Thursday, January 25, 2007

Problems for 1-26-07

Uh oh!! The original post appears to have been deleted. I also thought I lost all your comments but now they're back! I may have inadvertently done this or Blogger is doing weird stuff...
I am working on a better way to display the problems and diagrams since I know these images are too small or fuzzy and difficult to copy/paste for use with your students. Pls be patient...

I am planning something new for Monday - a more extended challenge that you can give to your interested students. We used it as part of an online contest - stay tuned


Unknown said...

Pretty nice problems!

For (1), I assume that ratios of sides in well-known triangles can be used.

For (3), one way to avoid the blind use of the calculator is to say (6.yy-x)(6.yy -x) where .yy is some decimal other than 0.

For (5), for extra credit, you may also ask the students to speculate on the ratio between the two numbers that give the maximum product. Related: Ask them to find two numbers that add up to a fixed sum that give the highest product.

Unknown said...

2 is fun: Let's see: 10! = 2*3*4*5*6*7*8*9*10 = 2*3*2*2*5*2*3*7*2*2*2*9*2*5, get rid of the 2s = 3*5*3*7*9*5

4 is easy too: 0.5(36-x^2) is an upside down parabola 18-0.5x^2 and will be maximum when x = 0 → 18. Another way to look at it is trying to maximize the area of the rectangle with constant perimeter; that'll obviously be a square so 6*6.

3 should be straightforward: x^2 + 6x + 9 + x^2 - 6x + 9 = 2x^2+18; x+18=20.

1 I need to sketch, but as TC says, if you drop a perpendicular from C you get a 30-60-90 and 45-45-90 triangle. Let's call that xR3, then AC is 2x. BC is xR6. So the ratio is R6/4, assuming I didn't make any stupid mistakes.

6 I hope I'm not messing up my obviously the original triangle was a 30-60-90 since one smaller triangle has a 30 and 90 degree angle and the larger triangel has a 60 and a 90. So....the short segment will be x, so the perpendicular is xR3 since it's opposite the 60. This is the 30 of the larger tringle, and opposite that 60 is xR3R3 or 3x. So ratio is 3:1, right?

5 I have no idea how to do aside from trying all the possibilities.

Unknown said...

Although for 5 since within each number the digits must go from max to min, there are only 4 + 3 = 7 possibilities, I believe...I could just try each that cheating?

Darmok said...

Er, the above is me Darmok; I'm at work and forgot to sign in.

Anonymous said...

1) I dropped a perpendicular from C to AB to make the 30-60-90 and 45-45-90 triangles. Lets say the point where C meets AB is D.

In 30-60-90 triangle CAD, if we let AD be 1 unit, AC is the hypotenuse so 2 units, and CD is R3. In the 45-45-90 triangle CBD, CD is already R3, os BD is also R3 and BC is R2xR3 or R6. So I'm getting R6/2 as the ratio of BC to AC

2) Love this one.
10! = 10*9*8*7*6*5*4*3*2*1
= 5*2*3^2*2^3*7*3*2*5*2^2*3*2*1
dropping all the 2's to make it odd (and the 1 which doesn't contribute anything) we get
(ooh, ooh, Mr. Marain, are we allowed to use calculators?)
= 14175 (yeah, I used a calculator)

3) This is just straightforward algebraic expansion -- I'd say Algebra I not II. (3+x)^2+(3-x)^2 = 9 +6x + x^2 + 9 - 6x + x^2 = 18 + 2x^2. If 18 + 2x^2 = a + bx^2 then a=18, b=2, and a+b = 20

4) 1/2*(6+x)(6-x) = 1/2*(36 - x^2)
without graphing or plugging into a calculator: we know that x^2 is always non-negative (for real x), so 36 - x^2 and hence the entire expression is maximized when x^2 is 0, when the expression has value 18

5) Hmm, I'm coming up with 24 possible combinations of a 1-digit number times a 3-digit number, and another 12 possible combinations of a 2-digit number times a 2-digit number. (If you set up the problem as _ * _ _ _ or _ _ * _ _ there are 4*3*2*1 ways to arrange the digits. In the 2-digit case, you list each combination twice since 24 * 31 is the same as 31 * 24)

To maximize, I tried maximizing the 1 by 3 and the 2 by 2 cases separately. To maximize the 1 x 3 case, you want the 1-digit number to be the 4, and the 3-digit number to have its digits in decreasing order: 4 * 321 = 1284.
For the 2 by 2 case, you want 4 and 3 in the tens places, and the 2 in the ones place to be where it is multiplying the forty-something number and not the thirty-something number. So 41 * 32 = 1312. That's bigger, so I think that's the max.

