Japan Mathematical Olympiad

Take three points $P$, $Q$, $R$ inside of the cube in such a way that the quadrilaterals $KLMP$, $MNIQ$ and $IJKR$ are parallelograms. Since $IJ$ and $ML$ are parallel, so are $PK$ and $RK$, and we see that the points $P$, $R$, $K$ lie on a same straight line. Therefore, we conclude that the hexagon $IJKLMN$ is partitioned into 3 parallelograms and a triangle $PQR$ as indicated in the figure below.



We then have $$\begin{array}{rl}\Delta IKM& =\Delta IKR+\Delta KMP+\Delta MIQ+\Delta PQR\\ & =\frac{1}{2}(\square IJKR+\square KLMP+\square MNIQ)+\Delta PQR\\ & =\frac{1}{2}(a-\Delta PQR)+\Delta PQR=\frac{1}{2}(a+\Delta PQR).\end{array}$$

Here we are denoting $\Delta XYZ$ and $\square XYZW$ for the areas of the triangle $XYZ$ and of the quadrilateral $XYZW$, respectively. We note also that since $P$ and $R$ lie on the same side from $K$ on the line $PK$, $PR=|PK-RK|=|ML-IJ|$ holds. Similarly, we have $PQ=|NI-LK|$ and $QR=|JK-MN|$.

Next, take three points $P^{\prime }$, $Q^{\prime }$, $R^{\prime }$ inside of the cube so that the quadrilaterals $NIJ{P}^{\prime }$, $JKL{Q}^{\prime }$, $LMN{R}^{\prime }$ are parallelograms. Then, in the same way as above, we obtain that $\Delta JLN=\frac{1}{2}(a+\Delta P^{\prime }{Q}^{\prime }{R}^{\prime })$. Furthermore, we get $P^{\prime }{R}^{\prime }=|ML-IJ|=PR$, $P^{\prime }{Q}^{\prime }=|NI-LK|=PQ$ and $Q^{\prime }{R}^{\prime }=|JK-MN|=QR$. Hence, the triangles $PQR$ and $P^{\prime }{Q}^{\prime }{R}^{\prime }$ are congruent, and in particular, we have $\Delta PQR=\Delta P^{\prime }{Q}^{\prime }{R}^{\prime }$. Thus we obtain the fact that the answer we seek is $\Delta IKM+\Delta JLN-a=\Delta PQR$, and thus it is enough to find the area of the triangle $PQR$.

Let us now observe that the right triangles $EJI$ and $CML$ are similar since the corresponding sides are parallel. Therefore, we have $$\begin{array}{rl}EJ:EI:IJ& =(CM-EJ):(CL-EI):(LM-IJ)\\ & =6:((2012-GL)-(2012-AI)):(ML-IJ)=6:8:(ML-IJ).\end{array}$$ Since the triangle for which the ratios of the lengths of the three sides are $6:8:|ML-IJ|$ is a right triangle whose hypotenuse is the side with length $|ML-IJ|$, we get $|ML-IJ|=\sqrt{6^2+8^2}=10$. Thus, we get $PR=10$. Similarly, we obtain $PQ=|NI-LK|=\sqrt{4^2+8^2}=4\sqrt{5}$ and $QR=|JK-MN|=\sqrt{4^2+6^2}=2\sqrt{13}$.

Let $S$ be the foot of the perpendicular line drawn from $Q$ to $PR$ and let $x=RS$ (assume that if $S$ lies on the opposite side from $P$ with respect to $R$, then $x$ is negative). Then, from the Pythagorean theorem, we get $$(2\sqrt{13}{)}^2-x^2=Q{S}^2=(4\sqrt{5}{)}^2-(10-x{)}^2.$$ Solving for $x$ we obtain $x=\frac{18}{5}$, and then we get $QS=\frac{4\sqrt{61}}{5}$. Finally, we obtain $$\Delta PQR=\frac{1}{2}\cdot 10\cdot \frac{4\sqrt{61}}{5}=4\sqrt{61},$$ which is the desired answer for the problem.