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

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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, 06:21
LaxAvenger wrote:

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° ?


Yes, that is another way of interpreting Bunuel's solution above.
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In the figure above, triangle ABC is equilateral, and point P is  [#permalink]

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New post 19 Jun 2016, 19:41
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I found this explanation on beathegmat by gmatguruny. very helpful!!

The triangle is to be rotated clockwise.
For B to end up in A's position, the triangle has to be rotated 2/3 OF THE WAY AROUND.
Since one complete rotation = 360 degrees, 2/3 of the way around = (2/3) * 360 = 240.
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In the figure above, triangle ABC is equilateral, and point P is  [#permalink]

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New post 15 Sep 2016, 02:54
This is quite tricky/annoying question....

My approach was the following: I draw a cross with centre the centre of the triangle. Then is quite easy/obvious to see that if you do 3 * 90 clockwise you go above the point A. Hence from my drawing is obvious that: 1st turn 90 degrees + 2nd turn another 90 degree (Effectively now the triangle is upside down) + 3rd turn roughly 90/2 degrees which is approximately answer D) .....

I don't like much my approximation (although is fast and this matters a lot) although to be fair in a real GMAT environment I would never be able to come up with Bunuel's tactic. Actually I don't understand the topic that this specific question tests such that you should come up with Bunuel's solution.

imho the takeaway message for this is what is refereed to the following

zxcvbnmas wrote:
I found this explanation on beathegmat by gmatguruny. very helpful!!

The triangle is to be rotated clockwise.
For B to end up in A's position, the triangle has to be rotated 2/3 OF THE WAY AROUND.
Since one complete rotation = 360 degrees, 2/3 of the way around = (2/3) * 360 = 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 30 May 2017, 16:35
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Attached is a visual that should help.
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Screen Shot 2017-05-30 at 4.08.24 PM.png
Screen Shot 2017-05-30 at 4.08.24 PM.png [ 87.82 KiB | Viewed 895 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 04 Jul 2017, 10:54
I would not spend more than 40 seconds on this problem.
See figure below. The answer is somewhere between 180 and 270. Answer is D
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rotate.jpg
rotate.jpg [ 28.73 KiB | Viewed 807 times ]


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

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Re: In the figure above, triangle ABC is equilateral, and point P is &nbs [#permalink] 11 Jul 2018, 16:09

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