Perpendicular following tangent circles

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Perpendicular following tangent circles

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Source: China Team Selection Test 2016 Test 2 Day 2 Q6

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#1 h Mar 21, 2016, 5:53 AM • 8 Y

Y by rkm0959, anantmudgal09, Saki, Davi-8191, tenplusten, Kobayashi, Adventure10, Mango247

The diagonals of a cyclic quadrilateral $ABCD$ intersect at $P$ , and there exist a circle $\Gamma$ tangent to the extensions of $AB,BC,AD,DC$ at $X,Y,Z,T$ respectively. Circle $\Omega$ passes through points $A,B$ , and is externally tangent to circle $\Gamma$ at $S$ . Prove that $SP\perp ST$ .

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#2 h Mar 21, 2016, 6:42 AM • 10 Y

Y by High, pi37, Dukejukem, Davi-8191, Idio-logy, No16, guptaamitu1, Adventure10, Mango247, TheHimMan

Let $Q$ be the intersection of $XY$ and $ZT$ and $P'$ be the intersection of $YZ$ and $XT$ . By Pascal on $XYYZTT$ , $Q$ , $C$ , and $P'$ are collinear. By Pascal on $XXYZZT$ , $A$ , $Q$ , and $P'$ are collinear. Hence, $P'$ is on $AC$ . Similarly, we can show that $P'$ is on $BD$ , so $P'=P$ .

Now, invert about $P$ fixing $\Gamma$ , and denote by $K'$ the image of point $K$ under this inversion. Since line $ABX$ is tangent to $\Gamma$ , the circumcircle of $PA'B'T$ is tangent to $\Gamma$ . Because $ABCD$ is cyclic, $A'B'$ is parallel to line $CD$ , which is the tangent to $\Gamma$ at $T$ . This implies that $TA'=TB'$ . If $T^*$ is the antipode of $T$ with respect to $\Gamma$ , then clearly the circumcircle of $T^*A'B'$ is also tangent to $\Gamma$ at $T^*$ , so $T^*=S'$ . But this means that $\angle PST=\angle PXS'=TXT^*=90^\circ$ , as desired.

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#3 h Mar 21, 2016, 6:54 AM • 10 Y

Y by mjuk, CeuAzul, tofu_, No16, Arefe, enhanced, tiendung2006, guptaamitu1, Adventure10, Mango247

Let $J$ be the pole of $SX$ WRT $\Gamma$ and let $V$ $\equiv$ $AB$ $\cap$ $YZ$ . It's well-known that $P$ lies on $XT,$ $YZ$ and $(A,B;V,X)$ $=$ $-1$ , so from ${JX}^2$ $=$ $JA$ $\cdot$ $JB$ we get $J$ is the midpoint of $VX$ , hence $J$ is the circumcenter of $\triangle SVX$ . Since $TX,$ $YZ$ is parallel to the bisector of $\angle (AB,CD),$ $\angle (BC,DA)$ , respectively, so $\angle XPV$ $=$ $90^{\circ}$ $\Longrightarrow$ $P,$ $S,$ $V,$ $X$ lie on a circle with diameter $VX$ , hence we conclude that $\measuredangle TSP$ $=$ $\measuredangle XSP$ $-$ $\measuredangle XST$ $=$ $\measuredangle XVP$ $-$ $\measuredangle VXP$ $=$ $90^{\circ}$ .

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#4 h Mar 21, 2016, 7:10 AM • 5 Y

Y by mjuk, Dukejukem, No16, Adventure10, Mango247

Let $O$ and $K$ be the centers of $\odot(ABCD)$ and $\Gamma.$ $E \equiv AD \cap BC$ and $F \equiv AB \cap CD.$ Since $E(F,Y,Z,P)=-1,$ then the pole of $EF$ WRT $\Gamma$ is on $EP$ and similarly it must be on $FP$ $\Longrightarrow$ $P$ is the pole of $EF$ WRT $\Gamma$ $\Longrightarrow$ $X,T,P$ are collinear and $EF \perp PK$ at $Q,$ i.e. $K \in OP.$ Since the polar of $Q$ WRT $(O)$ passes through $P,$ then it follows that $P,Q$ are the limiting points of $\Gamma,(O).$ Thus if the common tangent of $\Gamma,\Omega$ cuts $AB$ at $M,$ we get $MS^2=MX^2=MA \cdot MB$ $\Longrightarrow$ $\odot(M,MS)$ is orthogonal to $(O)$ and $\Gamma$ $\Longrightarrow$ $P \in \odot(M,MS)$ $\Longrightarrow$ $\angle PST=\angle MSP+\angle MST =90^{\circ}-\angle SXT+\angle SXT=90^{\circ}.$

