Just to show you how hard the SATs were, here’s a question (sent by reader Bryan) that I got wrong. I guessed three, assuming that the circumference of the smaller circle (1/3 of the larger), would translate into rotations needed to get around the circumference of the larger circle. How I got a perfect score in 1966 on the math part of the SAT defies me!

At any rate, the correct answer was not one of the SAT’s choices, so nobody got the question right.

Stop the video at 55 seconds in, make your guess, and then watch on.

This trick will make you the hit of any cocktail party—that is, if the cocktail party is full of academics.

ADDENDUM: This gif shows you that a coin rotates twice going around a coin of equal radius. Follow the blue dot from where it starts: you’ll see that it’s gone around once when it’s halfway around the central coin.

No. The area is proportional to r^2. The diameter is proportional to r. In fact, the diameter of a circle is always 2 x r. Even the circumference, which I think is what you meant, is proportional to r. It’s 2 x π x r.

To be pedantic, because I am a bitter old (ok, late middle aged) man:

The diameter is proportional to r with a constant in a space of uniform curvature. In a plane, D=2r. If the curvature is no zero, then, depending on the postulate involved, the diameter may differ, and the circumference certainly will.

Which obviously doen’t apply to the SAT, but, given the answers provided, might as well. I am of an age that I sat this xam, and remember the question, and remember there being no answer, and being told that I was obviously mistaken as the test designers were professionals. Today, I would contact them. At that time, there was not way to resolve such issues. Not that I am bitter (I was the only one I didn’t get and it ruined my score)

How are you defining “diameter”? Given that a circle is the set of points at a given distance from the centre, the radius must be constant and therefore the diameter also given that it must pass through the centre.

A simple way to see the origin of the extra rotation (giving 4) is to think of the moon which is tidally locked to the Earth, so that the same side faces Earth. It clearly rotates once when moving once around the Earth. So the answer is 1+3=4.

There are 366.24 sidereal days (periods between a distant star crossing your observatory’s meridian) in a year.
As is well known and deeply confusing to anyone setting out to chart the wine-dark seas of celestial mechanics.

That joke doesn’t work in spoken form. The number “ten” is a base-ten quantity and isn’t used in other bases, certainly not in base-two.

You can write that joke just fine: “there are 10 types of people…” Because you can read “10” as being “ten” (if you assume base ten) or as “one zero” in which case it could be read as binary.

There are 10 types of people in the world:
People who understand binary,
People who don’t understand binary…
…And people who weren’t expecting a joke in base three!

My problem with your joke is that it doesn’t divide the set of all people into a partition. The intersection of some of subsets is non empty. In fact, I would argue that “people who don’t understand binary” is a subset of “people who weren’t expecting a joke in base 3”.

Except that the answer isn’t 3. If you cut the larger circle and laid it out straight, *then* the answer would be 3. However, the smaller circle makes an additional rotation as it goes ’round the bigger circle, hence the total number of rotations is 4.

Except that the answer isn’t 3. If you cut the larger circle and laid it out straight, *then* the answer would be 3. However, the smaller circle makes an additional rotation as it goes ’round the bigger circle, hence the total number of rotations is 4.

After having gone through a spell when this problem drove me crazy, I used it to drive my own students crazy. Good for all of us.
Looking at the extreme case of zero rolls around the big circle helped. Using coins, let the outer coin go around the other coin so it always presents the same face to the other coin’s center — so with a penny, keep Abe’s eyes always looking at the center of the other coin. (Just as the moon always — more or less — presents the same face to the earth.) In doing so, you’ll find you have to keep turning the small coin. It will have gone through one rotation in the course of doing one trip.

Do USian coins have serrated edges? Sorry – the proper term is “milled”, I think. Until a few years ago, this was the norm for UKian ones, but I don’t think any other than the 5p and £2 coins now have milling. I’d have to dig out the coin collection to check the £5 and £20 coins. And I’m wondering why we don’t have a £10 coin.
(I checked – the £5, £20, £50 and £100 coins are milled. I couldn’t find out for the 1kilo coins proposed in gold and silver, but I bet they’re milled too.)

“If you find it difficult to grasp how things work on a circle, you might also want to imagine what would happen on a square. When a circle rolling along the outer periphery of a square encounters the first corner, it will have to rotate an “extra” 90° to continue along the next side. This will happen again at each corner, and since 90° x 4 = 360°, this accounts for an additional full revolution.”

I like this a lot — and it leads to what I think is a nice, instructive generalization. Replace the square with a regular hexagon. Now it will turn through six 60-degree angles (the congruent exterior angles at each vertex). Keep replacing the polygons with even more-sided regular n-gons. This is approaching the limit of a circle, just what we have in the original problem.
Don’t really need the regular-polygon stipulation except to see the circle as the limit. With a little finagling don’t even need a convex polygon.

I also like this explanation. The whole problem is akin to the _Around the World in Eighty Day_ plot twist at the end, where Fogg realizes he’s gained a day.

I think the demo with the cut-outs is deceptive. Doesn’t the ‘Circle A’ circle need to spin until it’s base touches again. Picture them as gears instead. Correct me if I’m wrong…

OK, I was somewhat alerted to the fact that there is a trick by PCC(E)’s “the correct answer was not one of the SAT’s choices” but I immediately figured out the correct answer.

There’s several ways of looking at it, but one way is to imagine the radius of circle B as being 0. Then you are rolling circle A round a point. That tells you where the extra rotation comes from.

It is easy to visualize the extra turn if you put an imaginary mark on the circumference of the small circle and keep track of it mentally as you roll the small circle around the large one.

The start/end point of Circle A is the point that represents the intersection of the two circles. Mark that point on Circle A. Therefore, one full rotation of Circle A will be when that point is once again at the intersection of the two circles. That means there will be 3 rotations of Circle A. It should have nothing to do with the orientation of the text ‘Circle A’.

I want to add that the way it is described in the video is NOT how you determine gear ratios. The gear ratio in this example is 3:1, not 4:1. In a simple two-gear plane train, you take the number of teeth of the driving gear and the number of the driven gear, then divide. The number of teeth represents a full rotation of the gear (circle).

I skip-watched watched the rest of the video past 45 seconds, so didn’t see the reference to gears.

HOWEVER, in a gearbox with fixed-position shafts (like a manual gearbox) then 3:1 is indeed the right gear ratio. In an epicyclic box (as with most autos), the gears do effectively roll around each other and frankly, trying to comprehend what their ratio is then just messes with my head. 😉

Lots of confusion here. If the larger circle were replaced by a straight lie of the same length the answer would be 3. If it rolls with out slipping once around and closed figure (doesn’t have to be a circle) you have to add one rotation just for moving around the figure without rolling. 3+1=4 for any closed figure with 3 times the circumference.

Yes, I’d like to know that. And also how difficult it generally is for a test taker to successfully contest a given test question. (I suppose the test generators could have included the answer, “none of the above,” eh?)

As ‘rgbowman’ above points out, 3 is a perfectly reasonable answer. If you define a full rotation of the smaller coin as when the initial point touching circle B again touches it (in the video cardboard demo this corresponds to the point on the circumference directly below the text ‘circle A’ when it is pointed right way up, you will see it only touches the bigger circle 3 times). In the comments, this is called looking at it form the frame of reference of the smaller coin. The poster of the video admits this a question of frames of reference and 3 is correct in that case and the test makers admitted a mistake because the used the wrong word (revolution, rotation..).

