Circumcircles of two triangles are tangent

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Circumcircles of two triangles are tangent

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Source: Tuymaada 2018 Senior League/Problem 8, Junior League/Problem 8

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#1 h Jul 20, 2018, 3:53 PM • 3 Y

Y by nguyendangkhoa17112003, Adventure10, kiyoras_2001

Quadrilateral $ABCD$ with perpendicular diagonals is inscribed in a circle with centre $O$ . The tangents to this circle at $A$ and $C$ together with line $BD$ form the triangle $\Delta$ . Prove that the circumcircles of $BOD$ and $\Delta$ are tangent.

Additional information for Junior League

Proposed by A. Kuznetsov

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#3 h Jul 20, 2018, 4:38 PM • 2 Y

Y by Adventure10, Mango247

Here's How to Crack It
Let the tangents to $(ABCD)$ at $A$ and $C$ intersect at $K$ and let $H$ be the orthocenter of $\triangle{ABC}$ , let $KD$ intersect $(ABCD)$ at $D$ and $V$ , let $M$ be the midpoint of $AC$ and let $KA$ intersect $(ABCD)$ at $A$ and $R$ . Let $T$ be the intersection of $(AOC)$ and $(AOD)$ , for some well-know reasons $\angle{AVH}=90$ and $AV$ , $BC$ and $KT$ concur at a point $W$ and $W, D$ and $R$ are colinear, now let $AD$ intersect $AK$ and $CK$ at $X_1$ and $X_2$ , for some obvious reasons if $Y_1$ and $Y_2$ are the inverses of $X_1$ and $X_2$ wrt $(K,KA)$ , then $Y_1Y_2$ passes through $W$ , but clearly $WRMV$ is cyclic with diameter $WM$ and $(VMR)$ is the inverse of $(AOD)$ wrt $(K,KA)$ and hence the conclusion.

This post has been edited 2 times. Last edited by rmtf1111, Sep 12, 2018, 2:24 PM

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dangit

#4 h Jul 20, 2018, 6:50 PM • 2 Y

Y by Adventure10, Mango247

Another solution here: let the tangents at $a$ and $C$ to $\odot{ABC}$ meet at $T$ , $M\in{BD\cap{AT}}$ and $N\in{BD\cap{CT}}$ . $X\in{\odot{MNT}\cap{\odot{AOC}}}$ $L\in{XT\cap{BD}}$ $K\in{XO\cap{BD}}$ and finally $P\in{XT\cap{AC}}$ . (i know it hurts seeing all of this, but i thought i'd just write them all from the begining). First notice that $MNT$ is isosceles $XO$ and $XT$ are bisectors of $\angle{CXT}$ and $\angle{NXM}$ , also $m(\angle{TXO})=90$ since $TO$ is diameter in $\odot{AOC}$ . Then we have $(N,M;L,K)=-1$ and also $(P,AC\cap{XO};A,C)=-1$ which gives $(TP,T(AC\cap{XO});TA,TC)=-1$ but intersecting this with $BD$ gives $(L,T(AC\cap{XO})\cap{BD};M,N)=-1$ hence $T(AC\cap{XO})\cap{BD}=K$ , which is only possible if $OK$ passes through $AC\cap{BD}$ , which is our main claim, proved. Now, using the power of point $K$ we get $XBOD$ cyclic, and we are left to prove that $\odot{MNT}$ and $\odot{BOD}$ are tangent at $X$ , which is equivalent to $m(\angle{XNT})+m(\angle{DXO})=90$ but $\angle{XNT}\equiv{\angle{XCO}}$ and since $\angle{TCO}=90$ we are left to prove that $\angle{XCT}\equiv{\angle{XDO}}\iff{\angle{XOT}\equiv{\angle{XDO}}}\iff{\angle{TOB}\equiv{\angle{BDO}}\equiv{\angle{DBO}}}$ which is true since $OT\parallel{BD}$

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#5 h Jul 20, 2018, 9:25 PM • 2 Y

Y by Adventure10, Mango247

ACGNmath wrote:

Quadrilateral $ABCD$ with perpendicular diagonals is inscribed in a circle with centre $O$ . The tangents to this circle at $A$ and $C$ together with line $BD$ form the triangle $\Delta$ . Prove that the circumcircles of $BOD$ and $\Delta$ are tangent.

Proposed by A. Kuznetsov

Nice problem!

Suppose tangents to $\odot(DAB)$ at $D$ and $B$ meet at point $X$ . Let $\overline{XA}$ and $\overline{XC}$ meet $\odot(DOB)$ again at points $Y$ and $Z$ respectively. Let $S=\overline{AC} \cap \overline{BD}$ . Let $W$ be the midpoint of $\overline{AC}$ .

Claim. $\odot(WSY)$ is tangent to line $\overline{BD}$ .

