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

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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


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Attachment:
ABC.png
ABC.png [ 4.92 KiB | Viewed 29818 times ]
[Reveal] Spoiler: OA

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

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Walkabout wrote:
Image

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:

Image

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.

[Reveal] Spoiler:
Attachment:
ABC+.png
ABC+.png [ 7.19 KiB | Viewed 31989 times ]

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

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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 P is [#permalink]

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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 P is [#permalink]

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New post 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 :-D.. That gives me clarity :)

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

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New post 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 :-D.. That gives me clarity :)

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

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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 P is [#permalink]

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New post 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 P is [#permalink]

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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 P is [#permalink]

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New post 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 P is [#permalink]

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New post 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 P is [#permalink]

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New post 23 May 2014, 07:53
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 P is [#permalink]

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New post 23 May 2014, 10:30
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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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Re: In the figure above, triangle ABC is equilateral, and point P is [#permalink]

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New post 23 May 2014, 10:47
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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 25148 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 P is [#permalink]

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New post 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 P is [#permalink]

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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 P is [#permalink]

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New post 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
Picture1.png [ 29.21 KiB | Viewed 19898 times ]

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

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New post 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 P is [#permalink]

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New post 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 P is [#permalink]

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New post 31 Oct 2015, 03:16
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:

Image

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.

[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° ?

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Re: In the figure above, triangle ABC is equilateral, and point P is   [#permalink] 31 Oct 2015, 03:16

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