tag:blogger.com,1999:blog-8231784566931768362.post9135105030882574151..comments2021-06-16T05:56:38.112-04:00Comments on MathNotations: Two Geometry Problems To Sharpen The Mind - Never Too Late In the Year For That!Dave Marainhttp://www.blogger.com/profile/13321770881353644307noreply@blogger.comBlogger9125tag:blogger.com,1999:blog-8231784566931768362.post-54659307619093323362009-07-27T12:13:39.268-04:002009-07-27T12:13:39.268-04:00I belive student would spend time trying to find a...I belive student would spend time trying to find all the solutions by using a listing strategy. Few student wold take an abstract approach. For example in the first question what do we really need. we need two odd numbers that add to fifty, and we have our answer. there are 25 combinations that add to fifty, twelve of which are two even numbers, and thirteen are two odd numbers. I do not think most students would think of that initially.marcelhttps://www.blogger.com/profile/14622826687954248660noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-62644497141866360152009-06-08T22:00:08.558-04:002009-06-08T22:00:08.558-04:00Agreed, Eric, on all counts...
The bigger issue f...Agreed, Eric, on all counts...<br /><br />The bigger issue for me is whether there should be a place for these kinds of problems in a regular curriculum. I think you know how I feel but a natural question is how and when does one bring these into the classroom.Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-10557851259750332382009-06-08T20:08:54.553-04:002009-06-08T20:08:54.553-04:00Problems like these seems to be much more algebrai...Problems like these seems to be much more algebraic than geometric. Even though they speak of areas and perimeters, the restrictions on the sides to integers of specific parities mean that their geometric contents are minimal. Only the most basic techniques of geometry show up in their solutions. Last week's problem on the length of the tangent to a circle in the first quadrant was much more geometric.<br /><br />Whether the student counts the number of solutions or applies algebra does not really matter. I would prefer an algebraic solution, but the time pressure of the SAT makes counting plausible.Eric Jablownoreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-87781098998976615342009-06-07T18:12:13.135-04:002009-06-07T18:12:13.135-04:00Beautiful analysis, Eric, and some nice manipulati...Beautiful analysis, Eric, and some nice manipulations. <br /><br />I would agree with you about about the inequalities for those students who approach this problem analytically. However, when confronted with this type of question, most students I've worked with plug in values for the sides. They might start with the largest possible value of the equal sides or, equivalently, as I did, the smallest possible value of the 'base'. If they start with a base of 2, they have to remember to keep the other sides even. My gut feeling is that it's one or more of the following which would trip them up:<br />the even condition OR<br />the triangle inequality OR <br />the equilateral case.<br /><br />As an instructor, I would certainly want to show them the algebraic analysis as well. I also think it's a good exercise in simple number theory for them to understand why the 'base' has to be a multiple of 4.Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-69763204326917943362009-06-07T17:52:21.632-04:002009-06-07T17:52:21.632-04:00For part 2, let the triangle have sides a, a, and ...For part 2, let the triangle have sides a, a, and b. B the triangle inequality and the the sides being even, <br /><br />2 ≤ b ≤ 2a-2.<br /><br />So, 2a+2 ≤ 2a+b ≤ 4a - 2.<br /><br />2a+2 ≤ 96 ≤ 4a - 2.<br /><br />This is where students will have the most problems.<br /><br />2a + 2 ≤ 96 means a ≤ 47, and so a ≤ 46 by the problem conditions.<br /><br />96 ≤ 4a - 2 means 98 ≤ 4a, 24.5 ≤ a and so 26 ≤ a by the problem conditions.<br /><br /><br />26 ≤ a ≤ 46. This sort or inequality reversal always seems to give students fits. Since a must be even, there are 11 solutions in this range.Eric Jablownoreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-3807906536387096742009-06-06T18:52:01.470-04:002009-06-06T18:52:01.470-04:00I know that feeling and I'm retired!
Compared ...I know that feeling and I'm retired!<br />Compared to some of the errors I've made....<br />At least yours can be attributed to end of year brain freeze!<br />Seriously, when you get a different answer from mine, I assume I made the error!Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-74778905889556466872009-06-06T17:37:30.446-04:002009-06-06T17:37:30.446-04:00Holy Guacamole.
I actually had a triangle that wa...Holy Guacamole.<br /><br />I actually had a triangle that was 2,2,92 ...<br /><br />I am really wiped out. This year really needs to end and I've got two weeks left.Curmudgeonhttps://www.blogger.com/profile/04323026187622872114noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-20457678679265490302009-06-06T12:06:11.136-04:002009-06-06T12:06:11.136-04:00I may have messed up but I believe the "base&...I may have messed up but I believe the "base" of the triangle must be a multiple of 4. E.g., if the base were 2, then the 'legs' would be 47 and 47. The conditions required that all sides be <i>even</i> integers! Thus the 'base' could be any positive integer multiple of 4 up to and including 44, hence 11 possibilities. This includes the equilateral case.Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-44267550031561513252009-06-06T11:53:11.980-04:002009-06-06T11:53:11.980-04:00Now I'm concerned ...
#2) I got 22 and a pos...Now I'm concerned ... <br /><br />#2) I got 22 and a possible 23rd if you count the one special case as isosceles - what the heck am I missing? <br /><br />#1) I agree with the 13.Curmudgeonhttps://www.blogger.com/profile/04323026187622872114noreply@blogger.com