(a) Let $B_1$ and $C_1$ be the points that $B$ and $C$ are midpoints of $AB_1$ and $AC_1$ Here, $O_1$, the circumcenter of $\triangle AB_1C_1$, is the circumcenter of $\triangle KPQ$(trivial) We need to prove $180 - \angle CAQ - \angle QKC = \angle BTC + \angle APB$, and $\angle QKC = 180 - \angle BAC - \angle BKP$ Hence, we need to prove $\angle BAC + \angle BKP - \angle CAQ = \angle BTC + \angle APB$ $\angle BKP = \angle KPB$ so it is sufficient to prove $\angle KPA + \angle BAC = \angle CAQ + \angle BTC$ Let $B_2, C_2$ be the points that $TB$ and $TC$ meets the circumcircle of $KPQ$. By butterfly theorem, $LB_1B_2$ and $LC_1C_2$ are collinear where $L$ is the intersection of $(KPQ)$ and $BC$.(not $K$) Then $\angle BTC = \angle B_2LC_2 = \angle B_1LC_1 = \angle B_1KC_1$ (Since $BC \parallel B_1C_1$) Also, $\angle CAQ = \angle KC_1A$ so it is sufficient to prove that $180 - \angle QKB_1 = \angle BAC + \angle KPA$ This is trivial since $\angle BAC = 180 - \angle QKP$ and $\angle KPA = \angle B_1KP$ Therefore, $\angle AQC = \angle BTC + \angle APB$

Let $X = BM \cap (ABP), Y = CM \cap (ACQ)$, $S = (ABP) \cap (ACQ)$ Claim: $B, C, S, T$ are concyclic. Proof: $\angle YAC = \angle CQK = \angle CKQ = \angle ACB$, so $AY \parallel BC$. We get $AX \parallel BC$ i.s.w, so both $ABKX, AYKC$ are parallelogram. Let $M = AK \cap BX \cap CY$. Notice that since $\angle SBM = \angle SAX = \angle SCM$, we have $B, C, S, M$ are concylic. Let $O$ be the circumcenter of $(KPQ)$, and $R = AK \cap (ABC)$. Notice that $BO \perp PK \parallel AB \implies A, B, C, O$ are concyclic, and $OR \perp RA \implies TR = RK$. Hence $KM \cdot KT = KR \cdot KA = KB \cdot KC $ implies $B, C, M, T$ are concyclic. Combine two results, we have $B, C, S, T$ are concyclic. $\square$ Since $$\angle BTC + \angle APB = \angle BSC + \angle ASB = \angle ASC = \angle CQA,$$part (a) is proved. Let $\angle TBC = \angle CMK = \alpha, \angle TMB = \angle TCB = \beta, \angle MKC = \gamma$. Then, $$\frac{AP \cdot BT \cdot CQ}{AQ \cdot CT \cdot BP} = \frac{BM \cdot BT \cdot CK}{CM \cdot CT \cdot BK} = \frac{BM}{BK}\cdot \frac{BT}{CT}\cdot \frac{CK}{CM}$$$$=\frac{\sin \gamma}{\sin \beta}\cdot \frac{\sin \beta}{\sin \alpha}\cdot \frac{\sin \alpha}{\sin \gamma} = 1$$, so $AP \cdot BT \cdot CQ = AQ \cdot CT \cdot BP. \ \blacksquare$

My solution(Chinese):

VicKmath7 wrote:

好像是 $\angle APB = \angle CTK$ 吧 我考场里把题目等价成这个事实，不知道有没有分 (最终 21 0 0 21 6 0)

Let $O$ be a point on the circle $(ABC)$ such that $AO$ is its diameter and let $AK$ meet $(ABC)$ again at $S\neq A$. Since $KP\parallel AB$, then the line $BO$ is the perpendicular bisector of $KP$. Similarly, $CO$ is also the perpendicular bisector of $KQ$. Therefore, $O$ is the circumcenter of $\triangle KPQ$. Moreover, $\angle ASO=90^\circ$ since $AO$ is a diameter, this means $S$ is a midpoint of $TK$ since $TK$ is a chord of $(KPQ)$. Consider $$\frac{TS}{CS}=\frac{KS}{CS}=\frac{KB}{AB}=\frac{PB}{AB}$$Since $\angle TSC=\angle ABP$, we have $\triangle TSC\sim\triangle PBA$. Similarly, we also have $$\frac{TS}{BS}=\frac{KS}{BS}=\frac{KC}{AC}=\frac{QC}{AC}$$Since $\angle ACQ=\angle BST$, we have $\triangle TSB\sim\triangle QCA$ Therefore, $\angle BTC+\angle APB=\angle BTC+\angle CTS=\angle STB=\angle CQA$ and question 1 is solved. For question 2, we also use these pairs of similarities. Consider $$\frac{AP}{CT}=\frac{BP}{ST}\hspace{5mm}\text{and}\hspace{5mm}\frac{AQ}{CQ}=\frac{BT}{ST}$$Combine these equations, after cancelling $ST$, we can derive what we want to prove in question 2.

