(JBMO 2000 - Shortlist, Problem G6), treated in class

Let $A B C D$ be a quadrilateral such that $\widehat{D A B}=60^{\circ}, \widehat{A B C}=90^{\circ}$ and $\widehat{B C D}=120^{\circ}$. Let $M$ be the intersection point of the diagonals $[A C]$ and $[B D]$. Given that $B M=1$ and $D M=2$, calculate the area of the quadrilateral $A B C D$.
Since $A B C$ is a right triangle at $B$ and since $\widehat{D A B}+\widehat{B C D}=180^{\circ}$, the quadrilateral $A B C D$ is therefore cyclic, inscribed in a circle $\mathcal{C}$ with radius $[A C]$. Let $O$ be the midpoint of $[A C]$, i.e., the center of $\mathcal{C}$, and let $R$ be the radius of $\mathcal{C}$. By the central angle theorem, we have $\widehat{D O B}=2 \widehat{D A B}=120^{\circ}$, and by the Law of Cosines, it follows that $B D^{2}=R^{2}(2-2 \cos (\widehat{D O B}))$; since $\cos (\widehat{D O B})=-1 / 2$, we deduce that $9=B D^{2}=3 R^{2}$, hence $R=\sqrt{3}$.

Furthermore, using the power of point $M$ with respect to $\mathcal{C}$, we get: $M A \cdot M C=M B \cdot M D=2$, while $M A+M C=2 R=2 \sqrt{3}$. Since $\widehat{B C D}>\widehat{D A B}$, we know that $M A>M C$, from which we deduce that $M A=\sqrt{3}+1$ and $M C=\sqrt{3}-1$. It follows that $O M=O C-M C=1$; then, since $O M^{2}+O D^{2}=$ $1+3=4=M D^{2}$, we deduce that triangle $M O D$ is a right triangle at $O$, i.e., $(A C)=(O M) \perp(O D)$.

We deduce that the area of triangle $A C D$ is $R^{2}$. Moreover, since $M D=2 B M$, we also find that the area of triangle $A B C$ is $\frac{1}{2} R^{2}$. The area of quadrilateral $A B C D$ is therefore $\frac{3}{2} R^{2}=\frac{9}{2}$.

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