Deceptively simple geometry question on SAT test

  • #1

DaveC426913

Gold Member
22,386
6,079
TL;DR Summary
This question appeared on the American SAT test until recently removed. Can you solve it?
Fascinating, and utterly unintuitive.

This is a question that appeared on the American SAT test until it was recently removed. (citation: Veritasium, to which I will not link at this time.)

1701384747634.png


Every student ever has gotten it wrong, and that's because the SAT writers got it wrong too. The correct answer is not listed at all.

And I guarantee that, even knowing this, you will get it wrong too (unless you cheat, or unless you are a PF-regular - IOW, a super-genius at math).

I got it wrong, and I can hardly believe it even after having been shown the correct answer.

Feel free to post your answers using the spoiler tag.
 
  • Like
Likes phinds
Mathematics news on Phys.org
  • #2
4? The ratio of the circumferences plus 1 since the small circle rotates its zero-point orientation once in going around the larger circle...
 
  • Like
  • Wow
Likes Vanadium 50, pinball1970 and DaveC426913
  • #3
OK, we gve one super-genius PF regular. So far.

(Or - to cover all possible bases - a cheater)
 
  • Haha
Likes berkeman
  • #4
DaveC426913 said:
TL;DR Summary: This question appeared on the American SAT test until recently removed. Can you solve it?

you will get it wrong too (unless you cheat, or unless you are a PF-regular - IOW, a super-genius at math).
, or you've digested the difference between a solar day and a sidereal day: by the time the Earth makes 365 solar days, it rotates 366 times relative to the stars.

So, I'd say the answer is 4.
 
  • Like
Likes Kontilera, DaveC426913 and berkeman
  • #5
FYI. This has appeared somewhere before (probably on physicsforums) and it is also a standard thing taught in some engineering disciplines (I forget which).
I remember that I got it wrong the first time I saw it and it had to be explained to me.

UPDATE: I saw it before in the Youtube video that @DaveC426913 linked in post #15.
 
Last edited:
  • #6
Hill said:
, or you've digested the difference between a solar day and a sidereal day: by the time the Earth makes 365 solar days, it rotates 366 times relative to the stars.

So, I'd say the answer is 4.
Yeah, After having seen the solution, I too noticed the analogy.

Although my preferred analogy was actually taken from Verne's "Around the World In 80 Days."
 
  • Like
Likes Hill
  • #7
berkeman said:
4? The ratio of the circumferences plus 1 since the small circle rotates its zero-point orientation once in going around the larger circle...
OK, MAKE me feel stupid !

Good catch

@DaveC426913 I'm w/ you :smile:
 
  • #8
I'm getting dizzy... o0)
 
  • #9
berkeman said:
I'm getting dizzy... o0)
Well, at least you counted the right number of spins while you were getting dizzy :oldlaugh:
 
  • Like
Likes berkeman
  • #10
I first encountered this phenomena from Verne's book as well. Im sure many students went from this SAT confused, knowing the difference between a solar day and a sidereal day. Although it take some guts to accuse the SAT writers of being wrong, thus they probably forgot about it instead of making a fuzz about it. Im not sure its fair to accuse everybody of being wrong when the correct answer wasnt an option, but I guess it makes a good video title.
 
  • Like
Likes PeroK
  • #11
It's easier to see what's happening if we take the smaller circle to be half the radius. Or, make the circles the same size and label four points round the moving one.

The critical question is whether the answer changes if the circumference of the larger circle changes to a straight line of equal length.

This comes up in a number of physics problems.
 
  • Like
Likes Lnewqban and Kontilera
  • #12
I like to think of it this way:
i) Imagine a small arrow at the center of the smaller circle, intially pointing upwards. This will give a hint of the orientation of the small circle, which helps in order to count revolutions during the process.

ii) Now let the smaller circle slide around the larger circle, i.e. the point of the smaller circle that is initially touching the larger circle will continue to do so but slide along the circumference of the larger circle.