Most likely some minutes after the test centres closed and the sitters extruded into the corridors. (I assume the tests are synchronised across the time zones – that must make setting exams in Russia challenging, with 9 time zones to accommodate.) Of course, they might have been able to adjust the marking scheme to just ignore the question, as you can’t infer the intention of a “null” answer in a multiple choice test, unless that is defined in the marking scheme.

No, I think you’re right. The question asks how many times it “revolves”. It only revolves once. It might “rotate” four times, but it only revolves once.

Going by what Jerry wrote, he indicated ‘rotations’ not ‘revolutions’. So I assume I’m wrong.

Addendum – I didn’t watch the video, and I had a brain block that day, but I knew there needed to be an *extra rotation* (key phrase – to get back to its ORIGINAL position). I couldn’t vocalize my initial thought ‘however many rotations plus the extra one’. Dunce day.

THANKS! So Jerry incorrectly described the problem, when he said ‘rotations’. I think I might have been thinking of revolutions, but now I can’t even remember.

I figured it had to be four, because for each full circumference of the small circle that was unrolled, it was also rotating a third of the way around just from following the surface of the larger circle.

Then I saw no four in the multiple choice answers, and wondered what I did wrong, before finishing the video to see what exactly they meant by everyone getting it wrong.

I solved this only because of Jerry’s hint that none of the proffered answers was correct.

I viewed it as follows: imagine the larger coin is unwound into a straight line and then you roll the smaller coin along the line. That gives 3 rotations. But then because of how the smaller coin moves around a _circular_ larger coin, you get one more rotation (as the wikipedia illustration shows).

I also got 800 on the math SAT (in 1980) and on the math GRE (in 1988). I think you don’t have to get a perfect score to get an 800 though.

What do you mean by rotation? Let’s forget Einstein and look at it from the point of view of the fixed stars. Define a radius of the moon, say from its center to some well known geological object, call it X. Then let the moon move around the Earth without the radius CX changing position. Can the moon be said to rotate? I would say not.

‘Geo’ to me implies Earth. Which makes no sense in the context.

If you took a line of direction from the moon to a fixed star, and the moon went round earth without that line moving, then I’d say it doesn’t rotate.

But then, in a motor, connecting rod big ends go round the crankshaft (in a circle about the crankshaft centreline) with no net rotation – they ‘wobble’ one way then the other.

I wonder if the semantics of these kinds of words was what the test writer was going for – but it should not have been, as it was not the verbal section.

Sorry, not geological. I mean to a radius from the moon’s center to some fixed point on its surface, like a mountain or crater. Then keep this radius parallel to itself (relative to the “fixed” stars) as you move the moon around the Earth. The moon must seem to have rotated once according to this example, but that parallel axis tells me differently. Sorry to belabor this, but I don’t get it — geometrically.

You’re missing the fact that there is a point on the Moon’s surface which is, to a first approximation, always the closest point to the Earth. So, as the Moon moves around it’s orbit and that point stays at the sub-Terran point, then the rest of the Moon must rotate in an Earth-centred frame of reference.
to a second approximation, that sub-Terran point moves by around 60km through each orbit because neither the Earth nor the Moon, nor the gravitational potential of the Solar system are perfectly smooth.

A BRAIN BUSTER PLANET MERCURY FACT: A Mercury day is twice as long as a Mercury year.

DEFINITIONS:

One Mercury year = time for Mercury to orbit Sol

One Mercury day = time for Sol to return to the same place in the sky from the POV of a ground observer

NOTES:

Sidereal vs. Solar Day:

It used to be thought that Mercury is tidally locked to Sol, with the same face always aimed at Sol – in the same way as the Moon always shows the same face to Earth [with a slight wobble actually]

But, Mercury is now known to be in 3:2 orbital resonance with Sol. This means that the planet completes three rotations on its axis for every two orbits it makes around Sol – It takes 58.646 EARTH DAYS to complete a single rotation on its axis & 87.969 EARTH DAYS to orbit Sol.

While this might lead some to conclude that a single day on Mercury is about 58 EARTH DAYS – thus making the length of a day and year correspond to the same 3:2 ratio – this would be inaccurate. Due to its rapid orbital velocity and slow sidereal rotation, a Solar Day on Mercury (the time it takes for Sol to return to the same place in the sky) is actually 176 EARTH DAYS.

I didn’t read the post before reading the question, so I was wondering why what I thought was the correct answer wasn’t among the choices. I don’t remember any of the math courses I took, but I figured this out intuitively. I figured if I added the radii of the two coins, that would come to 4/4. So, if circle A is 1/3 of circle B, it’s also 1/4 of the total, and therefore would revolve once for each quarter, giving the answer of 4.

It seems a lot of people made it more complicated than that with equations. Does my solution not work in other cases where it’s this exact question but with different values? Because the logic I used led me to the correct answer.

Yes it does work, without equations. Only ‘maths’ you need is knowing that circumference is directly proportional to radius. (It’s got a Pi in it but that’s for both circles so cancels out).

And it works the same for any other ratio of diameters. e.g. 5:1 ratio will give 6 rotations.

Only ‘maths’ you need is knowing that circumference is directly proportional to radius. (It’s got a Pi in it but that’s for both circles so cancels out).

You can buy tape measures printed in “pi-feet” – I’ve seen them, but rarely. Their use case is in places like pipe stock-yards, where you throw the tape around the circumference of the pipe, shuggle it tight like a tailor being “flattering” (why am I thinking of Trump’s ridiculous dinner suit?) and read off the diameter of the pipe directly from the tape. Check the paint marks for the steel grade, and double-check on the serial numbers.)
They’re rare – tapes printed in decimal feet are much more common (feet, tenths of feet, hundredths of feet).

I did not look at any other comments, and haven’t looked at the video yet.

Well, 3 times certainly “seems” like the correct answer substituting for the r of the larger circle 3 times the radius of the smaller. I don’t understand why it is not.

But a shot in the dark. 4 times?

In my older age, I’m more patient and determined (and enjoy for its own sake) learning rudimentary calculus. Observing that increasing the radius to 3, as compared to 1, increases the area by a factor of 9, I wondered what happens if one integrates 2(Pi)r. One gets (Pi)r^2. And I chose the “limits” of 1 and 3. I got 8(Pi) as the answer. For the smaller circle, if r = 1, then the circumference was 2(Pi). 8(Pi) divided by 2(Pi) = 4 times the smaller circle makes a complete revolution around the circumference of the larger circle.

But this seems to be nothing but an act of faith on my part, and my rotator cuff aches from having to reach so far. Without a head’s up from PCC(E), I would have answered 3 without any doubt or hesitation. And as much time as I spent on this attempt, I would have been well on the way to a mediocre SAT score.

Now to look at the comments and face the reality of my modest math skills.

The video narrator says that “n + 1” is the answer. (Is that only for the specific situation where the radius of the smaller circle is 1/3 that of the larger?) But short of a mathematical proof, I have to accept that answer on faith, it seems. Unless a test taker memorized that factoid, it seems s/he’d have to do some kind of mathematics to get the answer.

Don’t need maths, just need to visualise it. But I was forewarned from PCC’s comments that there was something tricky about it – which turned out to be the extra rotation. But ‘four’ was not in the options listed as answers at 45 seconds in the video.

It’s a bit like: Sam is in a race. He overtakes the second-placed runner. What is his position now?

I call this sort of question ‘Hidden in plain sight’.

That’s fine, we can do all the mathematical derivations you want. But it’s just not necessary. (And, note, I didn’t do any experiments, other than a ‘thought experiment’ visualising the small circle rolling around the big one).

