Chasing After the Tangent

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Chasing After the Tangent

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Source: 2022 Taiwan TST Round 1 Mock Day 2 P5

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530 posts

USJL

#1 h Mar 19, 2022, 6:43 AM • 2 Y

Y by itslumi, Rounak_iitr

Let $H$ be the orthocenter of a given triangle $ABC$ . Let $BH$ and $AC$ meet at a point $E$ , and $CH$ and $AB$ meet at $F$ . Suppose that $X$ is a point on the line $BC$ . Also suppose that the circumcircle of triangle $BEX$ and the line $AB$ intersect again at $Y$ , and the circumcircle of triangle $CFX$ and the line $AC$ intersect again at $Z$ .
Show that the circumcircle of triangle $AYZ$ is tangent to the line $AH$ .

Proposed by usjl

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#2 h Mar 19, 2022, 7:43 AM • 1 Y

Y by inoxb198

Not sure if I'm doing anything wrong, since the solution looks so simple...
$[asy] size(10cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair D = foot(A,B,C); pair E = foot(B,A,C); pair F = foot(C,A,B); pair X = 0.6*B+0.4*C; pair H = orthocenter(A,B,C); path BXE = circumcircle(B,X,E); path CXF = circumcircle(C,X,F); pair P = intersectionpoints(BXE,CXF)[0]; pair Y = intersectionpoints(BXE,A--B)[0]; pair Z = intersectionpoints(CXF,10*A-9*C--10*C-9*A)[0]; draw(A--B--C--A--cycle); draw(BXE,blue); draw(CXF,blue); draw(C--F); draw(B--E); draw(P--X,dashed+red); draw(circumcircle(A,F,E),rgb(0.1,0.6,0.1)); draw(circumcircle(A,Y,Z)); draw(circumcircle(B,C,E),rgb(1.0,0.6,0.6)+dashed); draw(A--Z); draw(A--D); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$H$",H,dir(H)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$Z$",Z,dir(Z)); dot("$P$",P,dir(P)); [/asy]$

We use directed angles modulo $180^{\circ}$ .

Let $P$ be the intersection of $(BEX)$ and $(CFX)$ . By the radical axis theorem on $(BFEC), (BEX), (CFX)$ , we have $BE, CF, PX$ are concurrent, which means that $P, H, X$ are collinear.

Next, we have $P$ lies on $(AEHF)$ because
$\measuredangle FPH=\measuredangle FPX=\measuredangle FCX=\measuredangle FCD=\measuredangle FAD$ where $D$ is the foot of the perpendicular from $A$ to $BC$ .

Lastly, we have $A,Y,P,Z$ are concyclic because
$\measuredangle ZPY=\measuredangle ZPX+\measuredangle XPY=\measuredangle ZCX+\measuredangle XBY = \measuredangle ACB+\measuredangle CBA=\measuredangle CAB=\measuredangle ZAY.$
To finish,
$\measuredangle PYA = \measuredangle PYB = \measuredangle PEB = \measuredangle PEH = \measuredangle PAH$ which implies that $(AYZ)$ is tangent to $AH$ .

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USJL

#3 h Mar 19, 2022, 7:58 AM

Y by

ACGNmath wrote:

Not sure if I'm doing anything wrong, since the solution looks so simple...
$[asy] size(10cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair D = foot(A,B,C); pair E = foot(B,A,C); pair F = foot(C,A,B); pair X = 0.6*B+0.4*C; pair H = orthocenter(A,B,C); path BXE = circumcircle(B,X,E); path CXF = circumcircle(C,X,F); pair P = intersectionpoints(BXE,CXF)[0]; pair Y = intersectionpoints(BXE,A--B)[0]; pair Z = intersectionpoints(CXF,10*A-9*C--10*C-9*A)[0]; draw(A--B--C--A--cycle); draw(BXE,blue); draw(CXF,blue); draw(C--F); draw(B--E); draw(P--X,dashed+red); draw(circumcircle(A,F,E),rgb(0.1,0.6,0.1)); draw(circumcircle(A,Y,Z)); draw(circumcircle(B,C,E),rgb(1.0,0.6,0.6)+dashed); draw(A--Z); draw(A--D); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$H$",H,dir(H)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$Z$",Z,dir(Z)); dot("$P$",P,dir(P)); [/asy]$

We use directed angles modulo $180^{\circ}$ .

