## Friday, May 29, 2009

### Geometry Challenge for SAT Prep or Review for Final Exam

In the coordinate plane, what is the area of ΔPQR given the coordinates P(4.5,4.5), Q(8.5,8.5), and R(6,0)?

• Example of "Grid-In" or student-constructed response question on the SAT
• This question seems more difficult than it really is. Students often give up on questions near the end of a section. DON'T!!
• Hopefully you will view the utility of questions like this as I do:
Students become more competent problem solvers only when challenged with nonroutine problems which are not always to be found in the textbooks. Questions like these should become more routine in our texts and in our classes (not only honors!).

Solution (no explanation, details omitted):
(1/2)(6)(8.5) - (1/2)(6)(4.5) = 12

Discussion Points
• What are some problem-solving strategies we need to review with our students here? Draw a diagram for sure but what are some other general attack strategies students should employ in triangle area problems?
• Although advanced theorems could be used here, the actual solution given above is efficient and fairly basic. But what insights are needed to use that approach? What geometry or algebra standards are being tested here?
• I chose this problem because coordinate geometry problems connect many important ideas in geometry and algebra. Not to mention that they are becoming more common on standardized tests like SATs, ACTs and state assessments. Besides, I enjoyed writing the question! Sometimes I'll get the germ of an idea, re-work it many times and then the question takes on a life of its own.
• If you find an error in my work or want to share your thoughts, please add a comment!

## Tuesday, May 19, 2009

### Light Humor and More on a Math Puzzle

We all need a break from the seriousness of math, so here are some (actual) amusing headlines. Before I get nasty comments, remember these are real!

1. Teachers Strike Idle Kids
2. Eye Drops Off Shelf
3. Miners Refuse To Work After Death
4. Hospitals Are Sued By 7 Foot Doctors
5. British Left Waffles On Falkland Islands

Ok, enough of the silliness...

I received several emails and comments about the math puzzle from the other day. Essentially, they all indicated that the problem was readily solved using technology.

First, here is the solution:
123456789 = 10821 x 11409

Aniket S., a computer science major from Solapur University, India, sent me the solution and the code she used in C++ to find the factors.

Both Mike Croucher and Pat Ballew, who emailed me, used the powerful new computing engine Wolfram Alpha to find the factors. As Pat pointed out, "the times they are a changin'...

In fact, I found the prime factors of 123456789 on the TI-89:

factor (123456789)
3^2 x 3607 x 3803

Is there anything special about the number 123456789? If we randomly permute the digits to, say, 586421739, we obtain the prime factorization 3^2 x 107 x 608953, which cannot be rearranged to form two 5-digit factors due to the 6-digit prime. Can you find a permutation of 123456789 which can also be factored into two 5-digit numbers?

If we make one change to the original number we can solve the puzzle:
123456784 = 10406 x 11864.

Have fun computing...

## Monday, May 18, 2009

### A Puzzle To Start the Week

Number puzzles always intrigued me and, perhaps, they are one way we can invite our students into the wonderful and exciting world of mathematics. Oh, alright, maybe that's a bit of a stretch, but, I suspect that if you give the following famous puzzle to your students in Grades 5 and up, they will try it even if you don't offer food or a 10 point bonus! Yes, calculators are allowed but after a few minutes of frustration they will be begging for a hint.

(Oh, and if you give them this problem at the beginning of class, you may as well forget the lesson!)

Find two 5-digit numbers whose product is 123456789.

If you solve it, don't post your answer immediately. I will probably publish a hint or the answer in a day or so. You can always email me with your solution at "dmarain at gmail dot com."

HINT: Rather than pressing random numbers into the calculator as some would do, encourage them to find the prime factors of 123456789. It's easy to show that this number is divisible by 3 and 9, but find finding the other factors will be challenging. I'll post another hint if you request it...

COMMENT: This beautiful puzzle was invented by Y. Yamamoto and has intrigued many puzzle enthusiasts for awhile now. Is there some profound meaning behind the solution or is it just a curiosity? Perhaps we'll have to wait for Dan Brown's next book to unlock the mystery! I will probably post the answer if I don't get a response within 24 hours. Probably...

