Problem Solving

I searched the forum but didn't see anyone asking about this one.

I read the explanation but I still feel like this one would take 20 minutes on the test if you did it that way..... Does anyone know a faster way to do it? thnx

Question ?

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:



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.

Thanks for replying, and thanks for being psychic enough to even guess which question I meant.... lol. Great explanation, appreciate it.

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.

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 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.

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.

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.

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?

Wow! See what I found with a little googling!

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

This clarifies my doubts

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:

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:

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.

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.

Joseph_Alexander wrote:Wow! See what I found with a little googling!

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

This clarifies my doubts

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!

Thanks Mitch and Mathsbuddy for those beautiful explanations. Mitch special thanks for your efforts in putting up those images!