What math level is required?
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From: Ask me about System Design on 15 Jan 2010 04:16 On Jan 15, 12:54 am, "Ostap S. B. M. Bender Jr." <ostap_bender_1...(a)hotmail.com> wrote: > On Jan 14, 11:26 pm, Ask me about System Design <grpad...(a)gmail.com> > wrote: > > > > > > > On Jan 14, 11:08 pm, "Ostap S. B. M. Bender Jr." > > > <ostap_bender_1...(a)hotmail.com> wrote: > > > On Jan 14, 10:43 pm, "Peter Webb" > > > > <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote: > > > > "mike3" <mike4...(a)yahoo.com> wrote in message > > > > >news:d4f6fc15-c015-4833-b108-06d60390b18d(a)l30g2000yqb.googlegroups.com... > > > > > > Hi. > > > > > > I saw this: > > > > > >http://news.bbc.co.uk/2/hi/uk_news/education/6589301.stm > > > > > > What sort of math level would that require (to be able to do problems > > > > > like the one on that Chinese test)? > > > > > Well, I haven't tried to solve it, but not much I suspect. > > > > > Alternatively (and equivalently) you could restate the problem in the > > > > language of cartesian co-ordinates and vectors, and solve it algebraically > > > > rather than geometrically. This still only requires high school maths. > > > > You can do THAT in the 5 minutes or so allocated to this problem on > > > the exam?! You must be a walking supercomputer. > > > I did not see where it was supposed to take only > > 5 minutes. Having not practiced such problems in over > > a decade, it took me less than 20 minutes to do in my > > head, with iii) being the challenging part. In my > > heyday, I could have probably done it in 5 minutes with > > pencil and paper, and perhaps another minute to check > > my work. > > 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.- Hide quoted text - > > - Show quoted text - I think you are combining my reply with that of a different poster. I solve geometry problems using a variety of methods, including various instances of coordinatizing the diagram using helpful points of origin and axes. For this problem, finding right triangles, computing lengths, and eventually computing a dot product mentally were what I did. I even checked the answer against what the RCS published, and (except for not computing an inverse cosine to get the angle in part iii) my answers agreed with theirs. However, I can solve the problem on paper using mostly just analytic geometry with a single coordinate system. If you tell me where you got the figure of 5 minutes, I will expand on the approach I just mentioned, which is as close as I will go to "solving the problem algebraically" . I might even convince you that such an approach might take under ten minutes. Gerhard "Ask Me About System Design" Paseman, 2010.01.15
From: Henry on 15 Jan 2010 04:17 On 15 Jan, 07:06, "Ostap S. B. M. Bender Jr." <ostap_bender_1...(a)hotmail.com> wrote: > On Jan 14, 10:23 pm, mike3 <mike4...(a)yahoo.com> wrote: > > > Hi. > > > I saw this: > > >http://news.bbc.co.uk/2/hi/uk_news/education/6589301.stm > > > What sort of math level would that require (to be able to do problems > > like the one on that Chinese test)? > > That would require a much better translation level, because the phrase > "the foot of perpendicular is E" is highly ambiguous. I would guess it means E is the intersection of AC and BD, which we already know are perpendicular. Not that E appears in the three questions. I also don't see what "square" means in "square prism ABCD-A1B1C1D1" since ABCD and A1B1C1D1 are not squares, and nor is ABB1A1 or ADD1A1. I suppose it might mean A1B1C1D1 is congruent to and vertically above ABCD.
From: Ostap S. B. M. Bender Jr. on 15 Jan 2010 04:40 On Jan 15, 1:17 am, Henry <s...(a)btinternet.com> wrote: > On 15 Jan, 07:06, "Ostap S. B. M. Bender Jr." > > <ostap_bender_1...(a)hotmail.com> wrote: > > On Jan 14, 10:23 pm, mike3 <mike4...(a)yahoo.com> wrote: > > > > Hi. > > > > I saw this: > > > >http://news.bbc.co.uk/2/hi/uk_news/education/6589301.stm > > > > What sort of math level would that require (to be able to do problems > > > like the one on that Chinese test)? > > > That would require a much better translation level, because the phrase > > "the foot of perpendicular is E" is highly ambiguous. > > I would guess it means E is the intersection of AC and BD, which we > already know are perpendicular. Not that E appears in the three > questions. > > I also don't see what "square" means in "square prism ABCD-A1B1C1D1" > since ABCD and A1B1C1D1 are not squares, and nor is ABB1A1 or ADD1A1. > I suppose it might mean A1B1C1D1 is congruent to and vertically above > ABCD. > Well, not just congruent but equal, given that it is vertically above ABCD. In any case, that's how I interpreted it too: a "right" prism, meaning: it's side edges are vertical.
