APMO 2015 P1

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

aditya21

717 posts

aditya21

#1 h Mar 30, 2015, 8:17 AM • 8 Y

Y by MSTang, Davi-8191, Samujjal101, Epsilon-, Adventure10, ItsBesi, DEKT, Rounak_iitr

Let $ABC$ be a triangle, and let $D$ be a point on side $BC$ . A line through $D$ intersects side $AB$ at $X$ and ray $AC$ at $Y$ . The circumcircle of triangle $BXD$ intersects the circumcircle $\omega$ of triangle $ABC$ again at point $Z$ distinct from point $B$ . The lines $ZD$ and $ZY$ intersect $\omega$ again at $V$ and $W$ respectively.
Prove that $AB = V W$

Proposed by Warut Suksompong, Thailand

This post has been edited 1 time. Last edited by djmathman, Apr 5, 2015, 3:46 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

TelvCohl

2311 posts

TelvCohl

#2 h Mar 30, 2015, 8:25 AM • 27 Y

Y by Math_CYCR, AmirAlison, joey8189681, bobthesmartypants, fz0718, TheDarkPrince, Wizard_32, e_plus_pi, Plops, AlastorMoody, Vietjung, Why-, vralex, Mathasocean, hsiangshen, yoyo1234, Loxing, 606234, OlympusHero, NightFury101, ACBY, Mathlover_1, centslordm, Creeper1612, Adventure10, vrondoS, ehuseyinyigit

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

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

aditya21

717 posts

aditya21

#3 h Mar 30, 2015, 9:45 AM • 3 Y

Y by aansc1729, Adventure10, Mango247

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

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Steve14-6

9 posts

Steve14-6

#4 h Apr 2, 2015, 7:03 AM • 1 Y

Y by Adventure10

It's the most easier 1st problem in APMO

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

KudouShinichi

160 posts

KudouShinichi

#5 h Apr 2, 2015, 7:05 AM • 1 Y

Y by Adventure10

$\angle XZB=\angle XDB=\angle CDY$
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....

This post has been edited 1 time. Last edited by KudouShinichi, Apr 21, 2015, 7:13 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

djmathman

7936 posts

djmathman

#6 h Apr 5, 2015, 1:48 PM • 2 Y

Y by Adventure10, Mango247

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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

MathPanda1

1135 posts

MathPanda1

#7 h Apr 5, 2015, 3:39 PM • 2 Y

Y by Adventure10, Mango247

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?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Dukejukem

695 posts

Dukejukem

#9 h Apr 6, 2015, 4:20 AM • 3 Y

Y by sa2001, Adventure10, Mango247

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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Steve14-6

9 posts

Steve14-6

#10 h Apr 28, 2015, 5:28 AM • 2 Y

Y by Adventure10, Mango247

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

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

fprosk

681 posts

fprosk

#11 h Jun 14, 2015, 5:23 AM • 2 Y

Y by Samujjal101, Adventure10

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.

This post has been edited 2 times. Last edited by fprosk, Jun 14, 2015, 6:46 AM
Reason: More stuff

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

MSTang

6012 posts

MSTang

#12 h Aug 27, 2015, 7:59 PM • 1 Y

Y by Adventure10

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 $AB = VW.$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

hayoola

123 posts

hayoola

#13 h Aug 28, 2015, 4:16 AM • 2 Y

Y by Adventure10, Mango247

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

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

the_executioner

533 posts

the_executioner

#14 h Jan 1, 2016, 6:37 PM • 2 Y

Y by math90, Adventure10

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 $\LaTeX$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

amplreneo

947 posts

amplreneo

#15 h Mar 13, 2016, 1:11 AM • 2 Y

Y by Adventure10, Mango247

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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

emiliorosado

8 posts

emiliorosado

#20 h May 21, 2016, 4:59 PM • 1 Y

Y by Adventure10

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

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

peppapig_

279 posts

peppapig_

#54 h Aug 15, 2023, 4:27 PM

Y by

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.

