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Help on Geometry Problem - OG 13 - PS#75

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Help on Geometry Problem - OG 13 - PS#75

by amirhakimi » Thu Oct 31, 2013 6:04 am

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I can't understand this problem:

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

I already read the explanation but still no clues to handle this question type. Please help me understand this solution.

[spoiler]Since ABC is equilateral, the measure of <ACB is 60°. Therefore, the measure of <BCD is 180°- 60° = 120°. Rotating the figure clockwise about point P through an angle of 120° will produce the figure shown below.

Then rotating this figure clockwise about point P through an angle of 120° will produce the figure shown below.

In this figure, point B is in the position where points was in the original figure. The triangle was rotated clockwise about point P through 120° + 120° = 240°.
The correct answer is D.[/spoiler]
Last edited by amirhakimi on Thu Oct 31, 2013 7:43 am, edited 1 time in total.
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by GMATGuruNY » Thu Oct 31, 2013 6:28 am
amirhakimi wrote:I can't understand this problem:

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
The triangle is 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.

The correct answer is D.
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by mevicks » Thu Oct 31, 2013 6:56 am
Image
amirhakimi wrote: 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
Firstly, please use the [spoiler][/spoiler] tag to hide the answers; doing this allows others to solve the question without knowing the answer before hand. Secondly, for posting images use a service such as Imgur (myimgur.eden.fm) or postimg.cc

Now, coming back to the question:

For rotation and reflection problems an alternate faster solution is to actually rotate the diagram on your scratch pad (if you cannot visualize the rotated image at first)

Normal Situation:
Image

But we need the first and last triangle to be same so the last increment should be slightly less than 90°:
Image

Thus the total slightly less than 3*90° and D works.
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Help on Geometry Problem - OG 13 - PS#75

by Brent@GMATPrepNow » Thu Oct 31, 2013 7:39 am
amirhakimi wrote: 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
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
Last edited by Brent@GMATPrepNow on Mon Apr 16, 2018 1:17 pm, edited 1 time in total.
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by amirhakimi » Thu Oct 31, 2013 7:53 am
Thank you all for those hints. The lines that Brent posted helped a lot.
It was the first time that I post a problem. I didn't know about the [spoiler][/spoiler] tag. So, sorry about that.
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by mcdesty » Thu Jul 10, 2014 9:16 am
Half a circle(180 degrees) plus the 60 degrees for one angle of the equilateral triangle. See attachment.
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by Scott@TargetTestPrep » Tue Jun 23, 2015 3:04 pm
amirhakimi wrote:I can't understand this problem:

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

I already read the explanation but still no clues to handle this question type. Please help me understand this solution.

[spoiler]Since ABC is equilateral, the measure of <ACB is 60°. Therefore, the measure of <BCD is 180°- 60° = 120°. Rotating the figure clockwise about point P through an angle of 120° will produce the figure shown below.

Then rotating this figure clockwise about point P through an angle of 120° will produce the figure shown below.

In this figure, point B is in the position where points was in the original figure. The triangle was rotated clockwise about point P through 120° + 120° = 240°.
The correct answer is D.[/spoiler]
Solution:

To more easily solve this problem we can inscribe triangle ABC in a circle. Remember that a circle has a total of 360 degrees. So, if we were to rotate point B around the ENTIRE circle, it would rotate 360 degrees, returning to its original position. We also know that the inscribed equilateral triangle breaks the circle into 3 arcs of equal length. In other words, arc AB = arc BC = arc CA. We can take this one step further and say that each arc equals 360/3 = 120 degrees. So the degree measurement from B to C is 120 degrees, from C to A is 120 degrees and from A to B is 120 degrees. Let's see what happens when we rotate the triangle clockwise. Let's first rotate the triangle so that point B ends up where point C currently is. In order to get B to the position where C is, we have to rotate the triangle 120 degrees. Finally, we must rotate the triangle one more time so that point B is where point A initially was. Notice that to get point B to where point A initially was, we have to again rotate the triangle 120 degrees. Thus, in total, we have rotated the triangle 240 degrees to get point B to the position where point A originally was.

Answer: D

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by Matt@VeritasPrep » Tue Jun 23, 2015 5:54 pm
Inscribing the figure in a circle is the way to go - it works well for any inscribed polygon, not just a triangle.
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Re: Help on Geometry Problem - OG 13 - PS#75

by [email protected] » Wed Apr 21, 2021 8:09 pm
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

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