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

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Manager
Joined: 02 Dec 2012
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In the figure above, triangle ABC is equilateral, and point [#permalink]

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03 Dec 2012, 04:36
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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
Math Expert
Joined: 02 Sep 2009
Posts: 39713
In the figure above, triangle ABC is equilateral, and point [#permalink]

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03 Dec 2012, 04:40
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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:

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°.

[Reveal] Spoiler:
Attachment:

ABC+.png [ 7.19 KiB | Viewed 28136 times ]

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Re: In the figure above, triangle ABC is equilateral, and point [#permalink]

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02 Jun 2013, 06:42
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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]

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02 Jun 2013, 06:45
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Expert's post
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]

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02 Jun 2013, 07:26
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 .. That gives me clarity
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Re: In the figure above, triangle ABC is equilateral, and point [#permalink]

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12 Jun 2013, 19:15
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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 .. That gives me clarity
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Joined: 29 Aug 2012
Posts: 6
Re: In the figure above, triangle ABC is equilateral, and point [#permalink]

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12 Jul 2013, 07:06
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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:

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]

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15 Sep 2013, 14:24
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]

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16 Sep 2013, 01:57
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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]

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06 Dec 2013, 10:01
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]

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06 May 2014, 12:42
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]

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23 May 2014, 07:53
Bunuel 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°.

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.

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Posts: 39713
Re: In the figure above, triangle ABC is equilateral, and point [#permalink]

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23 May 2014, 10:30
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Expert's post
russ9 wrote:
Bunuel 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°.

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.

Let me ask you a question: how many degrees are in one revolution? Isn't it 360°?
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Re: In the figure above, triangle ABC is equilateral, and point [#permalink]

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23 May 2014, 10:47
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russ9 wrote:
Bunuel 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°.

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.

Kindly refer the diagram.
Attachment:

g2.png [ 18.24 KiB | Viewed 21832 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 $$angleCAB$$

Press Kudos if it helped.

Kudos is the best form of appreciation

Kindly check the post with the poll and revert with your replies
http://gmatclub.com/forum/10-straight-lines-no-two-of-which-are-parallel-and-no-three-171525.html
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Re: In the figure above, triangle ABC is equilateral, and point [#permalink]

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23 Aug 2014, 22:59
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]

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13 Dec 2014, 08:52
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Just think about it before you calculate and you can eliminate a ton of answers...

Must rotate > 180 as 180 would have B pointing directly down.

Now we are left with 240 or 270. If you think about where 270 would out B it would be above where A is currently.

So must be between 270 and 180....only answer left is 240.
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Re: In the figure above, triangle ABC is equilateral, and point [#permalink]

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13 Apr 2015, 11:49
Alternate Solution:
First, rotate the figure by 180 degrees clockwise, which will give us the attached figure.

Further, in order to bring B to A's position, we must rotate the figure by 60 degrees (AB has to be in place of AC, and this has to traverse 60 degrees clockwise).
Attachments

File comment: ABC rotate by 180deg

Picture1.png [ 29.21 KiB | Viewed 16585 times ]

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In the figure above, triangle ABC is equilateral, and point [#permalink]

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01 Jun 2015, 06:16
Hello All,
Clarifying some doubts for those who asked:
1. Since tri ABC is equilateral, all 3 angles must be 60 degs.
Due to symmetrical property of centroid point P, lines PC, PA will divide angle BCA and CAB into 30degs each,
so, inside triangle APC, angle APC = 180 - (30+30) = 120 degs.

Now we rotate clockwise, so we get
120 + 120 = 240 degs
Hope that helps

2. Trick: Total angle at a point is always 360 degrees (that is why, the central angle is 360 degs in Bunuel's explanation).

Kudos if you found it helpful
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Re: In the figure above, triangle ABC is equilateral, and point [#permalink]

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11 Jul 2015, 00:15
a circle contains 360 degrees, so we need 3 turns (3*120=360) to have point B on it's initial position..but here we make just 2 turns --> 2*120=240
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Re: In the figure above, triangle ABC is equilateral, and point [#permalink]

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31 Oct 2015, 03:16
Bunuel 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:

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°.

[Reveal] Spoiler:
Attachment:
ABC+.png

Can another possible solution be that all angles in the triangle have 60° and hence, the opposite sites of a circle drawn around it have to have double of that figure? -> 120° for each side. Now we need to turn from BC to CA, which means 120° + 120° = 240° ?
Re: In the figure above, triangle ABC is equilateral, and point   [#permalink] 31 Oct 2015, 03:16

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