BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

OG Diagnostic question #10 (star)

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 437
Joined: Sat Nov 22, 2008 5:06 am
Location: India
Thanked: 50 times
Followed by:1 members
GMAT Score:580

by beat_gmat_09 » Tue Nov 09, 2010 9:59 am
Question ?
Hope is the dream of a man awake
Top
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Wed Nov 10, 2010 4:31 am
An easy approach to Q10 in the Official Guide (the star question) is to plug in for the angle measurements. We're being asked to find the sum of the angles measurements of the 5 points of the star. The key is to plug in values that follow the rules of geometry:

Image

Let's start with the most unusual shape, the pentagon inside the star. For any polygon with n sides, the sum of the interior angles = (n-2)*180. Thus, the sum of the angles inside the pentagon = (5-2)*180 = 540. There are 5 angles inside the pentagon. To make the math easy, let's plug in 540/5 = 108 for each interior angle.

Each of the adjacent angles must be 180-108 = 72 (see the picture), because the sum of angles that form a straight line must be 180.

There are 5 triangles around the outside of the star. The sum of the angles inside each of these triangles must be 180. This forces each point of the star to be 180-72-72 = 36.

Since the star has 5 points, the sum of the angle measurements of all 5 points is 5*36 = 180.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Newbie | Next Rank: 10 Posts
Posts: 3
Joined: Tue Nov 09, 2010 9:08 am

by matthewross » Wed Nov 17, 2010 8:51 pm
Thanks for replying, and thanks for being psychic enough to even guess which question I meant.... lol. Great explanation, appreciate it.
Top
Senior | Next Rank: 100 Posts
Posts: 45
Joined: Wed May 14, 2014 11:26 am
Thanked: 1 times

by Joseph_Alexander » Wed Nov 19, 2014 9:06 am
GMATGuruNY wrote:Let's start with the most unusual shape, the pentagon inside the star. For any polygon with n sides, the sum of the interior angles = (n-2)*180. Thus, the sum of the angles inside the pentagon = (5-2)*180 = 540. There are 5 angles inside the pentagon. To make the math easy, let's plug in 540/5 = 108 for each interior angle.
Hi Mitch!

Here we are deriving 108 as the interior angle as a regular pentagon. In GMAT quant, we can't assume that the figures are drawn to scale unless the question specifically mentions it. So how can we assume that each of those angles are 108 degrees.

Am I missing something here? Please assist.
Top
Master | Next Rank: 500 Posts
Posts: 447
Joined: Fri Nov 08, 2013 7:25 am
Thanked: 25 times
Followed by:1 members

by Mathsbuddy » Wed Nov 19, 2014 9:43 am
GMATGuruNY wrote: Let's start with the most unusual shape, the pentagon inside the star. For any polygon with n sides, the sum of the interior angles = (n-2)*180. Thus, the sum of the angles inside the pentagon = (5-2)*180 = 540. There are 5 angles inside the pentagon. To make the math easy, let's plug in 540/5 = 108 for each interior angle.
Alternatively, the EXTERIOR angles always add up to 360 degrees.
Using the pentagon again: 360/5 = 72
To find an internal angle: 180 - 72 = 108 (angles on a straight line)
This might save some time.
Top
Master | Next Rank: 500 Posts
Posts: 447
Joined: Fri Nov 08, 2013 7:25 am
Thanked: 25 times
Followed by:1 members

by Mathsbuddy » Wed Nov 19, 2014 9:55 am
Mathsbuddy wrote:
GMATGuruNY wrote: Let's start with the most unusual shape, the pentagon inside the star. For any polygon with n sides, the sum of the interior angles = (n-2)*180. Thus, the sum of the angles inside the pentagon = (5-2)*180 = 540. There are 5 angles inside the pentagon. To make the math easy, let's plug in 540/5 = 108 for each interior angle.
Alternatively, the EXTERIOR angles always add up to 360 degrees.
Using the pentagon again: 360/5 = 72
To find an internal angle: 180 - 72 = 108 (angles on a straight line)
This might save some time.
In fact, you could just find these external angles (360/N) and double them straight away to get the sum of the base angles of each external triangle. Then subtract from 180.

