Consider a quadrilateral with $\angle DAB=60^{\circ}$, $\angle ABC=90^{\circ}$ and $\angle BCD=120^{\circ}$. The diagonals $AC$ and $BD$ intersect at $M$. If $MB=1$ and $MD=2$, find the area of the quadrilateral $ABCD$.
Problem
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Tags: geometry, geometry proposed
Kouichi Nakagawa
21.11.2010 17:18
$AC$ is diameter. (Because $\angle ABC = \angle ADC = 90 ^\circ$) Let $O$ is circumcircle. So $\triangle OBD$ is isosceles triangle with $\angle BOD = 120^\circ$ and $OB=OD$. Let $R$ is radius. ($R = OA = OB = OC = OD$) So $2 \times \left(R \times \frac{\sqrt{3}}{2} \right) = R\sqrt{3} = BD = 3 \iff R = \sqrt{3}$. Now \[\left\{\begin{array}{l} MA \times MC = MB \times MD = 2 \\ MA + MC = 2R = 2\sqrt{3} \end{array}\right. \iff \left\{\begin{array}{l} MA = \sqrt{3} + 1 \\ MC = \sqrt{3} - 1 \end{array}\right.\] So $OM = OC - MC = \sqrt{3} - (\sqrt{3} - 1) = 1$. Therefore $\angle COD = \angle MOD = 90^\circ$. (Because $(OM, MD, DO) = (1, 2, \sqrt{3})$) Hence $\angle CAD = \frac{1}{2} \angle COD = 45^\circ$. So $\triangle DAC$ is isosceles right triangle with $\angle ADC = 90^\circ$ and $DA=DC$. Therefore $DA=DC = \frac{AC}{\sqrt{2}} = \sqrt{6}$. Thus $\triangle DAC + \triangle BAC = \left(\sqrt{6} \times \sqrt{6} \times \frac{1}{2}\right) + \left(\sqrt{6} \times \sqrt{6} \times \frac{1}{2}\right)\times \frac{1}{2} = 3 + \frac{3}{2} = \boxed{\frac{9}{2}}$.