## Tuesday, January 31, 2012

### this week's MGRE Math Beast Challenge problem

From here:

My answer will appear in the comments, but I'll tell you right now that I don't like the way this problem's been structured. As it's worded, I'd say the answer is "cannot be determined." Why? A couple reasons:

1. We can assume the figure isn't drawn to scale. The only "given" is that the round figure is indeed a circle with its center labeled. We can also safely assume that Points A, C, and D are all on the circle. It's also safe to assume that Point B is on the circle as well. Beyond that, what do we know for sure?

2. We can't assume-- since the problem doesn't specify this-- that the squarish-looking figure is indeed a square, which means we can't assume that the circle is properly inscribed within a square.

3. We also can't assume that Segment GF forms a 180-degree angle with the bottom of the putative square.

For those reasons, I think "cannot be determined" is the best answer, but I'm going to ignore the above concerns and charitably assume that the circle is inscribed in a square, thus making Point B the midpoint of that side of the square. With those assumptions in place, I believe the problem is easily soluble. Without them, however, "cannot be determined" is the only legitimate answer.

Here's how I would have presented the problem, so as to avoid any confusion about what we can and can't assume:

I'll be basing my answer off the above image, not off MGRE's.

_

#### 1 comment:

Kevin Kim said...

I know, from previous experience with similar problems, that angle ACD is a right angle. Since AC and CD are equal, we know this is an isosceles right triangle with legs of length 1/2, and whose hypotenuse must therefore be (√2)/2. (Triangle ACD is also half of a square inscribed inside circle E, but that fact may not be relevant here.)

If we rotate Triangle ACD such that Diameter AD is vertical, we see that we have another right triangle, which we can name either ADG (if we use the labels for the diameter) or GBL (if we use the tangent points B and L, since A and D now sit on top of those points).

We know that the hypotenuse of this right triangle, BG, is 4 times the length of AE. AE is half the diameter; the diameter measures (√2)/2, so AE measures (√2)/4. Four times that measurement is simply √2.

So for Triangle GBL, we know the following:

Hypotenuse BG = √2
Leg BL = (√2)/2
Leg GL = ?

We use the Pythagorean Theorem to solve for BL:

BL^2 + GL^2 = BG^2

(1/2) + x^2 = 2

x^2 = 1.5

So the length of GL is (√1.5), but the problem is asking for the length of GF. Since Segment FL is equal to the radius, i.e., the length of Segment AE-- which is (√2)/4-- then the length of GF is

GL - FL.

In other words, (√1.5) - (√2)/4, which is:

((2√6) - (√2))/4

--and that's the fourth selection!

Again, I'd say the correct answer, with the problem as written by MGRE, is "cannot be determined."