Let $ \alpha = \hat{(AB,CD)}$ and H be the projection of O on CD. We have: $ S_{ACBD} = \frac{1}{2} AB.CD.\sin{\alpha} = CD \sin{\alpha}$ (1) We need to find $ alpha$ in $ (0,\pi)$ such that it will minimize (1). Now: $ OH = OP \sin{\alpha}$ then $ CD = 2 \sqrt{1 - OH \sin^2{\alpha}}$ Now just use analysis to find the mimimum of the function $ f(x) = \sqrt{1 - OH \sin^2{\alpha}}\sin{\alpha}$

mathangel wrote: Let $ \alpha = \hat{(AB,CD)}$ and H be the projection of O on CD. We have: $ S_{ACBD} = \frac {1}{2} AB.CD.\sin{\alpha} = CD \sin{\alpha}$ (1) We need to find $ alpha$ in $ (0,\pi)$ such that it will minimize (1). Now: $ OH = OP \sin{\alpha}$ then $ CD = 2 \sqrt {1 - OH \sin^2{\alpha}}$ Now just use analysis to find the mimimum of the function $ f(x) = \sqrt {1 - OH \sin^2{\alpha}}\sin{\alpha}$ And what will you get after taking the derivative to 0? I wonder if it will be anything constructible. Would be nice to know if the derivative equation is reasonably doable, but I stop the moment I see a shadow of unnatural irrationality (other than root of N(atural)). The fact that this problem was given at some math competition hints that it is not a straightforward problem. And if it were up to calculus, then it would not be in the geometry department. This is a purely geometric problem and has a fine point to it. Thank you. M.T.

OK, how would you solve it then?

This is a rough problem because there are two different cases that produce different equality cases... WLOG $ P$ is on $ AO$. Let $ PO=p$. Let $ C'$ and $ D'$ be the projections of $ C$ and $ D$ onto $ AB$, respectively. Call $ CC'=c$ and $ DD'=d$. The area of the quadrilateral is $ c+d$. Let $ M$ be the midpoint of $ CD$ (so $ CD\perp OM$). Then it follows that if $ OM=x$, then $ \frac{d}{x}=\frac{PD}{p}$ and $ \frac{c}{x}=\frac{CP}{p}$. Hence $ c+d=\frac{x}{p}\cdot CD$. Note that $ CD=2\sqrt{1-x^2}$. Hence \[ [ABCD]=\frac{2}{p}\cdot x\cdot \sqrt{1-x^2}\] where $ x\le p$. Case 1: $ p\ge \frac{1}{\sqrt{2}}$. Note \[ [ABCD]=\frac{2}{p}\cdot x\cdot \sqrt{1-x^2}=\frac{2}{p}\cdot \sqrt{x^2\cdot(1-x^2)}\le \frac{2}{p}\cdot \frac{(x^2)+(1-x^2)}{2}=\frac{1}{p}\] by the AM-GM inequality. Equality occurs when $ 1-x^2=x^2\implies x=\frac{1}{\sqrt{2}}\le p$. Case 2: $ p<\frac{1}{\sqrt{2}}$. Note \[ [ABCD]= \frac{2}{p}\cdot x\cdot \sqrt{1-x^2}\le \frac{2}{p}\cdot p\cdot \sqrt{1-p^2}\iff \frac{4}{p^2}\cdot (p^2-x^2)(1-x^2-p^2)\ge 0\] but $ x^2+p^2\le p^2+p^2=2p^2<1$, so the inequality above is true. Equality occurs when $ x=p$. That is when $ CD\perp AB$.

rem wrote: Let $ AB$ be a fixed diameter of a given circle $ w = C(O,1)$ and let $ P$ be a fixed point on $ (AB)$. A mobile line through $ P$ meets the circle $ w$ at $ C$ and $ D$, so a convex quadrilateral $ ACBD$ is formed. Find the maximum possible area of this quadrilateral. Proof (similarly with Altheman's). Suppose w.l.o.g. that $ P\in (OB)$ . Denote $ \phi = m(\widehat {OPD})$ and the distance $ x$ of $ O$ to $ CD$ . Observe that $ CD = 2\sqrt {1 - x^2}$ and $ \sin\phi = \frac xp$ . Since $ S\equiv [ACBD] =$ $ \frac 12\cdot AB\cdot CD\cdot\sin\phi$ , then $ S$ is max. $ \Longleftrightarrow$ $ x\cdot\sqrt {1 - x^2}$ is max. $ \Longleftrightarrow$ $ x^2\cdot (1 - x^2)$ is max., where $ x\in [0,p]$ . But $ x^2 + (1 - x^2) = 1$ (constant). Observe that $ x^2 = 1 - x^2$ $ \Longleftrightarrow$ $ x = \frac {\sqrt 2}{2}$ . In conclusion, $ x_{\mathrm {max}} = \{\begin{array}{ccc} \frac {\sqrt 2}{2} & \mathrm {for} & \frac {\sqrt 2}{2}\le p \\ \\ p & \mathrm {for} & p < \frac {\sqrt 2}{2}\end{array}$ and $ S_{\mathrm {max}} = \{\begin{array}{ccc} \frac 1p & \mathrm {for} & \frac {\sqrt 2}{2}\le p \\ \\ 2\sqrt {1-p^2} & \mathrm {for} & p < \frac {\sqrt 2}{2}\end{array}$