Trig let’s goo

Angle chasing let's goo [asy][asy] size(9cm); import olympiad; import geometry; defaultpen(fontsize(10pt)); pair A = dir(230); pair B = dir(310); pair C = dir(30); pair D = dir(170); pair M = (B+C)/2; pair N = (A+D)/2; pair T = (C+D)/2; pair X = 2 * foot(circumcenter(A,B,T), C,D) - T; pair C_prime = reflect((0,0), (A+B)/2) * C; pair D_prime = reflect((0,0), (A+B)/2) * D; pair Q = extension(D, C_prime, A, B); pair P = extension(C, D_prime, A, B); fill(A--D--Q--cycle, 0.5*paleblue+0.5*white); fill(B--D--C--cycle, 0.5*paleblue+0.5*white); markscalefactor = 0.015; draw(anglemark(A,Q,D), linewidth(0.7)); draw(anglemark(C,A,D), linewidth(0.7)); draw(anglemark(C,B,D), linewidth(0.7)); draw(A--D--Q--cycle, blue+linewidth(0.7)); draw(B--D--C--cycle, blue+linewidth(0.7)); draw(B--T, blue+linewidth(0.7)); draw(Q--N, blue+linewidth(0.7)); draw(B--P--C--A, black); draw(A--B--C--D--cycle, linewidth(1)); draw(circle((0,0),1), gray+linewidth(0.8)); draw(circumcircle(A,B,T), red+linewidth(1)); dot("$A$", A, dir(-119)); dot("$B$", B, dir(-62)); dot("$C$", C, dir(29)); dot("$D$", D, dir(164)); dot("$M$", M, dir(-10)); dot("$N$", N, dir(166)); dot("$T$", T, dir(98)); dot("$X$", X, dir(100)); dot("$Q$", Q, dir(-145)); dot("$P$", P, dir(-36)); [/asy][/asy] Let $T$ be the midpoint of $CD$, and let $\odot(ABT)$ intersects $CD$ again at $X$. We claim that $X$ is the concurrency point. It suffices to show that $XA$ is tangent to $\odot(NQA)$. Notice that since $\measuredangle AQD = \measuredangle CAD = \measuredangle CBD$ and $\measuredangle QAD = \measuredangle BAD = \measuredangle BCD$, it follows that $\triangle DQA\sim\triangle DBC$, implying that $\measuredangle NQA = \measuredangle TBC$. Now, the problem follows from \begin{align*} \measuredangle XAN &= \measuredangle XAB + \measuredangle BAD \\ &= \measuredangle XTB + \measuredangle BCD \\ &= \measuredangle BCD + \measuredangle DTB \\ &= \measuredangle CBT \\ &= \measuredangle AQN, \end{align*}done.

Nice problem! My solution is similar with #3

Maybe easy for G3 since only angle chasing... and similarities This can be little bit generalized such as choosing $M,N$ such that $AN : ND = BM : MC$ The solution is the same with below Let $AA \cap CD = A_1$, $BB \cap CD = B_1$, and the midpoint of $CD$ as $L$ Firstly, $\begin{cases} \angle DQB = \angle DAC \\ \angle DCA = \angle DBQ \end{cases}$ so $\triangle DQB \sim \triangle DAC$ Here, $\begin{cases} \angle DQA = \angle DAC = \angle DBC \\ \angle QDA = \angle QDC - \angle ADC = \angle QDC - \angle QDB = \angle BDC \end{cases}$ so $\triangle QDA \sim \triangle BDC$ Then $\triangle NQA \sim \triangle LBC$ since $\begin{cases} \angle QDA = \angle BDC \\ QD : BD = AD : CD = AN : CL \end{cases}$ Hence, $\angle BLC = \angle QNA$ and $\angle QNA = \angle A_1AB$ so $A_1, L, A, B$ are concyclic In the same way, $B_1, L, A,B$ are concyclic Therefore, $A_1 = (ABL) \cap CD(\neq L \, \, \text{if} \, \, CD \not\parallel AB) = B_1$ Hence, $CD$, the tangent of circle $ANQ$ at point $A$, and the tangent to circle $BMP$ at point $B$ are concurrent

Trig that doesn't involve finding the concurrency point: Let the tangent from $A$ intersect $CD$ at $T_1$, and likewise construct $T_2$. We show $\frac{DT_1}{CT_1} = \frac{DT_2}{CT_2}$, which implies the problem since both $T_1,T_2$ lie in the interior of $CD$. Compute the first ratio as $\frac{AD}{AC}\frac{\sin DAT_1}{\sin CAT_1}$, angle chasing gives $\angle DAT_1 = \angle AQN, \angle AQD = \angle DAC $, thus we can compute $1 = \frac{\sin AQN}{\sin DQN} \frac{AQ}{QD} =\frac{\sin DAT_1}{\sin CAT_1} \frac{AQ}{QD} $, thus we can write the original ratio as $\frac{DQ}{QA}\frac{AD}{AC} = \frac{BD \cdot AD}{AC \cdot BC}$ since $\triangle DQA \sim \triangle DBC$ by angle chasing. By symmetry, we are done.

Let the mid-point of $CD$ be $S$, the intersection of the tangent of circle $ANQ$ at point $A$, and the tangent to circle $BMP$ at point $B$ is $T$ Notice that $\angle DAC=\angle DBC=\angle CPQ$,$\angle ADC=\angle BPC$ Which means that $\triangle DAC\sim \triangle BPC$ Hence $\triangle DAS \sim \triangle BPM$ Similarly $\triangle SBC \sim \triangle NQA$ Consider that $\angle SBT=\angle SBC-\angle TBC=\angle DAT-\angle DAS=\angle SAT$ Which means that $S,A,B,T$are cyclic Finally, $\angle TBA=\angle ABC-\angle TBC=180^{\circ}-\angle ABC+\angle TBC=\angle BMP=\angle DSA$ So $D-S-T-C$

Work on $\mathbb{RP}^2$. Let points $Q',P'$ be such that $QAQ'D,PBP'C$ are harmonic quadrilaterals. Since $\measuredangle DAQ'=\measuredangle DQQ'=\measuredangle NQA$ line $AQ'$ is tangent to $\odot (ANQ)$, and similarly $BP'$ is tangent to $\odot (BMP).$ But $$(AC,AD,AB,AQ')=(ADQQ')_{\odot (ADQ)}=-1=(CBPP')_{\odot (BCP)}=(BC,BD,BA,BP'),$$therefore $AQ',BP'$ are concur on $CD$ as desired.