Let $P$ be the intersection of $(BEX)$ and $(CFX)$ . By the radical axis theorem on $(BFEC), (BEX), (CFX)$ , we have $BE, CF, PX$ are concurrent, which means that $P, H, X$ are collinear.

Next, we have $P$ lies on $(AEHF)$ because
$\measuredangle FPH=\measuredangle FPX=\measuredangle FCX=\measuredangle FCD=\measuredangle FAD$ where $D$ is the foot of the perpendicular from $A$ to $BC$ .

Lastly, we have $A,Y,P,Z$ are concyclic because
$\measuredangle ZPY=\measuredangle ZPX+\measuredangle XPY=\measuredangle ZCX+\measuredangle XBY = \measuredangle ACB+\measuredangle CBA=\measuredangle CAB=\measuredangle ZAY.$
To finish,
$\measuredangle PYA = \measuredangle PYB = \measuredangle PEB = \measuredangle PEH = \measuredangle PAH$ which implies that $(AYZ)$ is tangent to $AH$ .

Or this problem is really simple itself

Yeah I think the point $P$ you have is crucial, but with that point there are so many ways to finish up the problem.

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352 posts

#4 h Mar 19, 2022, 8:09 AM

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Here's a more braindead solution. Also, the first part of this solution might not be necessary.

We begin by showing that the desired conclusion holds when $X$ is the midpoint of $\overline{BC}$ . Since $X$ is the center of $(BFEC)$ in this case, we know that $\overline{XY}$ bisects $\angle BYE$ by the trillium theorem, and that, as a consequence, $\overline{XY}$ is the perpendicular bisector of $\overline{EF}$ . By symmetry, $Z$ also lies on the perpendicular bisector of $\overline{EF}$ , and a two-second angle chase finishes (consider the midpoint of $\overline{EF}$ ).

Now, it suffices to show that the signed ratio $\tfrac{AY}{AZ}$ is independent of the position of $X$ . To do this, let $T\equiv\overline{AC}\cap (BXE)\neq E$ and $S\equiv\overline{AB}\cap(CXF)\neq F$ . By power of a point, we can obtain $CT = \tfrac{CX\cdot a}{CE}$ , $BS = \tfrac{BX\cdot a}{BF}$ ,
$AY = \frac{AE \cdot (b - CT)}{c} = \frac{AE \cdot (b - \frac{CX\cdot a}{CE})}{c},$ and
$AZ = \frac{AF \cdot (BS - c)}{b} = \frac{AF \cdot (\frac{BX\cdot a}{BF} - c)}{b}.$ Dividing the two yields
$\frac{AY}{AZ} = \frac{\frac{AE \cdot (b - \frac{CX\cdot a}{CE})}{c}}{\frac{AF \cdot (\frac{BX\cdot a}{BF} - c)}{b}} = \frac{AE\cdot BF\cdot b}{AF\cdot CE \cdot c},$ where the last equality comes from the substituting $BX = a - CX$ and doing some standard manipulations. Since this expression contains no $X$ , we're done.

This post has been edited 7 times. Last edited by Blast_S1, Mar 19, 2022, 8:15 AM

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#6 h Mar 19, 2022, 8:19 AM

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Great problem! Here's a length bash solution (i.e. PoP, Pythagorean theorems and a bit of trig)
Let $(BXE)$ intersect $AC$ at $Y'$ and let $(CXF)$ intersect $AB$ at $Z'$ . It's easy to see that $\frac {AZ'}{AZ}=\frac {AC}{AF}=\frac {AB}{AE} =\frac {AY'}{AY}$ , so $YZZ'Y'$ is cyclic (probably this could have been proven by radical axis, but I saw it that way). Note that $AH$ is tangent to $(AYZ)$ $\iff$ $Y'Z' \perp BC$ . We can prove the last one using $BZ'^2-CZ'^2=BY'^2-CY'^2$ . We calculate only the first one and we will see that it's symmetrical expression. I will put only a sketch: using $BZ'.BF=BX.BC$ , Pythagorean theorem for $CFZ'$ and $BF=a.cos \beta$ , $CF=a.sin \beta$ we obtain the first expression is $a(BX-CX)$ , so the second is $-a(CX-BX)$ and we're done.
Edit: this is similar to the above one - it again includes points $Y'$ and $Z'$

This post has been edited 5 times. Last edited by VicKmath7, Mar 19, 2022, 8:23 AM

