Source: Official Guide for the GMAT 13th Ed. Problem Solving; #75 Official Guide for the GMAT 2015 14th Ed. Problem Solving; #75

1

In the figure above, triangle ABC is

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 position where point A is now?

3 Explanations

1

hichem elkateb

how do you know that is two third ?!! can you please explain it with more details ?
Thank you :D

Dec 8, 2015 • Comment

Sam Kinsman, Magoosh Tutor

Hi Hichem,

Try to think of it like this. We'll draw a circle around the triangle, like Jose suggested below. The center of the circle is point P.

Note that all three sides of the triangle have the same length, so therefore all three corners of the triangle are equally spaced out along the circle. This means that if we start at one corner, and travel along the circle until we get to the next corner, we will have covered 1/3 of the circle's circumference.

In this problem, we're starting at point B. We travel along the circle to point C, and keep going along the circle until we get to point A. So we've covered 2/3 of the circle's circumference.

So we know that we need to rotate the triangle by 2/3 of a full turn. A full turn is 360 degrees, so the answer is 2/3 of 360 degrees = 240 degrees.

Dec 10, 2015 • Reply

Christina Lee

Had the same question so thank you!

Jan 5, 2019 • Reply

1

Let´s circumscribe the triangle:

If we draw a circle around the triangle, and then draw three lines from the center of the circle P to each of the three points of the triangle (A,B,C), we will obtain a kind of "Mercedes logo".

Each of the resulting fractions has 120º, since 360 / 3 = 120

Since we have to go over two portions (portion BPC and CPA):

120 * 2 = 240

Jan 25, 2015 • Comment

2

Gravatar Mike McGarry, Magoosh Tutor

Dec 27, 2013 • Comment

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