It is currently Fri Jan 18, 2019 6:40 am

 All times are UTC - 5 hours [ DST ]

 Page 1 of 1 [ 6 posts ]
 Print view Previous topic | Next topic
Author Message
 Post subject: GMAT GeometryPosted: Mon Jun 07, 2010 12:48 am

Joined: Sun May 30, 2010 3:15 am
Posts: 424

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?

Top

 Post subject: Re: Math (test 1, question 9): geometry (normal)Posted: Mon Jun 07, 2010 1:15 am

Joined: Sun May 30, 2010 2:23 am
Posts: 498
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.

Top

 Post subject: Re: Math (test 1, question 9): geometry, polygonPosted: Mon Jun 07, 2010 2:57 pm

Joined: Sun May 30, 2010 3:15 am
Posts: 424

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

----------

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)

Last edited by questioner on Tue Jun 08, 2010 4:03 pm, edited 1 time in total.

Top

 Post subject: Re: Math (test 1, question 9): geometry, polygonPosted: Mon Jun 07, 2010 3:05 pm

Joined: Sun May 30, 2010 2:23 am
Posts: 498
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:

Top

 Post subject: Re: Math (test 1, question 9): geometry, polygonPosted: Sat Jun 19, 2010 2:05 am

Joined: Sun May 30, 2010 3:15 am
Posts: 424

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".

Top

 Post subject: Re: Math (test 1, question 9): geometry, polygonPosted: Sat Jun 19, 2010 2:11 am

Joined: Sun May 30, 2010 2:23 am
Posts: 498
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 b-point position when n = 4.

So when point b is at original a-point position then point c is at original b-point position. The answer is choice (C).

When n = 4 the situation is following:

Top

 Display posts from previous: All posts1 day7 days2 weeks1 month3 months6 months1 year Sort by AuthorPost timeSubject AscendingDescending
 Page 1 of 1 [ 6 posts ]

 All times are UTC - 5 hours [ DST ]

#### Who is online

Users browsing this forum: No registered users and 3 guests

 You cannot post new topics in this forumYou cannot reply to topics in this forumYou cannot edit your posts in this forumYou cannot delete your posts in this forumYou cannot post attachments in this forum

Search for:
 Jump to:  Select a forum ------------------ GMAT    GMAT: Quantitative Section (Math)    GMAT: Verbal Section    GMAT: Integrated Reasoning    GMAT: General Questions GRE    GRE: Quantitative Reasoning (Math)    GRE: Verbal Reasoning    GRE: General Questions General questions    Other questions