## 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.

Totally_clueless said...

Hi Dave,

The units do not work out in the 2nd problem (r^3/2)/(ab) does not have length units.

After some calculation, I arrive at r^3/ab for the length. I need to double check, however.

And, of course, thanks for a very nice problem.

Cheers,

TC

Dave Marain said...

tc--
Nice to hear from you! You're absolutely right - it should be r^3 not r^(3/2). Hopefully, I caught all my errors on the contest!

I will change it in the post...

Eric Jablow said...

How many people do you think will provide solutions using only similar triangles and the fact that tangents are perpendicular to radii? Neither trigonometry, nor coordinates, nor the Pythagorean law are required to solve the problem

Dave Marain said...

Eric,
I take it your question was rhetorical!
Students struggle with similarity for a myriad of reasons.
Those chord-tangent-secant theorems are memorized for the test and soon forgotten.

Eric Jablow said...

Well, the simplest proof requires only similar triangles and the fact that tangents are orthogonal to radii. Let C be the point (a, b), O the origin, and A the foot of the perpendicular from C to the x-axis. Then, the triangles AOC, CAP, and OQP are all similar. The rest is immediate.

Dave Marain said...

Agreed, Eric, but most students (and others!) would struggle with what you find "immediate"! The fact is that our geometry texts do not have too many problems involving 3 similar triangles. Student I have seen students struggle mightily with TWO similar triangle problems, never mind THREE!

Actually, I used a slightly different approach involving areas and the "altitude on hypotenuse theorem", all of which is related to similar triangles of course:

Assume the radius of the circle is 1 for this problem.

Let B be the foot of the perpendicular from C to OQ.
Then CB = a and CA = b.

First look at triangle OPQ. From areas, we have
(PQ)(1) = (OP)(OQ).

Using areas in triangle OQC:
(OQ)(a) = (1)(QC)
Therefore, OQ = QC/a.

Similarly,in triangle OPC:
(OP)(b) = (1)(PC)
Therefore, OP = PC/b.

Hence, PQ = (OP)(OQ) = (QC/a)(PC/b)=
(QC)(PC)/(ab).
But (QC)(PC) = radius ^2 = 1 from the altitude on hypotenuse theorem in triangle OPQ.
Thus, PQ = 1/ab.

You probably have a more direct and easier set of proportions than this, but this is how I saw it.

I believe many students (if they get this at all)would approach the problem using the Pythagorean Theorem and have expressions involving a^2 + b^2 = r^2 in their derivation.

Eric Jablow said...

Of course, I made a typo in the process. CAP is similar to the other two, but the important one is that

AOC, COP, and OQP are similar.

a:b:r :: r:CP:OP :: OQ : OP : PQ

CP = rb/a, OP = r²/a from the first and second set of ratios.

PQ= r³/(ab) from the first and third set of ratios.

Note that CP and OQ are unimportant.

Dave Marain said...

Nice, Eric!
I wish I could have your conciseness...

Eric Jablow said...

It takes work to be concise, and the typical instruction to "show all work" militates against that. Perhaps examiners should request that test-takers should "show all relevant work" instead. I know we've all suffered from students dumping all the information they know so they can hide what they do not know.

One way to get to my solution is to ask the student:

1. Given the points O, P, Q, and C, find all the similarity relations among the triangles they form.

2. Given the points O, P, Q, C, and A, find all the similarity relations among the triangles they form.

3. Can the relationships in 1 solve the problem?

4. Can the relationships in 2 solve the problem?

5. Which relationships in 2 are necessary?

6. Construct a minimal proof of the problem.

7. Why is it important to be able to construct a minimal proof? How did mathematicians find their ways to discovering mathematics?