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#5 h Mar 21, 2016, 10:15 AM • 2 Y

Y by Adventure10, Mango247

Let $AD\cap BC=E, AB\cap CD=F, XY\cap TZ=G, XZ\cap TY=H, YZ\cap AB=Q$ . Note that polar of $H$ wrt $\Gamma$ is $AC$ , polar of $G$ is $BD$ , hence polar of $GH$ must be $P=XT\cap YZ$ . $YZ,XT$ are parallel to the internal angle bisectors of $\angle AEB,\angle AFB$ , which are also parallel to the internal bisectors of $\angle APB,\angle APD$ since $ABCD$ is cyclic $\implies PZ\perp PX$ , and $PQ$ is the angle bisector of $\angle APB\implies (A,B;Q,X)=-1$ .

Let the tangent at $S$ to the two circles meet $AB$ at $M$ . $MX^2=MS^2=MA\times MB$ , so $M$ is the midpoint of $QX$ , combining with $MS=MX\implies QS\perp SX$ . Hence $Q,P,S,X$ lie on a circle with diameter $QX\implies \angle PSQ=\angle PXQ=\angle TSX\implies \angle PST=\angle QSX= 90^{\circ}$ .

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#6 h Mar 21, 2016, 11:19 PM • 4 Y

Y by kapilpavase, guptaamitu1, Adventure10, Mango247

Let $O$ be the center of $\Gamma$ and let $U$ be the antipode of $T$ WRT $\Gamma.$ Set $A', B'$ as the midpoints of $\overline{XY}, \overline{XZ}$ respectively. It's enough to show that $U, P, S$ are collinear.

Since $A, C, P$ are collinear, their polars WRT $\Gamma$ are concurrent. Hence, the polar of $P$ passes through $XZ \cap YT.$ Similarly, the polar of $P$ passes through $XY \cap ZT.$ Thus, by Brokard's Theorem, $P \equiv XT \cap YZ.$

On the other hand, since $ABCD, AXOY, CZOT$ are cyclic, we have
$\begin{align*} \angle XOY = 180^{\circ} - \angle XAY = 180^{\circ} - \angle BAD = \angle BCD = \angle ZCT = 180^{\circ} - \angle ZOT = \angle ZOU. \end{align*}$ It follows that $XYZU$ is an isoceles trapezoid. In particular, $XU \parallel YZ$ and consequently $XT \perp YZ.$
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Meanwhile, inversion in $\Gamma$ sends $\odot(ABS) \mapsto \odot(A'B'S).$ Therefore, $\odot(A'B'S)$ is tangent to $\Gamma$ as well. Hence, $K \equiv XX \cap SS \cap A'B'$ is the radical center of $\Gamma, \odot(A'B'S), \odot(A'B'X).$ Clearly $KS = KX$ by equal tangents; also $KX = KP$ because $A'B'$ is the perpendicular bisector of $\overline{XP}.$ Therefore, $K$ is the circumcenter of $\triangle SXP.$ Thus, if $J$ is the reflection of $X$ in $K$ , we have $(JP \parallel XU) \perp XP.$ By the converse of Reim's Theorem for $XXJ$ and $UPS$ , it follows that $U, P, S$ are collinear, as desired.
_________________________________________________________________________________________________
Remark: This problem is essentially 2011 ISL G4, with $\triangle XYZ$ as the reference triangle.

This post has been edited 3 times. Last edited by Dukejukem, Mar 22, 2016, 2:51 PM

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suli

#7 h Mar 22, 2016, 5:16 AM • 4 Y

Y by High, RAMUGAUSS, Adventure10, Mango247

Projective KO:

1. Lemma: $P$ is $YZ$ intersect $TX$ .
Proof.
1a. With respect to $\Gamma$ , the pole of $P$ is the line through the polars of $BD$ and $AC$ .
1b. But the polar of $BD$ is the intersection of the poles of $B$ and $D$ , or $XY$ and $TZ$ .
1c. Similarly $AC$ polar is intersection of $XZ$ and $YT$ .
1d. Thus pole of $P$ is line through aforementioned intersections. But by Brokard's theorem the polar of this line is also the intersection of $YZ$ and $TX$ .

