Mongolian Mathematical Olympiad

Let the distance from the store to home be $1000$ meters.

Let $v$ be John's speed, so the dog's speed is $4v$.

Let $t$ be the time it takes John to walk home. Then $vt=1000$, so $t=\frac{1000}{v}$.

In time $t$, the dog can travel $4v\cdot t=4v\cdot \frac{1000}{v}=4000$ meters.

But the dog first runs home ($1000$ meters), then turns around and meets John somewhere on the way.

Let the meeting point be $x$ meters from the store. The dog runs $1000$ meters home, then turns back and meets John at $x$ meters from the store.

Let $t_1$ be the time until the dog meets John after turning back. The dog runs $1000$ meters home in $\frac{1000}{4v}=250$ meters per $v$ seconds, so $250$ seconds if $v=1$ m/s.

Let $t_2$ be the time the dog spends running back toward John. During this time, John is still walking toward home.

Let $d$ be the distance from the store to the meeting point. John covers $d$ meters in $\frac{d}{v}$ seconds. The dog covers $1000$ meters in $\frac{1000}{4v}$ seconds, then turns back and covers $(1000-d)$ meters in $\frac{1000-d}{4v}$ seconds.

The total time for John to reach the meeting point is $\frac{d}{v}$.

The total time for the dog to reach the meeting point is $\frac{1000}{4v}+\frac{1000-d}{4v}$.

Set these equal:

$$\frac{d}{v}=\frac{1000}{4v}+\frac{1000-d}{4v}$$

$$\frac{d}{v}=\frac{1000+1000-d}{4v}$$

$$\frac{d}{v}=\frac{2000-d}{4v}$$

Multiply both sides by $4v$:

$$4d=2000-d$$

$$5d=2000$$

$$d=400$$

So the meeting point is $400$ meters from the store.

The dog runs $1000$ meters home, then $600$ meters back to meet John.

Total distance: $1000+600=1600$ meters.

Answer: $1600$ meters.