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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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members

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

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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 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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Source: Beat The GMAT — Problem Solving | Fri Mar 20, 2020 7:36 pm

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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.

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.

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

Brent Hanneson - Creator of GMATPrepNow.com

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