My solution: Since $ Z $ is the Miquel point of complete quadrilateral $ \{ AB, BC, CA, XY \} $ , so $ Z \in \odot (CDY) \Longrightarrow \angle VZW=180^{\circ}- \angle ACB \Longrightarrow AB=VW $ . Q.E.D

nice problem ! and nice solution telv my solution = by cyclic quad. $AVZB,BXDZ$ we get $\angle BZV+\angle XAV=180$ and $\angle BZV=\angle AXD$ and hence $\angle AXD+\angle XAV=180$ and thus $AV$ is parallel to $XD$ now by cyclic quad. $ABZC,BXDZ$ we get $\angle ZDY=\angle XBZ=\angle ZCY$ and hence $ZDCY$ is a cyclic quad. also note that triangles $YCZ,YWA$ are similar and hence $\angle YZC=\angle CAW$ now by angle chase we get $\angle XDB=\angle YDC=\angle YZC=\angle CAW=\angle CBW$ and thus $XD$ is parallel to $BW$ or $AV$ is parallel to $BW$ hence $AVWB$ is a isosceles trapezium and thus we get $AB=VW$ as desired we are done

It's the most easier 1st problem in APMO

$ \angle XZB=\angle XDB=\angle CDY $ so $\angle AZX=\angle AYX \Longrightarrow AXZY$ cyclic so $\angle BAZ=\angle BCZ=\angle DYZ $ so $DCYZ$ cyclic so$ \angle CYD=\angle CZD=\angle AZX $ so$ \angle AZB=\angle VZW $ hence Proved Please check if it is right or wrong....

Note that $Z$ is the center of spiral similarity sending $\triangle ZXD$ to $\triangle ZAC$, so $\triangle ZXA\sim\triangle ADC\implies \angle XZA=\angle DZC$. But remark that \[\angle XZA=\angle AZB-\angle XZB=\angle ACB-\angle XDB=\angle ACB-\angle CDY=\angle CYD.\] Thus $ZDCY$ is cyclic, so $\angle ACB=\angle VZW\implies AB=VW$ as desired.

I used directed angles to prove $C,D,Z,$ and $Y$ are cyclic and then the directed angles $VZW$ and $BCA$ are equal. Would they take marks off for using directed angles even though you don't need to? Since the directed angles $VZW$ and $BCA$ are equal, $A,B,C,V,W,Z$ lie on $\omega$, and $AB$ and $VW$ subtend the angles $BCA$ and $VZW$ on $\omega$, then $AB=VW$. Would that reasoning be correct?

Using the two given cyclic quadrilaterals, we find that \[\measuredangle BAV = \measuredangle BZV = \measuredangle BZD = \measuredangle BXD,\] where the angles are directed. It follows that $VA \parallel DX.$ Meanwhile, by Pascal's Theorem on cyclic hexagon $ACBWZV$, it follows that $Y, D, BW \cap VA$ are collinear. But since $VA \parallel YD$, $BW$ must be parallel to $YD$ as well. Hence $VA \parallel BW$, so so since $A, B, W, Z$ are concyclic, they must form an isoceles trapezoid. Thus, $AB = VW$, as desired.

please how can I get a marks scheme for APMO2015???

First, note that quadrilaterals BXDZ and BACZ are cyclic. Now, we have <ZXY=<ZXD=<ZBD=<ZBC=<ZAC=<ZAY, so quadrilateral AXZY is cyclic. Now, denote <AYX by a and <AXY by b. Then by triangle AXY we have a+b=180-<XAY=180-<BAC. Now, by cyclic quad XBZD we have <XZV=<XZD=<XBD=<ABC. Also, by cyclic quad AXZY we have <AZX=<AYX=a and <AZY=<AXY=b. Now we have <XZW=<XZY=<XZA+<AZY=a+b. Finally, we have <VZW=<XZW-<XZV=a+b-<ABC=180-<BAC-<ABC=<ACB. This implies that arc VW and arc AB are equal, meaning that AB=VW. EDIT: I've been told this solution only covers one case. In my diagram I had an acute triangle and D closer to C than B. What other cases should I be worried about? Is it just the other case of <ACB obtuse? Edit2: Aaaand I just realized i didnt even need to do the whole a and b thing, it jusy follows from the opposite angles of a cyclic quad being supplementary.