This post has been edited 1 time. Last edited by peppapig_, Sep 3, 2023, 3:08 AM
Reason: Finishing not FInishing

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Math4Life7

1703 posts

Math4Life7

#55 h Sep 30, 2023, 4:25 PM

Y by

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. $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

shendrew7

792 posts

shendrew7

#56 h Oct 8, 2023, 9:18 PM

Y by

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 $AB = VW$ . $\blacksquare$

$[asy] size(300); pair A, B, C, D, X, Y, W, V; A = dir(120); B = dir(210); C = dir(330); D = .3B + .7C; X = .4A + .6B; Y = extension(A, C, X, D); pair [] Z = intersectionpoints(circumcircle(B, D, X), circumcircle(A, B, C)); V = IP(D--4D-3Z[1], circumcircle(A, B, C)); W = IP(Y--Z[1], circumcircle(A, B, C)); filldraw(Z[1]--D--X--cycle^^Z[1]--C--A--cycle, palegreen+linewidth(.5)); draw(circumcircle(A, B, C)^^circumcircle(B, D, X)); draw(A--B--Z[1]--C--cycle^^B--C--Y--Z[1]--V^^Y--X--Z[1]--A); draw(circumcircle(C, Y, Z[1]), blue); draw(Z[1]--Y--D--cycle^^Z[1]--A--B--cycle, red+linewidth(.7)); dot("$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$", W, dir(270)); [/asy]$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

abeot

123 posts

abeot

#57 h Oct 24, 2023, 8:21 PM • 1 Y

Y by centslordm

$Z$ 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. $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

joshualiu315

2513 posts

joshualiu315

#58 h Oct 28, 2023, 5:27 AM

Y by

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. $\square$

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.862013699401459, xmax = 24.643652945574914, ymin = -7.24172292270231, ymax = 13.388846403326983; /* image dimensions */ /* draw figures */ draw((5,12)--(0,0), linewidth(1)); draw((0,0)--(14,0), linewidth(1)); draw((14,0)--(5,12), linewidth(1)); draw((xmin, -0.24601752222343568*xmin + 2.4601752222343567)--(xmax, -0.24601752222343568*xmax + 2.4601752222343567), linewidth(1) + red); /* line */ draw((5,12)--(14.90504532247097,-1.2067270966279613), linewidth(1)); draw(circle((7,4.125), 8.125), linewidth(1)); draw(circle((5,-0.7739138974401139), 5.059539773601048), linewidth(1)); draw((9.113027864267206,-3.720427856071993)--(14.90504532247097,-1.2067270966279613), linewidth(1)); draw((9.113027864267206,-3.720427856071993)--(12.425358937734558,10.173231592518905), linewidth(1)); draw((14,0)--(9.113027864267206,-3.720427856071993), linewidth(1)); draw((0,0)--(9.113027864267206,-3.720427856071993), linewidth(1)); /* dots and labels */ dot((0,0),dotstyle); label("$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$", (11.388385430202264,-3.5783018465482685), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

brainfertilzer

1831 posts

brainfertilzer

#59 h Nov 25, 2023, 5:43 PM • 1 Y

Y by megarnie

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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

ItsBesi

136 posts

ItsBesi

#63 h Nov 28, 2023, 7:21 PM

Y by

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 $\omega$
$\angle WZC=\angle WBC=x$ $\implies$ $\angle WBC=x$

$\angle VBW=\angle VBC+\angle CBW=x+z=\gamma$ $\implies$
$\angle VBW=\gamma=\angle ACB$ $\implies$
$\angle VBW=\angle ACB$ $\implies$

$\widehat{\frac{VW}{2}}=\widehat{\frac{AB}{2}}$ $\implies$

$\widehat{VW}=\widehat{AB}$ $\implies$

$VW=AB$

This post has been edited 1 time. Last edited by ItsBesi, Nov 28, 2023, 9:14 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

bjump

987 posts

bjump

#64 h Dec 12, 2023, 2:28 AM

Y by

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 K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Inconsistent

1455 posts

Inconsistent

#65 h Mar 2, 2024, 5:09 AM

Y by

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

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

dolphinday

1318 posts

dolphinday

#66 h May 29, 2024, 12:20 AM

Y by

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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

fearsum_fyz

48 posts

fearsum_fyz

#67 h Oct 18, 2024, 5:47 PM

Y by

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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

LeYohan

29 posts

LeYohan

#68 h Dec 26, 2024, 4:11 AM

Y by

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 K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Likeminded2017

391 posts

Likeminded2017

#69 h Dec 26, 2024, 8:04 AM

Y by

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

This post has been edited 1 time. Last edited by Likeminded2017, Dec 26, 2024, 8:07 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Maximilian113

506 posts

Maximilian113

#70 h Jan 1, 2025, 7:13 PM

Y by

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

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

ethan2011

225 posts

ethan2011

#71 h Yesterday at 1:03 AM

Y by

Notice that $Z$ is the Miquel Point. This implies that $DCZY$ is concyclic, which implies that $\angle DCY\cong \angle DZY$ , or $\angle VZW\cong \angle ACB$ , implying that $VW=AB$ .

Z K Y