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In the figure above, triangle ABC is equilateral, and point [#permalink]
03 Dec 2012, 03:36

5

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

60% (01:48) correct
40% (00:46) wrong based on 604 sessions

Attachment:

ABC.png [ 4.92 KiB | Viewed 8284 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?

Re: In the figure above, triangle ABC is equilateral, and point [#permalink]
03 Dec 2012, 03:40

7

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

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:

ABC+.png [ 7.19 KiB | Viewed 8267 times ]

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]
12 Jul 2013, 06:06

1

This post was BOOKMARKED

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]
16 Sep 2013, 00:57

1

This post received KUDOS

Expert's post

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? _________________

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

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

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

Re: In the figure above, triangle ABC is equilateral, and point [#permalink]
23 May 2014, 09:30

Expert's post

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°? _________________

Re: In the figure above, triangle ABC is equilateral, and point [#permalink]
23 May 2014, 09:47

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 3883 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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