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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

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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

by BTGmoderatorDC » Fri Mar 20, 2020 7:36 pm

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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


OA D

Source: Official Guide
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Re: In the figure above, triangle ABC is equilateral, and point P is equidistant from vertices A, B, and C. If triangle

by Brent@GMATPrepNow » Sat Mar 21, 2020 5:42 am
BTGmoderatorDC wrote: ↑
Fri Mar 20, 2020 7:36 pm
 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


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.
Image
Aside, we know that each angle is 120º since all three (equivalent) angles must add to 360.º

Then draw a circle so that the triangles vertices are on the circle.
Image

From here, we can see that . ..
Image
. . . 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
Brent Hanneson - Creator of GMATPrepNow.com
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