Cute problem Let the tangent to $(ANQ)$ of $A$ intersect $(AQD)$ at $X$ and define the tangent to $B$ of $(BMP)$ hit $(BPC)$ at $Y$. Claim: $QAXD$ and $BPCY$ are harmonic. Proof: Angle chase to get \[\measuredangle AQM = \measuredangle XAD = \measuredangle XQD\]so $QX$ is a symmedian and the claim follows. $BOCY$ is the same thing. Now, let $AX$ hit $CD$ at $X’$, $BY$ hit $CD$ at $Y’$, and $AB\cap CD=E$. Then $-1=(AD;QX)\stackrel{A}{=}(CD;X’E)$ and similarly $-1=(CB;YP)\stackrel{B}{=}(CD;Y’E)$ so $X’=Y’$ as desired. Remark. That moment when $ABCD$ cyclic doesn’t matter

Let $X$ be the midpoint of $CD$. Notice that $\angle DBC = \angle AQD$ and $\angle QAD = \angle BCD$, so we receive $\triangle QAD$ is similar to $\triangle BCD$. Similarly, we get $\triangle PBC$ is similar to $\triangle ADC$. Now this means that $\angle NQA = \angle XBC$ and $\angle MPB = \angle XAD$ because of corresponding midpoints. Notice that if such a concurrency point were to exist, they must lie on $(ABX)$. We are motivated to introduce $(ABX) \cap CD = Z$. It suffices to prove that $\angle XBC = \angle ZAD$. Notice that \begin{align*} \angle XBC \\ = 180 - (\angle ZAB + 180 - \angle BAD) \\ = \angle BAD - \angle ZAB \\ = \angle ZAD \end{align*}

Let the tangent to $(BPM)$ at $B$ intersect $(BPC)$ at $R$, and define $S$ similarly for $Q$. Then $\angle RPC = \angle RBC = \angle BPM$, so $\overline{PR}$ is the $P$-symmedian in $\triangle PBC$ and hence $BPCR$ is harmonic. Similarly, $QASD$ is also harmonic. Now let $T = \overline{AB} \cap \overline{CD}$, and let $\overline{BR}$ and $\overline{AS}$ intersect $\overline{CD}$ at $P_1$ and $P_2$. Then \[ (T, P_1, D, C) \stackrel{B}{=} (P, R; B, C) = -1 = (Q, S; A, D) \stackrel{A}{=} (T, P_2; D, C), \]so $P_1 = P_2$ and we are done.

[asy][asy] import olympiad; size(8cm); pair A = (0,0); pair Q = (-10,0); pair D = (-4,-7); pair C = (75/14,-15); pair B = (11,0); pair N = (-2,-7/2); pair L = (-107/17,-147/34); pair K = (-25/17,70/17); pair O = (-92/13,133/26); pair X = (-195/16,0); pair Z = (-45/44,-105/11); draw(X--B--C--X); draw(Q--D--A); draw(Q--O, blue); draw(L--K, blue); draw(N--A, blue); draw(A--Z); draw(C--K); draw(circumcircle(A,D,Q)); draw(circumcircle(A,N,Q), red+linewidth(1)); dot("$A$", A, dir(50)); dot("$Q$", Q, dir(140)); dot("$D$", D, dir(260)); dot("$C$", C, dir(250)); dot("$B$", B, dir(60)); dot("$N$", N, dir(130)); dot("$L$", L, dir(30)); dot("$K$", K, dir(50)); dot("$O$", O, dir(120)); dot("$X$", X, dir(90)); dot("$Z_{A}$", Z, dir(220)); [/asy][/asy] Let $X$ be the intersection of $AB$ and $CD, O$ be the point on circle $AQN$ such that $OQ \parallel AD, K$ and $L$ be the intersections of $AC$ and $QD,$ respectively, with circle $AQN, Z_A$ be the point on $CD$ such that $AZ_A$ is tangent to circle $AQN$ and $Z_B$ to the point on $CD$ such that $BZ_B$ is tangent to circle $BPM.$ Suppose $P_{\infty}$ is the point at infinity on line $AD.$ Then we get $-1 = (A,D;N,P_{\infty}) \stackrel{Q}{=} (A,L;N,O).$ Since $KA$ is tangent to circle $AQD$ we get $\angle KAQ=\angle QDA = 180-\angle AQL - \angle QAN.$ Using arc lengths on circle $AQN$ we get ${KQ}+{AL}+{QN}=360$ so ${LN}={KA}$ and $LK \parallel NA.$ Therefore we have $O,K,A$ are reflections of $Q,L,N$ respectively over the perpendicular bisector of $AN,$ so $(A,L;N,O)=(N,K;A,Q) \stackrel{A}{=} (D,C;Z_A,X) = -1.$ Similarly we find $(D,C;Z_B,X) = -1,$ so $Z_A=Z_B$ and we are done.

[asy][asy] import olympiad; size(8cm); pair N = (0,0); pair D = (10,10); pair A = (20,20); pair Q = (8,20); pair G = (12,0); pair H = (24,0); pair C = (23,5); pair Z = (402/29,135/29); pair X = (49/3,20); draw(A--Q, green); draw(N--H, green); draw(A--N); draw(Q--N, blue); draw(A--G, blue); draw(Q--G, red); draw(A--H, red); draw(circumcircle(A,C,D)); dot("$N'$", N, dir(200)); dot("$D'$", D, dir(160)); dot("$A$", A, dir(50)); dot("$Q'$", Q, dir(140)); dot("$G$", G, dir(270)); dot("$H$", H, dir(290)); dot("$C'$", C, dir(330)); dot("$Z_A'$", Z, dir(20)); dot("$X'$", X, dir(100)); [/asy][/asy] Let $X$ be the intersection of $AB$ and $CD,$ and let $Z_A$ be the point on $CD$ such that $AZ_A$ is tangent to circle $AQN,$ and define $Z_B$ similarly. Let $C',D',N',Q',X',Z_A'$ be the images of points $C,D,N,Q,X,Z_A$ respectively under an inversion at $A.$ Let the line through $N'$ parallel to $AQ'$ intersect lines $AZ_A'$ and $AC'$ at points $G,H,$ respectively. Since $D'$ is the midpoint of $AN'$ we have that $D'$ is the center of parallelogram $AQ'N'G$ so $G$ is on $Q'D'$ and $AQ'GH$ is a parallelogram, so $N'G=GH=AQ'.$ If $P_{\infty}$ is the point at infinity on line $GH,$ we have $-1=(N',H;G,P_{\infty}) \stackrel{A}{=} (D',C';Z_A',X')=(D,C;Z_A,X).$ By symmetry we find $(D,C;Z_B,X)=-1$ as well, so $Z_A=Z_B.$