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#7 h Mar 19, 2022, 11:14 AM

Y by

Almost the same as #1 but anyways,
Let $G=\overline{HX} \cap (BEX)$ then by POP $GH \cdot HX = BH \cdot HE = CH \cdot HF$ so $G \in (CFX)$ .
$\measuredangle YGZ = \measuredangle XGZ - \measuredangle XGY = \measuredangle XCA - \measuredangle XBA = \measuredangle YAZ \implies G \in (AYZ).$ $\measuredangle HGE = \measuredangle XGE = \measuredangle XBE = \measuredangle HAE$ so $G \in \odot(AH)$ and finally to finish off, note that
$\measuredangle GZA = \measuredangle GZC = \measuredangle GXB = \measuredangle GEB = \measuredangle GEH = \measuredangle GAH.$

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L567

#8 h Mar 19, 2022, 1:21 PM

Y by

Nothing new I guess, Solved with Aayuu

By Miquel's theorem, $(BEX), (CFX), (AYZ)$ concur at a point, say $P$ . Then $\angle AZY = \angle APY = \angle EPA - \angle EPY = \angle APE - \angle ABE$ , so to show that $\angle BAH = \angle AZY$ , it suffices to show that $\angle APE = \angle AFE$ (or $P \in (AEF)$ )since then $\angle AZY = \angle AFE - \angle ABE = C - (90 - A) = 90 - B = \angle BAH$ .

Note that by radax, $PX$ goes through $H$ , so $\angle EPH = \angle EPX = \angle EBC = \angle EAH$ so $P \in (AEF)$ , so we're done. $\blacksquare$

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andywu

#9 h Feb 7, 2023, 2:55 AM

Y by

https://hackmd.io/@sine/TWTST_22_1_M5

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#10 h Feb 8, 2023, 5:37 AM

Y by

A common method,
We can know that $P$ , $X$ ,and $H$ are collinear because
$FH\cdot HC=HE\cdot BH$ That is, H is on the root axis of the $(PXB)$ and $(PXC)$ ,we have
$PH\cdot HX=HE\cdot HB=AH\cdot HD$ which means that $P$ , $A$ , $F$ , $H$ , $E$ lies on a same circle.
Consider $\angle ZPY=\angle ZPX-\angle YPX=\angle EHC=\angle FAC$ ,
so $A,P,Y,Z$ lies on a same circle.
At last, $\angle YZA=\angle APY=\angle APX-\angle YPX=\angle BAH$ which implies that $(AYZ)$ is tangent to $AH$ .

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#11 h Nov 22, 2023, 12:08 PM • 1 Y

Y by starchan

Let $K,X=(BXE)\cap (CXF)$ .

Claim: $AYKZ$ cyclic.
Proof: $180^\circ - \angle AZK = \angle KZC=\angle KXC= \angle BYK = \angle AYK$ .

Claim: $K \in (AFHE)$ .
Proof: $\angle KFH = \angle KFC = \angle KXC = 180^\circ - \angle KXB = 180^\circ - \angle KEB = 180^\circ - \angle KEH$ .

Now $\angle AZY = \angle AKY = \angle YKE - \angle AKE = 180^\circ - \angle YBE - (180^\circ - \angle AFE) = 90^\circ + A - (180^\circ - C) = 90^\circ - B$ .

Let $T$ be a point on line $AH$ on the opposite side of $A$ from $H$ . Then $\angle TAY = \angle HAF = 90^\circ - B = \angle AZY \implies AT$ tangent to $(AYZ) \implies AH$ tangent to $(AYZ)$ .

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#12 h Nov 22, 2023, 2:52 PM

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Let $(BEX)$ intersect $\overline{AC}$ at $P$ and $(CFX)$ intersect $\overline{AB}$ at $Q$ . I claim that $PQYZ$ is cyclic. Indeed, $\measuredangle PYQ=\measuredangle PYB=\measuredangle PEB=90^\circ$ and $\measuredangle PZQ=\measuredangle CFQ=90^\circ$ , as desired. Furthermore, we actually have $90^\circ=\measuredangle PEB=\measuredangle PXB$ and likewise $90^\circ=\measuredangle QFC=\measuredangle QXC$ , so $\overline{PQ}$ is the perpendicular to $\overline{BC}$ at $X$ .

