How Small?
On a real coordinate plane, let be the intersection point of the diagonals of a rectangle whose vertices have all integer coordinates. If none of the vertices lie strictly in the quadrant, what is the smallest possible area of the rectangle?
Clarification: Vertices can lie on the axes.
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1 solution
It's easy to find a rectangle of area 14: The vertices are respectively (5,0) (4,-1) (-2,7) (-3,6)
To show that this is the smallest rectangle, note first that since vertices can not be in the 1st quadrant, there must be a vertex in the 2nd quadrant (or the positive y-axis) and another in the 4th quadrant (or the positive x-axis), and the segment connecting these two vertices serve as the long side of the rectangle (and the extended line of this segment put point (1,3) on the lower semi-hyperplane). (If it's the short side, the rectangle would have a much larger area, so we don't have to consider it).
If at least one of the vertices lies not on the axis, the long side of the rectangle has a length of at least 1 + 4 2 . (one of the vertices is at least 1 unit away from the axis, plus the shortest length for a path from the y-axis to the x-axis in the 1st quadrant).
While if both vertices lies on the axis, the long side of the rectangle has a length of at least 4 2 + 5 2 = 4 1 . (the shortest length for a path from the y-axis to the x-axis in the 1st quadrant without touching (1,3)).
Both scenario considered, the shortest long side would have a length of at least 4 1 and therefore the short side would be at most 1 4 / 4 1 < 5 . The inequality means that, the only possible length for the short side is 2 , where the vertices making the short side lie on the diagonal of a unit square.
On the other hand, the property of rectangles require that the two vertices are of the same distance to point (1,3). Therefore, they have to lie on different side of x+y=4. It's easy to see that the smallest possible rectangle we could draw under these restrictions are the rectangle specified above.