is so nice, like bro what! Alright, it might lead to a solution (maybe? ), but still I consider it as misleading. Anyways, here is my solution. [asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions (otherwise throws error due to the missing xmin) are done using bubu-asy.py. This adds back the dps, real xmin coordinates, changes the linewidth, the fontsize and adds the directions to the labellings. */ pair A = (-109.61704,78.69024); pair B = (-16.32261,78.97165); pair D = (-129.90808,30.64640); pair P = (31.77006,79.11672); pair C = (-29.22183,-27.44024); pair M = (-22.77222,25.76570); pair Y = (-212.64762,78.37946); pair X = (-98.60899,12.58972); pair K = (-42.61658,57.75982); import graph; size(15cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); pen ffxfqq = rgb(1,0.49803,0); draw(arc(P,15.71684,-179.82717,-163.98087)--P--cycle, linewidth(0.75) + blue); draw(arc(P,15.71684,-135.63258,-119.78628)--P--cycle, linewidth(0.75) + blue); draw(circle((-62.82441,30.62082), 67.08368), linewidth(0.5)); draw(circle((7.89564,22.04815), 61.86120), linewidth(0.5) + ffxfqq); draw(circle((7.81419,49.05191), 38.44185), linewidth(0.5) + blue); draw(D--A, linewidth(0.5)); draw(B--C, linewidth(0.5)); draw(Y--P, linewidth(0.5)); draw(Y--C, linewidth(0.5)); draw(A--X, linewidth(0.5) + red); draw(X--B, linewidth(0.5) + blue); draw(D--B, linewidth(0.5) + ffxfqq); draw(K--P, linewidth(0.5) + red); draw(P--M, linewidth(0.5) + red); draw(C--P, linewidth(0.5)); dot("$A$", A, NW); dot("$B$", B, N); dot("$D$", D, SW); dot("$P$", P, NE); dot("$C$", C, dir(270)); dot("$M$", M, W); dot("$Y$", Y, NW); dot("$X$", X, SW); dot("$K$", K, W); [/asy][/asy] Let $K$ be the intersection point of the tangent at $B$ to $\odot(BMP)$, with $\odot(BCP)$. Define $L$ for $A$ also similarly. Also let $Y=AB\cap CD$. Now, we have that,\[\measuredangle KPC=\measuredangle KBC=\measuredangle KBM=\measuredangle BMP.\]This gives us that $PK$ is the $P$-symmedian in $\triangle PBC$. This gives us that $(P,K;B,C)=-1$. Now, \[-1=(P,K;B,C)\overset{B}{=}(Y,BK\cap BC;D,C).\]Similarly we also get $(Y,AL\cap BC;D,C)$ which gives $AL\cap BC\equiv BK\cap BC$ and we are done.

Let $X$ be the point where the tangent to $(ANQ)$ at $A$ intersects $CD$ and $Y$ be the point where the tangent to $(BMP)$ at $B$ intersects $CD$. Note that both $X$ and $Y$ lie within $\overline{CD}$, so it suffices to show that $\tfrac{XD}{XC} = \tfrac{YD}{YC}$. We start with the first ratio. By angle chasing, we we find that $\angle DAX = \angle NQA$, $\angle XAC = \angle DQN$, and $\angle QDA = \angle BAC$. Therefore, from the law of sines, we find that $$\frac{XD}{XC} = \frac{\sin \angle DAX}{\sin \angle XAC} \cdot \frac{\sin \angle ACD}{\sin \angle ADC} = \frac{\sin \angle NQA}{\sin \angle DQN} \cdot \frac{\sin \angle ACD}{\sin \angle ADC} = \frac{\sin \angle QAD}{\sin \angle ADQ} \cdot \frac{\sin \angle ACD}{\sin \angle ADC} = \frac{\sin \angle BAD}{\sin \angle CAB} \cdot \frac{\sin \angle ACD}{\sin \angle ADC}. $$Analogous calculations show that $\tfrac{YC}{YD} = \tfrac{\sin \angle ABC}{\sin \angle ABD} \cdot \tfrac{\sin \angle CDB}{\sin \angle BCD} = \tfrac{\sin \angle ADC}{\sin \angle ACD} \cdot \tfrac{\sin \angle CAB}{\sin \angle BAD}$. Then $\tfrac{XD}{XC} \cdot \tfrac{YC}{YD} = 1$, so $\tfrac{XD}{XC} = \tfrac{YD}{YC}$, as needed.

Stared at this problem for 6 hours and then the solution came to me in 2 seconds no ggb Let the tangent to $(PBM)$ at $B$ intersect $(BCP)$ at $E$ and tangent to $(QAN)$ at $A$ meet $(AQD)$ at $F$ $$\angle EPC=\angle EBC=\angle BPM \implies \text{ AE,AM are isogonal }$$Thus , $$(D,C;BE \cap CD,AB \cap CD)\stackrel{B}=(BC;EP)=-1$$Similarly $(D,C;AF \cap CD,AB \cap CD)=-1$.Thus $BE \cap CD \equiv AF \cap CD$ $\blacksquare$