So, angle at point of star = 180 - 2 x (360/N) = 180 - 720/N
Sum of such angles, S = N x (180 - 720/N) = 180N -720 = 180N - 180*4 = 180(N-4)

So the general formula is: S = 180(N-4)
[Note that a star cannot have N < 5]

If N = 5, then S = 180(5-4) = 180 x 1 = 180
If N = 6, S = 180 x 2 = 360
If N = 7, S = 180 x 3 = 540
etc.
Top
Master | Next Rank: 500 Posts
Posts: 447
Joined: Fri Nov 08, 2013 7:25 am
Thanked: 25 times
Followed by:1 members

by Mathsbuddy » Wed Nov 19, 2014 9:59 am
Joseph_Alexander wrote:
GMATGuruNY wrote:Let's start with the most unusual shape, the pentagon inside the star. For any polygon with n sides, the sum of the interior angles = (n-2)*180. Thus, the sum of the angles inside the pentagon = (5-2)*180 = 540. There are 5 angles inside the pentagon. To make the math easy, let's plug in 540/5 = 108 for each interior angle.
Hi Mitch!

Here we are deriving 108 as the interior angle as a regular pentagon. In GMAT quant, we can't assume that the figures are drawn to scale unless the question specifically mentions it. So how can we assume that each of those angles are 108 degrees.

Am I missing something here? Please assist.
I guess the question mentions EITHER a star with 5 points, OR asks what the smallest size star-point-angle there is for a regular star (as 5 is the minimum number points possible, then that would be the one to investigate).

In future, could the full question be published please. Thanks.
Top
Senior | Next Rank: 100 Posts
Posts: 45
Joined: Wed May 14, 2014 11:26 am
Thanked: 1 times

by Joseph_Alexander » Wed Nov 19, 2014 10:10 am
Mathsbuddy wrote: In future, could the full question be published please. Thanks.
Question is simple, wherever Mitch has taken the value as 36 in his figure, the question labels each of them with an alphabet and asks for the sum of each of those 5 alphabets.
Top
Senior | Next Rank: 100 Posts
Posts: 45
Joined: Wed May 14, 2014 11:26 am
Thanked: 1 times

by Joseph_Alexander » Wed Nov 19, 2014 10:13 am
Mathsbuddy wrote:
GMATGuruNY wrote: Let's start with the most unusual shape, the pentagon inside the star. For any polygon with n sides, the sum of the interior angles = (n-2)*180. Thus, the sum of the angles inside the pentagon = (5-2)*180 = 540. There are 5 angles inside the pentagon. To make the math easy, let's plug in 540/5 = 108 for each interior angle.
Alternatively, the EXTERIOR angles always add up to 360 degrees.
Using the pentagon again: 360/5 = 72
To find an internal angle: 180 - 72 = 108 (angles on a straight line)
This might save some time.
Mathsbuddy, I am unable to understand your response.
a) How can we assume that all the angles in the pentagon are equal as the pentagon need not be a regular pentagon?
b) If (a) above is true, how can the exterior angles be 360 degrees?
Top
Senior | Next Rank: 100 Posts
Posts: 45
Joined: Wed May 14, 2014 11:26 am
Thanked: 1 times

by Joseph_Alexander » Wed Nov 19, 2014 10:17 am
Wow! See what I found with a little googling! :P

A star is always regularly shaped!
1. The sum of the angles formed at the tips of the five pointed star is 180; the sum of the angles formed at the tips of the six pointed star is 360.
2. The formula for the sum of the angle measurements at the tips of an n-pointed star is f(n)=180(n)-720 where n is an integer greater than 4.

Read more here: https://mathforum.org/pom2/nov.98/winner.html :P

This clarifies my doubts :D
Top
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Wed Nov 19, 2014 11:23 am
Joseph_Alexander wrote:
GMATGuruNY wrote:Let's start with the most unusual shape, the pentagon inside the star. For any polygon with n sides, the sum of the interior angles = (n-2)*180. Thus, the sum of the angles inside the pentagon = (5-2)*180 = 540. There are 5 angles inside the pentagon. To make the math easy, let's plug in 540/5 = 108 for each interior angle.
Hi Mitch!