Here’s another one: Rotate the big circle once anticlockwise, holding the axis of the small circle fixed. The small circle will rotate three times clockwise (because the edges are in contact, and one circumference is three times the other). Now glue the circles together at their contact point and rotate the big circle once clockwise, with the small circle ‘riding’ round it; the small circle will also do one clockwise rotation. Total rotation of big circle: Zero (back at starting point). Total rotations of small circle: Four.

I could get all trig/calculus and calculate the rate of rotation (relative to a fixed azimuth) of the small circle relative to the rotation of its centrepoint about the centre of the big circle (which I’m sure would turn out to be 4 times) but quite honestly, I’m vegging out in front of the TV and I just can’t raise the energy.

Actually you can use that same trick for epicyclic gears.

Step 1: First rotate the input shaft AND the whole gearbox including the house over +360 (“glue everything together”): obviously all wheels have made 1 rotation now.

Step 2: Keep the input shaft (with the planet wheel carrier) fixed, unglue, and turn back the ring over -360. Because the planet wheel carrier is now fixed, it is easy to see what the sun wheel (output shaft) will do.

Then add the rotations of step 1 and step 2.

Thus the gear ratio (for speed up) is 1 + d_ring / d_sun (the diameter of the planet wheels doesn’t matter)

“..we can do all the mathematical derivations you want. But it’s just not necessary. (And, note, I didn’t do any experiments, other than a ‘thought experiment’ visualising…”

It is simply a fact of mathematics that visualizing geometric (and also kinematic) behaviour IS PART OF MATHEMATICS.

We find it frustrating that many people (not you I’m sure–I’m just criticizing a choice of words) think that math involves nothing but facts about numbers.

While I’m at it, there is an obverse aspect also, where facts that have everything to do with an arbitrary human choice of base 10 (and now sometimes base 2), i.e. some peculiarity about digits, has anything mathematical to do with numbers, when it has only to do with names for numbers!! That is, linguistics, not mathematics.

Concur. Watching the gif, it looks like it traces out some second or third cousin to an ellipsoid, kinda like a vertical cross section of an apple lying on its side.

Wait, what’s the trick in your question? If Sam just overtook the second-placed runner (and assuming there’s no trick here that his position has changed further since then and your asking of the question), Sam is now in second place, right?

I assume by “second-placed,” you mean the runner who is currently is second place. It’s not exactly how we would say it here.

The trick is that many people instinctively assume that, since he’s ahead of the (erstwhile) second-placed runner, he must be first. Without, of course, reflecting that his overtaking has changed the order.

In the case of the circles, the missing piece of the puzzle is that the small circle has rotated an extra time in going around the big circle.

There are hundreds of them: Port Arthur has six-foot tides. A ship ties up at the wharf at high tide and drops a rope ladder over the side. There are eight rungs showing above the water. Now the tide goes out. If the rungs are spaced nine inches apart, how many rungs are showing at low tide?

I don’t understand why someone wouldn’t account for the fourth rotation, but perhaps that’s because of the way I came to my conclusion and because of the way my brain works. Conversely, I often have trouble with things that even very below-average people don’t. For example, when screwing in something, people say, “lefty loosie, righty tighty.” It probably took me until age 25 or so to make sense of this because turning each way seems both left and right to me. Screwing to the right also seems like left to me because the bottom of the screw is rotating leftward. It took me a long time to figure out that I had to always look at it from only the top of the screw or the way only the top of my hand was turning. Before I figured out a way to compensate, I would always try to screw things in and out the wrong way (you could do a lot of innuendo work with that sentence!).

It’s because I have a learning disability involving the processing of visual information. I’m in the 26th percentile when it comes to spatial reasoning! Kinda crazy, huh? Have you ever taken an IQ test? You know the part where you have to take blocks that are half red and half white split down the middle diagonally, and form the picture they give you? I could only get past the second one, and even the second took me a significant amount of time. Oddly enough, I was still able to play sports that involve split-second reaction times and decision-making, like ice hockey and tennis.

I don’t know if my brain managed to compensate for it’s impaired processing of visual information or if it’s just a coincidence, but I score in the 99th percentile in everything else. When it comes to spatial reasoning and reading speed, I’m far below average (though I would obviously be above average if they tested reading speed/comprehension of, say, highly complex studies or something). I imagine my brain must be compensating in some way, considering how quickly I can process the speed and movement of a hockey puck and the positions, movements, and speeds of all the players on the ice.

I chuckled empathetically with your left-right confusion, BJ. My parents teased me for years about my youthful comment “It depends on which way you’re facing…”

Incidentally, I always have trouble with those IQ test questions that have a series of figures, then you have to choose the last figure in the series. I always read far too much complexity into the figures, at least I think I do. They look like visual gobbledegook to me.

Like this one:

Wtf???

If I can ever make anything out of it, I usually find several different alternatives for the final one depending which features one gives significance to.

@merilee: Right?!? My dad would always say something like, “now screw it in. Lefty loosie, righty tighty, remember? What are you doing?” And when I’d explain it to him, he’d just shake his head and say, “I’ll never get how your brain works.” He’s said that in many different situations…

@infiniteimprobibilit: Not as far as I know. My brain just doesn’t process most visual information properly. It’s also one of the reasons I haven’t read many books: it takes too long.

I think the problem with left / right rotation is knowing the direction from which one rotates – so if you imagine your field of view, I think “left” assumes rotating from the top of the view – almost overhead. Not from the hips. Thus, the right arm would arc over to the left, in order to rotate an object like a wrench to the left, effecting a loosening if a nut… (a right handed nut, in fact)… facing the nut, that is… not the hex head…. is anyone still reading this?…. I’d give you a $10 bill if I could.

I think I discussed this here, perhaps with Sturtevant.

The youthful Merilee has it exactly correct, as does BJ; these quick descriptions are simply incomplete, and I have seen many of the brightest of mathematics students complaining about poorly written texts in exactly that sense.

As an illustration, suppose at a beer party someone draws a circle clockwise on the transparent glass table, and says he/she has done that. But the big boozer who had passed out, fell on the floor under the table looking up, says “No, that was counterclockwise!”.

Oh, and the answer to your last riddle is obviously still eight, as the entire boat rises and lowers with the tide, so the ladder doesn’t sink any further below the water.

Yes, but it’s surprising how many people get tied up with working out how many nine-inch rungs there are in six feet and don’t notice the hidden trick.

On reflection, it looks to me like the SAT question setter intended it as a very simple question (answer: three, since circumference – which is what is in contact – is directly proportional to radius), and completely missed the hidden trap in it.

What could test takers who saw the ‘right’ answer do? Write it in?

Here might be another way to look at this, though it would be easier to look at with a drawing. Also, for the sake of simplicity, let’s make the ratio 4:1 rather than 3:1 (so the temptation in this version is to answer four rotations rather than 4 + 1 = 5).
Imagine the outer circle is a penny with the initial position with the coin on top and Abe looking off to the right at a very distant object, say a star, that is so distant that Abe is looking at it whenever his line of sight is off to the right along any line parallel to his initial line of vision. Also imagine a set of perpendicular axes through the center of the big circle with a vertical axis going through the initial point of contact on the penny. Call that point on the penny X. The axes would divide the figure into four quadrants.
Start the coin rolling. Whenever Abe sees the star he would say — correctly — that the coin has made one rotation from his last view of the star. When is his first new sight of the star? It would come somewhere IN the first quadrant, BEFORE X has reached the boundary between the first and second quadrants. His second view would be in the 2nd quadrant, his 3rd in the third quadrant and his 4th in the fourth quadrant. His fifth view would come just as X arrived back at its starting point for its fourth contact with the big circle. Abe would say — honestly — that he had made five rotations.