Direct all angles modulo $180^\circ.$ We show that quadrilateral $CDZY$ is cyclic. For the proof, let $Y' \neq D$ be the intersection of line $DX$ and the circumcircle of $\triangle CDZ$; we show that $Y = Y'.$ Using cyclic quadrilaterals $ABZC,$ $CDZY',$ and $BZDX,$ we get \[\begin{aligned} \angle ACY' &= \angle ACZ + \angle ZCY' \\ &= \angle ABZ + \angle ZDY' \\ &= \angle XBZ + \angle ZDX \\ &= \angle XDZ + \angle ZDX = 0 \end{aligned}\] so $A, C, Y'$ are collinear, and hence $Y'$ is the intersection of $AC$ and $XD.$ Therefore, $Y = Y',$ so $CDZY$ is cyclic. Then \[\angle AWB = \angle ACB = \angle YCB = \angle YCD = \angle YZD = \angle WZV = \angle WAV,\] so $AV \parallel BW$. Thus, $AVWB$ is a cyclic trapezoid, and hence an isosceles trapezoid, so $AB = VW.$

YAZ =ZBC = ZXY so YAXZ is cyclic so ZYA =ZXB =ZDB so YZDC is cyclic so WZD = YCD so WV = AB

fprosk wrote: First, note that quadrilaterals BXDZ and BACZ are cyclic. Now, we have <ZXY=<ZXD=<ZBD=<ZBC=<ZAC=<ZAY, so quadrilateral AXZY is cyclic. Now, denote <AYX by a and <AXY by b. Then by triangle AXY we have a+b=180-<XAY=180-<BAC. Now, by cyclic quad XBZD we have <XZV=<XZD=<XBD=<ABC. Also, by cyclic quad AXZY we have <AZX=<AYX=a and <AZY=<AXY=b. Now we have <XZW=<XZY=<XZA+<AZY=a+b. Finally, we have <VZW=<XZW-<XZV=a+b-<ABC=180-<BAC-<ABC=<ACB. This implies that arc VW and arc AB are equal, meaning that AB=VW. Your solution looks little untidy. Guess you should start using $\LaTeX$

It suffices to prove that $YZDC$ is cyclic because then $\angle VZW = \angle ACB \implies VW = AB$. Note that $Z$ is the center of spiral similarity mapping $AX$ to $CD$. Thus, we have $\angle AZX = \angle DZC$. We have $\angle DXZ = \angle DBZ = \angle YAZ$ so $AXZY$ is cyclic. This implies that $\angle AYD = \angle AZX = \angle DZC$ so $YZDC$ is cyclic as desired.

Note that the problem follows from the fact that the circuncircles of the triangles $ABC$, $AYX$, $XDB$, and $YDC$ are concurrent in $Z$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(14.36cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 0.14, xmax = 14.5, ymin = -3.56, ymax = 5.36; /* image dimensions */ /* draw figures */ draw((5.88,4.34)--(11.28,0.52)); draw((5.88,4.34)--(3.44,-2.06)); draw((3.44,-2.06)--(8.37419958069258,2.5755847410656187)); draw((11.28,0.52)--(4.606541738479059,0.9997816091254008)); draw(circle((8.004893472953015,1.6170221868969346), 3.453951424390603)); draw(circle((5.311710563152486,0.8915353477981152), 3.4949765866910085)); draw(circle((5.43378421854007,-1.067867409858131), 2.2269940697094484)); draw(circle((8.87950610939717,0.20826203620683048), 2.420651044015102)); /* dots and labels */ dot((5.88,4.34),dotstyle); label("$A$", (5.96,4.54), NE * labelscalefactor); dot((11.28,0.52),dotstyle); label("$B$", (11.46,0.42), NE * labelscalefactor); dot((3.44,-2.06),dotstyle); label("$Y$", (3.1,-2.42), NE * labelscalefactor); dot((8.37419958069258,2.5755847410656187),dotstyle); label("$X$", (8.46,2.78), NE * labelscalefactor); dot((4.606541738479059,0.9997816091254008),dotstyle); label("$C$", (4.2,1.02), NE * labelscalefactor); dot((6.548293392750314,0.8601813154396127),linewidth(3.pt) + dotstyle); label("$D$", (6.96,0.92), NE * labelscalefactor); dot((6.548293392750314,0.8601813154396127),linewidth(3.pt) + dotstyle); dot((11.28,0.52),linewidth(3.pt) + dotstyle); dot((7.535295283349836,-1.804857118743648),linewidth(3.pt) + dotstyle); label("$Z$", (7.58,-2.2), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Note that $Z$ is the miquel point of complete quadrilateral $ACDBXY$. Thus, $AXZY$ is cyclic and $$\angle BZD=\angle BXD = \angle AXY=\angle AZY \quad (*).$$But $$\angle BAV = \angle BZV = \angle BZD \stackrel{(*)}{=} \angle AZY = 180^{\circ}-\angle AZW = \angle ABW\implies AV\parallel BZ$$which means $AB=VW$, done.

