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From: Ask me about System Design on 15 Jan 2010 13:36 On Jan 15, 2:17 am, "Ostap S. B. M. Bender Jr." <ostap_bender_1...(a)hotmail.com> wrote: > On Jan 15, 2:09 am, "Peter Webb" > > <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote: > > You solve trivial geometry problems in your head by "restating the > > problem in the language of cartesian co-ordinates and vectors, and > > solve it algebraically rather than geometrically?" > > > What for? Wouldn't straightforward geometry be much easier on your > > head than this weird algebraic approach? > > > OK, you got me intrigued. So, describe to me in detail how to solve > > this problem algebraically please. > > > ________________________________ > > Well, I never really learned geometry until I did co-ordinate geometry, > > To me - it is the other way. We were taught Euclid geometry in 6th > grade as a separate subject, and that was the first time I was > introduced to the idea of axioms and rigorous proofs. I was very > impressed. Analytic geometry is something that I did very little of, > except for the usual ellipse, hyperbola and parabola. > > > > > and > > I suck at plane geometry, so I use the tools I know. I was also drawn to > > this because the problem is built around a square prism, which imposes a > > natural set of co-ordinates. An algebraic proof using vectors could exactly > > follow a purely geometric proof, just using a different notation. And no, I > > couldn't be bothered doing it for you. > > Let's see if "Ask me about System Design" can do it. OK then, here is an answer using a coordinate system. (By the way, you can use my new nickname for short: Mr. SDfae .) I start by picking D for the origin, and aligning axes with the three edges of the square prism = right prism. I use a scale suggested by the lengths of the problem. We get DA in the x direction, DC in the y, and DD1 in the z direction. Using s3 for sqrt(3), I coordinatize thusly: D = (0,0,0), A = (2,0,0), C=(0,2s3,0), D1=(0,0,s3). Now I will need the slope of AC in the x-y plane. This is slope AC = -2s3/2 . Since BD is perpendicular to AC in the x-y plane, this gives the slope of BD as 1/s3, so DB goes in the direction (s3,1,0). A1 is at (2,0,s3), so A1C has direction (-2, 2s3,-s3). i) It should now be clear that A1C dot DB is 0, so the two lines are perpendicular. Now to get the planes A1BD and BDC1, I will compute the normals, and compare those to get perpendicularity. I only need direction, so I will not use normals, just the cross products. ii) DA1 cross DBdir, is (-2,-3, s3), while DBdir cross DC1 is (-6, 3, -s3). The dot product of these vectors is 0, which gives that the two planes are perpendicular. Finally, the angle between AD and BC1 will be the most challenging, primarily because of coordinatizing B algebraically. B has coordinates (s3t,t,0) for some positive t (because B is not D) and is distance 2 away from (2,0,0). Thus t^2 + 3t^2 -4s3t +4 = 4, so t = s3. Then B = (3, s3,0), BC1 has direction (-3,s3,s3), and AD has direction (-2,0,0). Their cosine is their dot product divided by the product of their lengths, or 6/(2 * sqrt(15)), or sqrt(3/5). The angle is then inverse cosine of sqrt(.6), which I am told is 39 degrees and some number of minutes. If I am willing to accept some error, I could use interpolation between cos(45 degrees) = sqrt(1/2) and cosine (30 degrees) = sqrt(3/4) and say the angle is near 37.5 degrees, since sqrt(3/5) is near the average of sqrt(3/6) and sqrt(3/4). How is that for an algebraic solution? Gerhard "Ask Me About System Design" Paseman, 2010.01.15
From: Ostap S. B. M. Bender Jr. on 15 Jan 2010 21:38 On Jan 15, 10:36 am, Ask me about System Design <grpad...(a)gmail.com> wrote: > On Jan 15, 2:17 am, "Ostap S. B. M. Bender Jr." > > > > <ostap_bender_1...(a)hotmail.com> wrote: > > On Jan 15, 2:09 am, "Peter Webb" > > > <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote: > > > You solve trivial geometry problems in your head by "restating the > > > problem in the language of cartesian co-ordinates and vectors, and > > > solve it algebraically rather than geometrically?" > > > > What for? Wouldn't straightforward geometry be much easier on your > > > head than this weird algebraic approach? > > > > OK, you got me intrigued. So, describe to me in detail how to solve > > > this problem algebraically please. > > > > ________________________________ > > > Well, I never really learned geometry until I did co-ordinate geometry, > > > To me - it is the other way. We were taught Euclid geometry in 6th > > grade as a separate subject, and that was the first time I was > > introduced to the idea of axioms and rigorous proofs. I was very > > impressed. Analytic geometry is something that I did very little of, > > except for the usual ellipse, hyperbola and parabola. > > > > and > > > I suck at plane