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.

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 ...


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.

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.

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 ...


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.