Is this problem supposed to be hard? Why is everyone using projective though? Back to doing geo again coz sharygin MarkBcc168 wrote: Angle chasing let's goo [asy][asy] size(9cm); import olympiad; import geometry; defaultpen(fontsize(10pt)); pair A = dir(230); pair B = dir(310); pair C = dir(30); pair D = dir(170); pair M = (B+C)/2; pair N = (A+D)/2; pair T = (C+D)/2; pair X = 2 * foot(circumcenter(A,B,T), C,D) - T; pair C_prime = reflect((0,0), (A+B)/2) * C; pair D_prime = reflect((0,0), (A+B)/2) * D; pair Q = extension(D, C_prime, A, B); pair P = extension(C, D_prime, A, B); fill(A--D--Q--cycle, 0.5*paleblue+0.5*white); fill(B--D--C--cycle, 0.5*paleblue+0.5*white); markscalefactor = 0.015; draw(anglemark(A,Q,D), linewidth(0.7)); draw(anglemark(C,A,D), linewidth(0.7)); draw(anglemark(C,B,D), linewidth(0.7)); draw(A--D--Q--cycle, blue+linewidth(0.7)); draw(B--D--C--cycle, blue+linewidth(0.7)); draw(B--T, blue+linewidth(0.7)); draw(Q--N, blue+linewidth(0.7)); draw(B--P--C--A, black); draw(A--B--C--D--cycle, linewidth(1)); draw(circle((0,0),1), gray+linewidth(0.8)); draw(circumcircle(A,B,T), red+linewidth(1)); dot("$A$", A, dir(-119)); dot("$B$", B, dir(-62)); dot("$C$", C, dir(29)); dot("$D$", D, dir(164)); dot("$M$", M, dir(-10)); dot("$N$", N, dir(166)); dot("$T$", T, dir(98)); dot("$X$", X, dir(100)); dot("$Q$", Q, dir(-145)); dot("$P$", P, dir(-36)); [/asy][/asy] Let $T$ be the midpoint of $CD$, and let $\odot(ABT)$ intersects $CD$ again at $X$. We claim that $X$ is the concurrency point. It suffices to show that $XA$ is tangent to $\odot(NQA)$. Notice that since $\measuredangle AQD = \measuredangle CAD = \measuredangle CBD$ and $\measuredangle QAD = \measuredangle BAD = \measuredangle BCD$, it follows that $\triangle DQA\sim\triangle DBC$, implying that $\measuredangle NQA = \measuredangle TBC$. Now, the problem follows from \begin{align*} \measuredangle XAN &= \measuredangle XAB + \measuredangle BAD \\ &= \measuredangle XTB + \measuredangle BCD \\ &= \measuredangle BCD + \measuredangle DTB \\ &= \measuredangle CBT \\ &= \measuredangle AQN, \end{align*}done. This is basically my solution, except that I didn't realise it was $\odot(ABT) \cap CD$, which made me use ratio lemma, except that I didn't really do it and just saw the sol and confirmed my answer xD

[asy][asy] size(15cm); pair A, B, C, D, M, N, P, Q, X, Z; A=dir(-150); B=dir(-30); C=dir(15); D=dir(135); M=(B+C)/2; N=(A+D)/2; X=(C+D)/2; P=B-(M-B)*(D-A)/(X-D); Q=A-(N-A)*(C-B)/(X-C); Z=intersectionpoints(circumcircle(A,B,X),C--D)[0]; draw(circumcircle(B,C,P)); draw(circumcircle(A,D,Q)); draw(circumcircle(B,M,P)); draw(circumcircle(A,N,Q)); draw(A--B--C--D--A--P--Q--A--C--B--D--A--Z--B--X--A); draw(circle((0,0),1)); label("$A$", A, S); label("$B$", B, S); label("$C$", C, NNE); label("$D$", D, dir(90)); label("$M$", M, NW); label("$N$", N, SW); label("$P$", P, SE); label("$Q$", Q, SW); label("$X$", X, dir(90)); label("$Z$", Z, dir(90)); [/asy][/asy] Let $X$ be the midpoint of $CD$. We have $$\angle CPB=\angle CBD=\angle CAD$$and $$\angle CBP=180^{\circ}-\angle ABC=\angle CDA$$so $\triangle CPB\sim\triangle CAD$. Therefore, $\triangle MPB\sim\triangle XAD$, so if the tangents to the circumcircle of $ANQ$ at $A$ and the circumcircle of $BMP$ at $B$ intersect at $Z$, then $\angle ZBC=\angle MPB=\angle XAD$. Similarly, $\angle ZAD=\angle XBC$, so $\angle XBZ=\angle XAZ$. Therefore, we get $ABZX$ is cyclic. Now, we have \begin{align*} \angle DXA+\angle AXZ&=\angle BMP+180^{\circ}-\angle ZBA\\ &=\angle BMP+\angle PBZ\\ &=180^{\circ}. \end{align*}Therefore, $Z$ lies on $CD$.

Let $O$ be the midpoint of $CD$, and $P\neq O$ be on $CD$ such that $A$, $B$, $O$, $P$ are concyclic. Since \[\angle DQA = \angle DAC = \angle DBC\]\[\angle DAQ = \angle DAB = \angle DCB\]we have $\triangle DQAN\sim\triangle DBCO$. Now, \[\angle PAN=\angle BAD-\angle BAP=\angle BOP-\angle BCD=\angle OBC=\angle AQN\]so $AP$ tangent to $(ANQ)$, and we're done.

very surprising configuration but also how is $AQ=BP$ literally useless (even in trig solutions) and unrelated to everything else Let $G$ be the midpoint of $\overline{CD}$ and $X$ be the intersection of the tangents to $(ANQ)$ and $(BMP)$. Observe that $\measuredangle CBP=\measuredangle CBA=\measuredangle CDA$ and $\measuredangle BPC=\measuredangle DBC=\measuredangle DAC$, hence $\triangle CDA \sim \triangle CBP$. Likewise, $\triangle DCB \sim \triangle DAQ$. This implies that $\triangle AQN \sim \triangle CBG$ and $\triangle BPM \sim \triangle DAG$. Then, $$\measuredangle AGB=\measuredangle DGA+\measuredangle BGC=\measuredangle BMP+\measuredangle QNA=\measuredangle XBA+\measuredangle BAX=\measuredangle AXB,$$hence $ABGX$ is cyclic. To finish, note that $\measuredangle AGD=\measuredangle PMB=\measuredangle ABX=\measuredangle AGX$, hence $G,X,D$ are collinear, so $X$ lies on $\overline{CD}$. $\blacksquare$ Remark: i did this problem at mop but when i went home and decided to do some writeups i realized i had mostly forgotten my solution and i couldnt seem to find out where i had written it down so here we are about one month late. adding the midpoint is motivated after you get the two similarities, since they relate $\overline{DA}$ and $\overline{BC}$ to $\overline{CD}$ and we have the midpoints of the first two segments. Then you squint at the diagram and notice something looks concyclic