6) ooh, more 30-60-90 triangles.
Let's call the original triangle ABC where C is the 90-degree angle. Let's call the point where the altitude from C crosses the hypotenuse point D. And let's let ACD be the 30-degree piece of that chopped up angle C, and DCB is then the 60-degree piece.

Let AD be 1 unit. then AC is 2 and CD is R3. Then in triangle BCD, since CD is R3, CB is 2R3 and BD is R3R3 or 3.
The requested ratio is the ratio of side BC to AC which is 2R3:2 -> R3:1

Anonymous said...

Why on Earth are you using Word to type mathematics?

Dave Marain said...

'Unofficial' answers to be posted in an hour or so...
These problems were fairly straightforward but remember my goal is to provide some challenge for students in their current curriculum; many geometry students are getting to the Pythagorean Theorem and special right triangles soon so i wanted to give several of these...
Great comments thus far!

good question, eric! I'm working on a mac; all i have is Word and Equation Editor; no LaTex. Word at least allows me to make equations and draw diagrams but it's very crude...
Pls give me ideas here! I've used Math Type but Eqn Editor is a stripped down version so it's ok. The problem is that the equations and drawings require me to store these as images (.png files). If I took the time to use more html it would be better but I'm under a lot of time pressure...

Anonymous said...

Some alternatives:

(3) The equation is true for all x, so set x=1, RHS becomes a+b, which is what we want.

(4) (6+x)(6-x) is symmetric about the y-axis, and should have a maximum rather than a minimum since the coefficient of x^2 is -ve. The max is at x=0 due to symmetry.

Dave Marain said...

Answers and comments to 1-26-07 problems:

1) Rad(6):2
2) (3^4)(5^2)(7) in factored form
3) 20
4) 18
5) 1312
6) 3:1

By the time i post these, everyone pretty much knows the correct answers and several excellent explanations have already been given. However, I have a few thoughts and suggestions for extensions. Remember none of these are in the same league as jonathan's 'puzzles' like the recent quadrilateral problem he posed. I defer to him when it comes to the more advanced thinkng problems. Here, I want teachers to believe that these questions are reachable for many students. Of course, his 'what if' scenarios are awesome so I'll keep looking in to be challenged!

Ok, here's what I did with my 'skills' class with number 5. First, they struggled with this for quite a while and finally some got 1312, but, of course, they weren't sure they had the biggest. Another approach would be to give them the 'answer' and have them find the factors to produce. Believe it or not, I've tried that too and they still struggled. My role is to structure this by requiring they must record every attempt on paper before keying it in. They fight this of course, but this organization is critical for their development. As far as reasoning is concerned, if I leave them alone to explore, it'll mainly be trial and error and little thinking. Therefore I ask a lot of questions to get them thinking. Some stubbornly want to stay with a 3-digit by a 1-digit.
Ok, here's where we move to the next level (remember, 'skills' class): I ask for the 'winning' factors: 32 x 41 and these are written on the board. We discuss why it make sense for the 2 highest numbers to be in the tens' place and that seems easy to them and we notice that the other combination 31 x 42 is close but 10 less. No one is commenting about how in 32 x 41, the numbers are 'closer'.
Here's where we look for patterns and begin to generalize. i ask the group to do the same if the digits 6,7,8,9 are used. One young lady, K.C., suggests 86 x 97 while the others are banging away on the calculator not even considering if there might be a pattern. K.C.'s product beats everyone so now they listen to her explanation. She lines up 6,7,8,9 directly under 1,2,3,4 and says," just use the new numbers in the same place as before." The others are impressed so I asked C.D. to give me any 4 other single digits and we verify that the prduct is maximized using a similar pattern.

The period is closing so I pose the challenge problem: Use the 5 digits 1,2,3,4 and 5 and do the same thing. A lot more combinations now. The first one to email me a correct answer with the factors would earn 2 bonus points. I'm still waiting for that email! The next day, i TOLD them the largest product is 22,412, find the factors.

Enough for now, I'm nodding off at the keyboard!
The next round for Monday may not go up on Fri. We'll see...

Anonymous said...

Oops, I read the problem wrong for #6. I was comparing the legs of the big triangle, rather than the segments that the hypotenuse was divided into. I was thinking that "segments" was an odd way to describe the legs, and missed the fact that the "segments" in question were "defined" in the previous sentence. Well, at least my picture is right. :) My geometry is rusty, so I like working problems at this level to un-rust it. The highest level kids I am working with is middle school Mathcounts kids, so this is about where I need to be (and be able to get them to, ideally).