Join BW, AW and BV. AB = (BWsinACB)/sinBAW. VW = ( BWsinVZW)/sinBAW. So it suffices to show that ACB = VZW, and this follows from the fact that Z is the Miquel point of AXDC.

We have $<BXD=180-<BZD=180-<BZV=<BAV$, hence $XY \| AV$. Consider the hexagon inscribed in $\omega$, $BWZVAC$, applying pascal's theorem we get $BW \cap VA$, $WZ\cap AC=Y$ and $BC \cap ZV=D$ are collinear, but $AV$ and $YD$ are parallel, then they only meet at the infinite, thus $BW \cap VA=\infty \Rightarrow BW \|VA \Rightarrow BWVA$ is a cyclic cuadrilateral with parallel sides, then it is an isosceles trapezoid, $\Rightarrow AB=VZ$ $\square$

Reim's Theorem and its converse: $\implies$ $DY\|XD\|AV \implies$ $DYZC$ cyclic $\implies \angle DYZ= \angle DCZ=\angle BWZ \implies BW\|XY\|AV$ So $VAWB$ is an isosceles trapezoid and we're done .

Easy Angle Chase! Claim- CDZY is cyclic Proof-$\angle ZDY=\angle XBZ=180^{\circ}-\angle AVZ=180^{\circ}-\angle ACZ=\angle ZCY$ Now $\angle VAC=\angle VZC=\angle DYC \implies AV \parallel XY$. But, $\angle BWZ=\angle BCZ=\angle DYZ=\angle XYW \implies XY \parallel BW$ So $ABWV$ is cyclic trapezium, hence, we are done!

Notice how proving $AB=VW$ is equivalent to showing that $AVWB$ is an isosceles trapezoid, or that $AV\parallel BW$. Because \[\measuredangle DXB=\measuredangle DZB= \measuredangle VZB=\measuredangle VAB\]this tells us that $XY\parallel AV$. It suffices to show $BW\parallel XY$ now. Let's prove that $YCDZ$ is cyclic first, though, as shown below: \[\measuredangle ZDY=\measuredangle ZDX=\measuredangle ZBX=\measuredangle ZBA=\measuredangle ZVA=\measuredangle ZCA=\measuredangle ZCY.\] Now, we can actually show $BW\parallel XY$. Notice how \[\measuredangle ZWB=\measuredangle ZCB=\measuredangle ZYD=\measuredangle ZYX\]and our proof is complete.

1. $ZDCY$ is cyclic. This follows immediately from the fact that $Z$ is the Miquel point of the complete quadrilateral $\{AB, BC, AC, XY\}$, meaning that $Z$ must lie on the circumcircle of $CDY$. 2. Finishing. By (1), we have that \[\angle YZW = 180 - \angle DCY = \angle ACB\]which implies that chords $AB$ and $VW$ subtend (?, idk the word) arcs of the same measure, therefore $AB=VW$, finishing the problem.