I think the notion of a direct 1-to-1 mapping of each point of the circumferences is false because that would mean a transformation from a circle to a straight line and back. I think that is non trivial. The gif of the unit circle that illustrates where “2pi” comes from cannot be extended to two circles willy nilly – because a transformation is necessary.

Hoping to hear the more mathematically enlightened help me out here.

I got it, finally, by watching the addendum GIF, with two equal circles, for some time.

Draw a straight line with a length equal to the circumference of the (small) circle. Place the small circle at the left and and define the point of contact on the circle at C, the left end of the line. Now roll the circle to the midpoint of the line. The point C is now at the top of the circle and diametrically opposed to the new contact point. The circle has made a half rotation and so will rotate once as it rolls to the end the (straight) line. Now leave the circle attached to the center of the line with point C on the side opposite the line and drag the line by one end, leaving the other end fixed, so as to attach the two ends of the line and turn it into a circle too. The small circle has now been dragged around so that point C is on the downward side of the “small” (they’re really the same size) circle. The small circle has not moved relative to the line, but in space it has made a full rotation, the extra half of which is due just to the curving of the line, which drags the circle with it.

So we see that equal circles imply two rotations. It remains to be shown that n+1 is always the answer, rather than 2n. We can leave that as an exercise for the interested reader. 😉

I kinda see the resemblance. The twist in the tail of that did depend on the date line, or at least the one-day difference in dates caused by circumnavigation.

It’s really a relativity thing. If you were some tiny creature who lived on the circumference of the larger circle, you wouldn’t see that curved line or the extra rotation, any more than we sense the curvature of the surface of the Earth.

Please excuse the length; I could show you all this with diagrams much faster.

Also, after (b) below, I am moving the small circle to the right and downwards, not to the left. So those who do it the other way must either adjust or change a couple of my details. To repeat, the small circle here goes clockwise around the big one.

I think that what is fundamentally confusing is this:

(a) Each point on the circumference of the small circle, as it rolls around the big circle, actually only ‘touches down’ on the big circle just 3 times.

And yet,
(b) the small circle does rotate 4 times.

So, for this reason I think, the correct answer of 4 can be quite confusing.

There is yet another, I think simpler way of explaining (a) and (b) above, in particular a way to see (b), the solution:

Draw 3 radii of the big circle, equally spaced, the first being straight up from the centre to where the small circle touches that big one. So the second is obtained from rotating the first radius ⅓ the way round the big circle, i.e. rotating it 120 degrees; and the third big radius another 120 degrees, i.e. 240 degrees in total. Notice that another 120, i.e. 360 total, gets you back to the first radius.

Now roll the little one round, stopping to think three times.

First stop when the little circle touches the big one at the end of that second radius above. The length travelled is the circumference length of the small circle, so every point on the small circle has touched once. Now set out again and stop when the small one touches at the end of the third long radius. Again the little circle has now had each of its points touch a second time. And finally a third time, so we have now established (a).

Now go back and look at that motion again, this time seeing what happens to a small radius we draw for the small circle, from its centre down to where it touches the big one in the diagram. Notice that radius is directly lined up with each of our big radii at the three points. The first time we stopped, that little radius points leftwards and up, actually having rotated completely one and one-third times. Need to think a bit there, but it’s pretty obvious, and really we’re finished:

Why? Well the other two rolls are clearly symmetric at least in the sense that they have the little circle rotating also one and one-third times (480 degrees if you prefer expressing it that way). But three copies of that number added together give our magic number 4! (Or 3 x 480 = 1440 degrees, which is also 4 x 360 of course).

And the ‘same’ argument (essentially) gives the (n+1) answer in the general case mentioned.

Actually, when n=1 at the end there, that argument is harder to follow than is all the cases of bigger “n”s. That’s Jerry’s addendum I just noticed, and the wiki video sort of illustrates exactly that argument by showing the small radius ‘doing its thing’ that I described. But they go counterclockwise. Must be commies or something!!

Also ‘gears’ is a kind of illustration of the same thing happening, a kind of conservation law as follows. Look at a pair of gears with 3 to 1 gear ratio. The little gear goes round 3 times and the big one 1 time. But 3+1=4 according to the latest calculations, so that number 4 is ‘conserved’ in some sense between the two kinematic motions.

Somebody named priyaranjanblogs has “liked” every comment I have made on this page. Clearly to get me to go to his blog. Anybody else have this problem? Is he a real member?

I got about 100 emails from him this morning, over 3 or 4 original posts. I wrote to Jerry and presumably he has blockec the guy, as the emails from p have stopped coming. Very annoying. Maybe a bit?

## 129 Comments

9!

diameter is proportional to r^2.

ra = 1/3rb. ra^2 = 1/9 rb^2

diameter a is 1/9 of diameter b.

No. The area is proportional to r^2. The diameter is proportional to r. In fact, the diameter of a circle is always 2 x r. Even the circumference, which I think is what you meant, is proportional to r. It’s 2 x π x r.

To be pedantic, because I am a bitter old (ok, late middle aged) man:

The diameter is proportional to r with a constant in a space of uniform curvature. In a plane, D=2r. If the curvature is no zero, then, depending on the postulate involved, the diameter may differ, and the circumference certainly will.

Which obviously doen’t apply to the SAT, but, given the answers provided, might as well. I am of an age that I sat this xam, and remember the question, and remember there being no answer, and being told that I was obviously mistaken as the test designers were professionals. Today, I would contact them. At that time, there was not way to resolve such issues. Not that I am bitter (I was the only one I didn’t get and it ruined my score)

How are you defining “diameter”? Given that a circle is the set of points at a given distance from the centre, the radius must be constant and therefore the diameter also given that it must pass through the centre.

I thought it was three as well.

I rotated two identical bowls about each other but they slipped so I couldn’t get an accurate experiment.

Note that there is some discussion on Talwalker’s comments drawing distinctions between the words used – e.g. rotate, revolve, revolutions, etc.

A simple way to see the origin of the extra rotation (giving 4) is to think of the moon which is tidally locked to the Earth, so that the same side faces Earth. It clearly rotates once when moving once around the Earth. So the answer is 1+3=4.

After reading the websites posted here, this is the correct answer. I am surprised and to be honest it hurt my brain a little to think about it.

Thank you, my head hurts a little bit less after your example. So are there really 365.24 days in a year?

Solar days, yes (356.242…). Revolutions of the earth relative to the fixed stars, no.

There are 366.24 sidereal days (periods between a distant star crossing your observatory’s meridian) in a year.

As is well known and deeply confusing to anyone setting out to chart the wine-dark seas of celestial mechanics.

This is all way over my head. I’ll just stick to my standard math joke for cocktail parties:

There are ten types of people: those who get binary, and those who don’t.

That joke doesn’t work in spoken form. The number “ten” is a base-ten quantity and isn’t used in other bases, certainly not in base-two.

You can write that joke just fine: “there are 10 types of people…” Because you can read “10” as being “ten” (if you assume base ten) or as “one zero” in which case it could be read as binary.

There are 10 types of people in the world:

People who understand binary,

People who don’t understand binary…

…And people who weren’t expecting a joke in base three!

😎

I see what you did there.