2. $YZ$ intersect $AB$ at $M$ such that $A, B; M, X$ harmonic.
Proof.
2a. $BY$ and $BX$ are tangents, so $ZA, BY, ZB, ZX$ form harmonic bundle.
2b. Thus $ZA, ZM, ZB, ZX$ are harmonic, so $A, B; M, X$ harmonic.

3. $\angle ZPX = 90^\circ$ by easy angle chasing because $ABCD$ is cyclic, so $\angle A + \angle D = 180^\circ$ .
4. Because the two circles are tangent at $S$ , easy angle chasing show that $SX$ is external angle bisector of $ABS$ .
5. Thus by famous lemma $\angle MSX = 90^\circ$ .

6. $MPSX$ is cyclic.
7. $PST$ and $MSX$ are spirally similar, so $\angle PST = \angle MSX = 90^\circ$ .

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6870 posts

#8 h Apr 2, 2016, 12:03 AM • 6 Y

Y by anantmudgal09, A576, myh2910, HolyMath, guptaamitu1, Adventure10

Solution with Danielle Wang: Ignore $ABCD$ cyclic for now, and focus entirely on $\Gamma$ .

Let $Q$ be the inverse of $P$ with respect to $\Gamma$ . Since $P = AC \cap BD$ , it follows $P$ lies on the polars of $\overline{TY} \cap \overline{XZ}$ and $\overline{YZ} \cap \overline{TX}$ . By Brokard's Theorem, this implies $P = \overline{YZ} \cap \overline{XT}$ . Therefore $Q$ is the Miquel point of cyclic quadrilateral $YTXZ$ . Next let $\gamma$ be the circumcircle of $\triangle PQX$ , and $M$ its center. Thus $\gamma$ is orthogonal to $\Gamma$ . So if $W$ is the second intersection of $\Gamma$ and $\gamma$ , then $\overline{OW}$ and $\overline{OX}$ are tangents to $\gamma$ . Angle chasing, $\angle WPT = \frac12 \angle WMX = 90^{\circ} - \angle WOX = \angle WTX - 90^{\circ} \implies \angle PWT = 90^{\circ}.$ Thus we have shown that a circle centered at $M \in AB$ passes through $P$ , $W$ , $Q$ , $X$ with $\angle PWT = 90^{\circ}$ .

Now suppose $ABCD$ is cyclic, centered at $N$ with circumcircle $\omega$ . If $E = AB \cap CD$ , $F = BC \cap DA$ , then $Q \in EF$ , $PQ \perp EF$ , so $Q$ is the Miquel point of cyclic quadrilateral $ABCD$ . Consequently, $N$ , $P$ , $Q$ , $O$ collinear, and $P$ and $Q$ are inverses with respect to both $\omega$ and $\Gamma$ . Now the circle with diameter $PQ$ is orthogonal to both $\omega$ and $\Gamma$ , thus the midpoint $H$ of $PQ$ is on the radical axis of $\omega$ and $\Gamma$ . Thus from $HM \perp PQ$ , $M$ lies on this radical axis as well. Then $MA \cdot MB = MW^2$ , so $W = S$ and we're done.