Anonymous said...

The 5-digit challenge is quite interesting.

One interesting thing I noticed when playing with it is that
521 x 43 > 531 x 42 BUT
421 x 53 < 431 x 52
I don't know exactly what you mean by "skills" class but I'm guessing that these are not the most mathematically inclined kids, so asking them if they can see why might be a bit much, but it is interesting to look at. I found this unintuitive at first. I was also a little surprised that the pattern holds no matter which numbers you use, even if there is a gap, like 1,2,3,4,9. Intuitively, I want to optimize the hundreds place first, especially with that big gap, but that doesn't pan out. (If you use 5,6,7,8,9 the maximum product is an even number of thousands, which the students might find cool.)

Anonymous said...

oops, typed my name wrong above...

Anonymous said...

TeX is available for the Mac. XeTeX is wonderful; it works with fonts and unlocks all their capabilities.

Start with the TeX on Mac OS X site.

Then, use MetaPost for drawing figures. All these packages are available through the TeX on Mac OS X link, through the i-Installer program. I hate equation editors. They take me right out of the typing mood.

Anonymous said...

Wow, do you make those problems up yourself? I'm also an educator of sorts (a tutor), and I really admire your dedication to teaching math. I've started a blog about education yesterday, and I've just been surfing around for education-related blogs. Give me a visit at sometime, when you have time or if you feel like it. It's still a work in progress, but I hope to make it a great source of educational information, much like yours is for math. It'd be nice if we could bounce ideas off each other from time to time!

Anonymous said...

Anonymous, that's a great way to solve 3--I never thought of that!

Dave Marain said...

thanks for the info on TeX for OS X; I've been using Eqn Editor and Draw for so long, I'm actually pretty good at it but I know it's not the smart thing to do; i will have to find out the costs of these -- i'm assuming they don't give it away!

darmok et anon--
I agree -- some nice methods! Students do not appreciate the idea of 'true for all values of x'. Not enough attention is paid to the concept of an identity. Choosing x = 1 in a polynomial will always give the sum of the coefficients, a common math contest problem. I will have an alternate form of this soon. The other method of 'matching up the coefficients' is very powerful in math (e.g., partial fractions, method of undetermined coefficients in differential eqns). Students automatically assume that one can always equate coefficients but can they explain why it's true! Heres' an example to get your strong Alg 2 students to appreciate the beauty of this:
x^3 - 3x^2 +2x - 5 = ax^3 +bx^2 +cx + d -->
(a-1)x^3 + (b+3)x^2 + (c-2)x + (d+5) - 0
If that 3rd degree polynomial equals zero for ALL values of x, then ALL the coefficients must be ZERO or it would violate the Fundamental Theorem of Algebra which states there are AT MOST 3 'roots' for a 3rd degree polyn equation (subject to some restrictions). Students match up coefficients all the time without any thought about it. Would this argument work for non-polynomials!

i will check out your site; good luck! you're making me feel like a veteran when less than 30 days ago, nobody knew i existed; you're quickly figuring out the secret of spreading the word! if people like your content, they'll return...yes, i write my own problems! otherwise i would cite my sources and give credit to the authors...i love writing questions that are related to actual content being taught in the curriculum but are more thought-provoking; here's the plain truth as i see it:
Almost any math problem we teach can lead to deeper discussions in the class if the instructor asks more probing questions. The problem is that beyond middle school, math teachers tend to be content driven which by necessity leads to an emphasis on mechanical superficial thinking on the part of students; the way to combat this is to get through the content but ask questions that force reflection, questions that lead to disequilibrium and cause the student to think: "Wow, is that just a coincidence, or is there a reason for that!"

Anonymous said...

Those TeX programs are absolutely free. The licensing are typical open-source licenses; just give credit, and rename them if you change them. Just download them if you want, Oh, you could join the TeX Users' Group, as I have, but that's hardly a requirement.

I recommend the PracTeX Journal too.

Dave Marain said...

thanks eric!!
luv that open source code!!
now, how much do you charge per hour for tutorials and can i travel to your house!
seriously. i will look at it -- my problem is severe time-crunch; i'm barely able to find the time to get these problems in every day; my wife is very jealous of my affair with some woman named blogger!

Anonymous said...

