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In the figure above, triangle ABC is equilateral, and point

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In the figure above, triangle ABC is equilateral, and point [#permalink] New post 03 Dec 2012, 03:36
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Attachment:
ABC.png
ABC.png [ 4.92 KiB | Viewed 6636 times ]
In the figure above, triangle ABC is equilateral, and point P is equidistant from vertices A, B, and C. If triangle ABC is rotated clockwise about point P, what is the minimum number of degrees the triangle must be rotated so that point B will be in the position where point A is now?

(A) 60
(B) 120
(C) 180
(D) 240
(E) 270
[Reveal] Spoiler: OA
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Re: In the figure above, triangle ABC is equilateral, and point [#permalink] New post 03 Dec 2012, 03:40
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Walkabout wrote:
Attachment:
The attachment ABC.png is no longer available
In the figure above, triangle ABC is equilateral, and point P is equidistant from vertices A, B, and C. If triangle ABC is rotated clockwise about point P, what is the minimum number of degrees the triangle must be rotated so that point B will be in the position where point A is now?

(A) 60
(B) 120
(C) 180
(D) 240
(E) 270


Look at the diagram below:
Attachment:
ABC+.png
ABC+.png [ 7.19 KiB | Viewed 6629 times ]
Each of the three red central angles is 120° (360/3=120). Thus in order point B to be in the position where point A is, the triangle should be rotated clockwise by 120°+120°=240°.

Answer: D.
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Re: In the figure above, triangle ABC is equilateral, and point [#permalink] New post 02 Jun 2013, 05:42
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Look at the diagram below:Each of the three red central angles is 120°

Can you please explain this part? ( Which concept is applied here?)

and what is the level of this question?
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Re: In the figure above, triangle ABC is equilateral, and point [#permalink] New post 02 Jun 2013, 05:45
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pavan2185 wrote:
Look at the diagram below:Each of the three red central angles is 120°

Can you please explain this part? ( Which concept is applied here?)

and what is the level of this question?


Each read angle is equal and all three together constitute for 360°, thus each is equal to 360/3=120.

As for the difficulty: I'd say it's ~600.
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Re: In the figure above, triangle ABC is equilateral, and point [#permalink] New post 02 Jun 2013, 06:26
Bunuel wrote:
pavan2185 wrote:
Look at the diagram below:Each of the three red central angles is 120°

Can you please explain this part? ( Which concept is applied here?)

and what is the level of this question?


Each read angle is equal and all three together constitute for 360°, thus each is equal to 360/3=120.

As for the difficulty: I'd say it's ~600.


Thank you :-D.. That gives me clarity :)
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Re: In the figure above, triangle ABC is equilateral, and point [#permalink] New post 12 Jun 2013, 18:15
Is there any other way to solve this problem? I'm having trouble understanding the principles behind it. How did you know to use central angles?

Thanks!

pavan2185 wrote:
Bunuel wrote:
pavan2185 wrote:
Look at the diagram below:Each of the three red central angles is 120°

Can you please explain this part? ( Which concept is applied here?)

and what is the level of this question?


Each read angle is equal and all three together constitute for 360°, thus each is equal to 360/3=120.

As for the difficulty: I'd say it's ~600.


Thank you :-D.. That gives me clarity :)
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Re: In the figure above, triangle ABC is equilateral, and point [#permalink] New post 12 Jul 2013, 06:06
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Yes, I have an alternate way(equivalent to the one above)

Actually, for the purpose of understanding let that point P be Point O as in Fig:

Image

Now, since it is given that A,B,C are equidistant from O, so we can draw a circle with centre O, & since angle at centre is twice the angle made on the circle. angle(AOB)=angle(BOC)=angle(AOC) = 120'.
Now. we want to move point B clockwise first to position C's & then to A's position i.e 2 moves, thereby the change in angle at centre O is 120+120=240, which is the ans.

Bunuel, Pls correct if I am wrong.
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Re: In the figure above, triangle ABC is equilateral, and point [#permalink] New post 14 Jul 2013, 09:35
Yeah. 240 is the correct answer
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Re: In the figure above, triangle ABC is equilateral, and point [#permalink] New post 15 Sep 2013, 13:24
How can we assume that each sector is broken up in 1/3's of 360 - as in 120?

Once that fundamental point is clear, the rest is each, but how can we come to that conclusion?
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Re: In the figure above, triangle ABC is equilateral, and point [#permalink] New post 16 Sep 2013, 00:57
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russ9 wrote:
How can we assume that each sector is broken up in 1/3's of 360 - as in 120?