$[asy]size(16cm); pair O = origin; pair X = dir(-90); pair T = dir(210); pair Y = dir(165); pair Z = T/Y*(-X); pair A = conj(2/(X+Z)); pair B = conj(2/(Y+X)); pair C = conj(2/(Y+T)); pair D = conj(2/(T+Z)); pair E = conj(2/(T+X)); pair F = conj(2/(Y+Z)); draw(unitcircle, blue); pair P = extension(A, C, B, D); pair Q = extension(P, O, E, F); draw(A--B--C--D--cycle, orange); draw(A--C, orange); draw(B--D, orange); draw(F--C--E, orange); draw(D--Z, orange); draw(B--X, orange); draw(circumcircle(A, B, C), orange+dashed); draw(Y--Z--T--X--cycle, lightblue+1.5); draw(Y--P--T, lightblue+1.5); pair N = circumcenter(A, B, C); draw(N--O, heavygreen); pair H = midpoint(P--Q); draw(CP(H, P), deepgreen+dashed); pair M = circumcenter(P, Q, X); draw(H--M, heavygreen); draw(CP(M, X), lightolive); pair L = -T; pair W = foot(T, P, L); draw(P--W--T, mediumgreen); dot("$O$", O, dir(45)); dot("$X$", X, dir(-45)); dot("$T$", T, dir(T)); dot("$Y$", Y, dir(Y)); dot("$Z$", Z, dir(Z)); dot("$A$", A, dir(225)); dot("$B$", B, dir(-45)); dot("$C$", C, dir(170)); dot("$D$", D, dir(90)); dot("$E$", E, dir(-90)); dot("$F$", F, dir(F)); dot("$P$", P, dir(135)); dot("$Q$", Q, dir(Q)); dot("$N$", N, dir(N)); dot("$H$", H, dir(225)); dot("$M$", M, dir(-90)); dot("$W$", W, dir(45)); /* Source generated by TSQ */ [/asy]$

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#10 h Apr 6, 2016, 8:44 PM • 2 Y

Y by omarius, Adventure10

$(1)$ : $XT\cap YZ=P$ .

proof:

let $TZ\cap XY=R,XZ\cap TY=W$ .

apply pascal respectively to the hexagons:

$(TTXYYZ);(XXTZZY);(YYZXXT);(TTXZZY)$ to see that $XT\cap YZ\in AC\cap BD=P$ .
next let $G$ the intersection of the tangent at $P$ to $(ABP)$ and $AB$ .
$(2)$ : $S\in \Gamma(G,r=GP)$ .

proof:

on the one hand:

$XPG=XPB-GPB=180-AXT-XBD-BAC=ABD-BAC+180-AXT$

on the other hand $2AXT=180-BAD-ADC$ , equating the latters yellds $XPG=GXP$ ,

hence $GX^2=GP^2=GB.GA\rightarrow$ $G$ is on the radical axis of $(XYZ);(\Omega)$ i.e lays on the tangent of $(XYZ)$ at $S$ .

$(3)$ : $TSP=90$

proof:

$TSP=PSG+GST=90-1/2PGS+TXS=90-PXS+PXS=90$ which ends the proof.

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#12 h May 13, 2016, 10:33 PM • 2 Y

Y by Adventure10, Mango247

First by Pascal on $XXTYYZ$ and $TTXZZY$ , we get that $A$ , $C$ , and $XT\cap ZY$ are collinear. Applying Pascal again on $XXTZZY$ and $YYXTTZ$ gives that $B$ , $D$ , and $XT\cap ZY$ are collinear, so we conclude that $P\equiv XT\cap ZY$ . Then
$\angle ZYX=\frac{\angle ZOX}{2}=\frac{180^\circ - \angle ZBX}{2}=\frac{\angle ABC}{2}=\frac{180^\circ-\angle ADC}{2}=\frac{\angle TDY}{2}=\frac{180^{\circ}-\angle TOY}{2}=90^\circ-\angle TXY$ so that $\angle YPZ=90^\circ$ . If $F\equiv YZ\cap AB$ , then $(A,B;F,X)\stackrel{Y}{=}(D,B;P, XY\cap BD)\stackrel{X}{=}(DX\cap \odot(XYZ),X;T,Y)=-1$ which, combined with $\angle FPX=90^{\circ}$ , gives that $PF$ and $PX$ are the internal and external angle bisectors of $\angle APB$ . Then $\odot(FPX)$ is the $P$ -Apollonius circle in $\triangle APB$ , so its center $M$ lies on the common internal tangent of $\odot(ASB)$ and $\odot(XYZ)$ by the radical axis theorem, hence $S$ lies on $\odot(FPX)$ as well. Finally, $\angle SPT=\angle SPX=\angle SFX$ and $\angle STP=180^{\circ}-\angle STX=\angle SYX=\angle SXF$ whence $\angle PST=\angle FSX=90^\circ$ , as desired.

