Let $a$ be real number with $\sqrt{2}<a<2$, and let $ABCD$ be a convex cyclic quadrilateral whose circumcentre $O$ lies in its interior. The quadrilateral's circumcircle $\omega$ has radius $1$, and the longest and shortest sides of the quadrilateral have length $a$ and $\sqrt{4-a^2}$, respectively. Lines $L_A,L_B,L_C,L_D$ are tangent to $\omega$ at $A,B,C,D$, respectively. Let lines $L_A$ and $L_B$, $L_B$ and $L_C$,$L_C$ and $L_D$,$L_D$ and $L_A$ intersect at $A',B',C',D'$ respectively. Determine the minimum value of $\frac{S_{A'B'C'D'}}{S_{ABCD}}$.
Problem
Source: 2001 China National Olmpiad
Tags: geometry, circumcircle, ratio, trigonometry, function, cyclic quadrilateral, geometry proposed
sayantanchakraborty
20.10.2014 12:31
Consider the center $O$ and let the angles subtended by it on sides $AB,BC,CD,DA$ be $2\alpha,2\beta,2\gamma,2\theta$.Then note that $S_{ABCD}=\frac{1}{2}(sin2\alpha+2sin\beta+2sin\gamma+2sin\theta)$.Also note that $S_{A'B'C'D'}=\sum{AOBA'}=\frac{1}{2}\sum{AA'+A'B}=tan\alpha+tan\beta+tan\gamma+tan\theta$. Next note that atleast one of the sides is $\sqrt{4-a^2}$ and atleast another is $a$.Since the ratio is a symmetric expression without loss of generality we may assign $AB=\sqrt{4-a^2}$ and $CD=a$.By easy sine rule application we get $AB=2sin\alpha$ and $CD=2sin\gamma$.Then $sin^2\alpha+sin^2\gamma=1$ or $\alpha+\gamma=90^{\circ}$.Also by some manipulations we will get $tan\alpha=\frac{\sqrt{4-a^2}}{a}$ and $tan\gamma=\frac{a}{\sqrt{4-a^2}}$.Also we get $sin2\alpha=sin2\gamma=\frac{a\sqrt{4-a^2}}{2}$ Next note that $f(x)=tanx \implies f''(x)=2sec^2xtanx>0$ (I am considering only acute $x$).Thus this function is convex,so $tan\beta+tan\theta \ge 2tan\frac{\beta+\theta}{2}=2tan45^{\circ}=2$ Similarly we check the behaviour of $f(x)=sin2x$ where $x$ is acute.Note that $f''(x)=-4sin2x<0$ so this function is concave.So $sin2\beta+sin2\theta \le 2sin(\beta+\theta)=2sin90^{\circ}=2$ Thus $\frac{S_{A'B'C'D'}}{S_{ABCD}}=2\frac{tan\alpha+tan\beta+tan\gamma+tan\theta}{sin2\alpha+sin2\beta+sin2\gamma+sin2\theta} \ge 2\frac{2+\frac{a}{\sqrt{4-a^2}}+\frac{\sqrt{4-a^2}}{a}}{a\sqrt{4-a^2}+2}=\frac{4}{a\sqrt{4-a^2}}$ Equality is achieved when $\beta=\theta=45^{\circ}$ and $\alpha=sin^{-1}\frac{\sqrt{4-a^2}}{2}$ and $\gamma=sin^{-1}\frac{a}{2}$. Note that if we would have taken any pair of adjacent sides as $a$ and $\sqrt{4-a^2}$,then it would not have been a difficult task because one of $AC$ or $BD$ would have become the diameter.However there was no guarantee that in this case the ratio would be at its minimum.On the other hand the above is the general case which can be applied in any situation. So this is the required minimal value.