If any of your students solve it, email me at "dmarain at gmail dot com" and let me know if I can post their names.

## Saturday, May 16, 2009

### The "POWER" of Circles Part I - An Open-Ended Geometry Challenge and CONTEST Update

Well, registration for MathNotation's 2nd Online Contest has now closed. Not as many participants this time but we do have schools representing several states and one high school from Japan! The questions have been emailed to advisors but results will not be available for a couple of weeks. I also plan on publishing at least one of the contest problems in June. I'm still in the planning stages for running 2-3 of these contests for the 2009-10 school year. Stay tuned...

If you're interested in signing up for upcoming contests, just drop me an email at "dmarain at gmail dot com" and I will put you in my database.

For those of you who haven't been reading about these contests it's the team approach, open-endedness and multi-part nature of some of the problems which separates these contests from most others out there. In other words, these questions reflect the types of investigations I've been publishing since 2007.

Also, after I write 5-10 of these contests, I plan to publish these in a book with detailed solutions and comments. As my wife would say, "I'll believe that when I see it!"
**************************************************************************

Speaking of investigations, here is Part I of an open-ended geometry problem that starts out relatively simply but eventually will lead to deeper results. This investigation will review basic circle concepts involving tangents and secants but will connect to the more advanced ideas of power of a point and the inverse of a point in a circle in later posts. In the diagram, O is the center of the circle PT is a tangent segment, segment AB is a diameter.

(a) If the radius is 6 and PA = 4, show that the length of tangent segment PT is 8.

Note that (PA)(PB) = (4)(16) = 64 and, from part (a), (PT)2 also equals 64. This is a special case of the secant-tangent power theorem you may recall from geometry. Your job in the next part is to demonstrate a particular case of this theorem this using the above diagram.

(b) If the radius of the circle is r and PA = x, show that
(PA)(PB) = (PT)2.

Note: Using the secant-tangent power theorem here trivializes this problem. The idea is to demonstrate the result without using that theorem, in effect, proving a special case of this rule!

Click on Read More for further comments and a hint for part (b)...

(i) Neither of the above parts was intended to be highly challenging. Part (a) is definitely an SAT-type question. A review of geometry never hurts!

(ii) Here's a hint to get started on part (b):
From Pythagorean we know that (PT)2 = (PO)2 - r2. Factor this expression...

(iii) Note that my approach as always is to introduce or develop a theorem or concept such as the secant-tangent power theorem or "power of a point" by looking at particular or special cases (the secant in the diagram contains a diameter) and starting with numerical values rather than variables. This sequence (numerical values, particular case, generalization) forms the basis of the investigations I've been writing for this blog, but, more importantly the underlying foundation for the lessons I planned when I taught. I believed then and now that this approach may be time-consuming (both in planning and implementation) but the payoff is deeper conceptual understanding. Of course I needed to modify the presentation according to the backgrounds and ability levels of my students but I never assumed only my honors and AP classes were up to the challenge.

(iv) In Part II which I will post in a few days, we will discuss the "power of a point" and add a second circle, in which segment PT will be a radius. This will produce a another point, P', the inverse of point P.

## Monday, May 11, 2009

### A Sample Contest Problem (Open-Ended), Odds and Evens,...

First, some reminders and updates re the upcoming

• I am still accepting registration up to this Fri 5-15-09 and that may be extended. Just email me at "dmarain at gmail dot com" and I will email you the registration and information forms in short order!
• Because a few schools have expressed concern that some students are still taking AP's next week (makeups?) or their brains will be fried after this weeks AP's, I am willing to allow sponsors to administer the contest either the week of the 18th or the week of the 25th (after Memorial Day of course here in the US).
• Participating students should review their trig identities, infinite geometric series and probability. However, there are other questions or parts that do not involve these more advanced topics.
• Several questions are multi-part with later parts of increasing difficulty.
• At least one question requires a detailed explanation, i.e., showing one's method clearly.
ODDS and EVENS...
• Have you been keeping up with Burt's insightful comments, clear explanations and advocacy for balancing concept and procedure in our classrooms, K-12? Read Burt's comments to this post...
• I've been remiss in keeping up with all the carnivals. I will get caught up in a few days.
• Been thinking about the AP issues I raised in a recent post (from the NYT article)? I will have more to say about this, particularly based on the reader comments to that article. Link to the Times article from my post and skim through the 60 or so reader comments. Fascinating stuff...

