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# Pentagon problem

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Pentagon problem [#permalink]  03 Nov 2009, 14:56
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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
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Re: Pentagon problem [#permalink]  03 Nov 2009, 16:10
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arora2m 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

Any 3 points from 6 will make a triangle, since no 3 points are collinear, then:

6C3=20

TIPS ON RELATED ISSUES:
In a plane if there are n points of which no three are collinear, then
1. The number of straight lines that can be formed by joining them is nC2.
2. The number of triangles that can be formed by joining them is nC3.
3. The number of polygons with k sides that can be formed by joining them is nCk.

In a plane if there are n points out of which m points are collinear, then
1. The number of straight lines that can be formed by joining them is nC2 - mC2 + 1
2. The number of triangles that can be formed by joining them is nC3 - mC3
3. The number of polygons with k sides that can be formed by joining them is nCk -mCk
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Re: Pentagon problem [#permalink]  03 Nov 2009, 16:39
Bunuel - In this case I am a bit confused.
F is the center of the pentagon ABCDE ... So ABF form a triangle but not ACF, since a straight line drawn between AC will not pass through F.

I got 15 by doing 5 triangles formed with F as one of the points and then 5C3 thus 5+10 =15.

Please tell me what is wrong with my reasoning here.
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Re: Pentagon problem [#permalink]  03 Nov 2009, 16:55
arora2m wrote:
Bunuel - In this case I am a bit confused.
F is the center of the pentagon ABCDE ... So ABF form a triangle but not ACF, since a straight line drawn between AC will not pass through F.

I got 15 by doing 5 triangles formed with F as one of the points and then 5C3 thus 5+10 =15.

Please tell me what is wrong with my reasoning here.

acf does form a triangle. think of F immediately under A so drop a vertical line from a to f and then draw a line form a to c then connect c to f

the line from a to f stops at f and doesn't continue on to another letter. it angles over to c
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Re: Pentagon problem [#permalink]  03 Nov 2009, 17:01
arora2m wrote:
Bunuel - In this case I am a bit confused.
F is the center of the pentagon ABCDE ... So ABF form a triangle but not ACF, since a straight line drawn between AC will not pass through F.

I got 15 by doing 5 triangles formed with F as one of the points and then 5C3 thus 5+10 =15.

Please tell me what is wrong with my reasoning here.

Regular pentagon is a pentagon where all sides are equal. In such pentagon center is not collinear to any two vertex, so ANY three points (from 5 vertices and center point) WILL form the triangle.

You wrote that ACF won't form the triangle because they don't lie on straight line - that's not true. EXACTLY because these three points DON'T lie on the straight line they WILL form triangle. (In your other example you stated that ABF will form the triangle, but these three points also aren't collinear).

The question basically asks how many triangles can be formed from the six points on a plane with no three points being collinear.

Hope it's clear.
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Re: Pentagon problem [#permalink]  16 Apr 2010, 01:23
I also had similar question regarding total numbe rbeing 15 or 20..
the forum is amazing ..
and thanks Brunell
Re: Pentagon problem   [#permalink] 16 Apr 2010, 01:23
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