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

Question

https://gmatclub.com/forum/download/file.php?id=18813 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 ABC.png

Bunuel · Accepted Answer

Look at the diagram below: https://gmatclub.com/forum/download/file.php?id=18814 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. ABC+.png

nyknicks4412 · Answer

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.

pavan2185 · Answer

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?

Bunuel · Answer

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.

p111 · Answer

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:

https://3.bp.blogspot.com/-RnQkB9NvdO4/UTZAZVDZxtI/AAAAAAAAQ0Q/-iooDpud19k/s1600/picture3.jpg

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.

russ9 · Answer

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?

Bunuel · Answer

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?

dreamsfulfilled · Answer

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

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

zxcvbnmas · Answer

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.

mcelroytutoring · Answer

Attached is a visual that should help.

sasyaharry · Answer

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

BrentGMATPrepNow · Answer

It may help to add some lines to the diagram.

First add lines from the center to the 3 vertices. 
 https://s8.postimg.cc/e412m8uq9/clk1.jpg 
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. 
 https://s8.postimg.cc/4m5wzxyfl/clk2.jpg

From here, we can see that . .. 
 https://s8.postimg.cc/npun2jlw1/clk3.jpg 
. . . the triangle must be rotated clockwise 240º in order for point B to be in the position where point A is now.

Answer: D

Cheers,
Brent

EMPOWERgmatRichC · Answer

Hi All,

We’re given an EQUILATERAL triangle – with the three points of the triangle equidistant from the ‘center’ point P. We’re asked to rotate the triangle CLOCKWISE and we are asked how many degrees it would take for Point B (re: the top of the triangle) to get to where Point A is (the lower-left corner). This is a great ‘concept question’, meaning that you don’t have to do much math to answer it if you recognize the concepts involved.

Since we are rotating the shape, we are ultimately creating a pathway that is a CIRCLE, so we have to think in terms of the number of degrees associated with that shape (re: 360 degrees). Since we’re dealing with an EQUILATERAL triangle, we can essentially think in terms of breaking that circle into THIRDS (and 360/3 = 120 degrees per third).

Thus, when we rotate Point B to where Point C is, that will be a 120 degree ‘move’ and when we go from Point C to where Point A is, that will be another 120 degree ‘move’ – for a total of 120 + 120 = 240 degrees to get from Point B to where Point A is.

Final Answer:  E

GMAT Assassins aren’t born, they’re made,
Rich

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

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

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