Dave, take the plunge, find a "quick start" reference sheet, and just dive in and do it. TeX isn't really hard to learn, most of it is intuitive, and you save all the mouse-clicking of using an equation editor. My advice, to steal verbage from a series of Nike commercials: Just Do It. I think even at the start, it won't take any longer than what you are doing now, except for whatever overhead you have in installing software. Good luck!

Dave Marain said...

thanks mathmom -- i needed that kick!!
now pls call my wife and explain this to her!

Anonymous said...

Dave, do not spend so much time blogging that you get in trouble with your wife. If you had to go down to posting problems every other day, we'd all survive. ;-)

Anonymous said...

Actually, (ducking the bricks likely to get hurled my way), I would almost prefer fewer problems at a time. Three might be just enough to maintain a good discussion flow.

On another topic, is there MacTex or something like that?

Anonymous said...

jonathan, look at eric jablow's post above for information about TeX for Mac

Dave Marain said...

22,412 is the answer to the maximum product you can form from the digits 1,2,3,4 and 5 and, yes, the pattern works for any 5 digits you choose, a beautiful result...

Thanks for all the great advice, folks!
I checked out the TeX site and it doesn't look that difficult so maybe one can teach an old dog new...

jonathan, i tend to agree with you that less is more so I will probably cut back to 3-4

thanks mathmom, i needed to hear that before getting too burned out!

for mon 1-29 i plan on posting a different kind of problem - more sophisticated - that i used in a weekly online challenge i established a year ago; i would post the problem at 6 PM every Monday on our dept web page and students had 2-3 days to submit their solutions via email; these problems usually had several parts and i patterned some of them after AMC or AIME problems. I would announce the winners and post the best solutions online; it generated surprising traffic and the young lady who won the overall competition (after 2 months) is now in my BC Calc class.
i will still use an image file (.png) so it will appear tiny but you can click on it to magnify it (using Google's Picaso)
I will probably post it on Sat or Sun so look for it and let me know what you think...

Anonymous said...

One more request, while we're at it. I don't know you if you can post the files in a different format on here, but the .png format makes it hard to take a subset of the problems to use with my own students. I wonder if you could post .pdf or something with cut/paste ability? I get the impression that you'd like us to be able to use them with our students, but in any given set, only some will be appropriate for a particular group of students.

Not to make more work for you and get you in more trouble with the wife, of course. ;-) But while you're learning new tricks....

Anonymous said...

There are a few LaTeX to html programs. Most convert each equation to a PNG. Your Mileage May Vary, of course. Or, you could try a MediaWiki installation, like WikiPedia. That would require switching from Blogger, I think, so forget that.

Dave Marain said...

Blogger is only accepting my images in .png format and it definitely makes it harder to copy/paste. I may be able to get it into a compressed pdf but I'm not sure

I was wondering about the TeX to html format question; so I typeset it in TeX, then have to convert to html which makes it into .png all over again?

I'm thinking blogger was not my best option!
Too bad i can't export all this to wordpress or something similar; wordpress seems to have many built-in features that blogger doesn't and it might be more stable. i like that mouseover feature where you get a quick preview of another blog or whatever page you're planning to link to -- it's small but it works!
i'm wondering how much html you folks insert into your own pages...

Anonymous said...

Dave, if you have somewhere where you can host files for people to download, you could just post a link to the .doc or .pdf or whatever below your picture.

As to Wordpress, I just started, so I'm not sure what it can and can't do. You can import your blogger blog onto wordpress, but you'd have to re-import all your images by hand, which would be an issue for you. See wordpress' FAQ on the topic

p.s. I was wondering what happened to your blog last night! Glad you got it all back.

Anonymous said...

Dave,there are a few LaTeX→HTML programs. LaTeX2HTML,for example, is a Perl program that splits a LaTeX document, formats the text as HTML, turns each equation into a PNG, and includes the images in <IMG> tags. Cross-references are turned into <a> tags.

Essentially, you create the LaTeX document. You run pdflatex to turn it into a PDF you can print. You run latex2html to turn it into a HTML file and directory of PNGs you can post on your web site. How you publish that on blogger is up to you.

Alternately, some converters turn the equations into MathML, and then a browser with a MathML plugin shows the equation directly. Believe me, you do not want to create MathML by hand.

This page by Craig Small discusses alternatives. Also, look at the hyperref package. It works with the programs and allows liking in pdf output.

The UK TeX Users' Group discusses alternatives in its FAQ page. In any case, don't worry; you won't get one giant PNG for your entire document.