The inequality about areas
Source: Finnish Mathematics Competition 2009, Final Round - P4
November 16, 2011
inequalitiesgeometryrectanglecombinatorics unsolvedcombinatorics
Problem Statement
As in the picture below, the rectangle on the left hand side has been divided into four parts by line segments which are parallel to a side of the rectangle. The areas of the small rectangles are and . Similarly, the small rectangles on the right hand side have areas and . It is known that , , but .
[asy]
import graph; size(12cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-4.3,xmax=12.32,ymin=-10.68,ymax=6.3;
draw((0,3)--(0,0)); draw((3,0)--(0,0)); draw((3,0)--(3,3)); draw((0,3)--(3,3)); draw((2,0)--(2,3)); draw((0,2)--(3,2)); label("",(0.86,2.72),SE*lsf); label("",(2.38,2.7),SE*lsf); label("",(2.3,1.1),SE*lsf); label("",(0.82,1.14),SE*lsf); draw((5,2)--(11,2)); draw((5,2)--(5,0)); draw((11,0)--(5,0)); draw((11,2)--(11,0)); draw((8,0)--(8,2)); draw((5,1)--(11,1)); label("",(6.28,1.8),SE*lsf); label("",(9.44,1.82),SE*lsf); label("",(9.4,0.8),SE*lsf); label("",(6.3,0.86),SE*lsf);
dot((0,3),linewidth(1pt)+ds); dot((0,0),linewidth(1pt)+ds); dot((3,0),linewidth(1pt)+ds); dot((3,3),linewidth(1pt)+ds); dot((2,0),linewidth(1pt)+ds); dot((2,3),linewidth(1pt)+ds); dot((0,2),linewidth(1pt)+ds); dot((3,2),linewidth(1pt)+ds); dot((5,0),linewidth(1pt)+ds); dot((5,2),linewidth(1pt)+ds); dot((11,0),linewidth(1pt)+ds); dot((11,2),linewidth(1pt)+ds); dot((8,0),linewidth(1pt)+ds); dot((8,2),linewidth(1pt)+ds); dot((5,1),linewidth(1pt)+ds); dot((11,1),linewidth(1pt)+ds);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]
Prove that the big rectangle on the left hand side has area smaller or equal to the area of the big rectangle on the right hand side, i.e. .