A question with no right answer
On the American SAT, circle A, with a third of the radius of circle B, rolled around B. The choices were 3/2, 3, 6, 9/2 and 9. The setters meant 3, but the right answer is 4, which was not among the choices.
Two identical coins lie on a table. One is pinned down with a finger, and the other is rolled around it without slipping until it is back where it started. How many full turns does the moving coin make? Lock in your answer, then check it: roll the coin yourself, count on a computer and read the explanation.
Two identical coins lie on a table. One is pinned down with a finger, and the other is rolled around it without slipping. The rims are the same length, so it feels like exactly one turn. In fact the coin turns twice: once from rolling, and once more because it goes around a circle itself. The same "extra" turn hides in the sidereal day, in the Moon and in planetary gears.
You can check this puzzle in a minute with two coins from your pocket. But try to answer in your head first: it is more fun to find out whether your intuition matches what your eyes will see.
Two coins of the same size lie on a table touching each other. The lower one is pinned down with a finger and does not move. The upper one has an arrow drawn on it, pointing up.
The upper coin is rolled along the rim of the fixed one, like one gear wheel around another: the rims always touch and never slide against each other.
The coin is rolled until it has gone all the way around the fixed coin and is back in the same place, on top.
How many full turns has the moving coin made? That is, looking down at the table from above, how many times has its arrow gone round a full circle?
"The rims are the same length, so the coin rolls exactly its own length and makes one turn"
In fact there are two turns. Rolling along the rim really does give one turn. But the coin also travels around the fixed coin in a circle, and that lap turns it once more on its own, even if it did not roll at all.
Pick an answer with a button under the slider or type a number. Count only full turns of the arrow relative to the table.
Your answer will be compared with the right one here.
Hard to decide? Open the hints one at a time: the easy one first, then stronger ones.
Put two identical coins on a table and turn the moving one so its head is upright. Roll it halfway, to the far side. Which way is the head facing now?
Split the coin's turning into two parts: how much it turns from rolling along the rim, and how much from going around the fixed coin in a circle.
Watch the centre of the moving coin. What is the radius of the circle it travels, and how many times does the coin's rim fit along that path?
The next rounds count the coin's turns: you roll it yourself, the computer rolls coins of different sizes, and then comes the explanation. Answer round 1 first, or go on?
You chose to always show all rounds.
Drag the coin with a finger or the mouse, or move the slider. Pale copies stay behind after every quarter of the way. Then switch to a straight track exactly as long as the fixed coin's rim, and compare.
"Rim rolled out" shows how much of its rim the coin has used up in contact: along the line and around the coin it is the same, exactly one rim. But the turns relative to the table differ. The difference comes from the lap around the circle itself.
The computer replaces the fixed circle of radius R with a 3600-sided polygon and rolls a coin of radius r along it, outside or inside a ring. On the sides the coin rolls; at the corners it pivots about the vertex. Turns are counted the way a camera above the table would: from the arrow's angle. No formulas here, just rolling.
Outside, the dots lie on the line R/r + 1; inside, on R/r − 1; and none lies on the grey R/r that intuition suggests. At R/r = 3, as on the 1982 SAT (see the facts), the answer is 4. A coin bigger than the ring (R/r = ½) does not fit inside it.
Once you answer, the simulation code appears here so you can run it yourself.
The same simulation, short and with no libraries. Run the JavaScript in your browser console (F12) or in Node; the Python needs only the standard library.
// A coin of radius 1 rolls without slipping around a fixed coin of radius R, outside it and inside a ring. How many turns does the arrow on it make?
// The circle is replaced by a 3600-gon: on the sides the coin rolls, at the corners it pivots about the vertex.
const N = 3600, r = 1;
function roll(R, inside) {
const side = 2 * R * Math.sin(Math.PI / N); // a side of the polygon
// inside, the coin rolls along the polygon shifted in by r and never touches the corners
const track = inside ? 2 * (R * Math.cos(Math.PI / N) - r) * Math.tan(Math.PI / N) : side;
let angle = 0, prev = 0, turns = 0; // the arrow's angle relative to the table
for (let i = 0; i < N; i++) {
angle += (inside ? -1 : 1) * track / r; // rolled a distance track: a turn of track / r
if (!inside) angle += 2 * Math.PI / N; // the pivot about the vertex
// count turns as a camera above the table would: from the arrow's angle
const a = Math.atan2(Math.sin(angle), Math.cos(angle));
let d = a - prev; if (d > Math.PI) d -= 2 * Math.PI; if (d < -Math.PI) d += 2 * Math.PI;
turns += d / (2 * Math.PI); prev = a;
}
return Math.abs(turns);
}
for (const R of [1, 2, 3, 4, 6]) {
console.log('R/r =', R, 'outside', roll(R, false).toFixed(4), 'formula', R + 1,
'· inside', roll(R, true).toFixed(4), 'formula', R - 1);
}
# A coin of radius 1 rolls without slipping around a fixed coin of radius R, outside it and inside a ring. How many turns does the arrow on it make?
