China Mathematical Competition (Complementary Test)

(1) As shown in Fig. 1, by the Ptolemy inequality we have $$PA\times BC+PC\times AB\ge PB\times AC.$$ Therefore, $$\begin{array}{rl}f(P)& =PA\times BC+PC\times AB+PD\times CA\\ & \ge PB\times CA+PD\times CA\\ & =(PB+PD)\times CA.\end{array}$$ The equality holds if and only if $P$, $A$, $B$, $C$ lie on $\odot O$ and $P$ on $\widehat{AC}$. Furthermore, $PB+PD\ge BD$, and the equality holds if and only if $P$ lies on line $BD$. Combining the results above, we have $$f(P{)}_{\min }=BD\times CA.$$

(2) Denote $\angle ECB=\alpha$. Then $\angle ECA=2\alpha$. By the sine theorem we have $$\frac{AE}{AB}=\frac{\sin 2\alpha }{\sin 3\alpha }=\frac{\sqrt{3}}{2}.$$ That is to say, $\sqrt{3}\sin 3\alpha =2\sin 2\alpha$. By using trigonometric identities, $$\sqrt{3}(3\sin \alpha -4{\sin }^3\alpha )=4\sin \alpha \cos \alpha .$$ By simplification, $$3\sqrt{3}-4\sqrt{3}(1-{\cos }^2\alpha )-4\cos \alpha =0.$$ $$\text{That is to say, }4\sqrt{3}{\cos }^2\alpha -4\cos \alpha -\sqrt{3}=0.$$ $$\text{The solutions are }\cos \alpha =\frac{\sqrt{3}}{2}\text{ and }\cos \alpha =-\frac{1}{2\sqrt{3}}\text{ (discarded).}$$ $$\text{Therefore, }\alpha =30^{\circ }\text{ and }\angle ECA=60^{\circ }.$$ On the other hand, $$\frac{BC}{EC}=\sqrt{3}-1=\frac{\sin (\angle EAC-30^{\circ })}{\sin \angle EAC}.$$ That is to say, $$\frac{\sqrt{3}}{2}\sin \angle EAC-\frac{1}{2}\cos \angle EAC=(\sqrt{3}-1)\sin \angle EAC.$$ Then $$\frac{2-\sqrt{3}}{2}\sin \angle EAC=\frac{1}{2}\cos \angle EAC.$$ Therefore, $$\tan \angle EAC=\frac{1}{2-\sqrt{3}}=2+\sqrt{3}.$$ $$\text{We obtain }\angle EAC=75^{\circ }\text{ and }\angle AEC=45^{\circ }.$$ Since $△ADC$ is an isosceles triangle and $AC=\sqrt{2}$, we have $CD=1$. Furthermore, $△ABC$ is an isosceles triangle, so $BC=\sqrt{2}$ and $AB=2$. Then $$B{D}^2=A{B}^2+A{D}^2=4+1=5.$$ $$\text{We have }BD=\sqrt{5}.\text{ Therefore,}$$ f(P)_{\min} = BD \times CA = \sqrt{5} \times \sqrt{2} = \sqrt{10}. --- **Alternative solution.** (1) As shown in Fig. 2, the line $BD$ intercepts the circumscribed circle $O$ of $\triangle ABC$ at point $P_0$, which lies on $BD$ because $D$ is outside of $\odot O$. Through $A$, $C$, $D$ draw lines perpendicular to $P_0A$, $P_0C$, $P_0D$, respectively, and they constitute $\triangle A_1B_1C_1$ by intercepting with each other. It is easy to see that $P_0$ lies in $\triangle ABC$, as well as in $\triangle A_1B_1C_1$. Denoting the three inner angles of $\triangle ACD$ by $x$, $y$, $z$, respectively, we have Fig. 2 \angle AP_0C = 180^\circ - y = z + x. Furthermore, from $B_1C_1 \perp P_0A$ and $B_1A_1 \perp P_0C$, we have $\angle B_1 = y$. In a similar way, we have $\angle A_1 = x$ and $\angle C_1 = z$. It follows that $\triangle A_1B_1C_1 \sim \triangle ABC$. Now let B_1C_1 = \lambda BC, \quad C_1A_1 = \lambda CA, \quad A_1B_1 = \lambda AB. Then for any point $M$ on the plane, we have $$\begin{array}{rl}\lambda f(P_0)& =\lambda (P_{0}A\times BC+P_{0}D\times CA+P_{0}C\times AB)\\ & =P_{0}A\times B_1{C}_1+P_{0}D\times C_1{A}_1+P_{0}C\times A_1{B}_1\\ & =2S_{△A_1{B}_1{C}_1}\\ & \le MA\times B_1{C}_1+MD\times C_1{A}_1+MC\times A_1{B}_1\\ & =\lambda (MA\times BC+MD\times CA+MC\times AB)\\ & =\lambda f(M).\end{array}$$ That is to say, $f(P_0) \le f(M)$. Since $M$ is an arbitrary point, we know that $f(P)$ reaches the minimum at $P_0$, and at the same time $P_0A$, $P_0C$, $P_0D$ are concyclic. This completes the proof. (2) From (1) we know that the minimum of $f(P)$ is f(P_0) = \frac{2}{\lambda} S_{\triangle A_1 R_1 C_1} = 2\lambda S_{\triangle ABC}. In the same way as is shown in the proof of (2) in Solution I, we find that both $\triangle ADC$ and $\triangle ABC$ are right-angled isosceles triangles; so we have $CD = AD = \frac{AC}{\sqrt{2}} = 1$, $AB = \sqrt{2}AC = 2$, $S_{\triangle ABC} = 1$, BD = \sqrt{AB^2 + AD^2} = \sqrt{5}. Furthermore, since $\triangle A_1B_1C_1 \sim \triangle ABC$, $\angle AB_1B = \angle AB_1C + \angle BB_1C = 90^\circ$, $AB_1BD$ is a rectangle. It follows that $B_1C_1 = BD = \sqrt{5}$. Therefore $\lambda = \frac{\sqrt{5}}{\sqrt{2}}$ and f(P)_{\min} = 2 \times \frac{\sqrt{5}}{\sqrt{2}} \times 1 = \sqrt{10}.

