Chinese Mathematical Olympiad

Proof 1. (1) Let $A_1$ be the reflection of $A$ across $B$. By the given conditions, $KP{A}_{1}A$ forms an isosceles trapezoid, thus $\angle APB=\angle A_{1}KB$. Similarly, let $A_2$ be the reflection of $A$ across $C$. We have $\angle AQC=\angle A_{2}KC=\angle A_{2}KB$. Hence, the desired conclusion is equivalent to $\angle BTC=\angle A_{1}K{A}_2$. Perform a homothety centered at $A$ with a ratio of $1/2$ on $△A_{1}K{A}_2$, mapping $A_1$ to $B$, $A_2$ to $C$, and $K$ to the midpoint of $AK$ (denoted as $M$). The proposition then transforms to $\angle BTC=\angle BMC$, i.e., points $B,C,M$, and $T$ are concyclic. By the Power of a Point Theorem, this is equivalent to proving $KB\cdot KC=KM\cdot KT$. If we take $X$ as the midpoint of $KT$, it is equivalent to proving $KB\cdot KC=KA\cdot KX$, i.e., points $A,B,C$, and $X$ are concyclic. We note that the perpendicular bisector of segment $KP$ passes through point $B$ and is perpendicular to $AB$, and the perpendicular bisector of $KQ$ passes through $C$ and is perpendicular to $AC$. Hence, the circumcenter of $△PQK$ (denoted as $O$) is the antipodal point of $A$ on the unit circle. From $OX\perp KT$, we have $\angle OXA=\frac{\pi }{2}$, and thus $X$ lies on the circle with diameter $AO$ (the circumcircle of $△ABC$). Proof completed.



(2) We have $$\begin{array}{rl}\frac{BP}{CQ}& =\frac{BK}{CK}=\frac{S_{△BTK}}{S_{△CTK}}=\frac{BT}{CT}\cdot \frac{\sin \angle BTK}{\sin \angle CTK}=\frac{BT}{CT}\cdot \frac{\sin \angle MCK}{\sin \angle MBK}\\ & =\frac{BT}{CT}\cdot \frac{BM}{CM}=\frac{BT}{CT}\cdot \frac{K{A}_1}{K{A}_2}=\frac{BT}{CT}\cdot \frac{AP}{AQ}.\end{array}$$ Proof completed.

Proof 1 (Alternative). An alternative proof that points $B,C,M$, and $T$ are concyclic. Since $K,P,Q$, and $T$ are concyclic, by Ptolemy's Theorem, $KT\cdot PQ=PK\cdot QT+QK\cdot PT$. Also, by the Law of Sines, $\frac{PQ}{\sin \angle PKQ}=\frac{TQ}{\sin \angle TKQ}=\frac{PT}{\sin \angle PKT}$, hence $$KT=\frac{PK\cdot QT+QK\cdot PT}{PQ}=\frac{PK\cdot \sin \angle TKQ+QK\cdot \sin \angle PKT}{\sin \angle PKQ}.$$ Let $\angle ABC=\angle B$ and $\angle ACB=\angle C$. Noting that $KP=2BK\cos \angle B$ and $KQ=2CK\cos \angle C$, and utilizing the parallel relationships given in the problem, we obtain $$\begin{array}{rl}KT& =\frac{2BK\cos \angle B\cdot \sin \angle CAK+2CK\cos \angle C\cdot \sin \angle BAK}{\sin (\angle B+\angle C)}\\ & =\frac{2BK\cos \angle B\cdot \frac{KC}{KA}\sin \angle C+2CK\cos \angle C\cdot \frac{KB}{KA}\sin \angle B}{\sin (\angle B+\angle C)}\\ & =\frac{\cos \angle B\cdot \sin \angle C+\cos \angle C\cdot \sin \angle B}{\sin (\angle B+\angle C)}\cdot \frac{2KB\cdot KC}{KA}=\frac{KB\cdot KC}{KM},\end{array}$$ therefore $KM\cdot KT=KB\cdot KC$. By the Power of a Point Theorem, it follows that points $B,C,M$, and $T$ are concyclic. Proof completed.

