Does regular polygon X have more than 6 sides?

Question

Fresh GMAT Club Tests' Challenge Question:

(1) The ratio of the length of the longest line that can be drawn between two vertices of polygon X to the length of any one side of polygon X is greater than 2.
(2) The degree measure of an interior angle of polygon X is NOT an integer.

M36-139

pulkitaggi · Answer

IMO D.

Statement 1: As we move from Triangle to hexagon we can see that this ratio increases from 1 ----> 1.732, hence for ratio to be 2, the polygon should be the one with more than 6 sides

Statement 2: As we move from triangle to hexagon we can see that the interior angle will always be integer. But as soon as we have more than 6 sides the interior angle becomes non Integer.

Mathematically, any interior angle can be given by  , where N= number of sides of the regular polygon. All values of N from 3 to 6 are factor of 180.

GMATinsight · Answer

The regular polygon has 
Exterior angle = 360/n = 360/6 = 60 degrees
Interior angle = 180-Exterior = 180-(360/n) = 180-60 = 120 degrees
Length of Diagonal = 2*Length of side

Statement 1 : The ratio of the length of the longest line that can be drawn between two vertices of polygon X to the length of any one side of polygon X is greater than 2. 
Since the longest diagonal in a hexagon is twice the length of side therefore the polygon discussed in Question is definitely not a six sided polygon
SUFFICIENT

Statement 2 : The degree measure of an interior angle of polygon X is NOT an integer. 
But we have already seen that the degree measure of a six sided polygon is 120 hence it's definitely not a six sided polygon
SUFFICIENT

Answer: Option D

akela · Answer

The question asks "Does regular polygon X have  more than 6 sides? "

(1) 
Let's take a square. The length of the longest line is the diameter of the square. So the ratio of diameter to the side increases because the length of the diameter is constant, and the length of the side (chord) decreases.

As the question is concerned with a 6-sided polygon, let's take a regular hexagon - 6 sided regular polygon. 
The sum of its angles is (n-2)x180=720. So, each angle is 720/6=120 degrees. 
The length of the longest line is the diameter of the circumscribed circle, i.e. AD. 
Angle OAF = angle OFA= 60 degrees => triangle AOF is equilateral. So, AF is equal to the length of the radius => 1/2 of the diameter.
Per st.1, the ratio of diagonal is more than 2, so the polygon must have more than 6 angles.
Sufficient.

(2) 
Angle of a triangle = 180/3 = 60
Angle of a square = 360/4 = 90 
Pentagon: Sum of angles = (n-2) x 180 = 540. Angle of pentagon = 540/5 = 108
Angle of hexagon, as mentioned above, is 120 degrees.

Sufficient, we can stop here and move on.

The asnwer is D

Just to prove, the sum of angles of the 7 sided polygon (heptagon) = (7-3)x180=900. 900/7=128 4/7

Hence D

Bunuel · Answer

Par of  GMAT CLUB'S New Year's Quantitative Challenge Set

MusicCzar · Answer

This post, I think does not fit here. The question asks for whether the polygon has more than 6 sides, while this reply by GMATInsight proves that it is not a Hexagon/6 sided polygon.

chinmaysawant · Answer

For the 2nd option,

If the polygon has 12 sides,

interior angle = (n-2/n)*180 = (10/12)*180 = 150 which is an integer.

So only 1st option is correct according to me. 
Let me know if I am wrong.
Thanks!

souvikgmat1990 · Answer

I have a question. 
If it's a 10 sided poly then the interior angles are  =144 which is an integer so shouldn't A be the answer?

thyagi · Answer

Take n= 7 or 11..

You won't get an integral value for the interior angles.. infact all the values of n between 3 and 6 will give an integral value, and n is nothing but the number of sides of the polygon... so n > 6 will give you atleast one non integral value of the interior angles.. and that's all what you need.

Posted from my mobile device

Bunuel · Answer

Official Solution:

Does regular polygon X have more than 6 sides? 
 
(1) The ratio of the length of the longest line that can be drawn between two vertices of polygon X to the length of any one side of polygon X is greater than 2.
 
Consider regular polygons below:
 
 https://gmatclub.com/tests-beta/resources/import/img/M36-139.png 
 
Notice that when we increase the number of sides, the ratio of the longest diagonal to the side increases. So, more sides the larger the ratio.
 
Let's check what this ratio is for a 6-sided regular polygon, hexagon:
 
 https://gmatclub.com/tests-beta/resources/import/img/M36-139-2.png 
 
The degree measure of  \angle AOB = \frac{360°}{6}=60° . Since AO = BO, then  \angle ABO = \angle BAO = \frac{180° - 60°}{2}=60° . So, triangle ABO is equilateral. Thus, AB = AO, but AO is half the diagonal, so  diagonal:side=2:1 
 
Since we are given that the ratio for our polygon is more than 2 and we know that the ratio increases as we increase the number of sides, then our polygon must have more than 6 sides. Sufficient. 
 
(2) The degree measure of an interior angle of polygon X is NOT an integer.
 
The degree measure of an interior angle in a regular polygon is  \frac{(n-2)*180°}{n} , where  n  is the number of sides. 
 
So, we are told that  \frac{(n-2)*180°}{n}
eq integer . Notice that since 180 is divisible by 3, 4, 5, and 6, then for these values of  n ,  \frac{(n-2)*180°}{n}  will be an integer. Thus, the number of sides,  n , must be more than 6. Sufficient.

Answer: D

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Does regular polygon X have more than 6 sides?

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