geometry, so I use the tools I know. I was also drawn to > > > this because the problem is built around a square prism, which imposes a > > > natural set of co-ordinates. An algebraic proof using vectors could exactly > > > follow a purely geometric proof, just using a different notation. And no, I > > > couldn't be bothered doing it for you. > > > Let's see if "Ask me about System Design" can do it. > > OK then, here is an answer using a coordinate system. > > (By the way, you can use my new nickname for short: > Mr. SDfae .) > > I start by picking D for the origin, and aligning axes > with the three edges of the square prism = right prism. > I use a scale suggested by the lengths of the problem. > We get DA in the x direction, DC in the y, and DD1 in > the z direction. Using s3 for sqrt(3), I coordinatize > thusly: > D = (0,0,0), A = (2,0,0), C=(0,2s3,0), D1=(0,0,s3). > > Now I will need the slope of AC in the x-y plane. This > is slope AC = -2s3/2 . Since BD is perpendicular to AC > in the x-y plane, this gives the slope of BD as 1/s3, > so DB goes in the direction (s3,1,0). > > A1 is at (2,0,s3), so A1C has direction (-2, 2s3,-s3). > > i) It should now be clear that A1C dot DB is 0, so the > two lines are perpendicular. > > Now to get the planes A1BD and BDC1, I will compute the > normals, and compare those to get perpendicularity. I > only need direction, so I will not use normals, just the > cross products. > > ii) DA1 cross DBdir, is (-2,-3, s3), while > DBdir cross DC1 is (-6, 3, -s3). The dot product of > these vectors is 0, which gives that the two planes are > perpendicular. > > Finally, the angle between AD and BC1 will be the most > challenging, primarily because of coordinatizing B > algebraically. B has coordinates (s3t,t,0) for some > positive t (because B is not D) and is distance 2 away > from (2,0,0). Thus t^2 + 3t^2 -4s3t +4 = 4, so t = s3. > Then B = (3, s3,0), BC1 has direction (-3,s3,s3), and > AD has direction (-2,0,0). Their cosine is their dot > product divided by the product of their lengths, or > 6/(2 * sqrt(15)), or sqrt(3/5). The angle is then > inverse cosine of sqrt(.6), which I am told is > 39 degrees and some number of minutes. > > If I am willing to accept some error, I could use > interpolation between cos(45 degrees) = sqrt(1/2) and > cosine (30 degrees) = sqrt(3/4) and say the angle is > near 37.5 degrees, since sqrt(3/5) is near the average > of sqrt(3/6) and sqrt(3/4). > > How is that for an algebraic solution? > Very impressive. So, the answer was supposed to be in the 'arccos' form? Not an easy problem for a test for non-mathematicians.
From: Ask me about System Design on 16 Jan 2010 02:03 On Jan 15, 6:38 pm, "Ostap S. B. M. Bender Jr." <ostap_bender_1...(a)hotmail.com> wrote: > On Jan 15, 10:36 am, Ask me about System Design <grpad...(a)gmail.com> > wrote: > > > > > > > On Jan 15, 2:17 am, "Ostap S. B. M. Bender Jr." > > > <ostap_bender_1...(a)hotmail.com> wrote: > > > On Jan 15, 2:09 am, "Peter Webb" > > > > <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote: > > > > You solve trivial geometry problems in your head by "restating the > > > > problem in the language of cartesian co-ordinates and vectors, and > > > > solve it algebraically rather than geometrically?" > > > > > What for? Wouldn't straightforward geometry be much easier on your > > > > head than this weird algebraic approach? > > > > > OK, you got me intrigued. So, describe to me in detail how to solve > > > > this problem algebraically please. > > > > > ________________________________ > > > > Well, I never really learned geometry until I did co-ordinate geometry, > > > > To me - it is the other way. We were taught Euclid geometry in 6th > > > grade as a separate subject, and that was the first time I was > > > introduced to the idea of axioms and rigorous proofs. I was very > > > impressed. Analytic geometry is something that I did very little of, > > > except for the usual ellipse, hyperbola and parabola. > > > > > and > > > > I suck at plane geometry, so I use the tools I know. I was also drawn to > > > > this because the problem is built around a square prism, which imposes a > > > > natural set of co-ordinates. An algebraic proof using vectors could exactly > > > > follow a purely geometric proof, just using a different notation. And no, I > > > > couldn't be bothered doing it for you. > > > > Let's see if "Ask me about System Design" can do it. > > > OK then, here is an answer using a coordinate system. > > > (By the way, you can use my new nickname for short: > > Mr. SDfae .) > > > I start by picking D for the origin, and aligning axes > > with the three edges of the square prism = right prism. > > I use a scale suggested by the lengths of the problem. > > We get DA in the x direction, DC in the y, and DD1 in > > the z direction. Using s3 for sqrt(3), I coordinatize > > thusly: > > D = (0,0,0), A = (2,0,0), C=(0,2s3,0), D1=(0,0,s3). > > > Now I will need the slope of AC in the x-y plane. This > > is slope AC = -2s3/2 . Since BD is perpendicular to AC > > in the x-y plane, this gives the slope of BD as 1/s3, > > so DB goes in the direction (s3,1,0). > > > A1 is at (2,0,s3), so A1C has direction (-2, 2s3,-s3). > > > i) It should now be clear that A1C dot DB is 0, so the > > two lines are perpendicular. > > > Now to get the planes A1BD and BDC1, I will compute the > > normals, and compare those to get perpendicularity. I > > only need direction, so I will not use normals, just the > > cross products. > > > ii) DA1 cross DBdir, is (-2,-3, s3), while > > DBdir cross DC1 is (-6, 3, -s3). The dot product of > > these vectors is 0, which gives that the two planes are > > perpendicular. > > > Finally, the angle between AD and BC1 will be the most > > challenging, primarily because of coordinatizing B > > algebraically. B has coordinates (s3t,t,0) for some > > positive t (because B is not D) and is distance 2 away > > from (2,0,0). Thus t^2 + 3t^2 -4s3t +4 = 4, so t = s3. > > Then B = (3, s3,0), BC1 has direction (-3,s3,s3), and > > AD has direction (-2,0,0). Their cosine is their dot > > product divided by the product of their lengths, or > > 6/(2 * sqrt(15)), or sqrt(3/5). The angle is then > > inverse cosine of sqrt(.6), which I am told is > > 39 degrees and some number of minutes. > > > If I am willing to accept some error, I could use > > interpolation between cos(45 degrees) = sqrt(1/2) and > > cosine (30 degrees) = sqrt(3/4) and say the angle is > > near 37.5 degrees, since sqrt(3/5) is near the average > > of sqrt(3/6) and sqrt(3/4). > > > How is that for an algebraic solution? > > Very impressive. > > So, the answer was supposed to be in the 'arccos' form? Not an easy > problem for a test for non-mathematicians.- Hide quoted text - > > - Show quoted text - You can check the press releases for the Royal Chemical Society for 2007 to see what answer was preferred, as well as how they worked the problem. I did not compute or estimate the inverse cosine orignally, as I was used to giving answers of the form "alpha, where cos(alpha) equals blah". My point was that the problem was not that hard, and that it could be solved by a good student of analytic geometry, not necessarily a mathematician. I think that was what the test was designed to do, show which applicants had medium to good skills in analytic geometry. I can imagine a student who earned B's instead of A's in math would struggle with it. Did I convince you that it could be done in under 10 minutes, and with a somewhat algebraic treatment? Also, I apologize earlier for misinterpeting an earlier post of yours. At the time, it seemed to me that you were replying to me about Peter, and not to Peter about Peter. Gerhard "Ask Me About System Design" Paseman, 2010.01.15
From: Ostap S. B. M. Bender Jr. on 16 Jan 2010 02:44 On Jan 15, 11:03 pm, Ask me about System Design <grpad...(a)gmail.com> wrote: > On Jan 15, 6:38 pm, "Ostap S. B. M. Bender Jr." > > > > <ostap_bender_1...(a)hotmail.com> wrote: > > On Jan 15, 10:36 am, Ask me about System Design <grpad...(a)gmail.com> > > wrote: > > > > On Jan 15, 2:17 am, "Ostap S. B. M. Bender Jr." > > > > <ostap_bender_1...(a)hotmail.com> wrote: > > > > On Jan 15, 2:09 am, "Peter Webb" > > > > > <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote: > > > > > You solve trivial geometry problems in your head by "restating the > > > > > problem in the language of cartesian co-ordinates and vectors, and > > > > > solve it algebraically rather than geometrically?" > > > > > > What for? Wouldn't straightforward geometry be much easier on your > > > > > head than this weird algebraic approach? > > > > > > OK, you got me intrigued. So, describe to me in detail how to solve > > > > > this problem algebraically please. > > > > > > ________________________________ > > > > > Well, I never really learned geometry until I did co-ordinate geometry, > > > > > To me - it is the other way. We were taught Euclid geometry in 6th > > > > grade as a separate subject, and that was the first time I was > > > > introduced to the idea of axioms and rigorous proofs. I was very > > > > impressed. Analytic geometry is something that I did very little of, > > > > except for the usual ellipse, hyperbola and parabola. > > > > > > and > > > > > I suck at plane geometry, so I use the tools I know. I was also drawn to > > > > > this because the problem is built around a square prism, which imposes a > > > > > natural set of co-ordinates. An algebraic proof using vectors could exactly > > > > > follow a purely geometric proof, just using a different notation. And no, I > > > > > couldn't be bothered doing it for you. > > > > > Let's see if "Ask me about System Design" can do it. > > > > OK then, here is an answer using a coordinate system. > > > > (By