This now implies that
$\measuredangle BAD=\measuredangle BQX=\measuredangle YQP=\measuredangle YZP=\measuredangle YZA,$ done. $\blacksquare$

Remark: Actually, the motivation to add $P$ and $Q$ came from trying to lengthbash and prove $\frac{BF}{CE}=\frac{AY}{AZ}$ . Trying to use power of a point to relate $\frac{AY}{AZ}$ to other things naturally motivates the addition of $P$ and $Q$ . Unfortunately a comedy of errors occurred, and instead of noticing that the PoP implies $PQYZ$ cyclic, I thought that it was sufficient to prove that $PQYZ$ was cyclic to finish the problem (mixed up which equations were true and which ones we wanted to prove), and then found the angle chase that actually finishes the problem anyways

This post has been edited 1 time. Last edited by IAmTheHazard, Nov 22, 2023, 2:58 PM

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#13 h Dec 6, 2023, 11:42 AM

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$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); 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 = -2.368091302636755, xmax = 12.009319165682804, ymin = -6.798044077134987, ymax = 4.09530499803227; /* image dimensions */ pen zzttff = rgb(0.6,0.2,1); pen ffvvqq = rgb(1,0.3333333333333333,0); pen ccqqqq = rgb(0.8,0,0); pen zzttqq = rgb(0.6,0.2,0); pen wwwwww = rgb(0.4,0.4,0.4); pen fuqqzz = rgb(0.9568627450980393,0,0.6); draw((1.8,1.45)--(0.54,-3.47)--(11.46,-3.51)--cycle, linewidth(0.8) + zzttff); /* draw figures */ draw((1.8,1.45)--(0.54,-3.47), linewidth(0.8) + zzttff); draw((0.54,-3.47)--(11.46,-3.51), linewidth(0.8) + zzttff); draw((11.46,-3.51)--(1.8,1.45), linewidth(0.8) + zzttff); draw((xmin, 1.9475806451612907*xmin-4.5216935483870975)--(xmax, 1.9475806451612907*xmax-4.5216935483870975), linewidth(0.4) + zzttff); /* line */ draw((xmin, -0.25609756097560976*xmin-0.5751219512195117)--(xmax, -0.25609756097560976*xmax-0.5751219512195117), linewidth(0.4) + zzttff); /* line */ draw(circle((2.5165882201548744,-1.7014158977192724), 2.652317989777082), linewidth(0.8) + dotted + ffvvqq); draw(circle((7.998697575650328,4.314438152539373), 8.555842850106744), linewidth(0.8) + linetype("2 2") + blue); draw((xmin, -0.5134575569358178*xmin + 2.3742236024844723)--(xmax, -0.5134575569358178*xmax + 2.3742236024844723), linewidth(0.4) + zzttff); /* line */ draw(circle((0.42713749421416547,1.455028800387494), 1.3728717160120598), linewidth(0.8) + linetype("4 4") + ccqqqq); draw(circle((6,-3.49), 5.46003662991376), linewidth(0.4) + linetype("0 3 4 3") + zzttqq); draw((xmin, 272.9999999999993*xmin-489.94999999999885)--(xmax, 272.9999999999993*xmax-489.94999999999885), linewidth(0.8) + ffvvqq); /* line */ draw((0.5593293590296825,0.08853618807578001)--(4.480166912652623,-3.4844328458338922), linewidth(0.8) + linetype("2 2") + wwwwww); draw(circle((1.7954509751416954,0.2081162136832247), 1.241892117836188), linewidth(0.8) + linetype("2 2") + fuqqzz); /* dots and labels */ dot((1.8,1.45),dotstyle); label("$A$", (1.8648878394332957,1.6080952380952362), NE * labelscalefactor); dot((0.54,-3.47),dotstyle); label("$B$", (0.6055411255411272,-3.319098780007873), NE * labelscalefactor); dot((11.46,-3.51),dotstyle); label("$C$", (11.530373868555689,-3.350582447855177), NE * labelscalefactor); dot((1.7909019502833936,-1.0337675726335516),linewidth(4pt) + dotstyle); label("$H$", (1.8491460055096436,-0.9105981896891008), NE * labelscalefactor); dot((2.802035801392842,0.9354971454546074),linewidth(4pt) + dotstyle); label("$E$", (2.8723652105470303,1.0571310507674125), NE * labelscalefactor); dot((1.2025038381018844,-0.8830802512212138),linewidth(4pt) + dotstyle); label("$F$", (0.9361196379378214,-1.083758362849274), NE * labelscalefactor); dot((4.480166912652623,-3.4844328458338922),dotstyle); label("$X$", (4.540999606454154,-3.334840613931525), NE * labelscalefactor); dot((1.633420022111853,0.7995448482462824),linewidth(4pt) + dotstyle); label("$Y$", (1.4398583234946887,0.8052617079889788), NE * labelscalefactor); dot((-0.3769586858431993,2.5677758883832578),linewidth(4pt) + dotstyle); label("$Z$", (-0.30748524203069505,2.6942817788272317), NE * labelscalefactor); dot((0.5593293590296825,0.08853618807578001),linewidth(4pt) + dotstyle); label("$D$", (0.6212829594647792,0.20707201889019877), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$
Let $D = (CDX) \cap (BEX)$
Claim 1: $D-H-X$
Proof
Claim 2: $D$ lies on $(AEFH)$
Proof
Claim 3: $D$ lies on $(AYZ)$
Proof
Claim 4: $90^\circ-\measuredangle B = \measuredangle ADY$
Proof
By Claim 4, we're done

