# Inequality in Rectangle

Here's a problem by Leo Giugiuc who posted it along with a solution at the CutTheKnotMath facebook page:

Point $I$ lies within the rectangle $ABCD.$ Lines $AI,$ $BI,$ $CI,$ $DI$ meet the sides of the rectangle in points $M,$ $N,$ $P,$ $Q.$

Prove that

$AI+BI+CI+DI\ge MI+NI+PI+QI.$

### Proof 1

Draw the lines through $I$ parallel to the sides of the rectangle as shown:

The two lines split $ABCD$ into four smaller rectangles sharing a corner at $I.$ At $I,$ each of the lines $AI,$ $BI,$ $CI,$ $DI$ exits exactly one of the small rectangles and enters exactly one of them. Thus, each of the small rectangles house exactly one of the segments $AI,$ $BI,$ $CI,$ $DI$ (which serve as the diagonals of the corresponding rectangles) and exactly one of the segments $MI,$ $NI,$ $PI,$ $QI.$ Since, the diagonal is the longest segment between the points of a rectangle,

$AI\ge PI,\\ BI\ge QI,\\ CI\ge MI,\\ DI\ge NI.$

Adding the four up proves the required inequality. The equality achieved only when four inequalities degenerate into identities, i.e., when $MI,$ $NI,$ $PI,$ $QI$ serve as the diagonals of the respective rectangles, meaning the case where they physically coincide with segments $AI,$ $BI,$ $CI,$ $DI.$ This only happens when $I$ is the center of the rectangle $ABCD.$

### Proof 2

This proof is by Emil Stoyanov.

If the point $I$ liesn $\Delta AOB,$ then the angles $API,$ $BQI,$ $CMI,$ $DNI$ are not less than $90^{\circ},$ implying $AI\ge PI$ (with equality only when $I$ lies on $OA),$ $BI\ge QI,$ $CI\ge MI,$ $DI\ge NI;$ and the proposition follows. The cases where $I$ lies in triangles $BOC,$ $COD,$ or $DOA$ are treated similarly.