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Regular pentagon with F at its center

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by Uva@90 » Thu Oct 17, 2013 6:42 pm
ABCDE is a regular pentagon with F at its center. How many different triangles can be formed by joining 3 of the points A,B,C,D,E and F?
(A) 10
(B) 15
(C) 20
(D) 25
(E) 30

OA C

I solved the above problem as
6C3 = 20.

But I feel like I am missing something else.
Experts please help me how to do this problem or what I did is correct.

Thanks in advance.

Regards,
Uva.
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by theCodeToGMAT » Thu Oct 17, 2013 7:18 pm
Total points available = 5+1

So, 6C3 = 20
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by mevicks » Thu Oct 17, 2013 7:27 pm
Uva@90 wrote:ABCDE is a regular pentagon with F at its center. How many different triangles can be formed by joining 3 of the points A,B,C,D,E and F?
(A) 10
(B) 15
(C) 20
(D) 25
(E) 30
Drawing the pentagon:
Image

We have 5 Green triangles
For each vertex there are 3 Yellow ones, thus for 5 vertices we have 5*3 = 15 yellow triangles.
[spoiler]Total = 20[/spoiler]

Combinatorics:
We have 6 total points and have to select three points to form a triangle.
Here the order of the points is not important (triangle AEF is the same as FEA) thus we can use the formula for Combination : Image
[spoiler]Answer : 6C3 = 20[/spoiler]
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by [email protected] » Thu Oct 17, 2013 10:38 pm
Hi Uva@90,

Your approach here is correctly. While a pentagon is a rarity on the GMAT, the question in this prompt is remarkably straight-forward (there are no limitations on the 3 points used to form any of the triangles). You could even sketch out a quick drawing of all the possibilities. If you're ever in doubt about your "math", then see if there's another way to confirm that your answer is correct.

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by ganeshrkamath » Thu Oct 17, 2013 11:23 pm
Uva@90 wrote:ABCDE is a regular pentagon with F at its center. How many different triangles can be formed by joining 3 of the points A,B,C,D,E and F?
(A) 10
(B) 15
(C) 20
(D) 25
(E) 30

OA C

I solved the above problem as
6C3 = 20.

But I feel like I am missing something else.
Experts please help me how to do this problem or what I did is correct.

Thanks in advance.

Regards,
Uva.
Any 3 unique points on a surface can form a triangle as long as they are not collinear.

In this particular case, no 3 points are collinear.
So a triangle can be formed by choosing any 3 points out of 6.
(This wouldn't be the case for a hexagon)

So total number of possible triangles = 6C3 = 20
Choose C

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by Uva@90 » Sat Oct 19, 2013 6:27 pm
ganeshrkamath wrote:
Uva@90 wrote:ABCDE is a regular pentagon with F at its center. How many different triangles can be formed by joining 3 of the points A,B,C,D,E and F?
(A) 10
(B) 15
(C) 20
(D) 25
(E) 30

OA C

I solved the above problem as
6C3 = 20.

But I feel like I am missing something else.
Experts please help me how to do this problem or what I did is correct.

Thanks in advance.

Regards,
Uva.
Any 3 unique points on a surface can form a triangle as long as they are not collinear.

In this particular case, no 3 points are collinear.
So a triangle can be formed by choosing any 3 points out of 6.
(This wouldn't be the case for a hexagon)

So total number of possible triangles = 6C3 = 20
Choose C

Cheers
Hi Ganesh,
Can you explain why this is not possible in Hexagon, Even in hexagon 3 points wont be in co-linear only right.

Thanks in advance.

Regards,
Uva.
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by ganeshrkamath » Sun Oct 20, 2013 6:41 am
Uva@90 wrote:
ganeshrkamath wrote:
Uva@90 wrote:ABCDE is a regular pentagon with F at its center. How many different triangles can be formed by joining 3 of the points A,B,C,D,E and F?
(A) 10
(B) 15
(C) 20
(D) 25
(E) 30

OA C

I solved the above problem as
6C3 = 20.

But I feel like I am missing something else.
Experts please help me how to do this problem or what I did is correct.

Thanks in advance.

Regards,
Uva.
Any 3 unique points on a surface can form a triangle as long as they are not collinear.

In this particular case, no 3 points are collinear.
So a triangle can be formed by choosing any 3 points out of 6.
(This wouldn't be the case for a hexagon)

So total number of possible triangles = 6C3 = 20
Choose C

Cheers
Hi Ganesh,
Can you explain why this is not possible in Hexagon, Even in hexagon 3 points wont be in co-linear only right.

Thanks in advance.

Regards,
Uva.
In a hexagon, the line joining the opposite vertices coincides with the center of the hexagon.
So these 3 points are collinear and cannot form a triangle.
You will have to remove such cases from 6C3.

Hope this helps,
Cheers
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by Uva@90 » Sun Oct 20, 2013 6:53 am
ganeshrkamath wrote:
In a hexagon, the line joining the opposite vertices coincides with the center of the hexagon.
So these 3 points are collinear and cannot form a triangle.
You will have to remove such cases from 6C3.

Hope this helps,
Cheers
Ah!! Yes I forgot the Center Point. That's why I am wondering how 3 points will be in co-linear.

Thanks Ganesh.

Regards,
Uva.
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by sahilchaudhary » Mon Oct 21, 2013 6:38 am
6C3 is correct.
If it were given that taking F is a compulsion, then 6C3 would be wrong.
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