Okay, this is kinda amazing that I solved this one on my own. Very cool problem and a nice application of using similar triangles. I found the main idea in like 20-25 minutes, and later was figuring out details. And yes, I did geogebra, but well... I didn't need geogebra to guess symmedians, so its fine really. Here's a short solution, pretty much same as everyone else. Solution: Let $X$ be the midpoint of $\overline{CD}$ and define $K$ and $L$ as second intersections of tangents of $\odot(QAN)$ and $\odot(BPM)$ at $A$ and $B$ respectively with $\odot(ABC)$. The major claim is to show that $BK$ and $AL$ are $B$ and $A$ symmedians of $\triangle BDC$ and $\triangle ADC$. [asy][asy] import graph; size(300); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.6) + fontsize(11); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.0, xmax = 2.0, ymin = -1.6, ymax = 1.6; /* image dimensions */ pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen ubqqys = rgb(0.29411764705882354,0.,0.5098039215686274); pen dcrutc = rgb(0.8627450980392157,0.0784313725490196,0.23529411764705882); draw((-0.15488924714484542,0.9879318403204257)--(0.5825410032404356,0.8128013161552009)--(0.9003580058024234,-0.435149929779936)--(-0.9135099046553886,-0.4068164870018209)--cycle, linewidth(0.6) + qqwuqq); /* draw figures */ draw(circle((0.,0.), 1.), linewidth(0.6) + green); draw((-0.15488924714484542,0.9879318403204257)--(0.5825410032404356,0.8128013161552009), linewidth(0.6) + qqwuqq); draw((0.5825410032404356,0.8128013161552009)--(0.9003580058024234,-0.435149929779936), linewidth(0.6) + qqwuqq); draw((0.9003580058024234,-0.435149929779936)--(-0.9135099046553886,-0.4068164870018209), linewidth(0.6) + qqwuqq); draw((-0.9135099046553886,-0.4068164870018209)--(-0.15488924714484542,0.9879318403204257), linewidth(0.6) + qqwuqq); draw(circle((1.0310883014813292,0.2625882977567536), 0.7098795985055039), linewidth(0.6) + ubqqys); draw((0.5825410032404356,0.8128013161552009)--(-0.9135099046553886,-0.4068164870018209), linewidth(0.6) + blue); draw(circle((-0.8579087183071474,0.46662703458172117), 0.875211447205526), linewidth(0.6) + ubqqys); draw((-0.15488924714484542,0.9879318403204257)--(0.9003580058024234,-0.435149929779936), linewidth(0.6) + blue); draw((-0.9135099046553886,-0.4068164870018209)--(-1.2514736160331192,1.2483570062455538), linewidth(0.6) + blue); draw((0.9003580058024234,-0.435149929779936)--(1.6791253721287096,0.5523761502300726), linewidth(0.6) + blue); draw(circle((-0.7628771155996663,0.8667810277998146), 0.619940938769687), linewidth(0.6) + dcrutc); draw(circle((1.1150660977053701,0.616197513463053), 0.5676583756696456), linewidth(0.6) + dcrutc); draw((-0.534199575900117,0.29055767665930243)--(-1.2514736160331192,1.2483570062455538), linewidth(0.6) + blue); draw((0.7414495045214295,0.18882569318763245)--(1.6791253721287096,0.5523761502300726), linewidth(0.6) + blue); draw((-0.006575949426482597,-0.42098320839087844)--(0.5825410032404356,0.8128013161552009), linewidth(0.6) + blue); draw((-1.2514736160331192,1.2483570062455538)--(-0.15488924714484542,0.9879318403204257), linewidth(0.6) + blue); draw((0.5825410032404356,0.8128013161552009)--(1.6791253721287096,0.5523761502300726), linewidth(0.6) + blue); draw((-0.15488924714484542,0.9879318403204257)--(0.2356258109427729,-0.9718438543395543), linewidth(0.6) + blue); draw((-0.08538187571909811,-0.9963483001936062)--(0.5825410032404356,0.8128013161552009), linewidth(0.6) + linetype("4 4") + blue); /* dots and labels */ dot((-0.15488924714484542,0.9879318403204257),linewidth(3pt)+dotstyle); label("$A$", (-0.12861172585944775,1.0530544782066107), NE * labelscalefactor); dot((0.5825410032404356,0.8128013161552009),linewidth(3.pt) + dotstyle); label("$B$", (0.5538185261959065,0.9599958074717898), NE * labelscalefactor); dot((0.9003580058024234,-0.435149929779936),linewidth(3.pt) + dotstyle); label("$C$", (0.8764219180766195,-0.5909820381085583), NE * labelscalefactor); dot((-0.9135099046553886,-0.4068164870018209),linewidth(3.pt) + dotstyle); label("$D$", (-1.0343827876783727,-0.5351468356676659), NE * labelscalefactor); dot((-1.2514736160331192,1.2483570062455538),linewidth(3.pt) + dotstyle); label("$Q$", (-1.3445783567944427,1.3322304904110733), NE * labelscalefactor); dot((1.6791253721287096,0.5523761502300726),linewidth(3.pt) + dotstyle); label("$P$", (1.701542131925366,0.6001689472971491), NE * labelscalefactor); dot((-0.534199575900117,0.29055767665930243),linewidth(3.pt) + dotstyle); label("$N$", (-0.60,0.14), NE * labelscalefactor); dot((0.7414495045214295,0.18882569318763245),linewidth(3.pt) + dotstyle); label("$M$", (0.7585476018125129,0.28376946679875803), NE * labelscalefactor); dot((-1.0627745434602023,0.3242054106502089),linewidth(3.pt) + dotstyle); label("$E$", (-1.2205001291480146,0.20932253021090133), NE * labelscalefactor); dot((1.3086247791158379,0.08255805810153108),linewidth(3.pt) + dotstyle); label("$F$", (1.3479191831330462,-0.06364957061123995), NE * labelscalefactor); dot((-0.006575949426482597,-0.42098320839087844),linewidth(3.pt) + dotstyle); label("$X$", (-0.08,-0.6095937722555225), NE * labelscalefactor); dot((0.2356258109427729,-0.9718438543395543),linewidth(3.pt) + dotstyle); label("$K$", (0.25603077984447925,-1.1121105942235554), NE * labelscalefactor); dot((-0.08538187571909811,-0.9963483001936062),linewidth(3.pt) + dotstyle); label("$L$", (-0.15342737138873336,-1.15), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Observe that $\triangle DQA \sim \triangle DBC$ and $\triangle CPB \sim \triangle CDA$. By symmetry, we only need to show the result for first, which follows from \[\measuredangle AQP = \measuredangle DAC \qquad \text{and} \qquad \measuredangle BCD = \measuredangle DAQ.\]Finally, since $BX$ and $QN$ are medians, we also get that $\triangle BXC \sim \triangle QNA$. Finally, \[\measuredangle DBK = \measuredangle DAK = \measuredangle NQA = \measuredangle XBC\]which shows that $BK$ is the $B-$symmedian in $\triangle BDC$. The other one follows by symmetry. We can now finish, say $AB \cap CD = T$, then \begin{align*} -1 = (K,B;C,D) \overset{A}{=} (AK \cap CD, T;C,D) \\ -1 = (L,A;C,D) \overset{B}{=} (BL \cap CD,T;C,D). \end{align*}From the above two equations, we can easily conclude that $AK \cap CD = BL \cap CD$ which is what had to be proven. $\blacksquare$

