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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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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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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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19 Jun 2016, 18:41
2
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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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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30 May 2017, 15:35
Top Contributor
Attached is a visual that should help.
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Screen Shot 2017-05-30 at 4.08.24 PM.png [ 87.82 KiB | Viewed 1646 times ]

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

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04 Jul 2017, 09:54
3
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 [ 28.73 KiB | Viewed 1603 times ]

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

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

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01 Apr 2019, 03:45
No illustration or notes are necessary for this one, I think.

I just mentally projected 90 degree angles rotating clockwise starting from B. This way you can determine that the final point lies between 180 degrees (B would be pointing towards the South) and 270 (B would be pointing westwards).

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

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13 Sep 2019, 14:21
For some reason I always take the angle at the vertex instead of the central angle in these vertex questions, producing an incorrect 180.

Now I realise we calculate the central angle as the object is rotated around this central point P
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In the figure above, triangle ABC is equilateral, and point P is  [#permalink]

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17 Dec 2019, 14:14
Hi - chetan2u Gladiator59 Bunuel, generis

Its clear $$\angle BPN$$ | $$\angle NPD$$ | $$\angle MPD$$ and $$\angle BPM$$ are all $$90^{\circ}$$ each (totaling up-to $$360^{\circ}$$)

Why can't I assume Line PA splits angle MPD into 2 halves to give me $$45^{\circ}$$ each ?

Thats what i did and thus got a sum of $$180^{\circ}$$ + $$45^{\circ}$$ = $$225^{\circ}$$ instead of $$240^{\circ}$$

I am not disagreeing with option D but was wondering about where is the gap in my logic
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File comment: Triangle

Attachment 1.PNG [ 61.69 KiB | Viewed 249 times ]

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

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30 Dec 2019, 20:06

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

Attachment:
ABC.png

For this question, is point P the centroid and centre of the triangle as it is equilateral? Thanks!
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Re: In the figure above, triangle ABC is equilateral, and point P is  [#permalink]

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30 Dec 2019, 20:14
Yellowp wrote:

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

Attachment:
ABC.png

For this question, is point P the centroid and centre of the triangle as it is equilateral? Thanks!

Yes. P is equidistant, hence the center

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

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21 Mar 2020, 05:43
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Top Contributor

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

Attachment:
ABC.png

It may help to add some lines to the diagram.

First add lines from the center to the 3 vertices.

Aside, we know that each angle is 120º since all three (equivalent) angles must add to 360º

Then draw a circle so that the triangles vertices are on the circle.

From here, we can see that . ..

. . . the triangle must be rotated clockwise 240º in order for point B to be in the position where point A is now.

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

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20 May 2020, 03:48
Bunuel is this approach conceptually correct? (attached file)
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IMG_2165.jpeg [ 37.34 KiB | Viewed 20 times ]

Re: In the figure above, triangle ABC is equilateral, and point P is   [#permalink] 20 May 2020, 03:48

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