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Alternative solution.

(1) We discuss the problem on the complex plane, and regard points $A$, $B$, $C$ as complex numbers. Then by the triangle inequality we have $$|\overrightarrow{PA}\cdot \overrightarrow{BC}|+|\overrightarrow{PC}\cdot \overrightarrow{AB}|\ge |\overrightarrow{PA}\cdot \overrightarrow{BC}+\overrightarrow{PC}\cdot \overrightarrow{AB}|.$$ That is to say, $$\begin{array}{rl}& |(A-P)(C-B)|+|(C-P)(B-A)|\\ \ge & |(A-P)(C-B)+(C-P)(B-A)|\\ =& |-P\times C-A\times B+C\times B+P\times A|\\ =& |(B-P)(C-A)|\\ =& |\overrightarrow{PB}\cdot \overrightarrow{AC}|.\end{array}\text{\hspace{1em}(1)}$$ Then $$\begin{array}{rl}& |\overrightarrow{PA}\cdot \overrightarrow{BC}|+|\overrightarrow{PC}\cdot \overrightarrow{AB}|+|\overrightarrow{PD}\cdot \overrightarrow{AC}|\\ \ge & |\overrightarrow{PB}\cdot \overrightarrow{AC}|+|\overrightarrow{PD}\cdot \overrightarrow{AC}|\\ =& (\overrightarrow{PB}+\overrightarrow{PD})\cdot \overrightarrow{AC}\\ =& |\overrightarrow{BD}|\cdot \overrightarrow{AC}|.\end{array}\text{\hspace{1em}(2)}$$ The equality in (1) holds only when the complex numbers $(A-P)(C-B)$ and $(C-P)(B-A)$ are in the same direction. This means that there is a real $\lambda >0$ such that $$(A-P)(C-B)=\lambda (C-P)(B-A).$$ That is to say, $$\frac{A-P}{C-P}=\lambda \frac{B-A}{C-B}.$$ Therefore, $$\arg \left(\frac{A-P}{C-P}\right)=\arg \left(\frac{B-A}{C-B}\right),$$ which means that the angle of rotation from $\overrightarrow{PC}$ to $\overrightarrow{PA}$ is equal to that from $\overrightarrow{BC}$ to $\overrightarrow{AB}$. It follows that $P$, $A$, $B$, $C$ are concyclic. The equality in (2) holds only when $B$, $P$, $D$ are collinear and $P$ lies on the segment $BD$. This means that $f(P)$ reaches the minimum when $P$ lies on the circumscribed circle of $△ABC$ and $P$, $A$, $B$, $C$ are concyclic.

(2) From (1) we know that $f(P{)}_{min}=BD\times CA$. The following steps are the same as those in the proof of (2) in Solution I.