Does $AQ = BP$ count as a jumpscare? Let $K = QD \cap PC$. Notice $\angle KQP = \angle EAC = \angle EBC = \angle CPQ$ so $\triangle KPQ$ is isoceles. Let $C', D'$ be the reflections of $C, D$ over the angle bisector of $\angle PKQ$ respectively, then $\angle DC'C = \angle DQP = \angle DAC$ so $(DD'CBAC')$ cyclic by symmetry. By symmetry, it follows that $AQ = BP$, but we don't even care! Why? Because once you have enough structure, ratio lemma kisses you goodnight It suffices to show that the condition's tangent at $A$ intersects $CD$ into the same ratios as the tangent at $B$. Hence by ratio lemma in $\triangle AQD, \triangle BPC, \triangle ACD, \triangle BDC$, it suffices to show that: $\frac{DQ}{AQ} \cdot \frac{AD}{AC} \cdot \frac{CP}{BP} \cdot \frac{BC}{BD} = 1$ Now by ratio lemma in $\triangle DBQ, \triangle CAP$ we have that $\frac{DQ}{DB} \cdot \frac{C'A}{AB} = \frac{QA}{AB}, \frac{CP}{CA} \cdot \frac{BD'}{AB} = \frac{BP}{AB}$, so substituting in we obtain: $\frac{AD}{BD'} \cdot \frac{BC}{C'A} = 1 \cdot 1 = 1$ by symmetry over the angle bisector of $\angle PKQ$, finishing.

Let $P$ be the midpoint of $CD$ and suppose that the tangent from $A$ to $(ANQ)$ meets $CD$ at $Q$. We claim that the tangent from $B$ to $(BMP)$ meets $CD$ at $Q$, too. Observe that a spiral similarity centered at $D$ maps $DQA$ to $DBC$, by some simple angle chasing. Thus $\angle DQN = \angle DBP$. Moreover, as $\angle DAQ = \angle NQA$, we obtain $\angle DQN = \angle QAC$. Letting $BD \cap AQ = X$ and $AC \cap BP = Y$, we note that $A, X, Y, B$ are concyclic. Thus $XY$ is parallel to $DC$, and thus $A, Q, P, B$ are concyclic. This process is entirely symmetric, so if we’d begun by letting $Q’$ be the tangent from $B$ to $(BMP)$, we’d have found that $A, Q’, P, B$ were concyclic. Thus $Q=Q’$ and so the three lines desired concur at $Q$.

Different perspective [asy][asy] import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.3) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -20.21411278121972, xmax = 11.28361707838571, ymin = -6.476743372939919, ymax = 9.63388721374956; /* image dimensions */ pen ffttww = rgb(1,0.2,0.4); /* draw figures */ draw(circle((-4.179971840262307,-0.7536982022803161), 3.1800576626014823), linewidth(1)); draw((-2.34,1.84)--(-4.971719771058439,2.3262209446374156), linewidth(1)); draw((-4.971719771058439,2.3262209446374156)--(-7.36,-0.74), linewidth(1)); draw((-7.36,-0.74)--(-1.0065685673333689,-0.5480810706149173), linewidth(1)); draw((-2.34,1.84)--(-1.0065685673333689,-0.5480810706149173), linewidth(1)); draw((-2.34,1.84)--(-7.36,-0.74), linewidth(1)); draw((-4.971719771058439,2.3262209446374156)--(-1.0065685673333689,-0.5480810706149173), linewidth(1)); draw((-1.1492306869269697,7.233767365857066)--(-2.34,1.84), linewidth(1)); draw((-1.1492306869269697,7.233767365857066)--(-4.971719771058439,2.3262209446374156), linewidth(1)); draw((-11.182489084131468,-5.6475464212196504)--(-1.0065685673333689,-0.5480810706149173), linewidth(1)); draw((-7.36,-0.74)--(2.985276659696818,4.576895175700756), linewidth(1)); draw((-1.0065685673333689,-0.5480810706149173)--(2.985276659696818,4.576895175700756), linewidth(1) + ffttww); draw((-7.36,-0.74)--(-11.182489084131468,-5.6475464212196504), linewidth(1)); draw((-3.6558598855292193,2.0831104723187077)--(-1.1492306869269697,7.233767365857066), linewidth(1)); draw((-4.183284283666684,-0.6440405353074586)--(-11.182489084131468,-5.6475464212196504), linewidth(1)); draw((0.45176013380082797,1.3242109466268193)--(-1.0065685673333689,-0.5480810706149173), linewidth(1) + ffttww); draw((-4.971719771058439,2.3262209446374156)--(-0.2774042167662705,0.388064938005951), linewidth(1)); draw((-7.36,-0.74)--(0.9893540461817245,2.014407052542919), linewidth(1)); draw((-4.971719771058439,2.3262209446374156)--(0.45176013380082797,1.3242109466268193), linewidth(1)); /* dots and labels */ dot((-4.971719771058439,2.3262209446374156),dotstyle); label("$A$", (-4.875248929221975,2.5673980492285757), NE * labelscalefactor); dot((-7.36,-0.74),dotstyle); label("$B$", (-7.262902264674453,-0.4955511790791545), NE * labelscalefactor); dot((-2.34,1.84),dotstyle); label("$C$", (-2.2464184891783368,2.085043840046256), NE * labelscalefactor); dot((-1.0065685673333689,-0.5480810706149173),dotstyle); label("$D$", (-0.9199444139269598,-0.3026094954062266), NE * labelscalefactor); dot((-3.6558598855292193,2.0831104723187077),linewidth(4pt) + dotstyle); label("$N$", (-3.548774853970598,2.2779855237191837), NE * labelscalefactor); dot((-4.183284283666684,-0.6440405353074586),linewidth(4pt) + dotstyle); label("$M$", (-4.0793644840711485,-0.44731575816092256), NE * labelscalefactor); dot((2.985276659696818,4.576895175700756),linewidth(4pt) + dotstyle); label("$Y$", (3.0835955222862874,4.76210970100813), NE * labelscalefactor); dot((0.45176013380082797,1.3242109466268193),linewidth(4pt) + dotstyle); label("$X$", (0.551235924079113,1.5062187890274723), NE * labelscalefactor); dot((-1.1492306869269697,7.233767365857066),linewidth(4pt) + dotstyle); label("$Q$", (-1.0646506766816555,7.4150578515108885), NE * labelscalefactor); dot((-11.182489084131468,-5.6475464212196504),dotstyle); label("$P$", (-11.09761822767389,-5.415564112738815), NE * labelscalefactor); dot((-0.2774042167662705,0.388064938005951),linewidth(4pt) + dotstyle); label("$K$", (-0.1722953896943654,0.5897457915810648), NE * labelscalefactor); dot((0.9893540461817245,2.014407052542919),linewidth(4pt) + dotstyle); label("$L$", (1.0818255541796638,2.205632392341836), NE * labelscalefactor); dot((-1.904353708674545,1.0597884073293715),linewidth(4pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let $X$ be a point such that $\triangle ADX\sim \triangle QCA$, and $Y$ be a point such that $\triangle BDY\sim \triangle PBD$ Let $K,L$ be the midpoint of $XD,YD$ respectively. By angle chasing, $XD//AB,YD//AB\rightarrow Y,L,X,K,D$ collinear Also it is easy to see that the tangent of circle $ANQ$ at point $A$ is $AK$, and the tangent to circle $BMP$ at point $B$ is $BL$. Let $BL\cap CD=F_1$, $AK\cap CD=F_2$ Hence, by menelaus theorem it is easy that $F_1=F_2$, so the problem is solved$\blacksquare$

