aime26-00

Proposed solution

Let's denote the distance from the school to the park as \( d \) miles. Let \( v \) be the walking speed of Patrick in miles per hour. Then, the speed of Tanya is \( v + 2 \) miles per hour, and the speed of Jose is \( v + 9 \) miles per hour (since he bicycled 7 miles per hour faster than Tanya ran).

Patrick started walking 2 hours before Tanya and 1 hour before Jose. Let \( t \) be the time in hours that Tanya and Jose took to reach the park after they started moving.

Since all three arrived at the park at the same time, we can set up the following equations based on their travel times and speeds:

1. Patrick's travel time: \( t + 2 \) hours
2. Tanya's travel time: \( t \) hours
3. Jose's travel time: \( t - 1 \) hours

Using the formula \( \text{distance} = \text{speed} \times \text{time} \), we can write the equations for each person's travel:

1. Patrick's equation: \( d = v(t + 2) \)
2. Tanya's equation: \( d = (v + 2)t \)
3. Jose's equation: \( d = (v + 9)(t - 1) \)

Since all three equations equal \( d \), we can set them equal to each other:

\[ v(t + 2) = (v + 2)t \]
\[ v(t + 2) = (v + 9)(t - 1) \]

First, solve the equation \( v(t + 2) = (v + 2)t \):

\[ v(t + 2) = vt + 2v \]
\[ vt + 2v = vt + 2t \]
\[ 2v = 2t \]
\[ t = v \]

Next, solve the equation \( v(t + 2) = (v + 9)(t - 1) \):

\[ v(t + 2) = (v + 9)(t - 1) \]
\[ vt + 2v = vt + 9t - v - 9 \]
\[ 2v = 9t - v - 9 \]
\[ 3v = 9t - 9 \]
\[ v = 3t - 3 \]

Substitute \( t = v \) into \( v = 3t - 3 \):

\[ v = 3v - 3 \]
\[ 2v = 3 \]
\[ v = \frac{3}{2} \]

Now, substitute \( v = \frac{3}{2} \) back into \( t = v \):

\[ t = \frac{3}{2} \]

Now we can find the distance \( d \) using any of the original equations. Using \( d = v(t + 2) \):

\[ d = \frac{3}{2} \left( \frac{3}{2} + 2 \right) \]
\[ d = \frac{3}{2} \times \frac{7}{2} \]
\[ d = \frac{21}{4} \]

Thus, the distance from the school to the park is \( \frac{21}{4} \) miles. The values \( m \) and \( n \) are 21 and 4, respectively, and they are relatively prime. Therefore, \( m + n = 21 + 4 = 25 \).

The final answer is:

\[
\boxed{25}
\]