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????
P is a polygon
This topic has expert replies
- Shalabh's Quants
- Master | Next Rank: 500 Posts
- Posts: 134
- Joined: Fri Apr 06, 2012 3:11 am
- Thanked: 35 times
- Followed by:5 members
Statement 1...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????
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 Jain,
e-GMAT Instructor
e-GMAT Instructor
-
- Master | Next Rank: 500 Posts
- Posts: 161
- Joined: Mon Apr 05, 2010 9:06 am
- Location: Mumbai
- Thanked: 37 times
Agreed that statement 2 is insufficient.Shalabh's Quants wrote:
Statement 2...
Imagine Rectangle. All angels are equal but one cannot have all 4 vertexes on a circle. Insuff.
C.
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.
Naveenan Ramachandran
4GMAT, Dadar(W) & Ghatkopar(W), Mumbai
4GMAT, Dadar(W) & Ghatkopar(W), Mumbai
- ronnie1985
- Legendary Member
- Posts: 626
- Joined: Fri Dec 23, 2011 2:50 am
- Location: Ahmedabad
- Thanked: 31 times
- Followed by:10 members
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.
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.
Follow your passion, Success as perceived by others shall follow you
- Stuart@KaplanGMAT
- GMAT Instructor
- Posts: 3225
- Joined: Tue Jan 08, 2008 2:40 pm
- Location: Toronto
- Thanked: 1710 times
- Followed by:614 members
- GMAT Score:800
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).4GMAT_Mumbai wrote:Agreed that statement 2 is insufficient.Shalabh's Quants wrote:
Statement 2...
Imagine Rectangle. All angels are equal but one cannot have all 4 vertexes on a circle. Insuff.
C.
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.
Any polygon with equal angles is, in fact, cyclical.
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course