Here are my solutions: Proof 1 (Spiral Similarity): Let $T$ denote the midpoint of $CD$ and $X = (ABT) \cap CD$. We claim that $X$ is the intersection of tangents at $A, B$ in $(AQN), (BPM)$ respectively. Notice that a spiral similarity with center at $D$ maps $QA$ to $BC$. Thus, $\triangle ADQ \sim \triangle CDB$. Now, notice that: $$\measuredangle XAD = \measuredangle DAB + \measuredangle BAX = \measuredangle DCB + \measuredangle BTC = \measuredangle TBC = \measuredangle NQA$$Hence, we conclude the given statement. Proof 2 (Harmonic Bundles): Let the tangent to $(AQN)$ at $A$ meet $(ADQ)$ at $R$ and similarly, we define $S$. Notice that, $QR$ is the $Q$ - symmedian of $\triangle AQD$. Thus, $AQDR$ is harmonic. Similarly, $BPCS$ is also harmonic. Notice that: $$-1 = (AD; RQ) \overset{A}{=} (CD; (AR \cap CD) (AB \cap CD))$$$$-1 = (BC; SP) \overset{B}{=} (CD; (BS \cap CD) (AB \cap CD))$$Thus $AR \cap BC = BS \cap BC$ and we conclude. Proof 3 (Angle-Chase): Note that: $$\angle DQA = \angle DAC = \angle DBC, \angle DAQ = angle DCB$$which implies $\triangle DAQ \sim \triangle DCB$. Let $T$ denote the midpoint of $CD$ and $X = (ABT) \cap CD$. We claim that $X$ is the intersection of tangents at $A, B$ in $(AQN), (BPM)$ respectively. Notice that: $$\angle DAX = \angle DAB -\angle XAB = \angle XTB - \angle XCB = \angle TBC = \angle NQA$$which implies $XA$ is tangent to $(AQN)$ at $A$ and we are done.

Let $H$ be the midpoints of $CD$, and $G$ be the intersection of the the tangents from $A$ and $B$ on $(AMQ),(BNP)$. We can say: \[\angle ADC=\angle CBP,\]\[\angle DAC=\angle BPC,\]so: \[\triangle DAC\sim \triangle BPC.\]Using a similar approach: \[\triangle BDC\sim \triangle QDA.\] As a result, we can say: \[\angle DAH=\angle GBC=\angle BPC,\]\[\angle CBH=\angle DAG=\angle DQA,\]so $HGBA$ is cyclic. Therefore: \[\angle HGB=\angle QAD+\angle DAH=180-\angle CGB,\]implying that $G$ lies on $CD$, as desired $\blacksquare$

xooks harmonics are >>>> Let the tangent to $(BMP)$, at $B$ intersect $CD$ at $X$, and let the tangents to $(ABC)$, at $C$, and $D$, intersect at $K$, and let $J= AK \cap (ABCD) \neq A$. Now note that $AK$ is the $A$-symmedian of $\triangle ADC$ $$\measuredangle PBX= \measuredangle PMB$$$$\measuredangle PBJ = \measuredangle PBM+ \measuredangle CBJ = \measuredangle PBM + \measuredangle CAJ$$By spiral similarity $\triangle ADC \sim \triangle PBC$ So by symmedian properties $\measuredangle MPB= \measuredangle CAJ$ $$\measuredangle MBP+ \measuredangle JAC+ \measuredangle PMB= \measuredangle MPB + \measuredangle JAC =0^{\circ}$$So $B,X,J$ are collinear, which lets us harmonic chase $$-1 = (A,J; C,D) = (AB \cap BC, X; C, D)$$Let $Y$ be the point where the tangent to $(BMP)$ at $B$ intersects $BC$, by similar reasons $$-1= (B, J; C, D)= (AB \cap BC, Y ; C, D)$$$$(AB \cap BC, Y ; C, D)=(AB \cap BC, X ; C, D) $$$$X=Y$$And we are finished

Let $L$ be the midpoint of $BC$ and $T$ be the point where $(ABL)$ intersects $CD$. We claim that $T$ is indeed the point of concurrency. Note that, basic angle chasing gives that $\triangle AQD \sim \triangle BCD$ and $\triangle BPC \sim \triangle ADC.$ So, \[\measuredangle TBC = \measuredangle TBP - \measuredangle CBP = \measuredangle TLA - \measuredangle CBP = \measuredangle PMC - \measuredangle CBP = \measuredangle MPB,\] which gives us that $TB$ is tangent to $(BMP).$ Similarly, \[\measuredangle TAN = \measuredangle TAD = \measuredangle TAL + \measuredangle LAD = \measuredangle MPB + \measuredangle TBL = \measuredangle CBT + \measuredangle TBL = \measuredangle CBL = \measuredangle AQN.\] Thus, $TA$ is tangent to $(ANQ)$, as we claimed.

