24th Korean Mathematical Olympiad Final Round

Observation 1. $A$, $B$, $C$, $Z$ are cyclic. $$\angle BAD=\angle BXD=\angle ZBC\text{ and }\angle DAC=\angle DYC=\angle BCZ.$$ Hence $\angle BAC=\angle BAD+\angle BCZ=180^{\circ }-\angle BZC$, that is, $\angle BAC+\angle BZC=180^{\circ }$.

Observation 2. Both $A$, $E$, $Z$, $Y$ and $A$, $X$, $Z$, $F$ are cyclic. By Observation 1, we have $$\angle AEY=\angle ABC=\angle AZC=\angle AZY.$$ Similarly, $\angle AFX=\angle AZX$.

Lemma 1. $ZE=ZF$ if and only if $\frac{BD}{CD}=\frac{AB}{AC}$.

Proof. We can easily see that $△AEZ$ is similar to $△ADC$, and symmetrically, $△AFZ$ is similar to $△ADB$. Then we have $$ZE=\frac{AE\cdot CD}{AD}\text{and}ZF=\frac{AF\cdot BD}{AD}.$$ Therefore, we have $ZE=ZF\iff \frac{BD}{CD}=\frac{AE}{AF}=\frac{AB}{AC}$.

Now let $Q$, $R$, $S$ be the feet of perpendiculars from $A$, $B$, $C$ to $BC$, $CA$, $AB$, respectively. Then $A$, $S$, $D$, $H$, $F$ are cyclic. Since $AD\cdot AP=AH\cdot AQ=AS\cdot AB$, $S$, $D$, $P$, $B$ are cyclic. Similarly, $R$, $D$, $P$, $C$ are cyclic. Since $AS:AP=AD:AB$, $△ASP$ is similar to $△ADB$. Now we have $$(1)PS=BD\frac{AP}{AB},RP=CD\frac{AP}{AC}.$$

Lemma 2. Let $P$ be a point on $BC$. Then $$BP=PC\iff PR=PS.$$ Proof. Suppose $P$ is a midpoint of $BC$. Then $P$, $Q$, $R$, $S$ are on the nine-point circle, so we have $$\angle PSR=\angle RQC=\angle A=\angle SQB=\angle PRS.$$ That is, $PR=PS$. Hence $P$ is the intersection of the perpendicular bisector of $SR$ and $BC$, so it is unique. Thus $PR=PS$ implies that $P$ is a midpoint of $BC$. $\square$

Now we are ready to prove that $BP=PC\iff ZE=ZF$.

Suppose $BP=PC$, then by Lemma 2, $PR=PS$. Then by (1), we have $\frac{BD}{CD}=\frac{AB}{AC}$. Then by Lemma 1, $ZE=ZF$.

Conversely, assume that $ZE=ZF$, then by Lemma 1, $\frac{BD}{CD}=\frac{AB}{AC}$. Then by (1), we have $PS=PR$. By Lemma 2, $BP=PC$. $\square$