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Ywgh1

#14 h Apr 10, 2024, 2:53 PM

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$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(47.630266781564124cm); 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 = -15.468156288163454, xmax = 32.16211049340067, ymin = -10.461960997973392, ymax = 11.838373744345812; /* image dimensions */ pen ffcqcb = rgb(1.,0.7529411764705882,0.796078431372549); pen qqttzz = rgb(0.,0.2,0.6); pen ttffcc = rgb(0.2,1.,0.8); pen cczzff = rgb(0.8,0.6,1.); pen ffzzcc = rgb(1.,0.6,0.8); pen ffwwzz = rgb(1.,0.4,0.6); pen qqzzff = rgb(0.,0.6,1.); pen ccwwff = rgb(0.8,0.4,1.); pen zzwwff = rgb(0.6,0.4,1.); pen ffttcc = rgb(1.,0.2,0.8); pen eqeqeq = rgb(0.8784313725490196,0.8784313725490196,0.8784313725490196); pen ccccff = rgb(0.8,0.8,1.); /* draw figures */ draw(circle((9.5,-0.12), 5.368769111990194), linewidth(2.) + ffcqcb); draw((4.795733333333338,-2.7071909297052112)--(14.09203126922559,-2.901533857481228), linewidth(2.) + qqttzz); draw((14.09203126922559,-2.901533857481228)--(7.940798182004243,5.017370092627035), linewidth(2.) + qqttzz); draw((7.940798182004243,5.017370092627035)--(4.795733333333338,-2.7071909297052112), linewidth(2.) + qqttzz); draw(circle((7.405173710533244,-0.17979309448498995), 3.632756377688577), linewidth(2.) + ttffcc); draw(circle((12.13886161551904,3.8147482347819444), 6.994520486710662), linewidth(2.) + ttffcc); draw(circle((5.577784426835647,5.066769866033755), 2.3635300600433706), linewidth(2.) + cczzff); draw((7.778009697422992,-2.7695366423782812)--(7.940798182004243,5.017370092627035), linewidth(2.) + ffzzcc); draw((4.795733333333338,-2.7071909297052112)--(10.499503092356784,1.7233738914677879), linewidth(2.) + ffzzcc); draw((14.09203126922559,-2.901533857481228)--(6.049786618615247,0.372876284651286), linewidth(2.) + ffwwzz); draw((7.940798182004243,5.017370092627035)--(6.114438150018684,7.368568537968901), linewidth(2.) + qqttzz); draw((6.049786618615247,0.372876284651286)--(10.499503092356784,1.7233738914677879), linewidth(2.) + qqttzz); draw((5.2291087165406065,2.729100172989231)--(7.303265868975786,3.4515336178153673), linewidth(2.) + ffcqcb); draw(circle((9.443882301279462,-2.80436239359322), 4.649164561683396), linewidth(2.) + qqzzff); draw((5.2291087165406065,2.729100172989231)--(6.049786618615247,0.372876284651286), linewidth(2.) + ccwwff); draw((9.906707826570713,-2.8140379513534004)--(5.2291087165406065,2.729100172989231), linewidth(2.) + linetype("2 2") + zzwwff); draw((5.2291087165406065,2.729100172989231)--(10.499503092356784,1.7233738914677879), linewidth(2.) + qqttzz); draw(circle((7.884680483283708,2.333007699033815), 2.6849489112918357), linewidth(2.) + ffttcc); draw((5.2291087165406065,2.729100172989231)--(6.114438150018684,7.368568537968901), linewidth(2.) + eqeqeq); draw((7.303265868975786,3.4515336178153673)--(7.828562784563172,-0.3513546945594052), linewidth(2.) + ccccff); /* dots and labels */ dot((9.5,-0.12),dotstyle); label("$A$", (9.611912056489656,0.18847898400258614), NE * labelscalefactor); dot((4.795733333333338,-2.7071909297052112),dotstyle); label("$B$", (4.926967782893184,-2.4038568473874613), NE * labelscalefactor); dot((14.09203126922559,-2.901533857481228),dotstyle); label("$C$", (14.203157444614199,-2.5912546183313205), NE * labelscalefactor); dot((7.940798182004243,5.017370092627035),dotstyle); label("$D$", (8.050263965290831,5.341917684958705), NE * labelscalefactor); dot((7.828562784563172,-0.3513546945594052),linewidth(4.pt) + dotstyle); label("$E$", (7.956565079818903,-0.09261767241320215), NE * labelscalefactor); dot((10.499503092356784,1.7233738914677879),linewidth(4.pt) + dotstyle); label("$F$", (10.611366834856904,1.9687578079692454), NE * labelscalefactor); dot((6.049786618615247,0.372876284651286),linewidth(4.pt) + dotstyle); label("$G$", (6.176286255852244,0.6257404495382568), NE * labelscalefactor); dot((7.778009697422992,-2.7695366423782812),linewidth(4.pt) + dotstyle); label("$H$", (7.89409915617095,-2.5287886946833673), NE * labelscalefactor); dot((9.906707826570713,-2.8140379513534004),dotstyle); label("$I$", (10.01794056020135,-2.4975557328593907), NE * labelscalefactor); dot((7.303265868975786,3.4515336178153673),linewidth(4.pt) + dotstyle); label("$J$", (7.425604728811303,3.6865707082879515), NE * labelscalefactor); dot((6.114438150018684,7.368568537968901),linewidth(4.pt) + dotstyle); label("$K$", (6.238752179500197,7.621923898108988), NE * labelscalefactor); dot((5.2291087165406065,2.729100172989231),linewidth(4.pt) + dotstyle); label("$L$", (5.364229248428855,2.9682125863364925), NE * labelscalefactor); dot((5.229108716540603,2.7291001729892286),linewidth(4.pt) + dotstyle); label("$M$", (5.364229248428855,2.9682125863364925), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