(1)Let $O$ be the circumcenter of $\triangle KPQ$, assume that $BP\cap CQ=X$, $BP, CQ \cap \odot O=U,V$. Consider $$\angle BCQ=\angle CKQ +\angle CQK=2C$$We have $AC$ is the internal angle bisector of $\angle BCQ$, similarly we have $AB$ is the internal angle bisector of $\angle UBC$, this means $A$ is the $X$-excenter. Consider $$\angle PXQ=180^\circ-\angle BXC=\angle BCX+\angle CBX=180^\circ-2\angle BKP+180^\circ-2\angle BKQ=2(180^\circ-\angle PKQ)=\angle POQ$$Which means $P,X,O,Q$ are concyclic. Hence $$\angle OXQ=\angle OPQ=\angle OQP=\angle OXB$$Showing $A,O,X$ are collinear. Then $\angle QOU=2\angle QPU=2\angle AOQ$ we have $AO$ bisects $\angle QOU$, combine $OP=OU$ we have $Q$ and $U$ are symmetric wrt $AO$. Similarly $P$ and $V$ are symmetruc wrt $AO$. Since $\angle ATU=\angle KPU=B$ we have points $A,T,B,U$ are concyclic. Similarly $A,T,C,V$ are concyclic. As a result we have $$\angle APB+\angle BTC=\angle APB+180^\circ-\angle ATB-\angle CTK=\angle APB+180^\circ-(180^\circ-\angle AUP)-\angle AVQ=\angle AUP=\angle AQC$$means part (1) is done. (2)Since $$\angle BTK=\angle BTC+\angle CTK=\angle CQA-\angle APB+\angle CTK=\angle CQA$$Finally we have $$\dfrac{AP}{AQ}\cdot\dfrac{BT}{CT}\cdot\dfrac{CQ}{BP}=\dfrac{AP}{AQ}\cdot\dfrac{BT}{CT}\cdot\dfrac{CK}{BK}=\dfrac{AP}{AU}\cdot\dfrac{BT}{CT}\cdot\dfrac{CT\sin\angle CTK}{BT\sin\angle BTK}=\dfrac{\sin\angle AUP}{\sin\angle APU}\cdot\dfrac{\sin\angle CTK}{\sin\angle BTK}=1.$$Which means $AP\cdot BT\cdot CQ=AQ\cdot CT\cdot BP$ and we are done.$\;\;\blacksquare$

Please finish this thing sin<KTC/sin<KTB=AP/AQ

Let $A'$ be the $A$ antipode in $(ABC)$, let $P'$, $Q'$, and $T'$ be the midpoints of $KP$, $KQ$, and $KT$, and let $K_b$ and $K_c$ be the reflection of $K$ about $B$ and $C$. An elementary angle chase gives that $P$ and $K_b$ are reflections across $AB$ and that $Q$ and $K_c$ are reflections about $AC$. Then by homothety $KP'Q'T'$ is cyclic but $K'P'Q'A'$ is also cyclic with diameter $AK'$ so we must have that $T'$ lies on the circle with diameter $PK'$. This means that $T'$ lies on the circle with diameter $AA'$ or $(ABC)$. Notice that $ABTK_c$ and $ACTK_b$ are both cyclic as $$KT\cdot KA=2 KT' \cdot KA=2 KB\cdot KC=KK_b\cdot KC= KB\cdot KK_c$$Now to finish the first part $$\angle CQA-\angle APB=\angle AK_cC-\angle AK_bB=\angle KTB-\angle KTC=\angle BTC$$ For the second, using the Ratio Lemma $$\frac{BT}{CT}=\frac{KB\cdot \sin(\angle KTC)}{KC\cdot\sin(\angle KTB)}=\frac{BP}{CQ}\cdot \frac{\sin(\angle KK_bA)}{\sin(\angle KK_cA)}=\frac{BP}{CQ}\cdot \frac{K_cA}{K_b A}=\frac{BP}{CQ}\cdot \frac{AQ}{AP}$$