The limit of the straight route
Swimming from the centre straight away from the teacher works only if he is less than π ≈ 3.14 times as fast: the pupil swims one radius and the teacher runs half the shore, which is π radii.
A pupil runs from his teacher and jumps into a round lake. The teacher can't swim, but he runs along the shore four times as fast as the pupil swims, and waits wherever the pupil comes out. Swimming from the middle straight away from him fails: he gets round half the lake in time. Is there any way for the pupil to reach the shore before the teacher? Lock in your answer, then test it in the game.
A pupil runs from his teacher and jumps into a round lake. The teacher runs along the shore four times as fast as the pupil swims, so the straight route doesn't save him. Escape looks impossible. But near the centre the pupil circles the middle of the lake faster than the teacher. That way he gets exactly opposite him, and from there it is 75 m to the shore while the teacher has to run 314 m.
This is a classic pursuit puzzle. Older collections use a duck and a fox, a girl in a rowing boat and a goblin, or a swimmer and a bear instead of a pupil and a teacher. The idea is the same: a slower player who can go anywhere on the lake against a faster one who is tied to the shore.
The lake is perfectly round with a radius of 100 m. The shore is 2π × 100 ≈ 628 m long, half of it 314 m. The pupil starts in the middle of the lake: he could swim there at leisure, since the teacher stays out of the water.
The pupil swims at 1 m/s. The teacher runs along the shore at 4 m/s, four times as fast. On land the pupil runs faster than the teacher: if he gets out where the teacher isn't, he is free.
From the centre to the shore is 100 m, so 100 s. The teacher has to reach the opposite side: half the shore, 314 m, in 314 / 4 = 78.5 s. He is there 21.5 s before the pupil.
Both can see each other all the time and play as well as they can. Is there a way for the pupil to reach the shore before the teacher for sure?
"He is four times faster and reaches even the farthest point in time. No chance"
In fact there is a chance. The teacher is faster in a straight line but not at circling the centre: near the middle of the lake the pupil turns faster. He gets exactly opposite the teacher and from there reaches the shore 3.5 s ahead.
The teacher runs four times as fast as the pupil swims and always heads for the point of the shore nearest to the pupil. Pick an answer before reading on.
A breakdown of your answer will appear here.
Hard to decide? Open the hints one at a time: the easy one first, then stronger ones.
Swimming straight away from the teacher won't work: you swim 100 m, he runs half the shore, 314 m, four times as fast. Look for a place in the lake where you are faster than him at something.
Circles near the centre are short. You swim around a circle of radius 10 m in 63 s, while the teacher needs 157 s to run around the lake. Which of you circles the centre faster?
On a circle of radius just under 25 m you can stay exactly opposite the teacher all the time. How far do you swim from there to the shore, and how far does he run?
Tap the lake or drag across it: the pupil swims there. The clock starts with your first move. The teacher (orange) always runs the short way to the point of the shore nearest to you. On the right you can see who would win if you swam straight to the shore right now. Stuck? Turn on the safe circle or watch the autopilot.
The safe circle has a radius of 100 / k metres. Inside it you go around the centre faster than the teacher.
The computer plays both sides step by step, a thousand steps per chase. The pupil tries three plans: swim straight from the centre, circle and dash to the shore, circle and swim along a tangent. Each time the teacher runs the short way. The dots show how many metres behind the teacher is when the pupil reaches the shore (zero means caught). The lines show the same from the formula in round 4.
| Teacher faster by | Straight | Dash | Tangent |
|---|
In the table: how many metres behind the teacher is when the pupil lands. The thresholds found by simulation are slightly below the exact ones: the simulated pupil circles 0.5% closer to the centre than the edge of the safe circle, and "caught" means the teacher is within half a metre.
The teacher is faster in metres per second. But what matters in this chase is something else: how fast each of them goes round the centre of the lake. Near the middle even a slow swimmer turns faster than a runner on the shore.
“In plain words” is hidden on all puzzles.
This is not the whole answer yet, only its first half: this is how the pupil gets opposite the teacher. The second half is in the next card.
Once the pupil is on the 25 m circle exactly opposite the teacher, he turns and swims straight for the shore. He has 75 m left, while the teacher has to run around half the lake. The pupil is ashore 3.5 s earlier. From the centre this fails: the pupil has the full 100 m to swim, and the teacher gets there in 78.5 s.