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Ywgh1

#15 h Apr 10, 2024, 2:54 PM

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$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10.630266781564124cm); 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 = -15.468156288163454, xmax = 32.16211049340067, ymin = -10.461960997973392, ymax = 11.838373744345812; /* image dimensions */ pen ffcqcb = rgb(1.,0.7529411764705882,0.796078431372549); pen qqttzz = rgb(0.,0.2,0.6); pen ttffcc = rgb(0.2,1.,0.8); pen cczzff = rgb(0.8,0.6,1.); pen ffzzcc = rgb(1.,0.6,0.8); pen ffwwzz = rgb(1.,0.4,0.6); pen qqzzff = rgb(0.,0.6,1.); pen ccwwff = rgb(0.8,0.4,1.); pen zzwwff = rgb(0.6,0.4,1.); pen ffttcc = rgb(1.,0.2,0.8); pen eqeqeq = rgb(0.8784313725490196,0.8784313725490196,0.8784313725490196); pen ccccff = rgb(0.8,0.8,1.); /* draw figures */ draw(circle((9.5,-0.12), 5.368769111990194), linewidth(1.) + ffcqcb); draw((4.795733333333338,-2.7071909297052112)--(14.09203126922559,-2.901533857481228), linewidth(1.) + qqttzz); draw((14.09203126922559,-2.901533857481228)--(7.940798182004243,5.017370092627035), linewidth(1.) + qqttzz); draw((7.940798182004243,5.017370092627035)--(4.795733333333338,-2.7071909297052112), linewidth(1.) + qqttzz); draw(circle((7.405173710533244,-0.17979309448498995), 3.632756377688577), linewidth(1.) + ttffcc); draw(circle((12.13886161551904,3.8147482347819444), 6.994520486710662), linewidth(1.) + ttffcc); draw(circle((5.577784426835647,5.066769866033755), 2.3635300600433706), linewidth(1.) + cczzff); draw((7.778009697422992,-2.7695366423782812)--(7.940798182004243,5.017370092627035), linewidth(1.) + ffzzcc); draw((4.795733333333338,-2.7071909297052112)--(10.499503092356784,1.7233738914677879), linewidth(1.) + ffzzcc); draw((14.09203126922559,-2.901533857481228)--(6.049786618615247,0.372876284651286), linewidth(1.) + ffwwzz); draw((7.940798182004243,5.017370092627035)--(6.114438150018684,7.368568537968901), linewidth(1.) + qqttzz); draw((6.049786618615247,0.372876284651286)--(10.499503092356784,1.7233738914677879), linewidth(1.) + qqttzz); draw((5.2291087165406065,2.729100172989231)--(7.303265868975786,3.4515336178153673), linewidth(1.) + ffcqcb); draw(circle((9.443882301279462,-2.80436239359322), 4.649164561683396), linewidth(1.) + qqzzff); draw((5.2291087165406065,2.729100172989231)--(6.049786618615247,0.372876284651286), linewidth(1.) + ccwwff); draw((9.906707826570713,-2.8140379513534004)--(5.2291087165406065,2.729100172989231), linewidth(1.) + linetype("2 