Paul (![[info]](http://l-stat.livejournal.com/img/userinfo.gif){kind=small} happyfunpaul) wrote,@ 2009-01-28 11:57:00 |
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ANSWER to that (unintentionally) tricky geometry problem
Oh right, I never got around to posting the answer to that geometry problem. In honor of just having entered the finished math grades into a file (though they were done back on Saturday, plus one student on Monday), I'll show it now. :-)
Choice "(d) BC and plane ADC" is correct; since segment BC is perpendicular to both AC and CD, it's perpendicular to plane ADC.
Choice "(b) AD and plane E" is incorrect (though for some reason, a lot of students chose it) because, since DH is perpendicular to AC, AD meets AC at an acute angle, which means it must meet plane E at an acute angle.
Choice "(c) AB and plane ADC" is incorrect. Since BC is perpendicular to AC, triangle ACB is a right triangle with the right angle at C, so angle BAC is acute, so AB meets plane ADC at an acute angle.
Choice "(a) DH and plane E" is the tricky one...
My first thought was: It's incorrect because you don't have enough information. DH is perpendicular to AC, so it might be that D is directly "over" H and DH meets the entire plane E at a right angle... or it might be the case that D is a little "further away" from us, in which case DH does not meet plane E perpendicularly.
My second thought was: It is correct, because, given the answer to (d), there's nowhere for D to be except directly "over" H. If you draw an "above view" of the diagram correctly with ACB being a right angle, it's clear.
My third thought, thanks toultimatepsi, was: But it's also given that AD is perpendicular to AB! There is no way to put "D" to make that true! "BC perp. to CD" (in conjunction with "BC perp. to AC") forces D to be above H, but "AB perp. to AD" implies D is "further" from us than CD. It can't be both! So the problem describes an impossible figure! Therefore, choice (a) can be neither "true" nor 'false"!
In the end, I marked choices (b) and (c) wrong, (a) with "ok" (no points taken off), and (d) correct with a half-point bonus.
The whole thing reminded me of a semi-famous SAT math problem. It appeared as a problem #6 on the old pencil-and-paper version of the SAT (implying that it must be easy, to appear so early on). It went something like this:
Given two pyramids: One is a tetrahedron with the four faces being equilateral triangles. The other has a square base but the other four faces are equilateral triangles of the same size as in the tetrahedron. Now, stick the two pyramids together, one triangular face to another, so that both faces are covered up "inside" the resulting polyhedron. How many (external) faces does the new polyhedron have?
(a) 4
(b) 5
(c) 6
(d) 7
The intended answer was "7"-- you start with 9 total faces, cover 2 up, therefore 7 remain. The vast majority of students indeed answered "7". I would've answered '7" myself.
However, one student protested that the answer should be "5". Turns out, he was right. Two pairs of faces, it turns out, wind up being coplanar, so you get a five-sided polyhedron (one square face, two triangular faces, and two "double-triangle" faces). But it's almost impossible to tell that unless you build a model; most people's mental pictures have the pairs of faces meet at angles instead of "lining up".
The testing service, as I recall, went back and marked everyone correct who had put "5", even though you know that the vast majority of the students did so for incorrect reasons. I felt the same way about students choosing answer (a) on my test, but what are you going to do? :-) So they get the points.
Oh right, I never got around to posting the answer to that geometry problem. In honor of just having entered the finished math grades into a file (though they were done back on Saturday, plus one student on Monday), I'll show it now. :-)
Choice "(d) BC and plane ADC" is correct; since segment BC is perpendicular to both AC and CD, it's perpendicular to plane ADC.
Choice "(b) AD and plane E" is incorrect (though for some reason, a lot of students chose it) because, since DH is perpendicular to AC, AD meets AC at an acute angle, which means it must meet plane E at an acute angle.
Choice "(c) AB and plane ADC" is incorrect. Since BC is perpendicular to AC, triangle ACB is a right triangle with the right angle at C, so angle BAC is acute, so AB meets plane ADC at an acute angle.
Choice "(a) DH and plane E" is the tricky one...
My first thought was: It's incorrect because you don't have enough information. DH is perpendicular to AC, so it might be that D is directly "over" H and DH meets the entire plane E at a right angle... or it might be the case that D is a little "further away" from us, in which case DH does not meet plane E perpendicularly.
My second thought was: It is correct, because, given the answer to (d), there's nowhere for D to be except directly "over" H. If you draw an "above view" of the diagram correctly with ACB being a right angle, it's clear.
My third thought, thanks to
ultimatepsi, was: But it's also given that AD is perpendicular to AB! There is no way to put "D" to make that true! "BC perp. to CD" (in conjunction with "BC perp. to AC") forces D to be above H, but "AB perp. to AD" implies D is "further" from us than CD. It can't be both! So the problem describes an impossible figure! Therefore, choice (a) can be neither "true" nor 'false"!In the end, I marked choices (b) and (c) wrong, (a) with "ok" (no points taken off), and (d) correct with a half-point bonus.
The whole thing reminded me of a semi-famous SAT math problem. It appeared as a problem #6 on the old pencil-and-paper version of the SAT (implying that it must be easy, to appear so early on). It went something like this:
Given two pyramids: One is a tetrahedron with the four faces being equilateral triangles. The other has a square base but the other four faces are equilateral triangles of the same size as in the tetrahedron. Now, stick the two pyramids together, one triangular face to another, so that both faces are covered up "inside" the resulting polyhedron. How many (external) faces does the new polyhedron have?
(a) 4
(b) 5
(c) 6
(d) 7
The intended answer was "7"-- you start with 9 total faces, cover 2 up, therefore 7 remain. The vast majority of students indeed answered "7". I would've answered '7" myself.
However, one student protested that the answer should be "5". Turns out, he was right. Two pairs of faces, it turns out, wind up being coplanar, so you get a five-sided polyhedron (one square face, two triangular faces, and two "double-triangle" faces). But it's almost impossible to tell that unless you build a model; most people's mental pictures have the pairs of faces meet at angles instead of "lining up".
The testing service, as I recall, went back and marked everyone correct who had put "5", even though you know that the vast majority of the students did so for incorrect reasons. I felt the same way about students choosing answer (a) on my test, but what are you going to do? :-) So they get the points.