Claim $AV \parallel XY$ Proof Specifically we prove that $\angle XAV + \angle AXY = 180$. Thiis is because BXDZ cyclic means that $\angle BZD = \angle AXY$. From $AVZB$ cyclic we get $\angle XAV + \angle BZD = 180$. The result follows. We can see now that $\angle XDV = \angle AXD$ (their arcs are equal). This means that \[ \angle XBZ = 180 - \angle AXD = 180 - \angle XDV = \angle DZB \]This means that we have $BZ \parallel XD$ or $BZ \parallel AV$. Since cyclic trapezoids have to be isosceles we have the result. $\blacksquare$

not realizing it was Miquel until I finish asy'ing my diagram... Equivalently we can show $\measuredangle ABC = \measuredangle VZW$, or $\measuredangle AZB = \measuredangle DZY$, where we utilize directed angles to eliminate configuration issues. Our solution proceeds with 2 spiral similarities: $\textcolor{blue}{\textbf{Claim 1:}}$ $\triangle ZDX \sim \triangle ZCA$ This follows from the following equal angles: \begin{align*} \measuredangle ZCA &= \measuredangle ZBA = \measuredangle ZDX \\ \measuredangle ZAC &= \measuredangle ZBC = \measuredangle ZXD.~{\color{blue} \blacksquare} \end{align*} $\textcolor{blue}{\textbf{Claim 2:}}$ $\triangle ZAB \sim \triangle ZYD$ Note that $DCYZ$ is cyclic as \[\measuredangle ZCA = \measuredangle ZDX \implies \measuredangle ZDY = \measuredangle ZCY.\] As a result, we see a spiral similarity nested between $\omega$ and the blue circle: \begin{align*} \measuredangle ZAB &= \measuredangle ZCB = \measuredangle ZYD \\ \measuredangle ZBA &= \measuredangle ZCA = \measuredangle ZCY = \measuredangle ZDY.~{\color{blue} \blacksquare} \end{align*} This similarity produces the desired equivalence $\measuredangle AZB = \measuredangle DZY$, and thus $AB = VW$. $\blacksquare$ [asy][asy] size(300); pair A, B, C, D, X, Y, W, V; A = dir(120); B = dir(210); C = dir(330); D = .3B + .7C; X = .4A + .6B; Y = extension(A, C, X, D); pair [] Z = intersectionpoints(circumcircle(B, D, X), circumcircle(A, B, C)); V = IP(D--4D-3Z[1], circumcircle(A, B, C)); W = IP(Y--Z[1], circumcircle(A, B, C)); filldraw(Z[1]--D--X--cycle^^Z[1]--C--A--cycle, palegreen+linewidth(.5)); draw(circumcircle(A, B, C)^^circumcircle(B, D, X)); draw(A--B--Z[1]--C--cycle^^B--C--Y--Z[1]--V^^Y--X--Z[1]--A); draw(circumcircle(C, Y, Z[1]), blue); draw(Z[1]--Y--D--cycle^^Z[1]--A--B--cycle, red+linewidth(.7)); dot("$A$", A, NW); dot("$B$", B, SW); dot("$C$", C, 2dir(0)); dot("$Z$", Z[1], dir(305)); dot("$Y$", Y, SE); dot("$X$", X, dir(150)); dot("$D$", D, 2dir(110)); dot("$V$", V, dir(0)); dot("$W$", W, dir(270)); [/asy][/asy]

$Z$ is the center of spiral similarity sending $XD$ to $AC$, so it must also be the center of the spiral similarity sending $DC$ to $XA$. Therefore $YCDZ$ is cyclic. Then (directing all angles modulo 180), \[ \angle VZW = \angle DZY = \angle DCY = \angle BCA \]which is enough to imply $AB = VW$, as desired. $\blacksquare$

