This Week's Work : Week 12 -- for Inquisitive Young Mathletes

Link to the online timed test on questions you mostly got wrong or not fast enough. 
(sent through e-mail)

Common Pythagorean Triples: 
3, 4, 5 and its derivatives 
5, 12, 13 
8, 15, 17 
7, 24, 25 (at least these for SAT I and II) 

9, 40, 41, (the rest for state and Nationals, so we'll learn them later)
11, 60, 61
12, 35, 37
13, 84, 85
20, 21, 29

Shoe string method in finding the area of any polygon

Heron's formula in finding the area of a triangle.

Don't mix up the "s" with the other "S" of finding

the area of an equilateral triangle -- proof and formula (You can also use 30-60-90 special right
triangle to get that.)
or
the area of a regular hexagon

In Heron's case, "s" stands for half of the perimeter.

Besides, I've noticed most of the questions, when given the sides, are best solved by using Pythagorean triples, especially in sprint round questions, so make sure to actively evaluate the question(s) at hand and use the most efficient strategy.

Here is the link to 2003 chapter #29 that most of you got wrong:

You don't need to use complementary counting for that specific question since it's equal cases either way. Make sure you understand why you need to times 3. (AA_, A_A, and _AA for team A to be chosen two out of three days).

From Mathcounts Mini: Area of irregular polygon

See if you can use shoestring method to get the same answer.
Second half is again on similar triangles, dimensional change and sometimes
Pythagorean triples.

From NOVA : Fractals - Hunting the Hidden Dimension

The Largest Rectangle Inscribed in Any Triangle

\(\Delta EHI\sim\Delta EFG\) \(\rightarrow\) \(\dfrac {a} {c}=\dfrac {d-b} {d}\)\(\rightarrow\) \(a=\dfrac {c\left( d-b\right) } {d}=\dfrac {-c\left( b-d\right) } {d}\)

We are going to find out what the largest area of a rectangle is with the side length a and b.
It can be shown that by substituting the side length "a" with the previous equation + completing the square that the largest area is half of the area of the triangle the rectangle is embedded.

\(a\times b=\dfrac {-c\left( b-d\right) \times b} {d}=\dfrac {-c\left( b^{2}-bd\right)} {d}= \dfrac {-c\left( b-\dfrac {1} {2}d\right) ^{2}+\dfrac {1} {4}dc} {d}\).

From there, you know that when \(b= \dfrac {1} {2}d\), it will give you the largest area, which is \(\dfrac {1} {4}dc\).

\(a=\dfrac {-c\left( b-d\right) } {d}= \dfrac {-c\left( \dfrac {1} {2}d-d\right) } {d}=\dfrac {c\left( d-\dfrac {1} {2}d\right) } {d}=\dfrac {1} {2}c\).

Thus, the maximum rectangle area occurs when the midpoints of two of the sides of the triangle were joined to make a side of the rectangle and its area is thus 50% or half of the area of the triangle or 1/4 of the base times height.