## Friday, December 16, 2011

### nouvelle révélation triangulaire

One of my students, BZ, took a look at the MGRE problem from two weeks ago-- the one about the right triangles-- and arrived at the correct answer, (C), within a minute. BZ's approach was far superior to MGRE's own approach (and, by extension, mine, since my approach was simply a truncated version of MGRE's). First, take a look at the problem again: BZ noticed something very quickly: Quantities A and B are both related to the area of the large triangle!

The area of a triangle is one-half its base times its height. Quantity A is abc, which can be restated this way:

c times twice the triangle's area (since ab = the area of a rectangle)

Quantity B, which is h(a2 + b2), can be rewritten as hc2. Notice that hc is also twice the area of the triangle in question, because h is the triangle's height and c is, in this orientation, its base. This means that the expression hc2 can be restated as

c times twice the triangle's area

Amazing. If we use the letter Q to represent "twice the triangle's area," we see that, when comparing Quantities A and B, the equation is

cQ = cQ

The quantities are equal.

This is so easy to see in hindsight, and now I'm wondering why MGRE didn't include this very simple and obvious approach in its own explanation.

_