It is clear, that even though the smaller circle did'nt rotate at all relative to the circumference of the larger circle, it did complete a full revolution from our outside perspective - the arrow attached at the center did point in every direction once, which means it turned 360 degrees. Thus, by merely following the path of the larger circle, one revolution was made.

If sliding is not allowed, but the smaller circle has to roll around the larger one, then it has to complete an additional 3 revolutions in order to also travel the distance.
Thus 3 + 1 = 4 revolutions.

It dont know if this helped anybody, but I find it to be a good way to visiualize what is happening.
 
  • Like
Likes julian and Hill
  • #13
Kontilera said:
It dont know if this helped anybody, but I find it to be a good way to visiualize what is happening.
That's how I was able to wrap my head around what's happening. I was able to look at it as an exercise in reference frames. From the point of view of the large circle, the small circle just rotates around it 3 times, but from the point of view of the small circle, it is making two kinds of motions, first around the large circle and second around itself (which I though of exactly as you did, by considering an arrow fixed to the small circle, which has to go through a full rotation even if you consider that the small circle is stationary and the big circle rotates around it).
 
  • Like
Likes Kontilera
  • #14
The mathematical approach for the general case of circles of radius ##R## and ##r## is:

Mark the point of initial contact on the small circle as C. When the small circle has rolled an angle ##\theta## round the big circle, the distance rolled is ##R\theta##. Which equals ##r\phi'##, where ##\phi'## is the angle the small circle has rotated relative to its centre.

But, the line joining the centres through the new point of contact is at an angle of ##\theta## to the original line - horizontal in this case. So, the point C has rotated a total angle of ##\phi = \theta + \phi'##. Using the above equation we have:
$$\phi = \phi' + \theta = \frac R r \theta + \theta$$And if ##\frac R r = n##, then ##\phi = (n+1)\theta##
 
  • Like
  • Wow
Likes pinball1970, Kontilera and Hill
  • #15
Kontilera said:
Although it take some guts to accuse the SAT writers of being wrong, thus they probably forgot about it instead of making a fuzz about it. Im not sure its fair to accuse everybody of being wrong when the correct answer wasnt an option, but I guess it makes a good video title.
You might want to read to the whole article / watch the whole video. These students brought it to the attention of the SAT eggheads, who "didn't care". The students had to make a fuss about it.I think the reason it concentrates on the students getting it wrong is because, as is belaboured in the article, it asserts that, if students do poorly on their SATs, their futures are effectively finished.

Whether or not that's objectively true, it's certainly indicative of the pressure that was put on students to do well. It is therefore kind of surprising that every student must have guessed at an answer that seemed rightish instead of actually working out the answer. And that no students complained that the correct answer wasn't available.
 
Last edited:
  • Like
Likes FactChecker and Kontilera
  • #16
berkeman said:
4? The ratio of the circumferences plus 1 since the small circle rotates its zero-point orientation once in going around the larger circle...
Damn it!
 
  • Haha
  • Like
Likes berkeman and phinds
  • #17
PeroK said:
The mathematical approach for the general case of circles of radius ##R## and ##r## is:

Mark the point of initial contact on the small circle as C. When the small circle has rolled an angle ##\theta## round the big circle, the distance rolled is ##R\theta##. Which equals ##r\phi'##, where ##\phi'## is the angle the small circle has rotated relative to its centre.

But, the line joining the centres through the new point of contact is at an angle of ##\theta## to the original line - horizontal in this case. So, the point C has rotated a total angle of ##\phi = \theta + \phi'##. Using the above equation we have:
$$\phi = \phi' + \theta = \frac R r \theta + \theta$$And if ##\frac R r = n##, then ##\phi = (n+1)\theta##
LaTex not working on your post.
 
  • #18
pinball1970 said:
LaTex not working on your post.
Looks fine to me...
 
  • #19
What's crazy is that he makes a physical demo of it by unwrapping the circle's circumference into a straight line segment, and shows the circumference is exactly three times the smaller cirlce's circumference.

1701446539751.png

Even with that demo - the answer it produces is still wrong.

