Iranian Mathematical Olympiad

At first we shall prove following lemma: Lemma. If point $A^{\prime }$ is antipode of vertex $A$ in circumcircle of triangle $ABC$, then $$S_{ABC}-S_{A^{\prime }BC}=\frac{1}{2}|BC{|}^2\cot \angle A,$$ where $S_{XYZ}$ denotes the area of triangle $XYZ$. Proof. Let $H$ be the orthocenter of triangle $ABC$. Notice that $$S_{ABC}-S_{A^{\prime }BC}=S_{ABC}-S_{HBC}=\frac{1}{2}BC\cdot AH=\frac{1}{2}|BC{|}^2\cot \angle A.$$

Now let $AB=a$, $AD=b$, $\angle APB=\angle P_1$, $\angle BPC=\angle P_2$, $\angle CPD=\angle P_3$ and $\angle DPA=\angle P_4$. From the lemma, it's easy to see that $$\begin{gathered} 2S_{ABCD}-S_{XYZT}=\frac{a^2}{2}(\cot \angle P_1+\cot \angle P_3)+\frac{b^2}{2}(\cot \angle P_2+\cot \angle P_4) \\ =\frac{a^2}{2}\left(\frac{\sin (\angle P_1+\angle P_3)}{\sin \angle P_1\cdot \sin \angle P_3}\right)+\frac{b^2}{2}\left(\frac{\sin (\angle P_2+\angle P_4)}{\sin \angle P_2\cdot \sin \angle P_4}\right) \end{gathered}$$ Without loss of generality, we can assume $\angle P_1+\angle P_3>180^{\circ }$ (If $\angle P_1+\angle P_3=180^{\circ }$ we are done) so $\sin (\angle P_1+\angle P_3)<0$. Let points $K,L,M$ and $N$ be the foot of the perpendicular lines from $P$ to $AB,BC,CD$ and $DA$, respectively. We need to prove that $$\iff \frac{a^2}{2}\left(\frac{\sin (\angle P_1+\angle P_3)}{\sin \angle P_1\cdot \sin \angle P_3}\right)+\frac{b^2}{2}\left(\frac{\sin (\angle P_2+\angle P_4)}{\sin \angle P_2\cdot \sin \angle P_4}\right)<0$$ $$\iff \frac{a^2}{\sin \angle P_1\cdot \sin \angle P_3}>\frac{b^2}{\sin \angle P_2\cdot \sin \angle P_4}$$ $$\iff \frac{AP}{\sin \angle ABP}\cdot \frac{CP}{\sin \angle CDP}>\frac{AP}{\sin \angle ADP}\cdot \frac{CP}{\sin \angle CBP}$$ $$\iff \sin \angle CBP\cdot \sin \angle ADP>\sin \angle ABP\cdot \sin \angle CDP$$ $$\iff BP\cdot \sin \angle CBP\cdot DP\cdot \sin \angle ADP>BP\cdot \sin \angle ABP\cdot DP\cdot \sin \angle CDP$$ $$\iff PL\cdot PN>PK\cdot PM$$ $$\iff \angle LKN+\angle LMN>\angle KLM+\angle KNM$$ $$\iff \angle APB+\angle CPD>\angle APD+\angle BPC$$ The latest assertion is true, so the proof is completed. ■