## Tuesday, February 7, 2012

### the answer to last week's MGRE Math Beast Challenge

You may recall that I wasn't particularly happy about the way MGRE designed last week's problem. As it turned out, I was right to think that they would pick (D) as the correct answer, even though I contend that (E) ("cannot be determined") is correct because of the lack of proper labeling.

For what it's worth, here's how MGRE arrived at (D):

It may help to first redraw the figure by simply rotating triangle ACD about the center of the circle so that AD will be vertical. This is acceptable, because we aren’t changing any lengths or angles except to create a right triangle ADG, as shown:

Now, let’s start with the one length we were given. Since AC = CD, triangle ACD is an equilateral right triangle, or a 45–45–90 triangle (referring to the angle measures). In an equilateral right triangle, the hypotenuse is √2 times the length of either other side, so AD = ((√2)/2).

In the figure, AD is the diameter of the circle, and AE is a radius of the circle. Thus, AE is half AD, or AE = (√2)/4. Also, the square side length equals the diameter, and DF is half a side of the square, so DF = (√2)/4, too.

The problem states that BG = 4AE, so BG = √2.

We now have two of the side lengths for the right triangle we created:

By Pythagorean Theorem,

.

We are looking for GF, which is simply GD – DF. Since DF = (√2)/4,

.