The above problem is bogus, it is not possible to have a correct diagram with the angles and side length expressions given in the problem. |
There is no way that most of those 35 problems are tough. Maybe you meant to say tedious, or boring, or time consuming, which explains why your kid had to spend that long on them. A tough assignment would have 5-10 problems, maximum. |
x=3.5 (assuming degrees), y=5, seems to work? |
Yes those are the right values algebraically, but after plugging in and labeling the resulting given angles and lengths in the diagram, it is not geometrically possible. |
“Coy excuses”??? My kid is at school and their homework is with them. No I’m not going to jump through hoops for you. |
Except when it’s part of the magnet program. Yes, there were 35 challenging problems. I don’t understand why you think you are better than everyone else. |
It's not that at all, it's more of you should try to prove your case when you make the above claim. Others asked you to give examples and you've refused. At this point in time your statement is misinformation and I don't want others to come to the wrong conclusion. I'm telling you that a 35 problem packet means that most of the questions are straightforward and routine. It's just quantity. |
This whole conversation is stupid. 1hr, 3hrs, 6hrs. Easy, hard, routine, challenging. Every student has a different level of ability, speed, preparation, interest, and other demands on their time. Instead of all these out of context opinions on the course, people need to see it and decide for themselves. |
Oh, I see. The side opposite B is too short to allow the angle at B. I was focused on a different reason why the problem is impossible to solve as stated. Anyway, it certainly fails to instill confidence in the quality of the magnet math program, unless the problem was designed to intentionally troll the students, and was explained during class. |
Highly unlikely it was intentionally designed that way, especially since it didn't ask students to think/notice anything and just demanded a proof. Much more likely that whoever assigned it probably thought it would be ok to just change numbers and have the algebra work but didn't check the geometry. |
Absolutely. I can only tell you my experience. My kid who has breezed through everything thrown at them to date took maybe 8 hours total on this packet. I’m not sure exactly because I didn’t time it and they had homework for other classes they were working on some of the time as well, but kid is not easily distracted and spent a considerable amount of time on what seemed to be an excessive number of questions none of which were quick to solve and certainly weren’t easy. Why PP feels the need to argue or that they have some special knowledge of the types of problems or the preparation of the kids, I don’t know. There’s no quantifiable definition of tough so PP is just trying to assert superiority by arguing about something that cannot be defined and trying to denigrate the work of kids who are putting in a lot of effort. FWIW, functions has more problems over more pages so I guess it’s an easier class by PPs reckoning. |
I think it's even worse than that. I'm pretty sure (my kid and I agree, but maybe we are wrong) that no matter what any of the *numbers* in the problem are, it is *never* possible to determine whether A is acute or obtuse in a problem with that configuration. For some choices of numbers, you can proof that A is a right angle, but for other numbers, if there is an acute solution, there is an obstuse solution also. This is a theorem of Geometry (SSA *non*-congruence.) |
Well the numbers in the problem are already predetermined by solving for the variables x and y in the problem. First solve for x (the only non-extraneous solution here is x = 7/2), then that allows you to conclude that the triangles are similar (because when you plug the above x value in the expressions, both <ABD and <ECD come out to the same value of 43 degrees, hence similarity by AA). Now, you can use the four side lengths that are given in terms of y... just write out a similarity condition and solve it to get y = 5. So now you can label all the expressions in the problems with their values, so those are determined. However, the diagram is not actually possible, (for example if you use Law of Sines on one of the triangles, say triangle CED so that you can find <ECD, you will get an equation that does not have a solution because sin<CED will be greater than 1. |
I mean that even if the expressions are different, leading to different values for x and y, and different side lengths and angles, and even if that does make a valid triangle, "prove A is obtuse" is still impossible, because either A is a right angle or else there are 2 valid triangles, one with A acute and one with A obtuse. |
Sorry I misunderstood what you meant, yes you are absolutely right, even if the expressions were changed to actually yield a configuration that is geometrically possible, there would be two possible configurations, one obtuse and one acute, so it's impossible to 'prove that <A is obtuse'. In this sense, this problem is quite problematic, no pun intended. I would also argue that this problems is just plain stupid because it doesn't really teach any meaningful geometry. No meaningful geometry problems try to constrain the angles with algebraic expressions the way it is set up here. So I'd say it's a waste of time for multiple reasons. |