Friday, February 2, 2007

Problems 2-2-07

Posted by Dave Marain at 5:34 AM

22 comments:

Anonymous said...

I'm going to tackle #3 since I'm working on my geometry these days...

Answers:
SM = 7/10
a) only special thing I can see about triangle PSM is that it is (similar to) a 7-24-25 right triangle
b) (b^2-a^2)/2*sqrt(a^2+b^2)

I'll post my reasoning later, when I have more time

February 2, 2007 at 11:18 AM
Unknown said...

Suggested problem:

Within a given area code, phone numbers assigned by the phone company are seven digits long. By numbering rules, the first digit of the seven-digit phone number cannot be 0 or 1. The first three digits also cannot be 555, which is reserved by the phone company for special purposes. How many phone numbers can be assigned within a given area code?

February 2, 2007 at 4:13 PM
Anonymous said...

Reasoning for #3

Triangle QPR has legs with lengths 3 and 4, so it is a 3-4-5 triangle.

Since triangles QPR and QSP both have a right angle and also share angle Q, they are similar triangles. The QP/QR = QS/QP and subbing in we get QS=9/5 (and similarly SP = 12/5)

Since M bisects QR, QM is 5/2 units long. SM, our desired length, is QM - QS = 7/10

a) I wasn't sure what you were looking for, but I thought it was interesting that all of the lengths were rational. Converting to a common denominator showed them to be in the ration 7-24-25, a less-well-known Pythagorean triple.

b) if we let QP be a and PR be be, then QM is half of the hypotenuse, that is sqrt(a^2+b^2)/2. QS is found using the similar triangles to be a^2/sqrt(a^2+b^2), so by converting to a common denominator of 2*sqrt(a^2+b^2) and subtracting we get that SM = (b^2 - a^2)/2*sqrt(a^2+b^2)

This assumes that a is the shorter leg of the triangle. If a is longer, you get a negative answer with this formula, since S would be on the other side of M. So we can either put absolute value around the solution, or specify that a is the shorter leg.

February 2, 2007 at 7:48 PM
Anonymous said...

TC, how about this one: How many different 7-letter "words" can be formed using the letters: TOTALLY

February 2, 2007 at 7:50 PM
three of clubs said...

my take on altitudes to the hypotenuse

1) note that we have a 3-4-5 right triangle; midpoint is 2.5
2) the area of the big triangle is 6
3) so the altitude is (2*6)/5 = 2.4
4) sqrt(9-2.4^2) = 1.8
5) and so SM = 2.5 - 1.8

I can of like doing it without the tricky similar triangles.

February 2, 2007 at 7:50 PM
three of clubs said...

Were you pointing towards right triangles where one leg and the hypotenuse are consecutive numbers and add up to a perfect square?

February 2, 2007 at 7:56 PM
Anonymous said...

Mathmom,

Ah! That gives me more memories of high school. 7!/2!/2! should do it.

TC

February 2, 2007 at 8:21 PM
Dave Marain said...

TC --
Nice combinatorial problem.
Here's my thinking (correct me if I'm wrong):
Ignoring the '555' restriction leads to 8*10^6; from this we have to remove all of the 555-#### combos which would come to 10^4

mathmom -- nice solution of my triangle problem! yup, 3-4-5 generated a 7-24-25 triple, a coincidence? For the general expression, I find it very useful to teach the 'area' approach since the similar triangle relationships, while important, are far more difficult for most students. By this I mean that a*b = c*h, where h is the altitude on the hypotenuse 'c'. Thus, h = ab/sqrt(a^2+b*2). The length of the median to the hypotenuse is half of the hypotenuse by a famous theorem I mentioned yesterday. From there, I used the Pyth Thm and simplified. Similar triangles would have been another approach. Yes, you would need absolute values in the result. Note the result if a = b!

Ok, so no one wants to comment on the crazy square root domain problem, huh? I'm sure you hate this as much as my students did! (Given as a bonus).

How about #1? Just another exhaustive list problem? I think this requires some nice logic in addition to making an organized list. As a start, I ask students to start organizing by the number of 9's in the number: three 9's, then two 9's, then one 9, then no 9's. It's still a pain but it reviews some nice combinatorial methods.

February 2, 2007 at 8:40 PM
Anonymous said...

On problem (1), after recently being introduced to Kakuro, I was wondering how it would be possible before realizing that repetition was allowed.

The area method that Dave and 3 of clubs point out is quite cool.

And yes, I did (2), but did not enjoy it.

