aime26-00

Proposed solution

I need to find the distance from school to park given the walking, running, and bicycling rates and timing information.

Let me define variables:
- Let $v$ = Patrick's walking speed (miles per hour)
- Then Tanya's running speed = $v + 2$
- And Jose's bicycling speed = $(v + 2) + 7 = v + 9$

Let $d$ = distance from school to park (in miles)

Now I'll analyze the timing:
- Patrick starts at time 0 and walks at speed $v$
- Tanya starts at time 1 and runs at speed $(v + 2)$
- Jose starts at time 2 and bicycles at speed $(v + 9)$

Since all three arrive at the park at the same time, I need to find when each person arrives.

For Patrick:
Time to travel distance $d$ = $\frac{d}{v}$

For Tanya:
She travels for $(\frac{d}{v} - 1)$ hours (since she started 1 hour later)
Distance she covers = $(v + 2)(\frac{d}{v} - 1) = d$
So: $(v + 2)(\frac{d}{v} - 1) = d$

For Jose:
He travels for $(\frac{d}{v} - 2)$ hours (since he started 2 hours later)
Distance he covers = $(v + 9)(\frac{d}{v} - 2) = d$
So: $(v + 9)(\frac{d}{v} - 2) = d$

Let me solve the equation from Tanya's journey:
$(v + 2)(\frac{d}{v} - 1) = d$

Expanding:
$(v + 2)\frac{d}{v} - (v + 2) = d$

$\frac{d(v + 2)}{v} - v - 2 = d$

$\frac{d(v + 2)}{v} - d = v + 2$

$d(\frac{v + 2}{v} - 1) = v + 2$

$d(\frac{v + 2 - v}{v}) = v + 2$

$d(\frac{2}{v}) = v + 2$

Therefore: $d = \frac{v(v + 2)}{2}$

Now let me solve the equation from Jose's journey:
$(v + 9)(\frac{d}{v} - 2) = d$

Expanding:
$(v + 9)\frac{d}{v} - 2(v + 9) = d$

$\frac{d(v + 9)}{v} - 2v - 18 = d$

$\frac{d(v + 9)}{v} - d = 2v + 18$

$d(\frac{v + 9}{v} - 1) = 2v + 18$

$d(\frac{v + 9 - v}{v}) = 2v + 18$

$d(\frac{9}{v}) = 2v + 18$

Therefore: $d = \frac{v(2v + 18)}{9}$

Now I can set the two expressions for $d$ equal:
$\frac{v(v + 2)}{2} = \frac{v(2v + 18)}{9}$

Since $v \neq 0$, I can divide both sides by $v$:
$\frac{v + 2}{2} = \frac{2v + 18}{9}$

Cross-multiplying:
$9(v + 2) = 2(2v + 18)$
$9v + 18 = 4v + 36$
$5v = 18$
$v = \frac{18}{5}$

Now I can find $d$:
$d = \frac{v(v + 2)}{2} = \frac{\frac{18}{5}(\frac{18}{5} + 2)}{2} = \frac{\frac{18}{5} \cdot \frac{28}{5}}{2} = \frac{18 \cdot 28}{2 \cdot 25} = \frac{504}{50} = \frac{252}{25}$

Let me verify this is correct by checking that all three arrive at the same time:

Patrick's travel time: $\frac{d}{v} = \frac{\frac{252}{25}}{\frac{18}{5}} = \frac{252}{25} \cdot \frac{5}{18} = \frac{252}{90} = \frac{14}{5}$ hours

Tanya's travel time: She starts at time 1, so her total time is $\frac{14}{5} - 1 = \frac{9}{5}$ hours
Distance she covers: $(v + 2) \cdot \frac{9}{5} = (\frac{18}{5} + 2) \cdot \frac{9}{5} = \frac{28}{5} \cdot \frac{9}{5} = \frac{252}{25}$ ✓

Jose's travel time: He starts at time 2, so his total time is $\frac{14}{5} - 2 = \frac{4}{5}$ hours
Distance he covers: $(v + 9) \cdot \frac{4}{5} = (\frac{18}{5} + 9) \cdot \frac{4}{5} = \frac{63}{5} \cdot \frac{4}{5} = \frac{252}{25}$ ✓

Therefore, $d = \frac{252}{25}$ miles.

Since $\gcd(252, 25) = 1$, we have $m = 252$ and $n = 25$.

Therefore, $m + n = 252 + 25 = 277$.

$\boxed{277}$