The punchline is actually the setup, as well. Neat twist.

cr

My problem with your joke is that it doesn’t divide the set of all people into a partition. The intersection of some of subsets is non empty. In fact, I would argue that “people who don’t understand binary” is a subset of “people who weren’t expecting a joke in base 3”.

Why do you hate fun?

How many members did the Jackson 5 have?

Six. They started counting at zero.

They would still have five members but they would be indexed from 0 to 4 inclusive.

You just got yourself a job at Google.

Mary’s father has 5 daughters:

1. Nana

2. Nene

3. Nini

4. Nono

What was the name of the 5th daughter?

😉

Is this a joke? Wouldn’t it be Nunu?… I don’t understand either way…

The first word in the riddle is… 🙂

that is amazing – but I wonder if it works when spoken out loud.

Mary.

I am dumb

Is Mary a boy or a girl? Are we talking gender or biological sex? Is Mary’s middle name Nunu?

Spoiler alert.

It seemed pretty obvious to me that the answer was the ratio of perimeters of the circles.

Since perimeter = 2 * r * pi, the ratio of perimeter is the same as the ratio of radius which is 3.

That’s the standard CS joke.

The standard math joke is

“There are three types of mathematicians; those that can count, and those that can’t”

Except that the answer isn’t 3. If you cut the larger circle and laid it out straight, *then* the answer would be 3. However, the smaller circle makes an additional rotation as it goes ’round the bigger circle, hence the total number of rotations is 4.

Except that the answer isn’t 3. If you cut the larger circle and laid it out straight, *then* the answer would be 3. However, the smaller circle makes an additional rotation as it goes ’round the bigger circle, hence the total number of rotations is 4.

There are two hard things in computer science: cache invalidation, naming things, and off-by-one errors.

If anybody can track down the original source of that joke, I’d be grateful.

After having gone through a spell when this problem drove me crazy, I used it to drive my own students crazy. Good for all of us.

Looking at the extreme case of zero rolls around the big circle helped. Using coins, let the outer coin go around the other coin so it always presents the same face to the other coin’s center — so with a penny, keep Abe’s eyes always looking at the center of the other coin. (Just as the moon always — more or less — presents the same face to the earth.) In doing so, you’ll find you have to keep turning the small coin. It will have gone through one rotation in the course of doing one trip.

Use gears.

Lego gears are good for this, as they come in ratios of 1:1, 2:1, 3:1, 5:1 3:2, 5:2, 5:3, and likely others at this point.

Or 3-d print gears.

No slipping.

Now try calculating the ratios in an epicyclic gearbox 🙂

Regular (manual) gearboxes, no trouble, but epicyclics mess with my head.

cr

That explains it perfectly for me. Thank you for that!

Do USian coins have serrated edges? Sorry – the proper term is “milled”, I think. Until a few years ago, this was the norm for UKian ones, but I don’t think any other than the 5p and £2 coins now have milling. I’d have to dig out the coin collection to check the £5 and £20 coins. And I’m wondering why we don’t have a £10 coin.

(I checked – the £5, £20, £50 and £100 coins are milled. I couldn’t find out for the 1kilo coins proposed in gold and silver, but I bet they’re milled too.)

I found this that I think best explains it.

“If you find it difficult to grasp how things work on a circle, you might also want to imagine what would happen on a square. When a circle rolling along the outer periphery of a square encounters the first corner, it will have to rotate an “extra” 90° to continue along the next side. This will happen again at each corner, and since 90° x 4 = 360°, this accounts for an additional full revolution.”

https://plus.maths.org/content/circles-rolling-circles

I like this a lot — and it leads to what I think is a nice, instructive generalization. Replace the square with a regular hexagon. Now it will turn through six 60-degree angles (the congruent exterior angles at each vertex). Keep replacing the polygons with even more-sided regular n-gons. This is approaching the limit of a circle, just what we have in the original problem.

Don’t really need the regular-polygon stipulation except to see the circle as the limit. With a little finagling don’t even need a convex polygon.

I also like this explanation. The whole problem is akin to the _Around the World in Eighty Day_ plot twist at the end, where Fogg realizes he’s gained a day.

I think the demo with the cut-outs is deceptive. Doesn’t the ‘Circle A’ circle need to spin until it’s base touches again. Picture them as gears instead. Correct me if I’m wrong…

OK, I was somewhat alerted to the fact that there is a trick by PCC(E)’s “the correct answer was not one of the SAT’s choices” but I immediately figured out the correct answer.

There’s several ways of looking at it, but one way is to imagine the radius of circle B as being 0. Then you are rolling circle A round a point. That tells you where the extra rotation comes from.

I like that observation.

It is easy to visualize the extra turn if you put an imaginary mark on the circumference of the small circle and keep track of it mentally as you roll the small circle around the large one.

…or you can just look at the addendum gif

That addendum gif really helped!

Where is this “addendum gif”?

I’m sorry, never mind … it’s not animatedly in the email…. is why I asked…

The start/end point of Circle A is the point that represents the intersection of the two circles. Mark that point on Circle A. Therefore, one full rotation of Circle A will be when that point is once again at the intersection of the two circles. That means there will be 3 rotations of Circle A. It should have nothing to do with the orientation of the text ‘Circle A’.

I want to add that the way it is described in the video is NOT how you determine gear ratios. The gear ratio in this example is 3:1, not 4:1. In a simple two-gear plane train, you take the number of teeth of the driving gear and the number of the driven gear, then divide. The number of teeth represents a full rotation of the gear (circle).

I skip-watched watched the rest of the video past 45 seconds, so didn’t see the reference to gears.

HOWEVER, in a gearbox with fixed-position shafts (like a manual gearbox) then 3:1 is indeed the right gear ratio. In an epicyclic box (as with most autos), the gears do effectively roll around each other and frankly, trying to comprehend what their ratio is then just messes with my head. 😉

cr

Epicyclic gears are the work of the devil.

Yeah. Also them Disraeli gears on racing bikes.

Could be a Jewish conspiracy 😉

Lots of confusion here. If the larger circle were replaced by a straight lie of the same length the answer would be 3. If it rolls with out slipping once around and closed figure (doesn’t have to be a circle) you have to add one rotation just for moving around the figure without rolling. 3+1=4 for any closed figure with 3 times the circumference.

9/2 because it’s clearly more than 3 and less than 6.

The real puzzle is why the SAT examiners didn’t have the right answer among the choices.

Yes, I’d like to know that. And also how difficult it generally is for a test taker to successfully contest a given test question. (I suppose the test generators could have included the answer, “none of the above,” eh?)

I think they just fell into the trap. Probably thought they were setting a very simple question.

Those who can’t do, teach. Those who can’t teach, design tests. Those who can’t design tests…

“Those who can’t design tests…”

… are truly woke. 🙂

As ‘rgbowman’ above points out, 3 is a perfectly reasonable answer. If you define a full rotation of the smaller coin as when the initial point touching circle B again touches it (in the video cardboard demo this corresponds to the point on the circumference directly below the text ‘circle A’ when it is pointed right way up, you will see it only touches the bigger circle 3 times). In the comments, this is called looking at it form the frame of reference of the smaller coin. The poster of the video admits this a question of frames of reference and 3 is correct in that case and the test makers admitted a mistake because the used the wrong word (revolution, rotation..).

So this is really more an issue of semantics.

I think Charles Sawiki has got it right. The answer is “4”.

Follow the link that Mark posted to see why.

Right. The same way we see the earth rotating 365 times per year but from a non-rotating Frame of reference it rotates 366 times per year.

Without too much analysis, 4.

