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    Strategies and techniques for approaching featured GMAT topics

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, 05:21
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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Re: In the figure above, triangle ABC is equilateral, and point P is  [#permalink]

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New post 09 Feb 2016, 06:27
nyknicks4412 wrote:
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.

That is how I solved it as well. The other explanations are too complicated for me.
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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, 18:41
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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, 01:54
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, 15:35
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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 1056 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, 09:54
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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
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In the figure above, triangle ABC is equilateral, and point P is  [#permalink]

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New post 03 Nov 2018, 04:09
bunuel wrote:
Let me ask you a question: how many degrees are in one revolution? Isn't it 360°?


This is precisely the part where I wasn't understanding! It didn't strike me that I don't have to look at the interior angles (I was taking it as 60 degrees and trying to calculate it) but now it makes complete sense when you asked "How many degrees are in one revolution?"

All hail Bunuel!!!
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In the figure above, triangle ABC is equilateral, and point P is &nbs [#permalink] 03 Nov 2018, 04:09

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