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GMAT Geometry http://www.800score.com/forum/viewtopic.php?f=3&t=3 
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Author:  questioner [ Mon Jun 07, 2010 12:48 am ]  
Post subject:  GMAT Geometry  
The operation above rotates the regular pentagon clockwise by (72n) degrees from its center. What replaces the initial b position when n = 4? A. a B. b C. c D. d E. e (C) Think of the polygon as you would a wheel being rotated at its center. A wheel sweeps a 360º angle everytime it returns to its initial positon. Since this is a regular pentagon, we can increment our rotations by dividing 360/5 = 72. So everytime the center spins 72º clockwise, the polygon moves by one vertex. So, when n = 4 the pentagon rotates by 4 vertices and the initial b position is replaced by c or answer choice (C). An alternative way to consider the problem is to realize that everytime n = 5, it returns to its original configuration. So, when n = 4, the position is occupied by the vertex that is one vertex away in the clockwise direction because it lags by one 72º turn.  "Since this is a regular pentagon, we can increment our rotations by dividing 360/5 = 72. So everytime the center spins 72º clockwise, the polygon moves by one vertex." Doesn't pentagon has 540 degrees?

Author:  Gennadiy [ Mon Jun 07, 2010 1:15 am ]  
Post subject:  Re: Math (test 1, question 9): geometry (normal)  
The sum of internal pentagon angles is 540 degrees indeed, but 360 degrees  is "full rotation angle" when a point on a circle returns to its own position. So in the statement "Since this is a regular pentagon, we can increment our rotations by dividing 360/5 = 72. So everytime the center spins 72º clockwise, the polygon moves by one vertex." we note and use the statement that it is a REGULAR pentagon. I'd like to bring your attention to the term "REGULAR", because regular polygons can be inscribed in a circle and have vertexes "evenly distributed" on this circle. Following the same logic it will be true for any REGULAR polygon of n sides that by turning (360/n) degrees it rotates by one vertex.

Author:  questioner [ Mon Jun 07, 2010 2:57 pm ] 
Post subject:  Re: Math (test 1, question 9): geometry, polygon 
The operation above rotates the regular pentagon clockwise by (72n) degrees from its center. What replaces the initial b position when n = 19? A. a B. b C. c D. d E. e Right Answer is (C).  Since the rotation is clockwise and the position b is 72 degrees short of returning to its original configuration, shouldn't its position be at a? (c would replace b at 21n) 
Author:  Gennadiy [ Mon Jun 07, 2010 3:05 pm ]  
Post subject:  Re: Math (test 1, question 9): geometry, polygon  
Imagine that we rotated the pentagon 20 × 72º clockwise. That makes it look the same: If we rotate it (1 × 72) more degrees clockwise then it will look: We see that in this case (n = 21) point a stands at original position of point b. If we rotate (1 × 72) less degrees clockwise than (20 × 72º) clockwise turn then in this case it will look: For better understanding, please, take a look at this animation:

Author:  questioner [ Sat Jun 19, 2010 2:05 am ] 
Post subject:  Re: Math (test 1, question 9): geometry, polygon 
The operation above rotates the regular pentagon clockwise by (72n) degrees from its center. What replaces the initial b position when n = 4? A. a B. b C. c D. d E. e The right answer is choice (C). (See explanation above in this topic).  This answer is incorrect, if the polygon turns "clockwise" it should be turning from "b" to "c" in the first 72 degrees. Another 72 degrees 3 times would then leave the original point in upper most angle. The correct answer should be "a". 
Author:  Gennadiy [ Sat Jun 19, 2010 2:11 am ] 
Post subject:  Re: Math (test 1, question 9): geometry, polygon 
You're right about what place the point b itself will be at when n = 4, but the question asks about what point will be at original bpoint position when n = 4. So when point b is at original apoint position then point c is at original bpoint position. The answer is choice (C). When n = 4 the situation is following: 
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