The other question is whether they caught this back in ’82 and before calculating test results, or some yrs later?

Most likely some minutes after the test centres closed and the sitters extruded into the corridors. (I assume the tests are synchronised across the time zones – that must make setting exams in Russia challenging, with 9 time zones to accommodate.) Of course, they might have been able to adjust the marking scheme to just ignore the question, as you can’t infer the intention of a “null” answer in a multiple choice test, unless that is defined in the marking scheme.

Based on the wording of the problem and what Jerry wrote, my answer is 1. (Didn’t watch the video!)

Didn’t read the question properly either! My answer is wrong.

Today is definitely a dunce day for me.

No, I think you’re right. The question asks how many times it “revolves”. It only revolves once. It might “rotate” four times, but it only revolves once.

Going by what Jerry wrote, he indicated ‘rotations’ not ‘revolutions’. So I assume I’m wrong.

Addendum – I didn’t watch the video, and I had a brain block that day, but I knew there needed to be an *extra rotation* (key phrase – to get back to its ORIGINAL position). I couldn’t vocalize my initial thought ‘however many rotations plus the extra one’. Dunce day.

THANKS! So Jerry incorrectly described the problem, when he said ‘rotations’. I think I might have been thinking of revolutions, but now I can’t even remember.

I figured it had to be four, because for each full circumference of the small circle that was unrolled, it was also rotating a third of the way around just from following the surface of the larger circle.

Then I saw no four in the multiple choice answers, and wondered what I did wrong, before finishing the video to see what exactly they meant by everyone getting it wrong.

I solved this only because of Jerry’s hint that none of the proffered answers was correct.

I viewed it as follows: imagine the larger coin is unwound into a straight line and then you roll the smaller coin along the line. That gives 3 rotations. But then because of how the smaller coin moves around a _circular_ larger coin, you get one more rotation (as the wikipedia illustration shows).

I also got 800 on the math SAT (in 1980) and on the math GRE (in 1988). I think you don’t have to get a perfect score to get an 800 though.

Brilliant!

What do you mean by rotation? Let’s forget Einstein and look at it from the point of view of the fixed stars. Define a radius of the moon, say from its center to some well known geological object, call it X. Then let the moon move around the Earth without the radius CX changing position. Can the moon be said to rotate? I would say not.

What do you mean ‘geological’?

‘Geo’ to me implies Earth. Which makes no sense in the context.

If you took a line of direction from the moon to a fixed star, and the moon went round earth without that line moving, then I’d say it doesn’t rotate.

But then, in a motor, connecting rod big ends go round the crankshaft (in a circle about the crankshaft centreline) with no net rotation – they ‘wobble’ one way then the other.

Depends how you define ‘rotation’ I guess.

cr

“Depends how you define ‘rotation’ I guess.”

I wonder if the semantics of these kinds of words was what the test writer was going for – but it should not have been, as it was not the verbal section.

Sorry, not geological. I mean to a radius from the moon’s center to some fixed point on its surface, like a mountain or crater. Then keep this radius parallel to itself (relative to the “fixed” stars) as you move the moon around the Earth. The moon must seem to have rotated once according to this example, but that parallel axis tells me differently. Sorry to belabor this, but I don’t get it — geometrically.

You’re missing the fact that there is a point on the Moon’s surface which is, to a first approximation,

alwaysthe closest point to the Earth. So, as the Moon moves around it’s orbit and that point stays at the sub-Terran point, then the rest of the Moon must rotate in an Earth-centred frame of reference.to a second approximation, that sub-Terran point moves by around 60km through each orbit because neither the Earth nor the Moon, nor the gravitational potential of the Solar system are perfectly smooth.

I know that. I was trying to make a thought experiment and blew it. Sorry for the confusion.

In that particular case I would say that the Moon has revolved once around the Earth, but has not rotated.

But again, that’s assuming particular meanings for ‘revolved’ and ‘rotated’ which other people may disagree with – and which I may have got wrong.

cr

A BRAIN BUSTER PLANET MERCURY FACT: A Mercury day is twice as long as a Mercury year.

DEFINITIONS:

One Mercury year = time for Mercury to orbit Sol

One Mercury day = time for Sol to return to the same place in the sky from the POV of a ground observer

NOTES:

Sidereal vs. Solar Day:

It used to be thought that Mercury is tidally locked to Sol, with the same face always aimed at Sol – in the same way as the Moon always shows the same face to Earth [with a slight wobble actually]

But, Mercury is now known to be in 3:2 orbital resonance with Sol. This means that the planet completes three rotations on its axis for every two orbits it makes around Sol – It takes 58.646 EARTH DAYS to complete a single rotation on its axis & 87.969 EARTH DAYS to orbit Sol.

While this might lead some to conclude that a single day on Mercury is about 58 EARTH DAYS – thus making the length of a day and year correspond to the same 3:2 ratio – this would be inaccurate. Due to its rapid orbital velocity and slow sidereal rotation, a Solar Day on Mercury (the time it takes for Sol to return to the same place in the sky) is actually 176 EARTH DAYS.

I didn’t read the post before reading the question, so I was wondering why what I thought was the correct answer wasn’t among the choices. I don’t remember any of the math courses I took, but I figured this out intuitively. I figured if I added the radii of the two coins, that would come to 4/4. So, if circle A is 1/3 of circle B, it’s also 1/4 of the total, and therefore would revolve once for each quarter, giving the answer of 4.

It seems a lot of people made it more complicated than that with equations. Does my solution not work in other cases where it’s this exact question but with different values? Because the logic I used led me to the correct answer.

Yes it does work, without equations. Only ‘maths’ you need is knowing that circumference is directly proportional to radius. (It’s got a Pi in it but that’s for both circles so cancels out).

And it works the same for any other ratio of diameters. e.g. 5:1 ratio will give 6 rotations.

cr

Yay! I em smartz!

You can buy tape measures printed in “pi-feet” – I’ve seen them, but rarely. Their use case is in places like pipe stock-yards, where you throw the tape around the circumference of the pipe, shuggle it tight like a tailor being “flattering” (why am I thinking of Trump’s ridiculous dinner suit?) and read off the diameter of the pipe directly from the tape. Check the paint marks for the steel grade, and double-check on the serial numbers.)

They’re rare – tapes printed in decimal feet are much more common (feet, tenths of feet, hundredths of feet).

I did not look at any other comments, and haven’t looked at the video yet.

Well, 3 times certainly “seems” like the correct answer substituting for the r of the larger circle 3 times the radius of the smaller. I don’t understand why it is not.

But a shot in the dark. 4 times?

In my older age, I’m more patient and determined (and enjoy for its own sake) learning rudimentary calculus. Observing that increasing the radius to 3, as compared to 1, increases the area by a factor of 9, I wondered what happens if one integrates 2(Pi)r. One gets (Pi)r^2. And I chose the “limits” of 1 and 3. I got 8(Pi) as the answer. For the smaller circle, if r = 1, then the circumference was 2(Pi). 8(Pi) divided by 2(Pi) = 4 times the smaller circle makes a complete revolution around the circumference of the larger circle.

But this seems to be nothing but an act of faith on my part, and my rotator cuff aches from having to reach so far. Without a head’s up from PCC(E), I would have answered 3 without any doubt or hesitation. And as much time as I spent on this attempt, I would have been well on the way to a mediocre SAT score.

Now to look at the comments and face the reality of my modest math skills.

Cheers!