During my school days, I found that a problem relating to a real-life situation motivated me more than an abstract one. The combinatorial problem, simple as it is and as Dave showed, was formulated in that vein. In another example, the infinite series problem I still remember goes like: A ball is dropped from 12 feet high and bounces a third of the height from which it drops. Find the total distance traveled by the ball.

TC

February 2, 2007 at 10:49 PM
Anonymous said...

I didn't think of the area approach. It's tidier than the similar triangles approach, I think.

The theorem you stated about the median to the hypotenuse being half the hypotenuse can also be inferred by inscribing the triangle in a circle and then using the fact that the hypotenuse will always be a diameter (based on the inscribed/central angle theorem). The the median to the hypotenuse and the two halves of the hyp. are all radii of the circle.

So, yeah, #1 is a "make an organized list" problem, but I was too lazy.

And #2 just looks messy. :)

Even though I have an undergrad degree in Math and Computer science and a masters in EECS, as a computer scientist, I don't use much real math, and hardly remember any of what I learned in high school, let alone beyond!

February 2, 2007 at 10:53 PM
Dave Marain said...

'Official' Answers (at least until someone corrects any careless errors!):

1) 35

2) x < 0 or x >= 4/7

3) 0.7

Comments:
1) There is a strong parallel between this question and the number of different 7-letter 'words' one can form from 'TOTALLY!':
(a) 9,9,9, 5: 4C1 or 4P1 = 4 since we can view this as selecting one slot from among 4 choices in which to place the '5' or, in a complementary manner, use 4C3 to select 3 slots out of 4 in which to place the 9's;
(b) 9,9,8,6: 4P2 or 4*3 = 12 ways to place the 8 and 6; OR simply 4!/2! since we are counting arrangements of 4 objects, 2 of which are identical
(c) 9,9,7,7: 4!/(2!2!) = 6, using the permutation method with duplications or 4C2 since we need only choose the 2 slots in which to place the 7's for example
(d) 9,8,8,7: 4!/2! = 12 (same argument as (b))
(e) 8,8,8,8: only 1 of course!
There are many other approaches to this but i wanted to sugges how some combinatorics can be introduced early on. Further each permutation or combination formula can be viewed in a variety of ways.
Note: Most students use these methods to select k objects from a set of n. In the above discussion, the objects being selected were the slots not hte digits! This is important for students to grasp.

More later...

February 2, 2007 at 10:57 PM
Dave Marain said...

sorry three of clubs--
i missed reading your comments about the 'area' method! very nice ...
the 3-4-5, 5-12-13, 7-24-25, 9-40-41, types of triples are a wonderful pattern for middle schoolers to investigate and then prove algebraically;
actually, I was also thinking about how often one triple generates another...
Here's what I mean:
Take 3-4-5 which can be generated by the well-known formula for primitive triples (m^2+n^2, m^2-n^2, 2mn where m>n and m,n are coprime integers)
Using m = 2,n=1:
5 = 2^2+ 1^2; 3 = 2^2-1^2; 4 = 2*2*1
Now how do we get 7-24-25?
Use m=4, n=3:
m^2+n^2 = 25; m^2-n^2 = 7; 2mn = 24
This stuff always blows my mind!

mathmom--
you remember far more from hs math than most of my students! would you say you remember best what made sense to you or do you just remember random formulas?

February 2, 2007 at 11:14 PM
Anonymous said...

Dave,

I fought with the domain question. I get x < 0 or x > 4/7.

The combinatorial problem I teach as a (difficult) exercise in my fall elective.

In general, the number of whole number solutions to a + b + c + d = 32 would be C(32-1+4,4-1)

(Consider a row of 32 boxes and 3 dividers)

(Example, simpler case, sum of 6, 6 boxes, 3 dividers: @@|@||@@@ corresponds to 2103, the addends being consecutive @'s)

So C(35,3), but we need to exclude some things. First of all, the smallest possible digit is 5 (permutations of 9995)
We modify our number to C(32 - 1 + 4 - 20,4 - 1) = C(15,3), but we still have some problems. 85(11)8 is not okay; we need to cap each digit at 9.

(briefly, we take this number, subtract the solutions with one digit ten or higher, add back in those solutions with two digits ten or higher, subtract those solutions with three digits 10 or higher... but since that's 0, we would be done)

but look, let's take 5555 and augment the digits altogether by 12. The augmentation (consider 2433 to stand for 7988) What do we have?

4440 (4 permutations)
4431 (12)
4422 (6)
4332 (12)
3333 (1)

That's it, 35.

I wish I had started there.

February 2, 2007 at 11:31 PM
Anonymous said...

