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

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Manager
Joined: 02 Dec 2012
Posts: 172
In the figure above, triangle ABC is equilateral, and point P is  [#permalink]

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03 Dec 2012, 03:36
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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

Attachment:

ABC.png [ 4.92 KiB | Viewed 68602 times ]
Math Expert
Joined: 02 Sep 2009
Posts: 64243
In the figure above, triangle ABC is equilateral, and point P is  [#permalink]

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03 Dec 2012, 03:40
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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

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°.

Attachment:

ABC+.png [ 7.19 KiB | Viewed 70947 times ]

_________________
Intern
Joined: 31 Jul 2014
Posts: 39
Concentration: Finance, Technology
Schools: Owen '17 (M$) Re: In the figure above, triangle ABC is equilateral, and point P is [#permalink] ### Show Tags 13 Dec 2014, 07:52 9 2 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. ##### General Discussion Manager Joined: 15 Apr 2013 Posts: 67 Location: India Concentration: Finance, General Management Schools: ISB '15 WE: Account Management (Other) Re: In the figure above, triangle ABC is equilateral, and point P is [#permalink] ### Show Tags 02 Jun 2013, 05:42 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? Math Expert Joined: 02 Sep 2009 Posts: 64243 Re: In the figure above, triangle ABC is equilateral, and point P is [#permalink] ### Show Tags 02 Jun 2013, 05:45 1 1 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. _________________ Manager Joined: 15 Apr 2013 Posts: 67 Location: India Concentration: Finance, General Management Schools: ISB '15 WE: Account Management (Other) Re: In the figure above, triangle ABC is equilateral, and point P is [#permalink] ### Show Tags 02 Jun 2013, 06:26 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 Manager Joined: 30 Jun 2012 Posts: 79 Location: United States GMAT 1: 510 Q34 V28 GMAT 2: 580 Q35 V35 GMAT 3: 640 Q34 V44 GMAT 4: 690 Q43 V42 GPA: 3.61 WE: Education (Education) Re: In the figure above, triangle ABC is equilateral, and point P is [#permalink] ### Show Tags 12 Jun 2013, 18:15 1 1 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 Intern Joined: 28 Aug 2012 Posts: 4 Re: In the figure above, triangle ABC is equilateral, and point P is [#permalink] ### Show Tags 12 Jul 2013, 06:06 1 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. Manager Joined: 15 Aug 2013 Posts: 223 Re: In the figure above, triangle ABC is equilateral, and point P is [#permalink] ### Show Tags 15 Sep 2013, 13:24 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? Math Expert Joined: 02 Sep 2009 Posts: 64243 Re: In the figure above, triangle ABC is equilateral, and point P is [#permalink] ### Show Tags 16 Sep 2013, 00:57 2 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? _________________ Intern Joined: 16 Sep 2013 Posts: 11 Re: In the figure above, triangle ABC is equilateral, and point P is [#permalink] ### Show Tags 06 Dec 2013, 09:01 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. Manager Joined: 12 Feb 2011 Posts: 75 Re: In the figure above, triangle ABC is equilateral, and point P is [#permalink] ### Show Tags 06 May 2014, 11:42 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. Manager Joined: 15 Aug 2013 Posts: 223 Re: In the figure above, triangle ABC is equilateral, and point P is [#permalink] ### Show Tags 23 May 2014, 06:53 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. Math Expert Joined: 02 Sep 2009 Posts: 64243 Re: In the figure above, triangle ABC is equilateral, and point P is [#permalink] ### Show Tags 23 May 2014, 09:30 1 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°? _________________ Intern Joined: 13 May 2014 Posts: 32 Concentration: General Management, Strategy Re: In the figure above, triangle ABC is equilateral, and point P is [#permalink] ### Show Tags 23 May 2014, 09:47 4 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 [ 18.24 KiB | Viewed 57383 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 NUS School Moderator Affiliations: Oracle certified java programmer , adobe certified developer Joined: 14 Jul 2013 Posts: 51 GMAT Date: 02-12-2015 GPA: 3.87 WE: Programming (Telecommunications) Re: In the figure above, triangle ABC is equilateral, and point P is [#permalink] ### Show Tags 23 Aug 2014, 21:59 yeah 240 it is, initially overlooked the clockwise part , silly mistake. Its clear now _________________ IF IT IS TO BE , IT IS UP TO ME Intern Joined: 09 Dec 2014 Posts: 5 Location: United States Re: In the figure above, triangle ABC is equilateral, and point P is [#permalink] ### Show Tags 13 Apr 2015, 10:49 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 52062 times ] Intern Joined: 15 Sep 2014 Posts: 9 Re: In the figure above, triangle ABC is equilateral, and point P is [#permalink] ### Show Tags 01 Jun 2015, 05:16 3 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 Current Student Joined: 10 Mar 2013 Posts: 449 Location: Germany Concentration: Finance, Entrepreneurship Schools: WHU MBA"20 (A$)
GMAT 1: 580 Q46 V24
GPA: 3.88
WE: Information Technology (Consulting)
Re: In the figure above, triangle ABC is equilateral, and point P is  [#permalink]

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10 Jul 2015, 23:15
1
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
Manager
Joined: 18 Aug 2014
Posts: 109
Location: Hong Kong
Schools: Mannheim
Re: In the figure above, triangle ABC is equilateral, and point P is  [#permalink]

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31 Oct 2015, 02:16
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°.

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 P is   [#permalink] 31 Oct 2015, 02:16

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