Proof 2. Let's consider the circumcircle of $△ABC$ as the unit circle in the complex plane. We use uppercase letters for points and the corresponding lowercase letters for their complex numbers. Let $X$ be the intersection point of the line segment $AK$ with the unit circle. Then $$\bar{k}=\frac{a+x-b-c}{ax-bc}.$$ Thus, $$k=\frac{\bar{a}+\bar{x}-\bar{b}-\bar{c}}{\bar{a}\bar{x}-\bar{b}\bar{c}}=\frac{\frac{1}{a}+\frac{1}{x}-\frac{1}{b}-\frac{1}{c}}{\frac{1}{ax}-\frac{1}{bc}}=\frac{bc(x+a)-ax(c+b)}{bc-ax}=\frac{(ax-ac-cx)b+axc}{ax-bc}.$$ Note the following geometric facts: the perpendicular bisector of $KP$ passes through $B$ and is perpendicular to $AB$, the perpendicular bisector of $KQ$ passes through $C$ and is perpendicular to $AC$. Therefore, the circumcenter of $△PQK$ is the diametrically opposite point of $A$ on the unit circle, denoted as $A_1$. From $\angle AX{A}_1=\frac{\pi }{2}$, we know $X$ is the midpoint of $TK$. Therefore, $$k=\frac{\frac{1}{a}+\frac{1}{x}-\frac{1}{b}-\frac{1}{c}}{\frac{1}{ax}-\frac{1}{bc}}=\frac{bc(x+a)-ax(c+b)}{bc-ax}=\frac{(ax-ac-cx)b+axc}{ax-bc}.$$

We observe the following geometric facts: The perpendicular bisector of the segment $KP$ is the line passing through point $B$ and perpendicular to $AB$, and the perpendicular bisector of $KQ$ is the line passing through $C$ and perpendicular to $AC$. Therefore, the circumcenter of $△PQK$ is the diametrical opposite point of $A$ on the unit circle, denoted as $A_1$. Thus, according to $\angle AX{A}_1=\frac{\pi }{2}$, we know that $X$ is the midpoint of $TK$. So $$\begin{array}{rl}t& =2x-k=2x-\frac{(ax-ac-cx)b+acx}{ax-bc}=\frac{2a{x}^2-(ax-ac+cx)b-axc}{ax-bc},\\ t-b& =\frac{2a{x}^2-(ax-ac+cx)b-axc-axb+b^{2}c}{ax-bc}=\frac{(b-x)(bc+ac-2ax)}{ax-bc},\\ t-c& =\frac{2a{x}^2-(ax-ac+cx)b-axc-axc+b{c}^2}{ax-bc}=\frac{(c-x)(ab+bc-2ax)}{ax-bc}.\end{array}$$ On the other hand, since $Q$ and $K$ are symmetric about the line $C{A}_1$ (noting that $a_1=-a$), we have $$\frac{q+a}{c+a}=\frac{\overline{k+a}}{c+a}=\frac{(\bar{k}+\bar{a})ac}{a+c}=\frac{ac\bar{k}+c}{a+c},$$ $$\text{thus }q=-a+c+ac\bar{k}=(c-a)+\frac{ac(a+x-b-c)}{ax-bc}.\text{ Hence,}$$ $$\begin{array}{rl}q-c& =\frac{-a(ax-bc)+ac(a+x-b-c)}{ax-bc}=\frac{-a^{2}x+a^{2}c+acx-a{c}^2}{ax-bc}=\frac{a(a-c)(c-x)}{ax-bc},\\ q-a& =\frac{(c-2a)(ax-bc)+ac(a+x-b-c)}{ax-bc}=\frac{abc-b{c}^2-2a^{2}x+2acx+a^{2}c-a{c}^2}{ax-bc}\\ & =\frac{(a-c)(ac+bc-2ax)}{ax-bc}.\end{array}$$ $$\text{Similarly, we have }p=-a+b+ab\bar{k}=(b-a)+\frac{ab(a+x-b-c)}{ax-bc}.\text{ Thus,}$$ $$\begin{array}{rl}p-b& =\frac{-a(ax-bc)+ab(a+x-b-c)}{ax-bc}=\frac{-a^{2}x+a^{2}b+abx-a{b}^2}{ax-bc}=\frac{a(a-b)(b-x)}{ax-bc},\\ p-a& =\frac{(b-2a)(ax-bc)+ab(a+x-b-c)}{ax-bc}=\frac{abc-b^{2}c-2a^{2}x+2abx+a^{2}b-a{b}^2}{ax-bc}\\ & =\frac{(a-b)(ab+bc-2ax)}{ax-bc}.\end{array}$$ Therefore, we arrive at $$\frac{t-c}{t-b}\cdot \frac{p-b}{p-a}=\frac{q-c}{q-a},$$ and by taking the argument of both sides of this equation, we obtain conclusion (1); by taking the modulus, we obtain conclusion (2).