the way, you can use my new nickname for short: > > > Mr. SDfae .) > > > > I start by picking D for the origin, and aligning axes > > > with the three edges of the square prism = right prism. > > > I use a scale suggested by the lengths of the problem. > > > We get DA in the x direction, DC in the y, and DD1 in > > > the z direction. Using s3 for sqrt(3), I coordinatize > > > thusly: > > > D = (0,0,0), A = (2,0,0), C=(0,2s3,0), D1=(0,0,s3). > > > > Now I will need the slope of AC in the x-y plane. This > > > is slope AC = -2s3/2 . Since BD is perpendicular to AC > > > in the x-y plane, this gives the slope of BD as 1/s3, > > > so DB goes in the direction (s3,1,0). > > > > A1 is at (2,0,s3), so A1C has direction (-2, 2s3,-s3). > > > > i) It should now be clear that A1C dot DB is 0, so the > > > two lines are perpendicular. > > > > Now to get the planes A1BD and BDC1, I will compute the > > > normals, and compare those to get perpendicularity. I > > > only need direction, so I will not use normals, just the > > > cross products. > > > > ii) DA1 cross DBdir, is (-2,-3, s3), while > > > DBdir cross DC1 is (-6, 3, -s3). The dot product of > > > these vectors is 0, which gives that the two planes are > > > perpendicular. > > > > Finally, the angle between AD and BC1 will be the most > > > challenging, primarily because of coordinatizing B > > > algebraically. B has coordinates (s3t,t,0) for some > > > positive t (because B is not D) and is distance 2 away > > > from (2,0,0). Thus t^2 + 3t^2 -4s3t +4 = 4, so t = s3. > > > Then B = (3, s3,0), BC1 has direction (-3,s3,s3), and > > > AD has direction (-2,0,0). Their cosine is their dot > > > product divided by the product of their lengths, or > > > 6/(2 * sqrt(15)), or sqrt(3/5). The angle is then > > > inverse cosine of sqrt(.6), which I am told is > > > 39 degrees and some number of minutes. > > > > If I am willing to accept some error, I could use > > > interpolation between cos(45 degrees) = sqrt(1/2) and > > > cosine (30 degrees) = sqrt(3/4) and say the angle is > > > near 37.5 degrees, since sqrt(3/5) is near the average > > > of sqrt(3/6) and sqrt(3/4). > > > > How is that for an algebraic solution? > > > Very impressive. > > > So, the answer was supposed to be in the 'arccos' form? Not an easy > > problem for a test for non-mathematicians.- Hide quoted text - > > > - Show quoted text - > > You can check the press releases for the Royal Chemical > Society for 2007 to see what answer was preferred, as > well as how they worked the problem. > Do you have the link? How many respondents answered correctly and collected the 500 quid prize? > > I did not compute > or estimate the inverse cosine orignally, as I was used > to giving answers of the form "alpha, where cos(alpha) > equals blah". > Isn't that the same as arccos(alpha)? > My point was that the problem was not that hard, and > that it could be solved by a good student of analytic > geometry, not necessarily a mathematician. > I have to say that as a student at the best math school in Russia and a winner of math Olympiads in both Russia and USA, I never studied much analytic geometry in high school, nor even as a math major at Harvard U. Few if any pure math undergraduates of graduate students in USA study analytic geometry. Like most, I was more into algebraic topology, Galois theory, functional analysis, etc. But if you tell me that you know that even non-math high school students in China are taught to be wizards at analytic geometry - well, I am no expert on China to refute. In solving this problem, I myself would never even think of using analytic geometry, and would use the classic high school Euclidian geometry and stereometry. And parts (i) and (ii) are very easy with that. But part (iii) seems difficult. > > I think > that was what the test was designed to do, show which > applicants had medium to good skills in analytic > geometry. I can imagine a student who earned B's > instead of A's in math would struggle with it. > > Did I convince you that it could be done in under 10 > minutes, and with a somewhat algebraic treatment? > By a typical chemistry-bound high school graduate? Not really. Not in 10 minutes, working from scratch.
From: victor_meldrew_666 on 16 Jan 2010 06:47
On 15 Jan, 18:36, Ask me about System Design <grpad...(a)gmail.com> wrote: > On Jan 15, 2:17 am, "Ostap S. B. M. Bender Jr." > > ii) DA1 cross DBdir, is (-2,-3, s3), while > DBdir cross DC1 is (-6, 3, -s3). The dot product of > these vectors is 0, which gives that the two planes are > perpendicular. For an alternative geometric argument, note that as A_1 E and E C_1 lie in the two planes and are both perpendicular to the intersection of the planes (the line BD) then the angle between the two planes is angle A_1 E C_1. As the triangles A_1 A E and A C C_1 are easily seen to be similar and right-angled then A_1 E C_1 is a right angle. |