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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7495
GMAT 1: 760 Q51 V42 GPA: 3.82
If n regular pentagons are tangent each other in points of a circle as  [#permalink]

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16 00:00

Difficulty:   75% (hard)

Question Stats: 51% (02:06) correct 49% (01:52) wrong based on 125 sessions

HideShow timer Statistics Attachment: pentagons.jpg [ 5.39 KiB | Viewed 2532 times ]

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.

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Posts: 109
Re: If n regular pentagons are tangent each other in points of a circle as  [#permalink]

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10
5
I'm going to assume that the question implies that the pentagons go all the way around the circle and connect back with each other forming a closed loop of pentagons. (If the drawing were to depict this, then it would either give the answer away or be very misleading...)

Imagine if the sides of the pentagon that cross the circle were extended down to the center of the circle. Then each pentagon would be in one sector of the circle. If we could determine the central angle of the sector, we would know how many pentagons it would take to complete the circle. See the diagram below.
Attachment: Pentagon circle.png [ 4.69 KiB | Viewed 2439 times ]

In a regular polygon, the measure of the interior angles is [180*(n-2)]/n, where n the number of sides of the polygon. For a pentagon, the measure of the interior angles is [180*(5-2)]/5 = 108.

If each interior angle is 108, then the exterior angles, y = 180-108 = 72

The measure of angle x is therefore 180-2*72 = 36

How many sectors of $$36^{\circ}$$ will it take to complete a circle?

360/36 = 10

Attachment: Pentagon circle 2.png [ 7.96 KiB | Viewed 2435 times ]

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Dave de Koos
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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7495
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: If n regular pentagons are tangent each other in points of a circle as  [#permalink]

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1
1
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

Attachment: pentagon.jpg [ 5.87 KiB | Viewed 2373 times ]

Just like the picture above, there is a regular pentagon ABCDE. If you extend two lines, it becomes like the picture above. However, since an interior angle of one regular pentagon is 180(5-2)/5=108, angle A=E=B=108 and angle A+E+B=324. Since sum of interior angles of a rectangle ABFE is 360, angle F is 360-324=36. Then, angle F is a central angle, which is 360/36=10. Therefore, the answer is C.
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If n regular pentagons are tangent each other in points of a circle as  [#permalink]

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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
Intern  G
Joined: 18 Nov 2013
Posts: 46
Re: If n regular pentagons are tangent each other in points of a circle as  [#permalink]

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MathRevolution wrote:
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! Re: If n regular pentagons are tangent each other in points of a circle as   [#permalink] 26 Dec 2018, 02:36
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