement of 2023 such that it is not in the same row as 2024 and 2025.
6. Hence, the desired probability is
   \[
   \frac{45 \cdot \binom{45}{2} \cdot 44 \cdot 45}{\binom{2025}{3}}
   \]
7. Simplifying the expression, we get:
   \[
   \frac{45 \cdot \frac{45 \cdot 44}{2} \cdot 44 \cdot 45}{\frac{2025 \cdot 2024 \cdot 2023}{6}} = \frac{45 \cdot 45 \cdot 44 \cdot 44 \cdot 45 \cdot 3 \cdot 2}{2025 \cdot 2024 \cdot 2023 \cdot 2}
   \]
8. Further simplifying, we have:
   \[
   \frac{45^3 \cdot 44^2 \cdot 3}{2025 \cdot 2024 \cdot 2023}
   \]
9. Since \(2025 = 45^2\), we can cancel out \(45^2\) from the numerator and denominator:
   \[
   \frac{45 \cdot 44^2 \cdot 3}{2024 \cdot 2023}
   \]
10. Simplifying further, we get:
    \[
    \frac{45 \cdot 44 \cdot 44 \cdot 3}{2024 \cdot 2023} = \frac{22 \cdot 45 \cdot 3}{23 \cdot 2023}
    \]
11. Since the greatest common divisor of the numerator and denominator is 1, the answer is:
    \[
    22 \cdot 45 \cdot 3 = 2970
    \]

The final answer is \(\boxed{2970}\)