vikram4689 wrote:Thanks Mr Smith but unfortunately i am not able to comprehend both of the responses. Please explain in simpler language.
My concern is how would one solve this problem on GMAT DAY Even OG explanation assumes lot of stuff, it seems that they trying to prove ans and not derive it. for proving 5 to be a valid case they take different numbers and for 10 they chose different approach
Mr Smith wrote:You can not follow that method because:
1) You might disregard the important condition of the polygon being complex if you blindly add the limits of the diagonals by apply the triangle rule to the inner triangles.
2) You assume that the minimum of the last side will happen when the diagonals are at minimum length too. That might not be the case
For example, in the polygon described in your question, there is no lower limit for the last side. PT can be tending to 0 and PQRS could be a valid 4 sided Convex Polygon- A quadrilateral.
Use common sense.
It is possible to form a quadrilateral with sides of 3, 2, 4 and 5, since the length of each side would be less than the sum of the lengths of the other three sides.
Thus, it must be possible to add a 5th side that is INFINITELY SMALL:
1. Select any vertex.
2. Separate -- just a little -- the two sides forming that vertex.
3. Connect the separated sides with a tiny 5th side.
Click on the link in my post above for an illustration of just such a polygon.
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