The video narrator says that “n + 1” is the answer. (Is that only for the specific situation where the radius of the smaller circle is 1/3 that of the larger?) But short of a mathematical proof, I have to accept that answer on faith, it seems. Unless a test taker memorized that factoid, it seems s/he’d have to do some kind of mathematics to get the answer.

Don’t need maths, just need to visualise it. But I was forewarned from PCC’s comments that there was something tricky about it – which turned out to be the extra rotation. But ‘four’ was not in the options listed as answers at 45 seconds in the video.

It’s a bit like: Sam is in a race. He overtakes the second-placed runner. What is his position now?

I call this sort of question ‘Hidden in plain sight’.

cr

But I WANT a mathematical derivation. As a former experimental physicist, I need for math and experiment to agree.

That’s fine, we can do all the mathematical derivations you want. But it’s just not necessary. (And, note, I didn’t do any experiments, other than a ‘thought experiment’ visualising the small circle rolling around the big one).

Here’s another one: Rotate the big circle once anticlockwise, holding the axis of the small circle fixed. The small circle will rotate three times clockwise (because the edges are in contact, and one circumference is three times the other). Now glue the circles together at their contact point and rotate the big circle once clockwise, with the small circle ‘riding’ round it; the small circle will also do one clockwise rotation. Total rotation of big circle: Zero (back at starting point). Total rotations of small circle: Four.

I could get all trig/calculus and calculate the rate of rotation (relative to a fixed azimuth) of the small circle relative to the rotation of its centrepoint about the centre of the big circle (which I’m sure would turn out to be 4 times) but quite honestly, I’m vegging out in front of the TV and I just can’t raise the energy.

cr

Actually you can use that same trick for epicyclic gears.

Step 1: First rotate the input shaft AND the whole gearbox including the house over +360 (“glue everything together”): obviously all wheels have made 1 rotation now.

Step 2: Keep the input shaft (with the planet wheel carrier) fixed, unglue, and turn back the ring over -360. Because the planet wheel carrier is now fixed, it is easy to see what the sun wheel (output shaft) will do.

Then add the rotations of step 1 and step 2.

Thus the gear ratio (for speed up) is 1 + d_ring / d_sun (the diameter of the planet wheels doesn’t matter)

“..we can do all the mathematical derivations you want. But it’s just not necessary. (And, note, I didn’t do any experiments, other than a ‘thought experiment’ visualising…”

It is simply a fact of mathematics that visualizing geometric (and also kinematic) behaviour IS PART OF MATHEMATICS.

We find it frustrating that many people (not you I’m sure–I’m just criticizing a choice of words) think that math involves nothing but facts about numbers.

While I’m at it, there is an obverse aspect also, where facts that have everything to do with an arbitrary human choice of base 10 (and now sometimes base 2), i.e. some peculiarity about digits, has anything mathematical to do with numbers, when it has only to do with names for numbers!! That is, linguistics, not mathematics.

Point taken. 🙂

cr

Concur. Watching the gif, it looks like it traces out some second or third cousin to an ellipsoid, kinda like a vertical cross section of an apple lying on its side.

Wait, what’s the trick in your question? If Sam just overtook the second-placed runner (and assuming there’s no trick here that his position has changed further since then and your asking of the question), Sam is now in second place, right?

I assume by “second-placed,” you mean the runner who is currently is second place. It’s not exactly how we would say it here.

The trick is that many people instinctively assume that, since he’s ahead of the (erstwhile) second-placed runner, he must be first. Without, of course, reflecting that his overtaking has changed the order.

In the case of the circles, the missing piece of the puzzle is that the small circle has rotated an extra time in going around the big circle.

There are hundreds of them: Port Arthur has six-foot tides. A ship ties up at the wharf at high tide and drops a rope ladder over the side. There are eight rungs showing above the water. Now the tide goes out. If the rungs are spaced nine inches apart, how many rungs are showing at low tide?

cr

I don’t understand why someone wouldn’t account for the fourth rotation, but perhaps that’s because of the way I came to my conclusion and because of the way my brain works. Conversely, I often have trouble with things that even very below-average people don’t. For example, when screwing in something, people say, “lefty loosie, righty tighty.” It probably took me until age 25 or so to make sense of this because turning each way seems both left and right to me. Screwing to the right also seems like left to me because the bottom of the screw is rotating leftward. It took me a long time to figure out that I had to always look at it from only the top of the screw or the way only the top of my hand was turning. Before I figured out a way to compensate, I would always try to screw things in and out the wrong way (you could do a lot of innuendo work with that sentence!).

It’s because I have a learning disability involving the processing of visual information. I’m in the 26th percentile when it comes to spatial reasoning! Kinda crazy, huh? Have you ever taken an IQ test? You know the part where you have to take blocks that are half red and half white split down the middle diagonally, and form the picture they give you? I could only get past the second one, and even the second took me a significant amount of time. Oddly enough, I was still able to play sports that involve split-second reaction times and decision-making, like ice hockey and tennis.

I don’t know if my brain managed to compensate for it’s impaired processing of visual information or if it’s just a coincidence, but I score in the 99th percentile in everything else. When it comes to spatial reasoning and reading speed, I’m far below average (though I would obviously be above average if they tested reading speed/comprehension of, say, highly complex studies or something). I imagine my brain must be compensating in some way, considering how quickly I can process the speed and movement of a hockey puck and the positions, movements, and speeds of all the players on the ice.

I chuckled empathetically with your left-right confusion, BJ. My parents teased me for years about my youthful comment “It depends on which way you’re facing…”

Are you dyslexic, BJ?

Incidentally, I always have trouble with those IQ test questions that have a series of figures, then you have to choose the last figure in the series. I always read far too much complexity into the figures, at least I think I do. They look like visual gobbledegook to me.

Like this one:

Wtf???

If I can ever make anything out of it, I usually find several different alternatives for the final one depending which features one gives significance to.

cr

Those usually stump me, too. Do you know the answer?

@merilee: Right?!? My dad would always say something like, “now screw it in. Lefty loosie, righty tighty, remember? What are you doing?” And when I’d explain it to him, he’d just shake his head and say, “I’ll never get how your brain works.” He’s said that in many different situations…

@infiniteimprobibilit: Not as far as I know. My brain just doesn’t process most visual information properly. It’s also one of the reasons I haven’t read many books: it takes too long.

I think the problem with left / right rotation is knowing the direction from which one rotates – so if you imagine your field of view, I think “left” assumes rotating from the top of the view – almost overhead. Not from the hips. Thus, the right arm would arc over to the left, in order to rotate an object like a wrench to the left, effecting a loosening if a nut… (a right handed nut, in fact)… facing the nut, that is… not the hex head…. is anyone still reading this?…. I’d give you a $10 bill if I could.

I think I discussed this here, perhaps with Sturtevant.

The youthful Merilee has it exactly correct, as does BJ; these quick descriptions are simply incomplete, and I have seen many of the brightest of mathematics students complaining about poorly written texts in exactly that sense.

As an illustration, suppose at a beer party someone draws a circle clockwise on the transparent glass table, and says he/she has done that. But the big boozer who had passed out, fell on the floor under the table looking up, says “No, that was counterclockwise!”.

A good problem on that left/right business is to explain why, since a mirror reverses your left and right, does it not turn you upside down?

A mirror doesn’t reverse left-right, it reverses fore-aft. It is an error in perception/intuition to interpret that as reversing left-right.

In that case, why is it that no one commits the “error in perception/intuition to interpret that as reversing” up/down?