From: Ostap S. B. M. Bender Jr. on 15 Jan 2010 04:53 On Jan 15, 1:16 am, Ask me about System Design <grpad...(a)gmail.com> wrote: > On Jan 15, 12:54 am, "Ostap S. B. M. Bender Jr." > <ostap_bender_1...(a)hotmail.com> wrote: > > On Jan 14, 11:26 pm, Ask me about System Design <grpad...(a)gmail.com> > > wrote: > > > > On Jan 14, 11:08 pm, "Ostap S. B. M. Bender Jr." > > > > <ostap_bender_1...(a)hotmail.com> wrote: > > > > On Jan 14, 10:43 pm, "Peter Webb" > > > > > <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote: > > > > > "mike3" <mike4...(a)yahoo.com> wrote in message > > > > > >news:d4f6fc15-c015-4833-b108-06d60390b18d(a)l30g2000yqb.googlegroups..com... > > > > > > > Hi. > > > > > > > I saw this: > > > > > > >http://news.bbc.co.uk/2/hi/uk_news/education/6589301.stm > > > > > > > What sort of math level would that require (to be able to do problems > > > > > > like the one on that Chinese test)? > > > > > > Well, I haven't tried to solve it, but not much I suspect. > > > > > > Alternatively (and equivalently) you could restate the problem in the > > > > > language of cartesian co-ordinates and vectors, and solve it algebraically > > > > > rather than geometrically. This still only requires high school maths. > > > > > You can do THAT in the 5 minutes or so allocated to this problem on > > > > the exam?! You must be a walking supercomputer. > > > > I did not see where it was supposed to take only > > > 5 minutes. Having not practiced such problems in over > > > a decade, it took me less than 20 minutes to do in my > > > head, with iii) being the challenging part. In my > > > heyday, I could have probably done it in 5 minutes with > > > pencil and paper, and perhaps another minute to check > > > my work. > > > 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. > > I think you are combining my reply with that of a > different poster. > No. I think the problem is that in replying to my post, you didn't read what I was replying to. I had asked the previous poster - Peter Webb - how he would solve this geometry problem using the analytic geometry methods. Did you read the previous message or did you allow Google Groups hide it from you? > > I solve geometry problems using a > variety of methods, including various instances of > coordinatizing the diagram using helpful points of > origin and axes. For this problem, finding right > triangles, computing lengths, and eventually computing > a dot product mentally were what I did. I even checked > the answer against what the RCS published, and (except > for not computing an inverse cosine to get the angle > in part iii) my answers agreed with theirs. > > However, I can solve the problem on paper using mostly > just analytic geometry with a single coordinate system. > Really? I would be intrigued to see such a solution. For the life of me, I can't see myself solving this problem using analytic geometry in less than 50 minutes at best. > > If you tell me where you got the figure of 5 minutes, > Well, it says: "a sample question from Chinese university entrance tests". My own knowledge of American and Russian college entrance tests tell me that the average time allocated to one individual problem cannot be more than, say, 5 or 10 minutes. > > I > will expand on the approach I just mentioned, which is > as close as I will go to "solving the problem > algebraically" . I might even convince you that such an > approach might take under ten minutes. > I would be impressed if you did.
From: Ostap S. B. M. Bender Jr. on 15 Jan 2010 05:09
On Jan 15, 1:40 am, "Ostap S. B. M. Bender Jr." <ostap_bender_1...(a)hotmail.com> wrote: > On Jan 15, 1:17 am, Henry <s...(a)btinternet.com> wrote: > > > > > On 15 Jan, 07:06, "Ostap S. B. M. Bender Jr." > > > <ostap_bender_1...(a)hotmail.com> wrote: > > > On Jan 14, 10:23 pm, mike3 <mike4...(a)yahoo.com> wrote: > > > > > Hi. > > > > > I saw this: > > > > >http://news.bbc.co.uk/2/hi/uk_news/education/6589301.stm > > > > > What sort of math level would that require (to be able to do problems > > > > like the one on that Chinese test)? > > > > That would require a much better translation level, because the phrase > > > "the foot of perpendicular is E" is highly ambiguous. > > > I would guess it means E is the intersection of AC and BD, which we > > already know are perpendicular. Not that E appears in the three > > questions. > > > I also don't see what "square" means in "square prism ABCD-A1B1C1D1" > > since ABCD and A1B1C1D1 are not squares, and nor is ABB1A1 or ADD1A1. > > I suppose it might mean A1B1C1D1 is congruent to and vertically above > > ABCD. > > Well, not just congruent but equal, given that it is vertically above > ABCD. > > In any case, that's how I interpreted it too: a "right" prism, > meaning: it's side edges are vertical. > I actually tried to solve this problem, and in one minute I did get part (i): Since we know that BD is _|_ to both AA1 and AC, it must be _|_ to the entire plane AA1C, and thus to A1C. That wasn't too bad. However, I didn't have to use any of the given lengths. Are parts (ii) and (iii) more difficult? (Yes, I have grown rusty and lazy with age) |