The Return of the WarmUp Challenges!

Just when you thought that MathNotations is on permanent hiatus or in hibernation, here are a couple of WarmUps/Problems of the Day/Test Prep/Challenges/// to consider for your students. 

Actually, I'm embarking on a new venture - an online tutoring website with live audio and video for OneOnOne math tutoring for Grades 6-14 (through Calculus II). In addition, I'm also working on setting up a small group (5-10 students) online SAT or ACT Course grouped by ability (a 600-800 SAT group, a 450-600 group, etc.).  If you're interested in getting more information about these before the official launch just contact me at dmarain at gmail dot com.


Update: Answers/comments are at the bottom...

1.   NOTE: ANGLE B IS A RIGHT ANGLE IN DIAGRAM BELOW - THANKS TO JONATHAN FOR CATCHING THAT OVERSIGHT!

2.   If 10^-1000 - 10^-997 is written as a decimal, answer the following:


(a) How many decimal places are there, i.e., how many digits to the right of the decimal point?
(b) One can show that the decimal digits end in a string of 9's. How many 9's?
(c) How many zeros are to the right of the decimal point and to the left of the string of 9's?

Notes:
(1) If we write the negative exponent expressions as rational numbers, this is perfectly appropriate for middle schoolers and, in fact, I think they need more of these experiences!
(2) The "Make It Simpler - Look for a Pattern" Strategy should be second nature to our youngsters, but when they see questions like these on the SATs, how many of our students really think of it!
(3) The fact that some calculators return a value of zero for the expression in the problem is a teachable moment - seize it!!
(4) See below for an algebraic approach.



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ANSWERS


1. 9√3


2. (a) 1000   (b) 3   (c) 997


An Algebraic Approach to #2:
First, students need to be familiar with the basic pattern:
10^-1 = 1/10 = .1 Note that there is one decimal digit.

10^-2 = 1/10² = 1/100 = .01  Note that there are two decimal places, etc.


10^-1000 - 10^-997 = 1/10¹⁰⁰⁰ - 1/10⁹⁹⁷
Using 10¹⁰⁰⁰ as the common denominator, we obtain
1/10¹⁰⁰⁰ - 10³/10¹⁰⁰⁰ =
-999/10¹⁰⁰⁰ from which the results follow (with some additional reasoning)...

Note: I could have worked directly with the exponent form by factoring out 10^-1000 but I chose rational form for the younger student.

Another View of sin(A-B) for a Special Case: An Investigation

[As always, don't forget to give proper attribution when using this in the classroom or elsewhere as indicated in the sidebar]

In a standard trig unit, students learn those wonderful formulas for the sin and cos of the sum and difference of angles. Many creative methods have been developed to derive these formulas and, depending on the ability of the group and teacher preference, these are demonstrated or not. Students are typically shown various mnemonics for recalling them on the big test, but, in this investigation, students will derive sin 15° using only 30-60-90 triangle ratios and the Pythagorean Theorem. We will then compare the result to that obtained by the traditional formulas for sin(45°-30°) or sin(60°-45°) and show equivalence by algebraic methods using radicals. This is not an attempt to develop a general approach to deriving sum/difference formulas, although readers are invited to try a generalization. You may recall other posts on this blog of a similar nature.

THE PROBLEM/INVESTIGATION
Refer to the triangle above. If the print is too small, click on the image to magnify.
∠A = 75° and ∠B = 15°

(a) In the triangle above, locate point D on side BC such that ∠CAD = 60° . Express the lengths of the sides of triangle CAD in terms of a.
[Note: We could avoid the variable a altogether and assign a value of 1 since this is a ratio problem.]
(b) Show that CB = a √3 +2a.
(c) Use the Pythagorean Theorem to show that AB = a √(8+4 √ 3)
(d) Verify the identity (√ 6 + √ 2)² = 8+4 √ 3. Use this to rewrite AB.
(e) Use above results to obtain an expression for sin 15°.
(f) Use the standard trig formula for sin(45°-30°) to obtain an expression for sin 15°.
(g) Show your results in (e) and (f) are equivalent.

An Instructional Aside
When introducing the formula for sin(A+B), for example, teachers sometimes motivate it effectively using numerical values or considering the special case A=B. Here's an alternative:
Consider the special case A+B = 90°
Ask students to verify the formula for sin(A+B) in this special case. Simple, but at least it's something slightly different to pique their curiosity.

Standard Disclaimer:
This investigation is not copied from some other source. As it is original and has not been edited by others, there's always the possibility of error. Please feel free to suggest corrections/edits/extensions...
You know I welcome your comments!

Problems 3-5 thru 3-6-07: Geometry and Reasoning

Note: The problem below was chosen for the March 23rd Carnival of Mathematics. If you'd like to see a different kind of mathematical challenge, I recommend you also visit the posting for 3-30-07 (Challenging Geometry: Circles Inscribed in Quadrilaterals, Right Triangles).



Today's questions involve well-known ideas from geometry. Similar questions have appeared on SATs and math contests.
Some suggestions: Use it for in-class enrichment or assign it for extra credit outside of class. The first part lends itelf to a fairly simply visual approach (cutting up the square and matching the pieces), but the second is more sophisticated. Encourage the visualization but require the analytical approach as well!
Reviews: 30-60-90, equilateral, areas, symmetry, etc.