2016 China Mathematical Olympiad Q2

Note that $\frac{AN}{AK}=\frac{\frac{AB*AD}{AB+DC}}{\frac{AB*AD}{BC+AD}}=\frac{AB+DC}{BC+AD}=\frac{BL}{BK}=\frac{CL}{CM}=\frac{DN}{DM}$. Suppose $AN\neq AK$. WLOG $AN>AK$ Then $\angle NKA>\angle KNA$. Similarly $\angle BKL > \angle BLK$, $\angle LMC>\angle MLC$, $\angle DMN>\angle DNM$. Since $\angle MNK = 180^o - \angle DNM - \angle KNA$ and so on, $\angle MNK + \angle MLK < \angle NML + \angle NKL$, which is patently false. Hence $AN = AK$ and so on, which implies $AB+DC = BC + AD$, which implies $ABCD$ is tangential. Now let $G, H, I, J$ be the tangency points of the inscribed circle of $ABCD$ on $AB, BC, CD, AD$ respectively. $AF-AE = FJ-EG = FH-EI = CF-CE$, which implies $TF-SE=VF-UE$. However, since $SE+TF = VF+UE = FE$, $TF=VF,UE=SE$. $\angle TVU = \pi - \angle FVT + \angle CVU = \pi + \angle AFB/2 - \angle DCB/2 = \pi - \angle FDE/2 = \pi - \frac{\angle A + \angle DEA}{2} = \pi - \angle UST$, which implies $SUVT$ concyclic, and we are done.

joyce_tan wrote: Note that $\frac{AN}{AK}=\frac{\frac{AB*AD}{AB+DC}}{\frac{AB*AD}{BC+AD}}=\frac{AB+DC}{BC+AD}=\frac{BL}{BK}=\frac{CL}{CM}=\frac{DN}{DM}$. Suppose $AN\neq AK$. WLOG $AN>AK$ Then $\angle NKA>\angle KNA$. Similarly $\angle BKL > \angle BLK$, $\angle LMC>\angle MLC$, $\angle DMN>\angle DNM$. Since $\angle MNK = 180^o - \angle DNM - \angle KNA$ and so on, $\angle MNK + \angle MLK < \angle NML + \angle NKL$, which is patently false. Hence $AN = AK$ and so on, which implies $AB+DC = BC + AD$, which implies $ABCD$ is tangential. Now let $G, H, I, J$ be the tangency points of the inscribed circle of $ABCD$ on $AB, BC, CD, AD$ respectively. $AF-AE = FJ-EG = FH-EI = CF-CE$, which implies $TF-SE=VF-UE$. However, since $SE+TF = VF+UE = FE$, $TF=VF,UE=SE$. $\angle TVU = \pi - \angle FVT + \angle CVU = \pi + \angle AFB/2 - \angle DCB/2 = \pi - \angle FDE/2 = \pi - \frac{\angle A + \angle DEA}{2} = \pi - \angle UST$, which implies $SUVT$ concyclic, and we are done. it's been a while from the answer but isn't the nominator and denominator here $$\frac{AN}{AK}=\frac{\frac{AB*AD}{AB+DC}}{\frac{AB*AD}{BC+AD}}=\frac{AB+DC}{BC+AD}$$ has been changed....? $\frac{AN}{AK}=\frac{\frac{AB+DC}{AB*AD}}{\frac{BC+AD}{AB*AD}}$ would be right...

It is much easier to show ABCD is tangential using the lemma below http://artofproblemsolving.com/community/c6h487126p2729541 here posted by Luis Gonzalez

Plus, <joyce_tan> wrote: WLOG $AN>AK$ Then $\angle NKA>\angle KNA$. Similarly $\angle BKL > \angle BLK$, $\angle LMC>\angle MLC$, $\angle DMN>\angle DNM$. You can't use WLOG unconditionally for several times. It is possible to "WLOG" $\angle NKA>\angle KNA$ but you can't know the other angles "similarly"

Anyone trig bashed this? for63434 wrote: Plus, <joyce_tan> wrote: WLOG $AN>AK$ Then $\angle NKA>\angle KNA$. Similarly $\angle BKL > \angle BLK$, $\angle LMC>\angle MLC$, $\angle DMN>\angle DNM$. You can't use WLOG unconditionally for several times. It is possible to "WLOG" $\angle NKA>\angle KNA$ but you can't know the other angles "similarly" But we also have the first line of the solution with ratios that gets rid of the necessity of WLOG unconditionally for several times

I really have lost my geo proficiency. I can't even solve P2 now.