The correct answer is obvious - once you make that paradigm shift to include the extra rotation.
 
  • Haha
Likes Frimus
  • #20
berkeman said:
Looks fine to me...
That's weird! Now it's fine and the reply text is fine!
 
  • #21
pinball1970 said:
That's weird! Now it's fine and the reply text is fine!
Latex often takes a short time to render. Refreshing is the solution.
 
  • Like
Likes pinball1970
  • #22
DaveC426913 said:
This is a question that appeared on the American SAT test until it was recently removed.
Not that it matters, but 1983 has not been "recently" for quite some time, lol.
 
  • Haha
  • Sad
Likes OmCheeto and FactChecker
  • #23
Given that I've watched both the video and read the wiki 'such and such paradox', am I allowed to state that '3' is a legitimate answer?
 
  • #24
gmax137 said:
Not that it matters, but 1983 has not been "recently" for quite some time, lol.
Well, 1983 was only about 20 years ago.(And anybody who has anything to say about that can shut their filthy millennial pie hole!)
 
  • Haha
  • Love
  • Wow
Likes berkeman, Bystander, gmax137 and 1 other person
  • #25
I hope I am not supposed to hide this somehow (I don't know how). I skipped the answers, but certainly being warned this is a trick question helped.

My answer is four and I am very confident on that, just based on basic principles. I bet others got this, too, but (like me) would have answered the SAT too carelessly and thus incorrectly with "three". IIRC, a trick like this is often presented in fun tests with general trick questions which anybody can take.

BTW, I took the SAT in 1982. Either in 1986 or 1988, right before I got my either my first or second BS degree, I took the GRE and scored higher on the math than on the SAT. That test should have been harder than the SAT. So, did I find a way to think better in all those years, or had I just memorized patterns and formulas more?
 
  • #26
I guessed 3 as well until you said it was wrong. I think my solution is a little different than these rotation based attempts.

if a circle does a full rotation then its center moves one full circumference (this is true even on flat land). The center of the circle is 4 small circle radii away from the center of the big circle, so it needs to travel 4x as long as the small circle circumference to trace out the circle it will draw
 
Last edited:
  • Like
Likes gmax137
  • #27
I missed this thread the first time around. A few comments:

@berkeman has almost the correct answer. But it's not "X?" (I replaced his answer by X to avoid spoilers) but "X!". Be confident!

23:56:04 anyone?

I am fairly certain that this is not the only problem that a standardized test got wrong in the last 40 years.

A similar question was asked on the AMC12 (then called the MAA) in the 1950's. I don't believe the wrong answer was given then, but I'll look it up when I het some time.
 
  • #28
Ididn't get it at first, then I started breaking it down.

First unwind the larger circumference into a straight line and roll the smaller along it to the end, and it rolls 3 times as expected.

Then, re-wrap the circumference around the larger, and the smaller along with it. It rotates an additional one time in this second motion.
I often find breaking the movements down, sort of like addition of vectors, works for my mind!
 
  • #29
I deduced the correct answer and then noted the "problem" with that. We've discussed intuition before - in this case, experience with trying to keep orbital spacecraft pointed the right direction (reaction wheels, etc.) made it an 'easy' problem for me (I have plenty of deficits). Without that experience, I'd have stepped in the predictable pothole.
 
  • #30
Vanadium 50 said:
I am fairly certain that this is not the only problem that a standardized test got wrong in the last 40 years.
There is another one that I recall from a different SAT test. Circa 1972. Restating from memory...

We have a regular octahedron (8 faces) and a regular tetrahedron (4 faces). The size of the triangular faces on both are identical. The two polyhedra are then joined, face to identical face.

How many faces does the resulting shape have?

a) 12
b) 11
c) 10
d) 7.

A simple counting argument yields a result of 10. However, as I recall, the correct answer is 7.

You have a total of 12 faces. You lose two for the two faces that are directly glued together. You lose another three because the three remaining triangular sides of the tetrahedron join with their three adjoining faces on the octahedron at straight angles. So the three pairs of equilateral triangles become three rhombi, costing you another three faces.
One student reported this and was awarded points for his answer. The students with the supposedly correct answer did not lose their points.