I thought this was perfect for G3; not as easy as everyone else claims it is. Let $X$ be the midpoint of $\overline{BC}$. As a spiral similarity at $C$ takes $CBR$ to $CDA$, it takes $BMR$ to $DXA$. Now set $E = \overline{BB} \cap \overline{CD}$. As $\angle CBE = \angle XAD$, $\angle ABE + \angle AXE = 180^\circ$, hence $ABEX$ is cyclic. But for $E' = \overline{AA} \cap \overline{CD}$, $ABE'X$ is also cyclic, so $E'=E$ and the result follows.

Lessgo $10$ minute solve. Define $X$ on $(ADQ)$ such that $QAXD$ is harmonic. Reflect $X$ about $N$ to a point $X'$. Claim: $\triangle XAX' \sim \triangle AQD$. Note that $$\frac{AX}{AX'} = \frac{XA}{XD} = \frac{QA}{QD}$$Also we have $$\angle X'AX = \angle DAX + \angle X'AD = \angle DAX + \angle ADX = \angle AQD$$The conclusion then follows. $\square$ Now note that by construction $N$ is the midpoint of $\overline{XX'}$. Hence we find $\angle DAX = \angle NQA$, which implies $\overline{AX}$ is tangent to $(ANQ)$. Now projecting, \begin{align*} -1 = (XQ, AD) \overset{A}{=} (\overline{AX} \cap \overline{CD}, \overline{AB} \cap \overline{CD};C, D) \end{align*}However we may repeat a symmetric argument for $B$. Namely let $Y$ be the point on $(BCP)$ such that $BYCP$ is harmonic. Then we find, \begin{align*} -1 = (\overline{BY} \cap \overline{CD}, \overline{AB} \cap \overline{CD}; C, D) \end{align*}Thus $\overline{AX} \cap \overline{CD} \equiv \overline{BY} \cap \overline{CD}$ and the result follows.

My solution, and my skills 1. Call G the tangent at A of (AQN) with BC. 2. To prove the problem, we need to show that GB is tangent to (BMP) or GBC = BPM. This is equivalent to sin GBC / sin (DBC - GBC) = sin BPM / sin (BPC - BPM) (since BPC = DBC) Or sin GBC / sin DBG = sin BPM / sin CPM 3. Firstly, we show the equal angles: +) DAG = AQN (because AG is tangent to AQN) +) GAC = DAC - DAG = DQA - AQN = DQN +) QDA = DAB - DQA = DAB - DAC = BAC 4. Then sin DAG · AD / sin GAC · AC = DG / GC <=> sin AQN / sin DQN × AD / AC = DG / GC 5. Note that sin AQN / sin DQN = QD / QA = sin QAD / sin QDA = sin DAB / sin BAC = BD / BC 6. So, DG / GC = BD / BC × AD / AC 7. On the other hand, sin DBG / sin GBC = DG / GC × BC / BD = AD / AC 8. Continuing with the above idea, we have +) BCP = ABC - BPC = ABD 9. Then sin CPM / sin BPM = PB / PC = sin ABD / sin ABC = AD / AC = sin DBG / sin GBC 10. So we have proved the problem.

Sketch: Let $Y$ be the midpoint of $CD$. We claim that $X=(ABY)\cap CD$ is the desired intersection point. This follows by angle chasing and similar triangles.

Let $S$ be the midpoint of $CD$, and let $T=(ABS)\cap CD$. We claim that $T$ is the desired point. The key claim is $AQD\sim CBD$, which follows by angles. From here, we have \[\angle TAD = \angle BAD - \angle BAT = \pi - \angle BCD - \angle BSC = \angle CBS = \angle AQN,\]so $TA$ is tangent to $(AQN)$, as desired.

Let $E$ be the midpoint of $CD$, and let $CD$ intersect $(ABM)$ again at $X$. We will prove that $X$ is the desired concurrency point. Claim: $\triangle DAQ \sim \triangle DCB$ and $\triangle CBP \sim \triangle CDA$. Proof. We have $\measuredangle QAD = \measuredangle CAD = \measuredangle CBE$ and $\measuredangle CAQ = \measuredangle DCB$, so $\triangle DAQ \sim \triangle DCB$. Similarly, $\triangle CBP \sim \triangle CDA$. $\square$ Now let $X'$ be the point where the tangent of $(ANQ)$ at $A$ meets $CD$. Then $\measuredangle AQN = \measuredangle DAX' = \measuredangle LBC$, so $$\measuredangle BAX' = \measuredangle BAD + \measuredangle DAX' = \measuredangle BCD + \measuredangle LBC = \measuredangle BLX'.$$Therefore, $AX'LB$ is cyclic, so $X'=X$ and $X$ lies on the tangent to $(ANQ)$ at $A$. Similarly, $X$ lies on the tangent to $(BMP)$ at $B$, so the three lines concur at $X$, as desired.

I thought people will use $AQ=BP$.

New solution + overcomplicated + Pascal's theorem is the best theorem Let the two tangents to $ANQ$ at $A$ and $BMP$ at $B$ meet at $R$ and let $S$ be the midpoint of $CD$. Let $AS \cap BC = S_a, BS \cap AD = S_b, AR \cap BC = R_a, BR \cap AD = R_b$. We want to show that $R$ lies on $CD$. Claim: Triangles $DAQ$ and $DCB$ are similar. Analogously, $CDA$ and $CBP$ are similar. Proof: $\angle CBD = \angle CAD = \angle AQD$ from tangency condition and the fact that $ABCD$ is cyclic, and $\angle QAD = \angle DCB$ because $ABCD$ is cyclic, done. Claim: $ABR_aS_b$ is cyclic and analogously, $ABS_aR_b$ is cyclic. Proof: Due to similarity of triangles $DCB$ and $DAQ$, their medians make same angles with corresponding sides, so $\angle SBC = \angle NQA$, but as $\angle NQA = \angle NAR = \angle S_bAR_a$ and $\angle SBC = \angle S_bBR_a$, the conclusion follows. Note that $S_bR_a, R_bS_a, CD$ are parallel from the fact that $ABR_aS_b, ABS_aR_b, ABCD$ are cyclic ( this is just two step angle chasing). Now we apply Pascal/Pappus theorem on hexagon $AS_aR_bBS_bR_a$, from which it follows that $AS_a \cap BS_b = S, S_aR_b \cap S_bR_a, R_bB \cap R_aA = R$ are collinear. But $S_aR_b \cap S_bR_a$ is the point at infinity of lines $CD, S_bR_a, S_aR_b$ so it lies on $CD$. As $S$ also lies on $CD$, $R$ must lie on $CD$, which is what we wanted.