<p>Its from the GRE Math Subject Test. You guys should try it out </p>
<p>I'm sure you guys would enjoy solving it if you are into math and stuff.</p>
<p>The answer is E btw</p>
<p>It was a cool problem. I made a square in the middle using the radiuses to solve it. How did you do it?</p>
<p>Also interesting that the ratio they ask for is the same as the ratio of shaded to total area in the first circle…after you see that, you can ignore the second circle.</p>
<p>I did it that exact same way An0maly. And keller, I noticed that the ratios were the same as well. I guess they added extras to throw you off, I dunno.</p>
<p>i am wondering where is the root2 from, under a certain circumstance,maybe triangle.</p>
<p>^It comes from Pythagorean theorem</p>
<p>OMG! OMG! (becomes deranged and excited)
I solved a problem just like this (in fact, it basically WAS this) yesterday on an AMC-10 (American Mathematics Competition for 10th grade and below)! Wow…</p>
<p>You said this was a difficult question for the GRE? Gee, I feel smart lol. :)</p>
<p>According to the GRE score guide, only 25% of the test takers answered this question correctly.</p>
<p>Not that it matters, but do you know if that’s 25% of test-takers or 25% of those who answered it? I suspect that it is the latter. And then remember that 20% would get it right if they all just guessed. So it IS a hard problem!</p>
<p>Please explain the solution!</p>
<p>I’m not a math genius or anything (I got 720 for my SATI math) but I think first you need to establish the ratio between the area of a circle and the area of each of the four small circles in it.</p>
<p>Let r = radius of each of the 4 circles</p>
<p>Now what is the radius of the big radius then?</p>
<p>Refer to figure 1, extend the radius from the center of the white circle, extending through the center of a shaded circle and stop at the outer point where they touch. </p>
<p>What do you see? The radius of the bigger circle is obviously 2r + (a small portion from its center to the first point that the radius tine cuts the shaded circle)</p>
<p>Our job now is to find the length of that small portion. It can be seen that the distance between the centre of the big circle and the small circle is r * root2. (I hope you can work this out). Thus, the length of the small portion would be r * root2 - r</p>
<p>Thus, the radius of the big circle is 2r + r * root2 - r = r (1 + root 2)</p>
<p>Move on to figure 2, now r = radius of the smallest circle</p>
<p>Hence, the radius of the middle-size circle (also shaded circle of figure 1) would be r (1 + root 2)</p>
<p>We have then,</p>
<p>Area of a smallest circle = pi<em>r^2 (a)
Area of a mid-sized circle = pi</em>[r (1 + root 2)]^2 (b)</p>
<p>Take 16 (a) divided by 4 (b), as there are 4 mid-sized circles in figure 1 and 16 smallest circles in figure 2. The answer is thus E.</p>
<p>Hope this wasn’t too confusing. Wow I’m really, really bored.</p>
<p>Edit: my 99th post, heh.</p>
<p>That was never a difficult problem, I’d say it just require your tolerance with square roots, and that’s all.</p>