GRE 0568 #54

Forum for the GRE subject test in mathematics.
jerseyj282
Posts: 4
Joined: Mon Apr 05, 2010 3:35 pm

GRE 0568 #54

Postby jerseyj282 » Mon Apr 05, 2010 4:48 pm

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

enork
Posts: 33
Joined: Fri Sep 18, 2009 3:16 am

Re: GRE 0568 #54

Postby enork » Tue Apr 06, 2010 8:46 pm

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.

jerseyj282
Posts: 4
Joined: Mon Apr 05, 2010 3:35 pm

Re: GRE 0568 #54

Postby jerseyj282 » Wed Apr 07, 2010 3:29 pm

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 :(

yoyobarn
Posts: 80
Joined: Sun Dec 19, 2010 7:01 am

Re: GRE 0568 #54

Postby yoyobarn » Sat Jan 21, 2012 3:09 am

Hi,

What would be the fastest way to do this question?


Is there a faster way than this:
First, Pythagoras Theorem gives
(2r)^2=(R-r)^2+(R-r)^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 squared-form? Thanks a lot.

Charles.Rambo
Posts: 5
Joined: Wed May 22, 2013 7:33 pm

Re: GRE 0568 #54

Postby Charles.Rambo » Wed May 22, 2013 7:41 pm

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/GRE-math-subject-test-68-solutions.pdf.

-Charles

User avatar
redcar777
Posts: 34
Joined: Sun Nov 04, 2012 11:59 am

Re: GRE 0568 #54

Postby redcar777 » Thu May 30, 2013 3:56 pm

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

DDswife
Posts: 58
Joined: Thu Aug 14, 2014 5:29 pm

Re: GRE 0568 #54

Postby DDswife » Fri Aug 15, 2014 11:17 pm

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.




Return to “Mathematics GRE Forum: The GRE Subject Test in Mathematics”



Who is online

Users browsing this forum: No registered users and 4 guests