Oh, and the answer to your last riddle is obviously still eight, as the entire boat rises and lowers with the tide, so the ladder doesn’t sink any further below the water.

Yes, but it’s surprising how many people get tied up with working out how many nine-inch rungs there are in six feet and don’t notice the hidden trick.

cr

Aaargh. I was sure it was 3:(6(pi)x/2(pi)x),

But I was wrong😖

Without reading any other replies, I would have said 4, but that’s not an option in the answers.

Three times due to the difference in diameters, plus another rotation due to the fact that it’s gone once round the large circle.

Imagine the large circle cut and straightened out into a line. Now roll the small circle along it – that’s three.

Alternatively, fix the contact point with the large circle so the small circle ‘slides’ around the large circle – that’s one rotation.

Now combine the two – four rotations of the small circle.

That’s my guess.

(I’m surprised they didn’t put Pi in there as a red herring. But Pi cancels out.)

cr

On reflection, it looks to me like the SAT question setter intended it as a very simple question (answer: three, since circumference – which is what is in contact – is directly proportional to radius), and completely missed the hidden trap in it.

What could test takers who saw the ‘right’ answer do? Write it in?

cr

Here might be another way to look at this, though it would be easier to look at with a drawing. Also, for the sake of simplicity, let’s make the ratio 4:1 rather than 3:1 (so the temptation in this version is to answer four rotations rather than 4 + 1 = 5).

Imagine the outer circle is a penny with the initial position with the coin on top and Abe looking off to the right at a very distant object, say a star, that is so distant that Abe is looking at it whenever his line of sight is off to the right along any line parallel to his initial line of vision. Also imagine a set of perpendicular axes through the center of the big circle with a vertical axis going through the initial point of contact on the penny. Call that point on the penny X. The axes would divide the figure into four quadrants.

Start the coin rolling. Whenever Abe sees the star he would say — correctly — that the coin has made one rotation from his last view of the star. When is his first new sight of the star? It would come somewhere IN the first quadrant, BEFORE X has reached the boundary between the first and second quadrants. His second view would be in the 2nd quadrant, his 3rd in the third quadrant and his 4th in the fourth quadrant. His fifth view would come just as X arrived back at its starting point for its fourth contact with the big circle. Abe would say — honestly — that he had made five rotations.

I should note that Michael Fisher’s sidereal comment got me thinking this way.

I think the counterintuitive solution has something to do with the intuition of unrolling the circumference from one circle onto the other.

Consider the gifs of the unit circles in this article :

https://en.m.wikipedia.org/wiki/Circumference

My flailing attempt to explain it :

I think the notion of a direct 1-to-1 mapping of each point of the circumferences is false because that would mean a transformation from a circle to a straight line and back. I think that is non trivial. The gif of the unit circle that illustrates where “2pi” comes from cannot be extended to two circles willy nilly – because a transformation is necessary.

Hoping to hear the more mathematically enlightened help me out here.

I got it, finally, by watching the addendum GIF, with two equal circles, for some time.

Draw a straight line with a length equal to the circumference of the (small) circle. Place the small circle at the left and and define the point of contact on the circle at C, the left end of the line. Now roll the circle to the midpoint of the line. The point C is now at the top of the circle and diametrically opposed to the new contact point. The circle has made a half rotation and so will rotate once as it rolls to the end the (straight) line. Now leave the circle attached to the center of the line with point C on the side opposite the line and drag the line by one end, leaving the other end fixed, so as to attach the two ends of the line and turn it into a circle too. The small circle has now been dragged around so that point C is on the downward side of the “small” (they’re really the same size) circle. The small circle has not moved relative to the line, but in space it has made a full rotation, the extra half of which is due just to the curving of the line, which drags the circle with it.

So we see that equal circles imply two rotations. It remains to be shown that n+1 is always the answer, rather than 2n. We can leave that as an exercise for the interested reader. 😉

We’ve got all this way, and nobody has mentioned the definitive work on matters like this by Verne, Fogg and Passepartout (Hetzel, 1873).

😀

I kinda see the resemblance. The twist in the tail of that did depend on the date line, or at least the one-day difference in dates caused by circumnavigation.

cr

That was of course a reply to the gravelinspector at #30.

Something much more tricky than SAT questions – persuading WP to put comments in the right places. 😦

cr

It’s really a relativity thing. If you were some tiny creature who lived on the circumference of the larger circle, you wouldn’t see that curved line or the extra rotation, any more than we sense the curvature of the surface of the Earth.

I actually got that!

Please excuse the length; I could show you all this with diagrams much faster.

Also, after (b) below, I am moving the small circle to the right and downwards, not to the left. So those who do it the other way must either adjust or change a couple of my details. To repeat, the small circle here goes clockwise around the big one.

I think that what is fundamentally confusing is this:

(a) Each point on the circumference of the small circle, as it rolls around the big circle, actually only ‘touches down’ on the big circle just 3 times.

And yet,

(b) the small circle does rotate 4 times.

So, for this reason I think, the correct answer of 4 can be quite confusing.

There is yet another, I think simpler way of explaining (a) and (b) above, in particular a way to see (b), the solution:

Draw 3 radii of the big circle, equally spaced, the first being straight up from the centre to where the small circle touches that big one. So the second is obtained from rotating the first radius ⅓ the way round the big circle, i.e. rotating it 120 degrees; and the third big radius another 120 degrees, i.e. 240 degrees in total. Notice that another 120, i.e. 360 total, gets you back to the first radius.

Now roll the little one round, stopping to think three times.

First stop when the little circle touches the big one at the end of that second radius above. The length travelled is the circumference length of the small circle, so every point on the small circle has touched once. Now set out again and stop when the small one touches at the end of the third long radius. Again the little circle has now had each of its points touch a second time. And finally a third time, so we have now established (a).

Now go back and look at that motion again, this time seeing what happens to a small radius we draw for the small circle, from its centre down to where it touches the big one in the diagram. Notice that radius is directly lined up with each of our big radii at the three points. The first time we stopped, that little radius points leftwards and up, actually having rotated completely one and one-third times. Need to think a bit there, but it’s pretty obvious, and really we’re finished:

Why? Well the other two rolls are clearly symmetric at least in the sense that they have the little circle rotating also one and one-third times (480 degrees if you prefer expressing it that way). But three copies of that number added together give our magic number 4! (Or 3 x 480 = 1440 degrees, which is also 4 x 360 of course).

And the ‘same’ argument (essentially) gives the (n+1) answer in the general case mentioned.

Actually, when n=1 at the end there, that argument is harder to follow than is all the cases of bigger “n”s. That’s Jerry’s addendum I just noticed, and the wiki video sort of illustrates exactly that argument by showing the small radius ‘doing its thing’ that I described. But they go counterclockwise. Must be commies or something!!

Also ‘gears’ is a kind of illustration of the same thing happening, a kind of conservation law as follows. Look at a pair of gears with 3 to 1 gear ratio. The little gear goes round 3 times and the big one 1 time. But 3+1=4 according to the latest calculations, so that number 4 is ‘conserved’ in some sense between the two kinematic motions.

But maybe it should be 3-1 not 3+1 if clockwise and counterclockwise are taken as negatives of each other (or 1-3).

Somebody named priyaranjanblogs has “liked” every comment I have made on this page. Clearly to get me to go to his blog. Anybody else have this problem? Is he a real member?

I got about 100 emails from him this morning, over 3 or 4 original posts. I wrote to Jerry and presumably he has blockec the guy, as the emails from p have stopped coming. Very annoying. Maybe a bit?

A bot, not a bit.