
Problem 10.2. A quadrangle with perpendicular diagonals $A C$ and $B D$ is inscribed in a circle with centre $O$ and radius 1 .

a.) Calculate the sum of the squares of the sides of the quadrangle.

b.) Find the area of $A B C D$ if a circle with centre $I$ is inscribed in it and $O I=\frac{1}{\sqrt{3}}$.


Solution: a.) Denote $\angle B A C=\alpha, \angle C A D=\beta$. It follows from the Sine Theorem for $\triangle A B C$ and $\triangle A B D$ that $B C=2 \sin \alpha, A D=$ $2 \sin \angle A B D=2 \sin (90 \mathrm{deg}-\alpha)=2 \cos \alpha$. Analogously $C D=2 \sin \beta$, $A B=2 \cos \beta$. Therefore $A B^{2}+B C^{2}+C D^{2}+D A^{2}=4 \sin ^{2} \alpha+$ $4 \cos ^{2} \alpha+4 \sin ^{2} \beta+4 \cos ^{2} \beta=8$.

b.) The statement of the problem implies that $A B+C D=$ $A D+B C$ or $2 \cos \beta+2 \sin \beta=2 \cos \alpha+2 \sin \alpha$, i. e. $\sin (\beta+45 \mathrm{deg})=$ $\sin (\alpha+45 \mathrm{deg})$. Therefore $\beta=\alpha$ or $\beta=90 \mathrm{deg}-\alpha$. We shall consider only the case $\beta=\alpha$ (the case $\beta=90 \mathrm{deg}-\alpha$ is settled in a similar fashion). Now $A D=A B, C D=B C$ and $\angle A B C=\angle A D C=90 \mathrm{deg}$. The points $O$ and $I$ lie on $A C$ and when $\alpha \geq 45 \mathrm{deg}, I$ lies on
the segment $A O$. In $\triangle A C D$ ( $D I$ is a bisector) determine $A I=$ $\frac{2 \cos \alpha}{\sin \alpha+\cos \alpha}$. Consequently $O I=A O-A I=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}=\frac{1}{\sqrt{3}}$. Raising both sides to the 2 nd power gives $\frac{1-\sin 2 \alpha}{1+\sin 2 \alpha}=\frac{1}{3}$. Hence $\sin 2 \alpha=\frac{1}{2}$. It then follows that the area of $A B C D$ is equal to $S=2 S_{A B C}=A B \cdot B C=4 \cos \alpha \sin \alpha=2 \sin 2 \alpha=1$.