Hope it's not bad form to reference my own blog here, but I just posted a link to a cool math problems site, and also a sample geometry problem from that site that has me stumped. Thought some of you might like to take a peek

February 2, 2007 at 11:36 PM
Dave Marain said...

mathmom--
did i miss the link to your blog? you should absolutely include it here! that's how people came to know i existed!

jonathan--
that was some analysis!
check x = 4/7 -- i think it lead to sqrt(0) which is ok...

folks, i've been communicating with the national math panel and i've asked them for permission to publish their replies to me; they haven't replied yet but i told them i would publish my emails to them no mater what; anyone interested?

February 2, 2007 at 11:52 PM
Anonymous said...

Dave, to get to my blog, just click on my name. :)

February 3, 2007 at 12:15 AM
Anonymous said...

Dave, what is the National Math Panel, and what have you been communicating with them about? I'm sure lots of people would be interested!

February 3, 2007 at 12:16 AM
Anonymous said...

Sorry, 3 comments in a row is probably bad form, but they all answer different questions. :)

To answer the question about what I remember from high school, mostly whatever I have used since then. I don't really recall "formulas" at all, other than maybe some area/volume formulas. But mostly I look up formulas if I need them. Google is my friend. I was great at calculus, but recall nothing. Algebra is probably what I remember the most of, because it does get used most.

Mostly I've been refreshing my memory (over the past 6 years) on the types of problems that come up in Mathcounts and similar competitions at the middle school level and below, because that is what I am now "teaching". (It's a volunteer gig, I'm not a "real" teacher.) We do a lot of combinatorics, but I never give them formulas to remember. I figure if they understand where the numbers are coming from, they don't have to memorize. We do a fair bit of number theory too. Lots of primes, factors, etc. I don't teach much geometry, as I am sticking a little closer to the "regular" middle school curriculum, but I like making sure I can do all the Mathcounts level geometry problems myself, at least. ;-) But we will hit the Pythagorean theorem on Monday, because it will come in handy at Mathcounts, and is pretty quick to learn well enough to use appropriately.

February 3, 2007 at 12:22 AM
Anonymous said...

I haven't had a chance to try these problems recently, though I've read about them all. I think my situation is similar to Mathmom--I don't recall many exact formulas except some of the area/volume ones as well. I also have a vague memory of the "form" of many of them, so (at least for the simple ones) they're not that hard to derive. For instance, I had to use the area of an equilateral triangle for one of your problems the other day; I just bisected it into 2 30-60-90 triangles. To be honest, I still sometimes get confused with the ratios for the 30-60-90 triangles (though I've used them enough for your problems that it's solid again). I've always remembered the three sides are 1, 2, and sqrt(3). This probably sounds silly, but I think of the 60-degree angle as the "bad" one and that's how I remember it's opposite the radical. Even then I'm sometimes hesitant but then I just look at the sine of 30 and 60 which is obviously 1/2 and sqrt(3)/2 and then draw the sides.

Most of what's left of my math ability these days, I think, is using some relatively basic principles and formulas and deriving more complex relationships. I suppose a bit of "intuition" guides as well.

February 3, 2007 at 1:34 AM
Anonymous said...

Darmok,

until I started teaching, I rederived my 30-60-90 ratios each time I needed them (from equilateral triangles). In fact, I show students how to do that in case they forget.

February 3, 2007 at 10:42 AM
Anonymous said...

Darmok, I too remember 1, 2, sqrt(3) and then have to remind myself that 2 is bigger than sqrt(3) so it goes on the hypotenuse.

The types of things I don't remember are like, 95% of trig (everything beyond SOHCAHTOA -- and I sort of remember the unit circle stuff) and most of calculus. Just haven't needed them. Geometry was equally rusty until I started "practicing" on Mathcounts problems.

I think I've forgotten all of the math I learned beyond high school, with the exception of a few cool things (like the proofs of the same/different cardinalities of the various sets of numbers, proof of infinitely many primes, etc.) What I've retained are intuition and problem solving skills. And at least an appreciation for what kinds of skills will really be important for kids moving forward.

February 3, 2007 at 11:04 AM
Anonymous said...

I'm sure I could still differentiate pretty well, and perhaps integrate very basic equations--no integration by parts or whatever it was called though! Integral of u dv = uv - int v du? Was that it? I'd never be able to see how to manipulate an equation to get it into the form. And differential equations are completely out the window.

More basic high school math, I though, I think I could derive for the most part. Given enough time, you can derive most of the theorems and such, especially if you remember that there's such a theorem and you don't actually need to prove it.

February 4, 2007 at 3:45 AM

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