# The circle is replaced by a 3600-gon: on the sides the coin rolls, at the corners it pivots about the vertex.
import math
N, r = 3600, 1.0
def roll(R, inside):
side = 2 * R * math.sin(math.pi / N) # a side of the polygon
# inside, the coin rolls along the polygon shifted in by r and never touches the corners
track = 2 * (R * math.cos(math.pi / N) - r) * math.tan(math.pi / N) if inside else side
angle = prev = turns = 0.0 # the arrow's angle relative to the table
for _ in range(N):
angle += (-1 if inside else 1) * track / r # rolled a distance track: a turn of track / r
if not inside:
angle += 2 * math.pi / N # the pivot about the vertex
# count turns as a camera above the table would: from the arrow's angle
a = math.atan2(math.sin(angle), math.cos(angle))
d = (a - prev + math.pi) % (2 * math.pi) - math.pi
turns += d / (2 * math.pi)
prev = a
return abs(turns)
for R in [1, 2, 3, 4, 6]:
print("R/r =", R, "outside", f"{roll(R, False):.4f}", "formula", R + 1,
"· inside", f"{roll(R, True):.4f}", "formula", R - 1)
Intuition counts only the rolling: how many times the coin's rim fits along the fixed rim. But the coin also travels in a circle, and anything that goes around a circle turns along with it.
“In plain words” is hidden on all puzzles.
Roll the coin to the far side, to the bottom. It has rolled out only half its rim, yet the arrow points up again: a whole turn is already done. So a full lap gives two.
Why? Half the rim turns the coin half a turn relative to the fixed one: the point that first touched it is now on the side facing away from it. But the fixed coin itself, seen from the moving one, is now on the other side, above instead of below. That is another half turn.
Roll the coin around a triangle. Along each side it rolls as on a straight line and turns exactly by how much of its rim the side takes up. At each corner the coin does not roll; it swings around the vertex like a door on its hinge. The three swings (the pink arcs) add up to exactly one full turn, because after going all the way around, the coin faces the shape with the same side again.
A circle is like a polygon with a huge number of tiny corners. You can't see the corners, but together they still give their one turn. That is exactly how the computer counts in round 3.
“Rigorous” is hidden on all puzzles.
Rolling along a line, the coin turns once each time its centre covers 2πr, the length of its rim. The same rule holds on a convex path: the angle turned equals the length of the centre's path divided by r. At the triangle's corners the centre follows a small arc of radius r, and the turn there is also "path / r".
Around a coin of radius R, the moving coin's centre travels a circle of radius R + r. Its length is 2π(R + r). Divide by the rim 2πr and you get (R + r)/r = R/r + 1 turns. For identical coins the centre travels a circle of radius 2r, twice the rim: 2 turns.
| R/r | centre's path | rims around | turns |
|---|---|---|---|
| 1 | 2π · 2r | 1 | 2 |
| 2 | 2π · 3r | 2 | 3 |
| 3 | 2π · 4r | 3 | 4 |
| ½ | 2π · 1.5r | ½ | 1.5 |
Now the coin rolls along the inside of a ring of radius R. Its centre travels a circle of radius R − r, so it makes (R − r)/r = R/r − 1 turns. The lap still gives one turn, but rolling along the inside spins the coin the opposite way, so the turns subtract.
Inside a ring twice the coin's size there is only 1 turn, even though the ring's rim equals two coin rims. And in a ring of the same size the coin just sits there: 1 − 1 = 0. Outside the lap adds a turn, inside it takes one away, while the rolling is the same in both cases, R/r.
Outside: the moving coin's centre travels a circle of radius R + r, and every 2πr of path gives one turn.
Inside a ring: the centre travels a circle of radius R − r, and the coin spins the other way.
The same as a sum of two parts: rolling along a rim of length 2πR, plus the line joining the two centres turning a full circle.
Any convex shape with perimeter P: rolling on the sides, swings at the corners that add up to 2π.
Intuition takes only the first term, R/r. It really does count how many times the coin's rim fits along the fixed rim. The second term depends on neither size nor shape: it is one full turn of the direction to the centre per lap.
“Deeper” is hidden on all puzzles.
"How many turns" only makes sense together with "relative to what". Imagine an ant sitting on the fixed coin and always facing the moving one. For the ant the moving coin turns exactly R/r times, just as intuition promises: the ant itself turns along with the line between the centres. A camera above the table stays put and sees that line's turning too, one per lap.
Angular velocities in different frames add up: turns relative to the table = turns relative to the line of centres + turns of the line itself. Intuition's mistake is not arithmetic; it quietly counts in the ant's frame while the question is about the table. Astronomers see the same one-turn difference between the sidereal and the solar day (round 5).
