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

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03 Dec 2012, 04:36

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Attachment:

ABC.png [ 4.92 KiB | Viewed 25520 times ]

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?

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°.

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

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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 [ 18.24 KiB | Viewed 20995 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.

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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

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?
_________________

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°?
_________________

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

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12 Jul 2013, 07:06

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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:

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.

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

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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.

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

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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.

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

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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.

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

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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).

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

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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).

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

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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 [#permalink]

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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:

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°.

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° ?

gmatclubot

Re: In the figure above, triangle ABC is equilateral, and point
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31 Oct 2015, 03:16

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