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Saki

#13 h Jun 6, 2016, 5:42 AM • 1 Y

Y by Adventure10

Ignore cyclicity of $ABCD$ , and move $T$ . then $BD\cap YZ\rightarrow D\rightarrow T\rightarrow C\rightarrow AC\cap YZ$ are homography, which is identity on $T=X,Y,Z$ , hence is identity for all $T\in \Gamma$ . So $P\in YZ$ , similarly $P\in XT$ .
\Define
$M$ := $YZ\cap AB$ , $U$ := $AD\cap BC$ , $J$ := $SS\cap AB$
\end Define;
Since $ABU$ and $YZX$ are perspective, by FT of projective geometry $(A,B;M,X)=-1$ .
So $JX^2=JS^2=JA\cdot JB$ implies that $J$ is midpoint of $MX$ and $\angle MSX=\pi/2$ .
Also $YZ\perp TX$ , hence $\angle MPX=\pi/2$ , which means $\angle PSM=\angle PXM=\angle TSX$ .
Therefore $\angle PST=\pi/2$ , which means $PS\perp ST$ , as desired.
\return;

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#15 h Nov 30, 2018, 9:53 AM • 2 Y

Y by karitoshi, Adventure10

Consider inversion wrt $\Gamma$ . Vertices of $ABCD$ go to midpoints of $XZ,XY,TY,ZT$ and still form a cyclic quadrilateral; since it's also a parallelogram, it must be a rectangle. Then $YZ \perp XT$ . Let $M,N$ be the midpoints of $XY,XZ$ . Then $(MSN)$
and $\Gamma$ are tangent.
$AC$ is a pole of $XZ \cap YT$ and $DB$ is a pole of $XY \cap ZT$ . Then by Brocard's thm their intersection, $P$ , also is an intersection of $XT$ and $YZ$ .
Introduce points $U$ and $V$ where the tangent from $X$ to $\Gamma$ meets $MN$ and $YZ$ . Then $US$ is also tangent to $\Gamma$ and $U$ is the midpoint of $XV$ . Now $UV=UX=US$ , so $\angle XSV=90^{\circ}=\angle XPV$ , so $XSPV$ is cyclic and $\angle PSV=\angle PXV=\angle XST$ . Finally, $\angle PST=\angle PSV+\angle VST=\angle XST+\angle VST=\angle VSX=90^{\circ}$ .

This post has been edited 3 times. Last edited by Mindstormer, Nov 30, 2018, 9:55 AM
Reason: Typoes

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#16 h Nov 6, 2019, 3:38 PM • 1 Y

Y by Adventure10

Let $Q$ be the inverse of $P$ wrt $\Gamma$ , let $O$ be the center of $\Gamma$ , and let $R_1=YT\cap XZ$ , $R_2=YX\cap TZ$ .

Claim. $Q$ is the Miquel point of quadrilateral $YTZX$ .
Proof. Because $YT$ is the polar of $C$ and $XZ$ is the polar of $A$ , we know that $AC$ is the polar of $YT\cap XZ=R_1$ . Similarly $BD$ is the polar of $R_2$ . Thus $P=AC\cap BD$ is the pole of $R_1R_2$ . This means that $Q$ lies on $R_1R_2$ and $OQ\perp R_1R_2$ , which implies that $Q$ is the Miquel point of cyclic quadrilateral $YTZX$ .

Consider the circumcircle of $PQX$ , and redefine $S$ as the other intersection between $\omega=(PQX)$ and $\Gamma$ .

Claim. $\angle PST=90^{\circ}.$
Proof.
$\begin{align*} \angle PST &= \angle PSX+\angle XST\\ &= 90^{\circ} -\angle XQR_2 +\angle XZT\\ &= 90^{\circ} -\angle XZR_2 +\angle XZT\\ &= 90^{\circ}. \end{align*}$ Next, observe that $\omega$ is orthogonal to $\Gamma$ . Let $O'$ be the center of $\omega$ . Then $O'$ lies on $AX$ and $O'S$ is tangent to $\Gamma$ because $\omega$ and $\Gamma$ are orthogonal. Hence, to prove that $(ABS)$ is tangent to $\Gamma$ , we only need $O'P^2=O'S^2=OA\cdot OB$ , which is equivalent to $\angle BPO'=\angle CAB$ .

Claim. $PO'\parallel CD$ , which implies $\angle BPO'=\angle CAB$ .
Proof.
$\begin{align*} \angle AO'P &= 180^{\circ} - \angle XO'P\\ &= 180^{\circ} - 2\angle XSP\\ &= 2\angle XST\\ &= \angle XOT\\ &= \measuredangle(AX, DT). \end{align*}$

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sa2001

#17 h Apr 10, 2020, 9:38 PM

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Can't believe I get to say this, but not that hard for China TST #6!