Notice that showing that $\angle ACB = \angle VZW$ would suffice to prove $AB=VW$. This happens if and only if $CDZY$ is cyclic. Notice that \[\angle ZCY = \pi-\angle ACZ = \pi-\angle AVZ = \angle XBZ = \pi-\angle XDZ = \angle ZDY\] proving that $CDZY$ is cyclic. $\square$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.862013699401459, xmax = 24.643652945574914, ymin = -7.24172292270231, ymax = 13.388846403326983; /* image dimensions */ /* draw figures */ draw((5,12)--(0,0), linewidth(1)); draw((0,0)--(14,0), linewidth(1)); draw((14,0)--(5,12), linewidth(1)); draw((xmin, -0.24601752222343568*xmin + 2.4601752222343567)--(xmax, -0.24601752222343568*xmax + 2.4601752222343567), linewidth(1) + red); /* line */ draw((5,12)--(14.90504532247097,-1.2067270966279613), linewidth(1)); draw(circle((7,4.125), 8.125), linewidth(1)); draw(circle((5,-0.7739138974401139), 5.059539773601048), linewidth(1)); draw((9.113027864267206,-3.720427856071993)--(14.90504532247097,-1.2067270966279613), linewidth(1)); draw((9.113027864267206,-3.720427856071993)--(12.425358937734558,10.173231592518905), linewidth(1)); draw((14,0)--(9.113027864267206,-3.720427856071993), linewidth(1)); draw((0,0)--(9.113027864267206,-3.720427856071993), linewidth(1)); /* dots and labels */ dot((0,0),dotstyle); label("$B$", (0.21000293624274426,0.203022867628873), NE * labelscalefactor); dot((14,0),dotstyle); label("$C$", (14.300266164704891,0.0103022867628873), NE * labelscalefactor); dot((5,12),dotstyle); label("$A$", (5.002904849298981,12.24833581405018), NE * labelscalefactor); dot((10,0),dotstyle); label("$D$", (10.095806762355219,0.2603022867628873), NE * labelscalefactor); dot((0.9297652799241796,2.231436671818031),dotstyle); label("$X$", (0.3224114076641886,2.540634105793081), NE * labelscalefactor); dot((14.90504532247097,-1.2067270966279613),linewidth(4pt) + dotstyle); label("$Y$", (15.012674636126334,-1.0069317013224501), NE * labelscalefactor); dot((9.113027864267206,-3.720427856071993),linewidth(4pt) + dotstyle); label("$Z$", (9.00874297069548,-4.4660549977314184), NE * labelscalefactor); dot((12.425358937734558,10.173231592518905),linewidth(4pt) + dotstyle); label("$V$", (12.52889601947907,10.37282951168388), NE * labelscalefactor); dot((11.287201826252467,-2.776849426130471),linewidth(4pt) + dotstyle); label("$W$", (11.388385430202264,-3.5783018465482685), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Solved with megarnie Observe that \begin{align*} AB = VW&\iff \widehat{VABW} = \widehat{AVWB} = 2(180^\circ - C)\\ &\iff \angle VZW = 180^\circ - C\\ &\iff \angle DZY = C = 180^\circ - \angle DCY \\ &\iff DCYZ \text{ cyclic}. \end{align*}But \[\angle ZDY = 180^\circ - \angle XDZ = \angle ABZ = 180^\circ - \angle ACZ = \angle ZCY,\]so $DCYZ$ is indeed cyclic and we are done.

