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

[BTGmoderatorDC]

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

OA D

Source: Official Guide

[Brent@GMATPrepNow]

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

OA D

Source: Official Guide

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.

Answer: D

Cheers,
Brent

[Scott@TargetTestPrep]

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

We have a triangle with 3 vertices. Thus, when one vertex is rotated clockwise to the next vertex (for example, B is rotated clockwise to C), the number of degrees is 360/3 = 120. So, in order for point B to be in the position where point A is now, it must be rotated by 120 x 2 = 240 degrees.

Answer: D