Deceptively simple geometry question on SAT test

[Vanadium 50]

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.

 

[gmax137]

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.

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?

 

[jbriggs444]

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.

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.

 

[Vanadium 50]

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.

 

[mpresic3]

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"

 

[mpresic3]

Actually d would be the "best" answer. 9/2 is closer to 4 than any of the other answers

 

[Vanadium 50]

Magellan's crew, in circum navigaing the globe saw one fewer sunsets

Didn't Phineas Fogg have the sane experience, as a plot point?

Perhaps this was really supposed to be on the English portion.

 

[PeroK]

Didn't Phineas Fogg have the sane experience, as a plot point?

Fogg travelled east and although 80 days had passed for him, upon his return only 79 days had passed in London.

 

[Vanadium 50]

I still like my answer in #34. I'm not very good in solid geometry...but I was that time.

If people still are having problems seeing the "+1" part of the answer, think about a coin going around a nail.

 

[berkeman]

If people still are having problems seeing the "+1" part of the answer, think about a coin going around a nail.

Says the guy with the hammer!

 

[gleem]

If people still are having problems seeing the "+1" part of the answer, think about a coin going around a nail.

I see said the blind man as he picked up his hammer and saw.

 

[bob012345]

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.)

View attachment 336410

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.

How can you first reach the starting point unless you are just starting?

 

[pellis]

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.

I knew I'd seen something like it before in relation to spinors: the rotation of two identical coins shown in the previously posted youtube video is one of a number of illustrations of the need for a 4-pi rotation identity. And there's also e.g. the Balinese Cup Trick and others described in https://en.wikipedia.org/wiki/Spinor.

But those only correspond to equal-radius cases, rather than the r/3 and others explored in this informative thread. Thank you.

 

[mathwonk]

when I read the geometry question, I wondered: "rotates with respect to what?" the answer is 4, (reasoning as explained by others here), if it means rotation with respect to a fixed axis on the page, but it is 3 if it means with respect to a little boy walking around the circle with a stick, rolling his hoop. so in my opinion, the problem is not entirely well posed, without more clear definitions.
just an alternate viewpoint, maybe unreasonable. (and of course it is zero, if it means with respect to the person rolling inside the tire, like scout in "to kill a mockingbird"!).

 

[DaveC426913]

when I read the geometry question, I wondered: "rotates with respect to what?" the answer is 4, (reasoning as explained by others here), if it means rotation with respect to a fixed axis on the page, but it is 3 if it means with respect to a little boy walking around the circle with a stick, rolling his hoop. so in my opinion, the problem is not entirely well posed, without more clear definitions.
just an alternate viewpoint, maybe unreasonable. (and of course it is zero, if it means with respect to the person rolling inside the tire, like scout in "to kill a mockingbird"!).

If the frame of reference has to, itself, rotate to rationalize a given answer, I'd say that's a pretty weak solution. I mean, that opens up an unlimited number of solutions: in the FoR of an observer rotating 23 times counter-clockwise, the small circle rotates, what? 27 times clockwise, right?(Although, note: it still adds up to four).

I think the problem, as stated, implies the frame of reference is that of the non-rotating, objective observer - not some arbitrary moving location within the setup.

 

[mathwonk]

of course you take the intelligent, reasonable, physicists point of view. but it seems the test makers took the other, as did the guy in the video who unfurled the circle. I'm just saying, when us clueless amateurs weigh in, all bets are off.

I had posted but deleted, an earlier comment that, as a mathematician with no talent for making reasonable assumptions, I am challenged to answer a question whose terms are left undefined, so I wanted a definition of the FOR (see how I picked up on that!?) that should be used.

I don't argue this is reasonable, just that with no specific guidance, some of us will go astray in this way. and of course I agree entirely it opens up an unlimited number of solutions; that's what I meant by saying the problem is not well posed.

I apologize for sharing the perspective of someone with little to no common sense!

 

[Gavran]

The motion of the circle A consists of two periodic motions with two different revolutions and these two revolutions can not be mixed with each other. The correct answer is 3.

 

[Gavran]

After including time instead of angle in the equations $$ x \left ( \theta \right ) = \left ( R + r \right ) \cos \theta – r \cos \left ( \frac { 3 r + r } { r } \theta \right ) $$ $$ y \left ( \theta \right ) = \left ( R + r \right ) \sin \theta – r \sin \left ( \frac { 3 r + r } { r } \theta \right ) $$ (https://en.wikipedia.org/wiki/Epicycloid) there will be $$ x _ 1 \left ( t \right ) = \left ( R + r \right ) \cos \left ( \frac { 2 \pi } { T } t \right ) – r \cos \left ( \frac { 2 \pi } { \frac { T } { 4 } } t \right ) $$ $$ y _ 1 \left ( t \right ) = \left ( R + r \right ) \sin \left ( \frac { 2 \pi } { T } t \right ) – r \sin \left ( \frac { 2 \pi } { \frac { T } { 4 } } t \right ) $$ and there are two periodic motions, the first one with a period of T and the second one with a period of T/4. During the one revolution of the first periodic motion there will be four revolutions of the second periodic motion. The correct answer seems to be 4.