I always speculated that the student that got it right played a lot of D&D.
 
  • Like
  • Haha
Likes e_jane and PeroK
  • #31
jbriggs444 said:
I always speculated that the student that got it right played a lot of D&D.
He might have, but not before the SAT, which was a few years before the game was published.

I see a plausible answer, and verified it by looking up the relevant dihedral angles. I guess the easiest way to see it is to know one fact: if I partially stellate a regular octahedron using 4 alternating sides, I get a tetrahedron again.
 
  • #32
Vanadium 50 said:
I am fairly certain that this is not the only problem that a standardized test got wrong in the last 40 years.
I just tried to find a list or collection of "known bad" SAT question-answers but I haven't had success with my googling.

Vanadium 50 said:
He might have, but not before the SAT, which was a few years before the game was published.
Or maybe @jbriggs444 "Circa 1972" is off by a few years?
 
  • #33
Vanadium 50 said:
He might have, but not before the SAT, which was a few years before the game was published.
Found a reference. My chronology was off by eight years and the question actually involved a square pyramid and a tetrahedron.
https://mathlair.allfunandgames.ca/saterrors.php said:
The first error to be caught was found when the question appeared on the October 1980 PSAT:

[diagram showing one pyramid ABCD with a triangular base and one with a square base]
44. In pyramids ABCD and EFGHI shown above, all faces except base FGHI are equilateral triangles of equal size. If face ABC were placed on face EFG so that the vertices of the triangles coincide, how many exposed faces would the resulting solid have?
(A) five
(B) six
(C) seven
(D) eight
(E) nine
The wanted answer for this question was (C) seven; the reasoning was that the two solids have a total of nine faces; when the pyramids are joined, two of the faces disappear, leaving seven. However, 17-year-old Daniel Lowen noticed that, when the solids are joined, faces ABD and EGH merge into a single face, as do faces ACD and EFI, so the resulting solid only has five faces (this can be difficult to see; if you want to prove it to yourself, you could make models out of paper or cardboard).

As a result of Lowen's successful challenge, ETS (Educational Testing Service) raised the scores of 240,000 test-takers by an average of one point. The question had been previously used on an SAT in 1979, and a the scores of people who had taken that test as juniors were raised; the scores of seniors weren't adjusted as presumably they would already have been accepted (or not) to college.
 
  • Like
Likes e_jane and gmax137
  • #34
After thinking about it, I see a high school level solution for @jbriggs444 problem.

Take a regular tetrahedron. Cut the four corners off to make a regular octahdron. Take one of the smaller tetrahedra and glue it back where you found it. Count the sides.
 
  • Like
Likes jbriggs444
  • #35
I did get this one right, but I think it was that I realized many years ago, Magellan's crew, in circum navigaing the globe saw one fewer sunsets. I thought about the voyage in a reference frame fixed to the stars and you can see that the number of times the Earth turns is one more than the number of sunsets along the voyage.

I should qualify this by saying, I can remember being a HS student, and I think I would have got this wrong if I encountered it back then. However, because of the nature of the test, the number of solutions required in the time allotted and so on, you are expecting to give the "easy" motivated answer.

Add to this that the correct answer is not one of the choices, what is a student to do.

I may be mistaken as it has been > 50 years since I took the SAT, but I think the instructions given, are:
"choose the BEST answer" In this contingency the answer 3 is closest to the true answer 4, so the obvious answer is also the correct one if constrained by the instructions of the test.

Many years ago, I worked to create an exam with a multiple choice section with a professor of the course. He told me, "always make one of your choices : none of the above"
 
  • Like
Likes PeroK

Suggested for: Deceptively simple geometry question on SAT test

Replies
4
Views
963
Replies
1
Views
855
Replies
38
Views
3K
Replies
1
Views
681
Replies
7
Views
983
Replies
2
Views
201
Replies
3
Views
1K
Replies
18
Views
1K
Back
Top