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[quote="MathRevolution"]
Attachment:
pentagons.jpg
If n regular pentagons are tangent each other in points of a circle as above figure, n=?

A. 8
B. 9
C. 10
D. 11
E. 12

imagine a regular decagon within the circle,
using 10 identical 36°-72°-72° triangles,
with the 36° (360°/10) angles in the center
and the 10 sides between the 72° angles forming the decagon
each of those 10 sides will be the base of 10 regular pentagons,
confirmed by the fact that the internal angle of a regular pentagon,
108°, is supplementary to 72°
n=10
C
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MathRevolution
Attachment:
pentagons.jpg
If n regular pentagons are tangent each other in points of a circle as above figure, n=?

A. 8
B. 9
C. 10
D. 11
E. 12


* A solution will be posted in two days.

I got the same problem in the GMAT exam. What a coincidence!
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MathRevolution
Attachment:
pentagons.jpg
If n regular pentagons are tangent each other in points of a circle as above figure, n=?

A. 8
B. 9
C. 10
D. 11
E. 12


* A solution will be posted in two days.

MathRevolution Bunuel chetan2u yashikaaggarwal

Hi, What if the side of the regular pentagon is smaller and when extended, doesn't merge at the centre?
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MathRevolution
Attachment:
pentagons.jpg
If n regular pentagons are tangent each other in points of a circle as above figure, n=?

A. 8
B. 9
C. 10
D. 11
E. 12


* A solution will be posted in two days.

MathRevolution Bunuel chetan2u yashikaaggarwal

Hi, What if the side of the regular pentagon is smaller and when extended, doesn't merge at the centre?
We are not considering side length here. The stated pentagon is regular that means it does have equal sides. And will have 108° as its one of the side angles.
And their extended forms will meet at center because Its mentioned that they are aligned with each other in form of tangent. Since they are alined there is no chance they will bisect each other and will definitely have a common point of meeting(center) in between circle.

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