Once that fundamental point is clear, the rest is each, but how can we come to that conclusion?


Triangle ABC is equilateral, and point P is equidistant from the vertices. Now, ask yourself why should any of the central angles be greater than the others?
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Re: In the figure above, triangle ABC is equilateral, and point [#permalink] New post 06 Dec 2013, 09:01
Instead of rotating clock-wise can we not rotate it anti-clock wise, since it is asking for the minimum number of degrees?
If so we get 120 as the answer.
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Re: In the figure above, triangle ABC is equilateral, and point [#permalink] New post 06 May 2014, 11:42
aleem681 wrote:
Instead of rotating clock-wise can we not rotate it anti-clock wise, since it is asking for the minimum number of degrees?
If so we get 120 as the answer.

No because the question specifically asks to rotate clockwise.
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Re: In the figure above, triangle ABC is equilateral, and point [#permalink] New post 23 May 2014, 06:53
Bunuel wrote:
Walkabout wrote:
Attachment:
ABC.png
In the figure above, triangle ABC is equilateral, and point P is equidistant from vertices A, B, and C. If triangle ABC is rotated clockwise about point P, what is the minimum number of degrees the triangle must be rotated so that point B will be in the position where point A is now?

(A) 60
(B) 120
(C) 180
(D) 240
(E) 270


Look at the diagram below:
Attachment:
ABC+.png
Each of the three red central angles is 120° (360/3=120). Thus in order point B to be in the position where point A is, the triangle should be rotated clockwise by 120°+120°=240°.

Answer: D.


Hi Bunuel,

How do you make the jump that all three areas total to 360?

When I solved for this, I visualized the total internal angles to equal 180 and 180/3 = 60, therefore it had to move 120.

Thanks for your help.
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Re: In the figure above, triangle ABC is equilateral, and point [#permalink] New post 23 May 2014, 09:30
Expert's post
russ9 wrote:
Bunuel wrote:
Walkabout wrote:
Attachment:
ABC.png
In the figure above, triangle ABC is equilateral, and point P is equidistant from vertices A, B, and C. If triangle ABC is rotated clockwise about point P, what is the minimum number of degrees the triangle must be rotated so that point B will be in the position where point A is now?

(A) 60
(B) 120
(C) 180
(D) 240
(E) 270


Look at the diagram below:
Attachment:
ABC+.png
Each of the three red central angles is 120° (360/3=120). Thus in order point B to be in the position where point A is, the triangle should be rotated clockwise by 120°+120°=240°.

Answer: D.


Hi Bunuel,

How do you make the jump that all three areas total to 360?

When I solved for this, I visualized the total internal angles to equal 180 and 180/3 = 60, therefore it had to move 120.

Thanks for your help.


Let me ask you a question: how many degrees are in one revolution? Isn't it 360°?
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: In the figure above, triangle ABC is equilateral, and point [#permalink] New post 23 May 2014, 09:47
russ9 wrote:
Bunuel wrote:
Walkabout wrote:
Attachment:
The attachment ABC.png is no longer available
In the figure above, triangle ABC is equilateral, and point P is equidistant from vertices A, B, and C. If triangle ABC is rotated clockwise about point P, what is the minimum number of degrees the triangle must be rotated so that point B will be in the position where point A is now?

(A) 60
(B) 120
(C) 180
(D) 240
(E) 270


Look at the diagram below:
Attachment:
The attachment ABC+.png is no longer available
Each of the three red central angles is 120° (360/3=120). Thus in order point B to be in the position where point A is, the triangle should be rotated clockwise by 120°+120°=240°.

Answer: D.


Hi Bunuel,

How do you make the jump that all three areas total to 360?

When I solved for this, I visualized the total internal angles to equal 180 and 180/3 = 60, therefore it had to move 120.

Thanks for your help.


Kindly refer the diagram.
Attachment:
g2.png
g2.png [ 18.24 KiB | Viewed 2248 times ]

The angle by which point B has to be rotated around P is angleBPC + angleCPA .

Please do not confuse it with the internal angles: angleBCA and angleCAB

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Re: In the figure above, triangle ABC is equilateral, and point [#permalink] New post 23 Aug 2014, 21:59
yeah 240 it is, initially overlooked the clockwise part , silly mistake. Its clear now :)
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Re: In the figure above, triangle ABC is equilateral, and point   [#permalink] 23 Aug 2014, 21:59
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