Starting with a convex pentagon P, we define the "middle polygon" Q, whose vertices are the middle points of the sides of the initial pentagon. The ratio between the areas of this polygons seem to always satisfy:
1/2 < area(Q)/area(P) < 3/4
The lower bound is easy to obtain. I don't see how to get the upper bound.
This problem is equivalent to the following one. Just forget about the middle polygon for a moment. Start with a convex pentagon and consider also all his 5 diagonals. You will obtain a central pentagon. Prove that the area of the new central pentagon is less than the sum of the areas of the five small triangles which have a side adjacent to the sides of this central polygon.
Best Answer
I used qepcad to compute that the intersection of the set of possible area ratios with the interval [1/2, 3/4] is (1/2, 3/4). Since the set of possible area ratios is the image of a connected space under a continuous function, and we know the set contains (1/2, 3/4), but not 1/2 or 3/4, it must equal (1/2, 3/4). Here is a log of the qepcad session.
======================================================= Quantifier Elimination in Elementary Algebra and Geometry by Partial Cylindrical Algebraic Decomposition Version B 1.53, 16 Jul 2009 by Hoon Hong (hhong@math.ncsu.edu) With contributions by: Christopher W. Brown, George E. Collins, Mark J. Encarnacion, Jeremy R. Johnson Werner Krandick, Richard Liska, Scott McCallum, Nicolas Robidoux, and Stanly Steinberg ======================================================= Enter an informal description between '[' and ']': [ area of middle pentagon ] Enter a variable list: (a,x1,y1,x2,y2) Enter the number of free variables: 1 Enter a prenex formula: (E x1)(E y1)(E x2)(E y2)[ a >= 1/2 /\ a <= 3/4 /\ x1 > 0 /\ y1 > 0 /\ 1 - x1 - y1 < 0 /\ x2 > 0 /\ x2 y1 + y2 - x1 y2 - y1 < 0 /\ x1 + x2 y1 - x2 - x1 y2 < 0 /\ a (1/2)(y1 + x1 y2 - x2 y1 + x2) = (1/8)(0 - 1 + x1 + 2 x2 + 2 y1 + y2 + 2 x1 y2 - 2 x2 y1) ]. ======================================================= Before Normalization > finish An equivalent quantifier-free formula: 2 a - 1 > 0 /\ 4 a - 3 < 0 ===================== The End ======================= ----------------------------------------------------------------------------- 12 Garbage collections, 473385670 Cells and 0 Arrays reclaimed, in 8158 milliseconds. 1345504 Cells in AVAIL, 40000000 Cells in SPACE. System time: 79624 milliseconds. System time after the initialization: 79028 milliseconds. -----------------------------------------------------------------------------