Selected Problems from Thailand Training Camp

Solution: Denote the area of $\Gamma$ by $[\Gamma ]$. Let $E$ be the intersection of $AC$ and $BD$. From any point $P^{\prime }$ on $AB$, we construct points $Q^{\prime },R^{\prime },S^{\prime }$ as follows: Let the line $P^{\prime }E$ intersect $CD$ at $R^{\prime }$. Let $\ell$ be the line perpendicular to $P^{\prime }{R}^{\prime }$ at $E$. Let $\ell$ intersect $BC,DA$ at $Q^{\prime },S^{\prime }$ respectively.

Define a continuous function $f$ from a point $P^{\prime }$ on $AB$ to the interval $[0,1]$ by $f(P^{\prime })=\frac{[P^{\prime }{Q}^{\prime }{R}^{\prime }{S}^{\prime }]}{[ABCD]}$ where $Q^{\prime },R^{\prime },S^{\prime }$ are constructed as above.

We see that $f(A)=f(B)=1$. If we can find a point $P^{\prime }$ on $AB$ such that $f(P^{\prime })\le \frac{1}{2}$, then by continuity there exists a point $P$ on $AB$ such that $f(P)=\frac{1}{2}$. Then the construction above will yield the points $P,Q,R,S$ as required.

Next, we prove that $f(P^{\prime })\le \frac{1}{2}$ if $E{P}^{\prime }$ bisects $\angle AEB$. We see that, in such case, $E{Q}^{\prime },E{R}^{\prime },E{S}^{\prime }$ bisect $\angle BEC,\angle CED,\angle DEA$ respectively.



Let $a,b,c$ be the lengths of $EA,EB,EC$ respectively. We have that the lengths of $E{P}^{\prime },E{Q}^{\prime }$ are $\frac{\sqrt{2ab}}{a+b}$, $\frac{\sqrt{2bc}}{b+c}$ respectively. It follows from the AM-GM inequality that $$\begin{gathered} (a+b)(b+c)(ab+bc)\ge 8\sqrt{ab}\sqrt{bc}\sqrt{ab\cdot bc} \\ =8ab\cdot bc, \end{gathered}$$ or $$\begin{array}{rl}[ABC]& =\frac{1}{2}(ab+bc)\\ & \ge 4\frac{ab}{a+b}\frac{bc}{b+c}=4[E{P}^{\prime }{Q}^{\prime }]\end{array}$$ Similarly, $$\begin{array}{rl}[ABCD]& =\frac{1}{2}([ABC]+[BCD]+[CDA]+[DAB])\\ & \ge \frac{1}{2}(4[E{P}^{\prime }{Q}^{\prime }]+4[E{Q}^{\prime }{R}^{\prime }]+4[E{R}^{\prime }{S}^{\prime }]+4[E{S}^{\prime }{P}^{\prime }])\\ & =2[P^{\prime }{Q}^{\prime }{R}^{\prime }{S}^{\prime }]\end{array}$$ So, $f(P^{\prime })\le \frac{1}{2}$ as required.