Japan Mathematical Olympiad

Let $M$ be the mid-point of the side $AD$ of the square, and let $N$ be the point of tangency of the line $CE$ to the circle $O$. Since $MD=MN$, and $\angle MDC=\angle MNC=90^{\circ }$, we have $△MNC\equiv △MDC$. Similarly, we have $△MNE\equiv △MAE$. From $△MNC\equiv △MDC$ we have $\angle CMN=\angle CMD$, which implies that $2\angle CMN=\angle DMN$. Similarly, we get $2\angle EMN=\angle AMN$ from $△MNE\equiv △MAE$. Consequently, we have $$\angle EMC=\angle EMN+\angle CMN=\frac{1}{2}(\angle DMN+\angle AMN)=90^{\circ }.$$ Hence, we have $\angle EMN=90^{\circ }-\angle CMN=\angle MNC$, which together with the fact that $\angle ENM=90^{\circ }=\angle MNC$ implies that $△EMN$ and $△MCN$ are similar triangles. Therefore, we have $EN:NM=NM:NC$. On the other hand, from the fact that $△MNC\equiv △MDC$, we have $NC=DC=1$, $NM=DM=\frac{1}{2}$, from which we obtain $EN=\frac{1}{4}$. From $△MNE\equiv △MAE$ we get $EA=EN=\frac{1}{4}$ so that we have $BE=\frac{3}{4}$, and therefore, the area of $△CBE=\frac{1}{2}\cdot 1\cdot \frac{3}{4}=\frac{3}{8}$.

Alternatively: We can argue in the same way as above to conclude that $△MNC\equiv △MDC$, $△MNE\equiv △MAE$. Then we can put $AE=t$ and get $EC=EN+NC=EA+DC=t+1$ and $BE=BA-AE=1-t$. By applying the Pythagorean theorem to the right triangle $CBE$, we get $(1+t{)}^2=(1-t{)}^2+1^2$. Solving for $t$ we get $t=\frac{1}{4}$, and we get the area of the $△CBE=\frac{3}{8}$ as above.