Here is an ugly solution Let $\angle BAC=\alpha$ , $\angle ABC=\beta$ , $\angle ACB=\gamma$ From $\square BXDZ-cyclic$ we get: $\angle XBD=\angle XZD=\beta$ From $\omega$ we get: $\angle ABC=\angle AZC=\beta$ $\implies$ $\angle XZD=\angle AZC=\beta$ $\angle XZD=\angle AZC$ $\implies$ $\angle XZA+\angle AZD=\angle AZV+\angle VZC$ $\angle XZA+\angle AZV=\angle AZV+ \angle VZC$ $\angle XZA=\angle VZC=x$ $\angle XZD=\angle XZA+\angle AZV$ $\beta=x+\angle AZV$ $\angle AZV=\beta-x$ From $\omega$ again we have: $\angle AZV=ACV=\beta-x$ $\implies$ $\angle ACV=\beta-x$ Let $\angle ZAC=y$ $\angle ZAC=\angle ZVC=\angle ZBC=\angle ZBD=\angle ZXD=\angle ZXY=y$ $\implies$ $\angle ZAC=\angle ZAY=ZXY$ $\implies$ $\angle ZAY=\angle ZXY=y$ $\implies$ $\square ZAXC-cyclic$ Let $\angle WAC=z$ From $\omega$ we get: $\angle WAC=\angle WBC=\angle WZC=z$ $\implies$ $\angle WZC=z$ From $\square ZAXC-cyclic$ we get: $\angle AXY=\angle AZY=\angle AZV+\angle YZX+\angle CZY=\beta-x+x+z=\beta+z$ $\implies$ $\angle AXY=\beta+z$ From triangle $\triangle DVC$ we get: $\angle DVC+\angle CDV+\angle DVC=180$ $\angle ZVC+\angle CDV+\angle BCV=180$ $y+\angle CDV+\angle BCA+\angle ACV=180$ $y+\angle CDV+\gamma+\beta-x=180$ $\implies$ $\angle CDV=\alpha+x-y$ $\angle CDV=\angle BDZ=\angle BXZ=\alpha+x-y$ $\implies$ $\angle BXZ=\alpha+x-y$ $\angle BXA=180$ $\angle BXA=\angle BXA+\angle XZY+\angle AXC$ $\implies$ $180=\alpha+x-y+y+\beta+z$ $\implies$ $180-\alpha-\beta=x+z$ $\implies$ $\gamma=x+z$ From $\omega$ $\angle WZC=\angle WBC=x$ $\implies$ $\angle WBC=x$ $\angle VBW=\angle VBC+\angle CBW=x+z=\gamma$ $\implies$ $\angle VBW=\gamma=\angle ACB$ $\implies$ $\angle VBW=\angle ACB$ $\implies$ $\widehat{\frac{VW}{2}}=\widehat{\frac{AB}{2}}$ $\implies$ $\widehat{VW}=\widehat{AB}$ $\implies$ $VW=AB$

Note that by reims $AV \parallel XD$ $$\angle VZC= \angle VAC= \angle CVD$$So $DCYZ$ is cyclic. $$\angle VZW = 180^{\circ}-\angle DCY=\angle ACB$$So $VW$ and $AB$ subtend equal arcs on $(ABC)$, we are done.

$Z$ is the miquel point of $AXBDYC$. Thus $(DZCY)$ so $\angle VZW = \angle DZY = \angle ACB$ so $AB = VW$.

Clearly $Z$ is the miquel point of $XACD$ so $Z \in (CDY)$ from which it follows that $\measuredangle VZW = \measuredangle BCA$, so $AB = VW$ since they subtend the same arc.

The problem is just an angle chase. We want to show that arcs $\widehat{AB}$ and $\widehat{VW}$ of the circumcircle have the same measure. Claim: $C, D, Y, Z$ are concyclic. Proof. $\measuredangle{ZDY} = \measuredangle{ZDX} = \measuredangle{ZBX} = \measuredangle{ZBA} = \measuredangle{ZCA} = \measuredangle{ZCY}$ Then $\measuredangle{VZW} = \measuredangle{DZY} = \measuredangle{DCY} = \measuredangle{BCA}$ as desired.

Since $X-D-Y$ we know the miquel point lies on $(ABC)$, so $ZDCY$ is cyclic which means $\angle VZW = \angle DZW = \angle C = \angle AZB$.

$Z$ is the Miquel Point of complete quadrilateral $AXCD$ so $\angle VZW=180-\angle DZY=180-\angle BZA=180-\angle BCA.$ Thus $AB=VW.$

Notice that $\angle ZDY = \angle ABZ = \angle ZCY,$ so $ZDCY$ is cyclic. Therefore, $\angle DYZ = \angle BCZ = \angle BWZ \implies BW \parallel XY.$ Also, $\angle AVZ = 180^\circ - \angle XBZ = \angle XDZ \implies XY \parallel AV.$ Hence, $BW \parallel AV$ so $AVWB$ is an isosceles trapezoid, and the desired result follows. QED