Japan Mathematical Olympiad

$[2+4\sqrt{2}]$ $$\text{Choose a point }P\text{ so that the quadrilateral }ABCP\text{ becomes a parallelogram. Then, we have }AP=PC=2.\text{ Since }M\text{ is the mid-point of }BC\text{ and }N\text{ is the mid-point of }BD,\text{ we have }PD=2MN=2\sqrt{2}.\text{ Since }P{C}^2+P{D}^2=2^2+(2\sqrt{2}{)}^2=12=(2\sqrt{3}{)}^2=C{D}^2,\text{ we see that the }△PCD\text{ is a right triangle with }CD\text{ as its hypotenuse. We also have from the law of cosines that}$$ \cos \angle APD = \frac{AP^2 + PD^2 - AD^2}{2 \cdot AP \cdot PD} = -\frac{\sqrt{2}}{2}, $$\text{from which we get that }\angle APD=135^{\circ }.\text{We separate two cases depending on whether or not the points }A\text{ and }C\text{ lie on the same side of the plane divided by the line }PD.\text{If }A\text{ and }C\text{ lie on the opposite sides of the plane divided by the line }PD,\text{ the quadrilateral becomes convex, and its area is the sum of the areas of the parallelogram }ABCP,\text{ the triangle }CDP\text{ and the triangle }DAP.\text{ Since }\angle APC=360^{\circ }-\angle APD-\angle CPD=135^{\circ },\text{ we get that the area of the quadrilateral is given by}$$ AP \cdot CP \cdot \sin \angle APC + \frac{1}{2} AP \cdot DP \cdot \sin \angle APD + \frac{1}{2} CP \cdot DP \cdot \sin \angle CPD = 2 + 4\sqrt{2}. $$ \text{On the other hand, if the points } A \text{ and } C \text{ lie on the same side of the plane divided by the line } PD, \text{ then we have } \angle BCP = 180^\circ - \angle CPA = 180^\circ - (\angle APD - \angle CPD) = 135^\circ \text{ and } \angle BCP + \angle PCD = 135^\circ + \angle PCD. \text{ Since } \angle CPD = 90^\circ \text{ and } PC < PD, \text{ we have } \angle PCD > 45^\circ \text{ and hence } \angle BCP + \angle PCD > 135^\circ + 45^\circ = 180^\circ. \text{ Therefore, the point } P \text{ and the point } C \text{ lie on the same side of the plane divided by the line } BD. \text{ Since the points } A \text{ and } P \text{ lie on the same side of the plane divided by the line } BD, \text{ the points } A \text{ and } C \text{ also lie on the same side of the plane divided by the line } BD. \text{ This means that the 4 given points } A, B, C, D \text{ traversed in this order will not form the 4 vertices of a convex quadrilateral.} \text{Thus, the desired area is } 2 + 4\sqrt{2}.