Here's a sample contest question that demonstrates showing all work. Some of you may recognize a similar question posted earlier on MathNotations.

(i) Consider the circle of radius 1 centered at (0,0). Let L be the line tangent to this circle at the point (a,b) in the first quadrant. If P and Q are the x- and y-intercepts of L, respectively, show that the length of segment PQ equals 1/(ab). All work must be shown clearly.

(ii) Same as part (i) except the radius of the circle is now r. Show that the length of segment PQ can be expressed as r3/(ab). All work must be shown clearly.

## Saturday, May 2, 2009

### A Tale of Two Equations: Balancing Procedures and Conceptual Understanding

WHAT WILL YOUR STUDENTS BE DOING AFTER THE AP'S?
TAKING MATHNOTATIONS 2ND ONLINE (FREE) MATH CONTEST!

UPDATE
-- Registration deadline extended until Fri May 15th!

The following is intended for all students in 2nd year algebra. Your stronger students should not find these overly challenging but there is more here than meets the eye. The purpose here is to demonstrate how we can review procedures AND develop deeper understanding of important mathematical ideas in the same lesson. The graphing calculator can be used to enhance the lesson by employing multiple representations (Rule of Four) to reinforce the essential ideas.

Note: Finding the solutions is only the tip of the iceberg. Understanding WHY one equation must have finitely many solutions and the other must have infinitely many solutions is the bigger idea here...

SOLVE EACH OF THE FOLLOWING

Equation 1:
(x-1)(x-2) = (1-x)(2-x)

Equation 2:
(x-1)(x-2)(x-3) = (1-x)(2-x)(3-x)

Click on Read More... for solutions and further discussion.

Equation 1: All real numbers
Equation 2: {1,2,3}

DISCUSSION

Would most of your students eliminate parentheses in the first equation and solve by traditional methods? Even though the left and right sides of the equation appear similar, it is reasonable to expect they will distribute and solve since that is what they're used to doing. This is fine and the standard procedure should be reviewed.

Assuming students will not make "careless" mechanical errors in distributing, they should obtain:
x2 - 3x + 2 = 2 - 3x + x2.
This generates a nice discussion of an "identical equation" or identity since the left and right sides are mathematically equivalent (if they recognize that!). The instructor may or may not want to continue the mechanical approach of moving all terms to one side producing 0 = 0 to reinforce that the equation is satisfied by all real numbers. Your stronger student will not have much difficulty with this.

Before moving on to the 2nd equation, we can develop a
deeper conceptual understanding by asking students to approach the problem another way. We know that some students will wonder about the form of the original equation. Could we have predicted that the two sides would be identical without removing parentheses? Could we also have determined by inspection that both x = 1 and x = 2 are solutions? Asking them to revisit the original equation to see this is critical. Now what about trying some other real number, say x = 5. This should strongly suggest that all real numbers will satisfy the equation. Using the graphing calculator will also drive this point home visually. Store the left side of the equation Y1 and the right side of the equation in Y2. Change the appearance of Y2 (make it bold for example) and have them observe on the viewscreen that the graphs are identical.

So, how come the 2nd equation only has 3 solutions! I'll leave that to my readers to elaborate on...
How can we generalize this?

These kinds of lessons seem to involve way too much overhead, stealing so much valuable time away from other content. BUT these are precisely the kinds of problems students are expected to grapple with in Japan and other countries. Do you really believe "Less is More?"