Hi everyone,
Can anyone help me with this question:
I can't include the figures for this, and also I don't know latex sorry.
54. The four shaded circles in Figure 1 above are congruent and each is tangent to the large circle and to two of the other shaded circles. Figure 2 is the result of replacing each of the shaded circles in Figure 1 by a figure that is geometrically similar to Figure 1. What is the ratio of the area of the shaded portion of Figure 2 to the area of the shaded portion of Figure 1?
A) 1/(2sqrt2)
B) 1/(1+sqrt2)
C) 4/(1+sqrt2)
D) (sqrt2/1+sqrt2)^2
E) (2/1+sqrt2)^2
Correct answer is E
GRE 0568 #54
Re: GRE 0568 #54
Basically with any tangent circle problem like this the first thing you want to do is draw all the radii to the relevant tangent points. In this case let r be the small radius in the second diagram. Drawing radii you find that a middle size circle has radius $$R = r(1 + \sqrt{2})$$. So the middle circle has area $$\pi r^2(1 + \sqrt{2})^2$$, and so the ratio of the area of 4 small circles to one middle one is $$4/(1 + \sqrt{2})^2$$ which is the value they're looking for.

 Posts: 4
 Joined: Mon Apr 05, 2010 3:35 pm
Re: GRE 0568 #54
Thanks enork,
was getting too bogged down with things like the big radius. Took depressingly long to figure out why the ratio of the radii was (1+sqrt2) also, doubt I would have time for one like this on the real test
was getting too bogged down with things like the big radius. Took depressingly long to figure out why the ratio of the radii was (1+sqrt2) also, doubt I would have time for one like this on the real test
Re: GRE 0568 #54
Hi,
What would be the fastest way to do this question?
Is there a faster way than this:
First, Pythagoras Theorem gives
$$(2r)^2=(Rr)^2+(Rr)^2$$
I then expanded it (?!) and used the quadratic formula (?!) and then took me a while to reject the "negative" answer. So $$R=r(1+\sqrt{2})$$.
Finally, when the answer came out, it had a different form from the options ($$\frac{4}{3+2\sqrt{2}}$$), so I had to spend seconds to evaluate the options, esp D and E. Overall took rather long (>5 min).
Anyone has a faster method, that preferably produces an answer in the squaredform? Thanks a lot.
What would be the fastest way to do this question?
Is there a faster way than this:
First, Pythagoras Theorem gives
$$(2r)^2=(Rr)^2+(Rr)^2$$
I then expanded it (?!) and used the quadratic formula (?!) and then took me a while to reject the "negative" answer. So $$R=r(1+\sqrt{2})$$.
Finally, when the answer came out, it had a different form from the options ($$\frac{4}{3+2\sqrt{2}}$$), so I had to spend seconds to evaluate the options, esp D and E. Overall took rather long (>5 min).
Anyone has a faster method, that preferably produces an answer in the squaredform? Thanks a lot.
 Charles.Rambo
 Posts: 9
 Joined: Wed May 22, 2013 7:33 pm
Re: GRE 0568 #54
Hey, I already responded to another feed on this website with the same question. But I want to get the word out that I've solved all of the GRE test 68 questions. You can view the solutions at http://rambotutoring.com/GREmathsubje ... utions.pdf.
Charles
Charles
Re: GRE 0568 #54
Yet another set of solutions can be found here: http://www.gigauploader.com/file/0770203183169331
Here's the post which describes the solutions:
viewtopic.php?f=1&t=1378
Here's the post which describes the solutions:
viewtopic.php?f=1&t=1378
Re: GRE 0568 #54
Consider the first quadrant of the big circle (radius r), a pair of axe x and y. The small circle is tangent to both. Let's say its radius is 1. Hence, the coordinates of its center are (1,1). The distance between it and the big circle's center is sqrt (2). So, r = 1+ sqrt (2)
The rest is easy.
The rest is easy.