Reply to How hard will Blair's Functions class be for a kid who currently finds Algebra 2 "easy"?

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Reply to "How hard will Blair's Functions class be for a kid who currently finds Algebra 2 "easy"?"

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[quote=Anonymous][quote=Anonymous][quote=Anonymous][quote=Anonymous][quote=Anonymous][quote=Anonymous][quote=Anonymous][quote=Anonymous][quote=Anonymous][quote=Anonymous][quote=Anonymous][quote=Anonymous][quote=Anonymous][quote=Anonymous][quote=Anonymous]Can one of you PPs share an example of what the problems are like? [/quote]

Why? Are you the parent of a current student?[/quote]

If I were a parent of a current student, I would post a picture of homework.

I am a parent of prospective students, so I'd like to know what the class is about, o see whether it would be a fit for my students, and also to see what to target if they don't go to Blair but want to pursue a similar course of study on their own. [/quote]

1) I’m not sharing my kid’s homework here (it’s completed)
2) you can’t post a photo without having it hosted somewhere which is usually not anonymous
3) I’m not sure how sharing a problem or two at the start of the year would give you an accurate overview of the breadth and depth of this program[/quote].

1. I presume that not all the homework always completed before you see it.
Also, often homework problems are losted digitally or are printed not on the same page as the student work, especially for hard problems that take several minutes of work.

2. https://imgbb.com/
Free hosting without login.

Example: (funny problem I saw in the  Precalculus C summer math packet at https://mbhs.edu/departments/magnet/Summerpackets.php , the closest thing I could find to an example of what Precalculus/Functions covers.)

[img]https://i.ibb.co/hdYMfvw/Screenshot-20230911-092644.png[/img]

That's

https://i.ibb.co/hdYMfvw/Screenshot-20230911-092644.png  , wrapped in the DCUM [ img ] tag 
(Long press or right -click on uploaded image, to get the address of the image.)

3. Commenters said that it's already 3+hrs to do a packet this week, so it seems representative enough, certainly better than the nothing already public (except for the summer packet which is the same normal MCPS Honors math packet).

Look, you are a free human being. You don't have to post anything if you don't want to. But the coy excuses are just weird.

Have a great day. [/quote]
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.[/quote]

x=3.5 (assuming degrees), y=5, seems to work?
[/quote]
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.[/quote]

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. [/quote]
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.[/quote]

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.)
[/quote]
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.[/quote]

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. [/quote]
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.[/quote]

It's perilously similar to the classic (probably apocryphal) V.I. Arnold story of how American geometry (math) classes and students don't understand geometry (math).

https://math.stackexchange.com/questions/1594740/v-i-arnold-says-russian-students-cant-solve-this-problem-but-american-student[/quote]

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