Here we are deriving 108 as the interior angle as a regular pentagon. In GMAT quant, we can't assume that the figures are drawn to scale unless the question specifically mentions it. So how can we assume that each of those angles are 108 degrees.

Am I missing something here? Please assist.
We can plug in ANY COMBINATION OF ANGLES that satisfies the following constraints:
1. Angles inside the pentagon must have a sum of 540
2. Angles that form a straight line must have a sum of 180
3. Angles inside a triangle must have a sum of 180

In my solution above, I made the angles inside the pentagon equal:
Image
In this case, v+w+x+y+z = 36+36+36+36+36 = 180.

But the angles inside the pentagon can be ANY COMBINATION THAT HAS A SUM OF 540:
Image
In this case, v+w+x+y+z = 45+60+15+25+35 = 180.

The same answer -- 180 -- is yielded in each case.
The reason is that each case satisfies all of the constraints in the problem.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Master | Next Rank: 500 Posts
Posts: 447
Joined: Fri Nov 08, 2013 7:25 am
Thanked: 25 times
Followed by:1 members

by Mathsbuddy » Thu Nov 20, 2014 12:46 am
Joseph_Alexander wrote:
Mathsbuddy wrote:
GMATGuruNY wrote: Let's start with the most unusual shape, the pentagon inside the star. For any polygon with n sides, the sum of the interior angles = (n-2)*180. Thus, the sum of the angles inside the pentagon = (5-2)*180 = 540. There are 5 angles inside the pentagon. To make the math easy, let's plug in 540/5 = 108 for each interior angle.
Alternatively, the EXTERIOR angles always add up to 360 degrees.
Using the pentagon again: 360/5 = 72
To find an internal angle: 180 - 72 = 108 (angles on a straight line)
This might save some time.
Mathsbuddy, I am unable to understand your response.
a) How can we assume that all the angles in the pentagon are equal as the pentagon need not be a regular pentagon?
b) If (a) above is true, how can the exterior angles be 360 degrees?
a) Without knowing the full question, I do not know whether the star is regular or not. However, the sum of all points will be the same anyway - by standard rules of geometry.
b) The exterior angles of ANY polygon (triangle, quadrilateral, pentagon, hexagon, heptagon, octagon, etc... any 2D shape with straight sides)ALWAYS add up to 360 degrees. Look here: https://www.geom.uiuc.edu/~dwiggins/conj09.html

I hope this helps.
Top
Master | Next Rank: 500 Posts
Posts: 447
Joined: Fri Nov 08, 2013 7:25 am
Thanked: 25 times
Followed by:1 members

by Mathsbuddy » Thu Nov 20, 2014 12:49 am
Joseph_Alexander wrote:Wow! See what I found with a little googling! :P

A star is always regularly shaped!
1. The sum of the angles formed at the tips of the five pointed star is 180; the sum of the angles formed at the tips of the six pointed star is 360.
2. The formula for the sum of the angle measurements at the tips of an n-pointed star is f(n)=180(n)-720 where n is an integer greater than 4.

Read more here: https://mathforum.org/pom2/nov.98/winner.html :P

This clarifies my doubts :D
I'm pleased you see it now. This matches what I had said earlier:

Sum = 180 x (N-4)

If N = 5, then S = 180(5-4) = 180 x 1 = 180
If N = 6, S = 180 x 2 = 360
If N = 7, S = 180 x 3 = 540
etc.
Good luck!
Top
Senior | Next Rank: 100 Posts
Posts: 45
Joined: Wed May 14, 2014 11:26 am
Thanked: 1 times

by Joseph_Alexander » Thu Nov 20, 2014 2:53 am
Thanks Mitch and Mathsbuddy for those beautiful explanations. Mitch special thanks for your efforts in putting up those images! :)
Top
Post new topic Post Reply
• Page 1 of 1