Why does "the lap" give exactly one turn for any convex shape? Because the tangent to a simple closed curve turns exactly 360° on the way around. For smooth curves Heinz Hopf proved this rigorously in 1935 (the turning tangent theorem, Umlaufsatz). For a polygon it is the familiar school fact that the exterior angles add up to 360°.
And "without slipping" matters. The ancient "Mechanics", ascribed to Aristotle, describes a wheel with two rims of different radii on one axle: in one turn the big rim rolls out its own length, yet the small rim covers the same distance, though it is shorter. The answer is that the small rim does not roll cleanly but partly slides. Galileo discussed this paradox in his "Two New Sciences" of 1638.
"Plus one turn per lap" does not depend on the shape, carries over to the sky and works in gearboxes. And a point on the coin's rim draws curves with names of their own.
A square, a triangle and a coin with the same perimeter, equal to the moving coin's rim, all give the same 2 turns: one from rolling along the perimeter, one from the lap. The shape only changes where the coin turns: in jerks at the corners, smoothly on a circle.
A point on the rim of a coin rolling outside a circle draws an epicycloid with R/r cusps. With R = r it is a cardioid, a "heart" with one cusp; with R = 2r a nephroid, a "kidney" with two, the curve light draws at the bottom of a cup. Inside a ring you get hypocycloids: with R = 3r a deltoid, with R = 4r an astroid, and with R = 2r the point moves along a straight line. This "Tusi couple" was described by the Persian astronomer Nasir al-Din al-Tusi in 1247.
The Earth is a coin "rolling" around the Sun. Relative to the distant stars it turns in 23 h 56 min 4 s. But in that time it moves along its orbit, and for the Sun to be overhead again it has to turn about 4 more minutes. Over a year these minutes add up to a whole turn: relative to the stars the Earth turns about 366.24 times, but there are only 365.24 solar days. The Earth spins the same way it goes around the Sun, so the lap takes one day away, as for a coin inside a ring.
The Moon always shows the Earth the same face, so relative to the Earth–Moon line it does not turn at all. Yet relative to the stars it turns once per lap, every 27.32 days. That is 13.37 turns a year, while full cycles of phases (new moon to new moon, 29.53 days) are one fewer, 12.37, because the Earth and Moon are themselves going around the Sun. Each time the difference of one comes from the same place: counting turns in different frames of reference.
Teeth are "no slipping" made of metal. In the middle is the sun gear, planet gears roll around it on a carrier, and an outer ring with teeth on the inside encloses everything. The numbers in the table can be checked with Willis's formula: (ωsun − ωcarrier) / (ωring − ωcarrier) = −zring/zsun, where z is the number of teeth.
| Device | What is held | What comes out |
|---|---|---|
| Two identical gears with an arm between their axles | one gear | The other one, carried around it on the arm, makes 2 turns: the coin puzzle with teeth. In general a planet makes 1 + zsun/zplanet turns per turn of the carrier. |
| A 24-tooth sun, a 72-tooth ring | the ring | The sun turns 1 + 72/24 = 4 times as fast as the carrier, not 3, even though the ring is three times bigger. It is the same "3 + 1" as on the 1982 SAT. |
| The Sturmey-Archer AW three-speed bicycle hub | the sun | In third gear the carrier drives the ring, and the wheel turns 4/3 as fast: 1 + zsun/zring = 1 + ⅓. In first gear it is the reverse, ¾. Second is direct, 1 : 1. |
| A car's open differential | one wheel (on the ground, with the other lifted) | The wheels are linked by ωleft + ωright = 2ωcase. If one wheel stands still, the other spins twice as fast as the differential case: one turn with the case and one more relative to it. |
On the American SAT, circle A, with a third of the radius of circle B, rolled around B. The choices were 3/2, 3, 6, 9/2 and 9. The setters meant 3, but the right answer is 4, which was not among the choices.
Three New York high-school students spotted the error independently and wrote to the organisers. The question was dropped and the maths section of about 300,000 test takers was rescored; scores could change by up to 10 points on the 200–800 scale. According to Education Week, it was the fourth error found in the SAT in two years.
Relative to the stars the Earth turns in 23 h 56 min 4 s. So every star rises about 4 minutes earlier each day than the day before, and over a year the sky makes one "extra" turn.
The Moon shows the Earth one face, yet relative to the stars it turns 13.37 times a year. There are only 12.37 new moons a year: the difference is exactly one, the same turn from the lap around the Sun.
Six coins are laid in a ring like petals, and a seventh is rolled around them. It makes 4 turns: by each coin it rolls along a 120° arc and turns ⅔ of a turn, and 6 × ⅔ = 4.
The bright curve at the bottom of a cup on a sunny day is a nephroid, the trace of a point on a coin rolling around one twice its size. The name "nephroid" ("kidney-like") was given to this curve by the English astronomer Richard Proctor in 1878.
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