A perfect example for demonstrating the power of Miquel Points. 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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] CLAIM 1. Let $G$ be the intersection of the angle bisector of $\angle DEA$ and $AD$, then $LN\|EG$. Proof. The key step is to consider the Miquel Point of $ABCD$. Then $$\frac{BK}{KA}=\frac{BC}{DA}=\frac{PB}{PA}$$Hence $PK$ is the angle bisector of $\angle BPA$, and the cyclic variants hold by symmetry. Therefore, $P$ is also the center of spiral similarity sending $MK$ to $DA$, hence $P,E,M,K$ are concyclic. Similarly $P,L,N,F$ are concyclic. Now suddenly the angles become chasable. We have $$\angle LNG=\angle FPL=\angle FPC+\frac{1}{2}\angle CPL=\angle EDA+\frac{1}{2}\angle CEB=\angle EGA$$$\blacksquare$ By symmetry $MK\|FH$ where $FH$ is the angle bisector of $\angle BFA$ CLAIM 2. $K,L,M,N$ concyclic implies $ABCD$ is a circumscribed quadrilateral (in fact after proving this claim it is easy to show that it is inscribed as well but it is not needed in the second part of the problem). Proof. We have $$\angle MNL=\angle DNL-\angle DNM=\angle DGE-\angle DNM$$and $$\angle MKL=\angle BKM-\angle BKL=\angle BHF-\angle BKL$$Since they are equal,$$\angle BKL-\angle DNM=\angle BHF-\angle DGE=\frac{1}{2}\angle BFD+\angle BAD-\frac{1}{2}\angle DEA-\angle BAD=\frac{1}{2}(\angle CDA-\angle CBA) \qquad(1)$$Rearraning gives $$\angle DNM+\frac{1}{2}\angle MDN=\angle BKL+\frac{1}{2}\angle LBK$$Let $B_1$ be intersection of the angle bisector of $\angle LBK$ and $LK$. Define $D_1$ similarly. Then From $(1)$ we have $$\angle MD_1D=\angle BB_1L$$ On the other hand, We have $\frac{BL}{BK}=\frac{NA}{AK}=\frac{DN}{DM}=\frac{CM}{CL}=k$ for some $k$. Suppose on the contrary that $k>1$, then $DN>DM$ and $BL>BK$, hence $\angle MD_1D>90^{\circ}>\angle BB_1L$, contradiction. Similarly $k<1$ is impossible. This implies $k=1$ so $DN=DM$, $CM=CL$, $NA=NK$ and $BL=BK$ so we are done. $\blacksquare$ [asy][asy] size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -3.9959790753015536, xmax = 4.556735100145964, ymin = -2.4376172871819906, ymax = 4.827591528520739; /* image dimensions */ /* draw figures */draw(circle((0,0), 1), linewidth(0.8)); draw((-2.9175509758486338,-0.43045231576364834)--(1.8866099483124603,-1.4016001707505017), linewidth(0.8)); draw((1.8866099483124603,-1.4016001707505017)--(-0.43054417783469734,2.91817360526708), linewidth(0.8)); draw((-0.43054417783469734,2.91817360526708)--(-1.178237903141253,-0.7820496469886266), linewidth(0.8)); draw((-2.9175509758486338,-0.43045231576364834)--(0.40849440731718795,1.3539889719243832), linewidth(0.8)); draw(circle((-0.6113106602005902,0.45415488584969743), 1.3240259920962445), linewidth(0.8)); draw((-2.9175509758486338,-0.43045231576364834)--(-0.43054417783469734,2.91817360526708), linewidth(0.8)); draw(circle((-1.3002819207681409,0.9658502047882521), 0.4658223465017571), linewidth(0.8)); /* dots and labels */dot((0,0),dotstyle); label("$I$", (0.027475173900907653,0.0760836532721184), NE * labelscalefactor); dot((0.8812259804908418,0.4726952203142682),dotstyle); label("$K$", (0.9088032475357325,0.5512344407969805), NE * labelscalefactor); dot((-0.47276298742979583,0.8811896264235382),dotstyle); label("$L$", (-0.44001189124452117,0.9574117269069432), NE * labelscalefactor); dot((-0.980189078650461,0.19806405553345718),dotstyle); label("$M$", (-0.9458175682871163,0.27534043513738315), NE * labelscalefactor); dot((-0.19813943397435527,-0.9801738441237463),linewidth(4pt) + dotstyle); label("$N$", (-0.16411788558492382,-0.9202002560542053), NE * labelscalefactor); dot((1.8866099483124603,-1.4016001707505017),linewidth(4pt) + dotstyle); label("$A$", (1.9204146016209227,-1.3417049869230346), NE * labelscalefactor); dot((0.40849440731718795,1.3539889719243832),linewidth(4pt) + dotstyle); label("$B$", (0.4413161823903037,1.4172350696729388), NE * labelscalefactor); dot((-0.8870665151689868,0.6589135491714202),linewidth(4pt) + dotstyle); label("$C$", (-0.8538528997339172,0.7198363331445122), NE * labelscalefactor); dot((-1.178237903141253,-0.7820496469886266),linewidth(4pt) + dotstyle); label("$D$", (-1.145074350152381,-0.7209434741889406), NE * labelscalefactor); dot((-0.43054417783469734,2.91817360526708),linewidth(4pt) + dotstyle); label("$E$", (-0.4016932793473549,2.9806344350773237), NE * labelscalefactor); dot((-2.9175509758486338,-0.43045231576364834),linewidth(4pt) + dotstyle); label("$F$", (-2.8847393302837308,-0.36841224473501066), NE * labelscalefactor); dot((0.1551505149394301,-1.0515904310141926),linewidth(4pt) + dotstyle); label("$G$", (0.1884133438690061,-0.9891737574691046), NE * labelscalefactor); dot((1.0515612864816217,0.15514621498661318),linewidth(4pt) + dotstyle); label("$H$", (1.0850688622626974,0.21403065610191707), NE * labelscalefactor); dot((-1.6742470174178203,1.2435921083393704),linewidth(4pt) + dotstyle); label("$J$", (-1.8731279761985407,1.3405978458786063), NE * labelscalefactor); dot((0.5554554428797828,1.0800156438854494),linewidth(4pt) + dotstyle); label("$S$", (0.5869269075995356,1.1413410640133415), NE * labelscalefactor); dot((-0.8736524208418742,-0.8436207605430356),linewidth(4pt) + dotstyle); label("$T$", (-0.846189177354484,-0.7822532532244066), NE * labelscalefactor); dot((-0.8436879441357878,0.8735875416820027),linewidth(4pt) + dotstyle); label("$U$", (-0.8155342878367509,0.9344205597686435), NE * labelscalefactor); dot((-1.0800583566244126,0.5553723852946328),linewidth(4pt) + dotstyle); label("$V$", (-1.0531096815991818,0.6202079422118798), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] CLAIM 3. If $ABCD$ is circumscribed then the incircle of $\triangle CEF$ and $\triangle AEF$. Proof. This is just length chasing. It suffices to show $$AE-AF=EC-EF$$Indeed, $AE-AF=EB+BA-DA-DF=EB+BC-CD-DF$, hence it suffices to show $$EB+BC-EC=CD+DF-CF\qquad(2)$$Notice that $L$ is the touching point of the $E$-excircle and $BC$, hence the left hand side of $(2)$ equals $CL$ while the right hand side of $(2)$ equals $CM$. Obviously they are equal so we are done. $\blacksquare$. Now let $Z$ be the intersection of $UV$ and $EF$. Then $(E,F;J,Z)=-1$. Therefore, $T,S,Z$ are collinear. As a result $$ZT\times ZS=ZJ^2=ZU\times ZV$$hence $S,T,U,V$ are concyclic as desired.