Why circle instead of heading straight for the 25 m ring? While the pupil swims out, the teacher runs too and is no longer opposite. Circling brings the pupil back to the point farthest from the teacher.
“Rigorous” is hidden on all puzzles.
The pupil is opposite the teacher at a distance r from the centre. He has 100 − r seconds to swim, and the teacher always needs 78.5 s. The pupil wins if r is more than 21.5 m. But he can stay opposite the teacher only inside the 25 m circle: beyond it the teacher turns faster. So the dash has to start from a narrow ring between 21.5 and 25 m.
| Teacher k times faster | Pupil | Teacher |
|---|
The safe circle has a radius of 100 / k metres, so the pupil swims for 100 − 100 / k seconds and the teacher runs for 314 / k seconds. The pupil wins while k − 1 < π, that is, while the teacher is less than 4.14 times as fast.
A lake of radius R, the pupil swims at speed v, the teacher runs at kv. Angular speed is speed divided by the radius of the circle you move on. The pupil turns faster than the teacher on circles of radius less than R/k.
A radial dash from the circle R/k with the teacher opposite: the pupil covers R − R/k, the teacher πR.
A dash along the tangent to the circle R/k, in the direction the teacher is running. The pupil swims R·sin θ, where cos θ = 1/k, and the teacher has to run π + θ radians.
For k = 4 and a 100 m lake: the dash gives 75 s against 78.5 s, the tangent 96.8 s against 111.5 s. The tangent is longer, but the teacher has to run more than half the shore.
“Deeper” is hidden on all puzzles.
The straight dash challenges the teacher to run half the shore. You can do better: don't turn sharply for the shore, keep going along the tangent to the safe circle, in the same direction the teacher is already running. Then he chases the pupil from behind and has to run more than half the shore. Turning back doesn't pay: the pupil switches to the mirror tangent, and the teacher's run is wasted. This path gives an escape while the teacher is less than 4.6033 times as fast.
Problems where one player flees, the other pursues, and both decide at every instant are called differential games. Rufus Isaacs began studying them at the RAND Corporation: his first report, "Games of Pursuit", came out in 1951, and the book "Differential Games" in 1965. The examples there are military: interception, evasion, aiming.
The lake puzzle appears in Martin Gardner's collections. The 4.6 limit for the tangent path is credited to Richard Guy, and IBM published the full solution with the exact 4.6033 in its Ponder This puzzle for May 2001.
First the maths: how the answer depends on the teacher's speed. Then the idea behind the escape. And finally, where such chases happen for real.
The pupil loses on speed but wins on angular speed near the centre. Mathematicians solve such problems by looking from the teacher's point of view: in a frame that turns with him, the teacher stands still and the lake spins. Then you can see that the pupil can steer the teacher into his worst position.
A squirrel hides from a dog the same way: it runs around a tree trunk, staying on the far side, and the dog, though faster, can't keep up on such a small circle.
A similar puzzle: a lion and a man in a round arena with equal speeds. For a long time it was thought that the lion, keeping on the same radius as the man, would catch him. In 1952 Abram Besicovitch showed the man can run forever: the lion gets arbitrarily close but never catches him. The proof is in "Littlewood's Miscellany".
At RAND Isaacs described the "homicidal chauffeur": a car much faster than a pedestrian but with a poor turning circle, against a slow pedestrian who can turn on the spot. It is the same "speed against agility" game as the lake. Similar models are used for intercepting aircraft and missiles: a faster but less agile pursuer against a slower but nimbler target.
Swimming from the centre straight away from the teacher works only if he is less than π ≈ 3.14 times as fast: the pupil swims one radius and the teacher runs half the shore, which is π radii.
On a circle of radius 25 m the pupil and the teacher complete a lap in the same 157 s. Inside it the pupil turns faster, even though he swims four times as slowly.
The dash from the safe circle gives the pupil 75 s against the teacher's 78.5 s. The margin is just 3.5 s, so the circling has to stay close to the edge of the 25 m circle.
The best path looks like the letter J: a semicircle near the centre, then a tangent. It works while the teacher is less than 4.6033 times as fast. IBM's Ponder This of 2001 named its goblin Jimmy after the shape of the path.
Rufus Isaacs's book "Differential Games", about games of pursuit and evasion, came out in 1965 and won the Lanchester Prize in operations research the same year.
In 2021 Quanta Magazine gave its readers this puzzle with a swimmer and a bear running 3.5 times as fast, and asked for the quickest path to the shore.
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