Let $DA \cap CB \equiv M$ , and $BA \cap CD \equiv L$ . Let $O$ be the center of $\Omega$ .

We claim that $P \equiv TX \cap YZ$ , and $TX \perp YZ$ .
Assume WLOG that $M$ is closer to $AB$ , and $L$ is closer to $AD$ . Then, $\angle TYZ = \angle TOZ/2 = \angle CDA/2$ . Similarly, $\angle XTY = \angle ABC/2$ . Thus $TX \perp YZ$ .
Now, wrt $\Omega$ , $B$ is pole of $YX$ , $D$ is pole of $ZT$ , so $BD$ is polar of $YX \cap ZT$ . Similarly, $AC$ is polar of $YT \cap ZX$ , so, by Brokard's theorem, $P \equiv BD \cap AC$ satisfies $P \equiv YZ \cap XT$ .

Invert about $\Omega$ . $A^*$ and $B^*$ are midpoints of $ZX, YX$ respectively, and $S^* \equiv S$ . As $X \neq S$ , and $S^*A^*B^*$ is tangent to $\Omega$ , $S$ is the $2011 G4$ point, wrt $XZY$ , with $X$ being the vertex at top. Note $P$ is foot of altitude from $X$ to $YZ$ . Let $PS$ meet $\Omega$ again in $X'$ . Then, $XX' \parallel ZY$ . Then $\angle TXX' = \angle TPY = 90$ , so $\angle TSX' = 90$ , so $ST \perp SP$ , and we're done.

This post has been edited 2 times. Last edited by sa2001, Apr 10, 2020, 9:39 PM
Reason: Typo

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626 posts

#18 h Sep 16, 2020, 1:56 AM • 1 Y

Y by Lcz

Claim: $P$ lies on lines $XT$ and $YZ$ .

Proof: Take a homography fixing $\Gamma$ and sending $XZYT$ to a rectangle; the claim becomes trivial (we don't need $ABCD$ cyclic).

Let the image of $ABCD$ after inversion about $\Gamma$ be $A'B'C'D'$ . Note this is the Varignon parallelogram of $XYTZ$ , and since it is cyclic, it must be a rectangle. This gives $YZ \parallel C'D' \perp C'B' \parallel XT$ so in fact $P$ is the foot from $X$ to $Z$ . Since $(A'B'S)$ is tangent to $\Gamma$ by 2011 G4 we have that if $PS$ intersects $\Gamma$ again at $X'$ , then $XX' \parallel YZ$ . Thus $90^\circ = \angle X'XT = \angle X'ST = \angle PST$ as desired.

This post has been edited 2 times. Last edited by VulcanForge, Sep 16, 2020, 1:59 AM

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#19 h Oct 25, 2020, 10:41 AM • 3 Y

Y by Afternonz, Tudor1505, R8kt

This is a very long and messy solution but this was what I really thought.

Let $\omega$ be the circumcircle of $\square ABCD$ . We fix $\omega$ and $\Gamma$ . By Poncelet porism, we can vary $D$ along $\omega$ and the conditions still hold.

Claim $P$ is a fixed point.