2") + zzwwff); draw((5.2291087165406065,2.729100172989231)--(10.499503092356784,1.7233738914677879), linewidth(1.) + qqttzz); draw(circle((7.884680483283708,2.333007699033815), 2.6849489112918357), linewidth(1.) + ffttcc); draw((5.2291087165406065,2.729100172989231)--(6.114438150018684,7.368568537968901), linewidth(1.) + eqeqeq); draw((7.303265868975786,3.4515336178153673)--(7.828562784563172,-0.3513546945594052), linewidth(1.) + ccccff); /* dots and labels */ dot((9.5,-0.12),dotstyle); label("$O$", (9.611912056489656,0.18847898400258614), NE * labelscalefactor); dot((4.795733333333338,-2.7071909297052112),dotstyle); label("$B$", (4.926967782893184,-2.4038568473874613), NE * labelscalefactor); dot((14.09203126922559,-2.901533857481228),dotstyle); label("$C$", (14.203157444614199,-2.5912546183313205), NE * labelscalefactor); dot((7.940798182004243,5.017370092627035),dotstyle); label("$A$", (8.050263965290831,5.341917684958705), NE * labelscalefactor); dot((7.828562784563172,-0.3513546945594052),linewidth(4.pt) + dotstyle); label("$H$", (7.956565079818903,-0.09261767241320215), NE * labelscalefactor); dot((10.499503092356784,1.7233738914677879),linewidth(4.pt) + dotstyle); label("$E$", (10.611366834856904,1.9687578079692454), NE * labelscalefactor); dot((6.049786618615247,0.372876284651286),linewidth(4.pt) + dotstyle); label("$F$", (6.176286255852244,0.6257404495382568), NE * labelscalefactor); dot((7.778009697422992,-2.7695366423782812),linewidth(4.pt) + dotstyle); label("$D$", (7.89409915617095,-2.5287886946833673), NE * labelscalefactor); dot((9.906707826570713,-2.8140379513534004),dotstyle); label("$X$", (10.01794056020135,-2.4975557328593907), NE * labelscalefactor); dot((7.303265868975786,3.4515336178153673),linewidth(4.pt) + dotstyle); label("$Y$", (7.425604728811303,3.6865707082879515), NE * labelscalefactor); dot((6.114438150018684,7.368568537968901),linewidth(4.pt) + dotstyle); label("$Z$", (6.238752179500197,7.621923898108988), NE * labelscalefactor); dot((5.2291087165406065,2.729100172989231),linewidth(4.pt) + dotstyle); label("$L$", (5.364229248428855,2.9682125863364925), NE * labelscalefactor); dot((5.229108716540603,2.7291001729892286),linewidth(4.pt) + dotstyle); label("$M$", (5.364229248428855,2.9682125863364925), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Sketch :

1) Show that $M-H-X$ collinear.

2) Show that $M \in (AEFH)$ .

3) Show that $M \in (AZY)$ .

This post has been edited 4 times. Last edited by Ywgh1, Apr 10, 2024, 3:04 PM

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