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P is a polygon

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by amsm25 » Fri Apr 27, 2012 3:06 am
P is a polygon. Is it possible to construct a circle such that each vertex of the polygon is a point on the circle?
1)All sides of the polygon are of equal length.
2)All Angles of the polygon have equal measures.



AO - B. Why can't it be D????
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by Shalabh's Quants » Fri Apr 27, 2012 4:22 am
amsm25 wrote:P is a polygon. Is it possible to construct a circle such that each vertex of the polygon is a point on the circle?
1)All sides of the polygon are of equal length.
2)All Angles of the polygon have equal measures.



AO - B. Why can't it be D????
Statement 1...

Imagine Rhombus. All sides are equal but one cannot have all 4 vertexes on a circle. Insuff.

Statement 2...

Imagine Rectangle. All angels are equal but one cannot have all 4 vertexes on a circle. Insuff.

So it should be C.
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by 4GMAT_Mumbai » Fri Apr 27, 2012 6:52 am
Shalabh's Quants wrote:
Statement 2...

Imagine Rectangle. All angels are equal but one cannot have all 4 vertexes on a circle. Insuff.

C.
Agreed that statement 2 is insufficient.

However, a rectangle may not be the right example. Even a rectangle is a cyclic quadrilateral - i.e., a circle can always be drawn touching the 4 vertices of a rectangle.

Any quadrilateral whose opposite angles are supplementary is a cyclic quadrilateral (works vice versa also).

Statement 2 could be proved to be insufficient with a hexagon which looks like this ...
Image

If a circle is drawn which passes through the 3 vertices on the left portion of the hexagon, then, the other 3 vertices will not lie on the circle. This is, in spite of, all the angles being equal.

I hope this helps. Thanks.
Naveenan Ramachandran
4GMAT, Dadar(W) & Ghatkopar(W), Mumbai
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by ronnie1985 » Fri Apr 27, 2012 7:46 am
Equal chords subtend equal angle at the centre of the circle in question.

If all the angles of a polygon are equal it implies it is a cyclic polygon as it is the corollary of the theorem stated in the first sentence.
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by Stuart@KaplanGMAT » Fri Apr 27, 2012 7:48 am
4GMAT_Mumbai wrote:
Shalabh's Quants wrote:
Statement 2...

Imagine Rectangle. All angels are equal but one cannot have all 4 vertexes on a circle. Insuff.

C.
Agreed that statement 2 is insufficient.

However, a rectangle may not be the right example. Even a rectangle is a cyclic quadrilateral - i.e., a circle can always be drawn touching the 4 vertices of a rectangle.

Any quadrilateral whose opposite angles are supplementary is a cyclic quadrilateral (works vice versa also).

Statement 2 could be proved to be insufficient with a hexagon which looks like this ...
Image

If a circle is drawn which passes through the 3 vertices on the left portion of the hexagon, then, the other 3 vertices will not lie on the circle. This is, in spite of, all the angles being equal.

I hope this helps. Thanks.
Umm.. not to be too picky, but the angles in your hexagon are NOT all equal (the 2 angles of the "pointy bits" are different from the other 4 angles).

Any polygon with equal angles is, in fact, cyclical.
Image

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