Let $\ell$ be one of the external tangents of $\Gamma$ and $\omega$ and touch $\omega$ at $X$ . Another tangent line from $X$ touches $\Gamma$ and intersect $\omega$ at $Z'$ and $Y$ respectively. By Poncelet porism, Another tangent line from $Y$ to $\Gamma$ is also tangent to $\omega$ and intersect $\ell$ at $Z$ . By DDIT, the pairs $(\overline{DA},\overline{DC}),(\overline{DX},\overline{DY}),(\overline{DZ},\overline{DZ'})$ . Hence, $AC,XY,Z_0Z_0'$ are concurrent ( $Z_0=AZ \cap \omega,Z_0'=AZ' \cap \omega$ ). Let $P' = XY \cap Z_0Z_0'$ . $(X,Y;P',Z') \overset{Z_0'}{=} (X,Y;Z_0,D)=-1$ . Therefore, $P'$ is fixed and $AC$ always passes through $P'$ . Similarly, $BD$ also passes though $P'$ . So, $P'=P$ . By symmetry, $P$ lies on the line $\ell_0$ passing through the centers of $\Gamma,\omega$ . But $(X,Y;P,Z')=-1$ . Polar of $P$ , $\ell_P$ , wrt. $\Gamma$ passes though $Z'$ and $Z$ and perpendicular to $\ell_0$ . Let $N$ be the midpoint of $Z'P$ . $NZ^2=NY \cdot NX$ . So radical axis of $\Gamma$ and $\omega$ is the midline of $\ell_P$ and $P$
$[asy] import geometry; size(300); pair X,Y,Z,M,Ix,Z1,A,Z0,Z01; X=dir(-40); Y=dir(220);Z=dir(90);M=dir(270); triangle t=triangle(X,Y,Z); Ix=excenter(t.BC); Z1=foot(Ix,X,Y); path omega = circle(M,length(M-X)); pair CCC = 2/3*X+1/3*Y; pair CCCC = 2*CCC-Z; A = IP(CCC--CCCC,omega); Z0=IP(Z--CCC,omega); pair[] Z01 = intersectionpoints(A--Z1,omega); clipdraw(omega,deepmagenta); draw(line(A,Z1),fuchsia); drawline(t,blue); draw(line(A,Z),fuchsia); clipdraw(excircle(t.BC),deepmagenta); dot("$D$",A,dir(-60)); dot("$X$",X,dir(60)); dot("$Y$",Y,dir(120)); dot("$Z$",Z,dir(0)); dot("$Z'$",Z1,dir(270)); dot("$Z_0'$",Z01[0],dir(240)); dot("$Z_0$",Z0,dir(60)); pair P = IP(Z0--Z01[0],X--Y); dot("$P$",P,dir(280)); draw(Z0--Z01[0],purple+dashed); draw(box((-2.3,-2), (2,1.5)), invisible); draw(line(Z1,Z),deepcyan); [/asy]$

Assume that $S'$ lies on $\Gamma$ and $\angle PS'T = 90^{\circ}$ . Then, we'll show that $\Omega$ is tangent to $\Gamma$ at $S'$ .

Since, $AB \cap CD = R$ lies on $\ell_P$ . So, the polar of $R$ , $TX$ , passes through $P$ . Let $T'$ be $T$ antipode wrt. $\Gamma$ . $X,S'$ are the foot from $T',T$ to $TP,T'P$ respectively. By Brokard theorem, $H$ , the orthocenter of $\triangle PT'Y$ lies on $\ell_P$ . Since $AB$ intersects the tangent line to $\Gamma$ at $S'$ is the intersection of tangent line at $S',X$ to $\Gamma$ is $M$ , the midpoint of $PH$ (angle chasing after we know that $H$ is orthocenter) lies on radical axis of $\omega,\Gamma$ . So, $MS'^2=MA \cdot MB$ and $\Omega$ is tangent to $\Gamma$ at $S'$ . $S'=S$ .

This post has been edited 1 time. Last edited by k12byda5h, Mar 12, 2023, 2:59 PM

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#20 h Jan 9, 2022, 3:52 AM • 2 Y

Y by centslordm, rama1728

This is actualy a nice thing but u have to focus on $\Gamma$ to win!.
Claim 1: $AC,BD,TX,YZ$ are concurrent at $P$
Proof: Take all the polars here w.r.t. $\Gamma$ (projective is back baby!) and now since $A,P,C$ are colinear we have $\mathcal P_A,\mathcal P_P, \mathcal P_C$ concurrent which means that $YT \cap ZX \in \mathcal P_P$ and since $B,P,D$ are colinear we have $\mathcal P_B, \mathcal P_P, \mathcal P_D$ concurrent which means that $ZT \cap XY \in \mathcal P_P$ but by Brokard the only point satisfying this conditions is $ZY \cap TX$ hence done!.
Finishing: The next move was made because of the cyclic quadrilaterals we get. We make an inversion with center $P$ and radius $\sqrt{PT \cdot PX}$ which means that $\Gamma$ is fixed after this transformation now let $A',B'$ be the inverses of $,B$ respectivily and since the line $ABX$ is tangent to $\Gamma$ we have $(PA'B'T)$ cyclic and tangent to $\Gamma$ and also $ABA'B'$ cyclic, now by angle chase:
$\angle BDC=\angle BAC=\angle A'B'B \implies DC \parallel A'B' \implies \angle TA'B'=\angle DTA'=\angle TB'A' \implies TA'=TB'$ Now let $U$ be the antipode of $T$ w.r.t. $\Gamma$ , since $TU \perp CD$ we have $TU \perp A'B'$ hence $TU$ is perpendicular bisector of $A'B'$ meaning that $UA'=UB'$ and also that the tangent from $U$ to $\Gamma$ is parallel to $A'B'$ meaning that $(UA'B')$ is tangent to $\Gamma$ BUT remember that $(ABS)$ was tangent to $\Gamma$ so by the inversion we have letting $S'$ the inverse of $S$ that $(A'B'S')$ tangent to $\Gamma$ so $S'=U$ and this means that $\angle PST=\angle UXT=90$ thus we are done :blush:

:blush:

This post has been edited 2 times. Last edited by MathLuis, Jan 10, 2022, 2:53 PM

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CyclicISLscelesTrapezoid

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CyclicISLscelesTrapezoid

#21 h Sep 22, 2022, 1:18 AM • 1 Y

Y by crazyeyemoody907

Sketch, with v4913 and crazyeyemoody907.

By Brianchon's on $AXBCTD$ and $AZDCYB$ , $\overline{XY}$ and $\overline{YZ}$ intersect at $P$ . By angle chasing, $\overline{XT}$ and $\overline{YZ}$ are perpendicular. Let $M$ and $N$ be the midpoints of $\overline{XZ}$ and $\overline{XY}$ , respectively, let $T'$ be the antipode of $T$ with respect to $\Gamma$ , and redefine $S$ as the second intersection of $\overline{PT'}$ with $\Gamma$ . By inversion about $\Gamma$ , it suffices to show that $\Gamma$ is tangent to the circumcircle of $SMN$ at $S$ . By angle chasing, $XYZT'$ is a cyclic isosceles trapezoid, so we are done by ISL 2011 G4.

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#22 h Aug 8, 2023, 1:30 AM

Y by

I'm disappointed that no complex bash has been posted yet.

Solution

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cursed_tangent1434

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cursed_tangent1434

#23 h Mar 30, 2025, 1:44 AM

Y by

Pretty easy for CHNTST standards actually. We start off with the following well-known claim.

Claim : Lines $\overline{AC}$ , $\overline{BD}$ , $\overline{XT}$ and $\overline{YZ}$ concur at $P$ .

Proof : By Pascal's Theorem on concyclic hexagons $TTXYYZ$ and $XXYZZT$ it follows that points $C=TT \cap YY$ , $A=XX \cap ZZ$ , $TX \cap YZ$ and $XY \cap ZT$ are all collinear. The first three imply that $P$ lies on $\overline{AC}$ . A similar argument shows that $P$ lies on $\overline{BD}$ , proving the claim.

Now, let $R = AB \cap YZ$ . We identify the tangency point $S$ with the following claim.

Claim : The tangency point $S$ is the second intersection of circle $(RPX)$ with $\Gamma$ .

Proof : First note that,
$\measuredangle ZPX = \measuredangle PZX + \measuredangle ZXP = \measuredangle YZX + \measuredangle ZXT = \measuredangle BYX + \measuredangle ZTD = 90^\circ$ Thus, $\triangle RPX$ is right-angle and hence its circumcenter $O$ is the midpoint of segment $RX$ . Let $S' = (RPX) \cap \Gamma$ . Then, $OX^2 = OS'^2$ which implies that $\overline{OS'}$ is tangent to $\Gamma$ . Further, let $W = BZ \cap \Gamma$ . Note,
$-1=(XY;WZ)_{\Gamma} \overset{Z}{=}(XR;BA)$ which since $O$ is the midpoint of segment $XR$ implies,
$OS'^2 = OR^2=OA \cdot OB$ so $\overline{OS'}$ is also tangent to $(ABS')$ which implies that $(ABS')$ is tangent to $\Gamma$ at $S'$ . Thus, $S' \equiv S$ which implies the claim.

The problem is now reduced to a straightforward angle chase. We observe,
$\measuredangle RSP = \measuredangle RXP = \measuredangle BXT = \measuredangle XST$ Thus,
$\measuredangle TSP = \measuredangle TSR + \measuredangle RSP = \measuredangle TSR + \measuredangle XST = \measuredangle XSR = \measuredangle XPR =90^\circ$ as desired.

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