(Proof) Note that $\angle OWA=\angle OYA=90^{\circ}$ so $AOYW$ is cyclic. Thus $\angle AYW=\angle AOW$ . Now we require $\angle SYX=\angle OAC$ . Let $A'$ be the point symmetric to $A$ in line $\overline{BD}$ . Let $R=\overline{XA} \cap \odot(ABCD)$ with $R \ne A$ ; note that $R$ lies on line $\overline{A'M}$ . Observe that $\overline{SY} \parallel \overline{A'M}$ as it is the midline in $\triangle AMA'$ . Thus $\angle SYX=\angle ARM$ . Now $\overline{RM} \cap \odot(ABCD) \overset{\text{def}}{:=} L$ then $\overline{AL} \parallel \overline{BD}$ so $\angle ARM=\angle ACL=\angle OAC$ and we're done. $\blacksquare$

Finally, observe that $\odot(WSZ), \odot(WSY)$ are both tangent to line $\overline{BD}$ . Hence $\odot(WYZ)$ touches line $\overline{BD}$ at point $S$ . Now we invert at $O$ ; clearly $\triangle WYZ$ maps to the triangle formed by $\overline{BD}, \overline{AA}, \overline{CC}$ ; hence the inverse point $T$ of $S$ in $\odot(ABCD)$ is the tangency point of $\Delta$ and $\odot(BOD)$ . $\blacksquare$

Remark. As always, "guessing" the tangency point is the hardest part of the proof. The intuition behind our candidate is as follows. We know $T \in \odot(BOD)$ . Swapping $A$ and $C$ preserves $\Delta$ so whatever way we use to define $T$ given $\triangle DAB$ must also work if we used $\triangle CBD$ instead.

Another idea is that we invert at $\odot(O)$ because (a) images of vertices of $\Delta$ are especially nice; (b) because "line-circle" tangency is convenient to prove as opposed to "circle-circle".

The inverse of $T$ must lie on $\overline{BD}$ . So we seek a point on $\overline{BD}$ symmetric with respect to $A$ and $C$ ; but depending on them. The most natural choice would then be $\overline{AC} \cap \overline{BD}$ ; which happens to work!

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#6 h Jul 21, 2018, 7:51 AM • 2 Y

Y by Adventure10, Mango247

Let $M$ be the Miquel point of the three sides of $\Delta$ and $AC$ , and let $X=AC\cap BD$ . Since $MX$ bisects $\angle AMC$ , hence $O,M,X$ are collinear. We get $OX\cdot OM=OB^2=OC^2$ , thus $M$ lies on $(OBC)$ . But $\Delta$ and $AOC$ are spirally similar about $M$ with a $90$ degree rotation, so $(\Delta)$ is orthogonal to $(AOC)$ which is in turn orthogonal to $(BOD)$ .

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#7 h Jul 21, 2018, 11:29 AM • 2 Y

Y by Adventure10, Mango247

DVDthe1st wrote:

Let $M$ be the Miquel point of the three sides of $\Delta$ and $AC$ , and let $X=AC\cap BD$ . Since $MX$ bisects $\angle AMC$ , hence $O,M,X$ are collinear. We get $OX\cdot OM=OB^2=OC^2$ , thus $M$ lies on $(OBC)$ . But $\Delta$ and $AOC$ are spirally similar about $M$ with a $90$ degree rotation, so $(\Delta)$ is orthogonal to $(AOC)$ which is in turn orthogonal to $(BOD)$ .

Why $MX$ bisect $\angle AMC$ ?

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#8 h Jul 21, 2018, 5:47 PM • 7 Y

Y by qweDota, nguyendangkhoa17112003, AlastorMoody, SenatorPauline, Adventure10, Mango247, Tastymooncake2

$[asy] size(10cm); pair O = origin; pair A = dir(120); pair B = dir(200); pair C = dir(240); pair D = dir(340); draw(unitcircle); draw(A--B--C--D--cycle); draw(A--C); draw(B--D); pair X = extension(A, A+dir(90)*A, C, C+dir(90)*C); pair Y = extension(A, X, B, D); pair Z = extension (C, X, B, D); pair E = extension(A, C, B, D); draw(Y--Z--B); draw(A--X--Y); draw(C--Z--X); draw(circumcircle(B,O,D),red); draw(circumcircle(A,O,C),heavycyan); draw(circumcircle(X,Y,Z),red); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$Z$", Z, dir(Z)); dot("$O$", O, dir(50)); dot("$E$", E, dir(E)); [/asy]$
$\text{Figure 1: The Big Picture}$
Above is the big picture. $E = AC \cap BD, X = AA \cap CC, Y = AX \cap BD, Z = CX \cap BD.$

Now we consider a smaller diagram:
$[asy] size(10cm); pair O = origin; pair A = dir(120); pair B = dir(200); pair C = dir(240); pair D = dir(340); draw(unitcircle); draw(A--B--C--D--cycle); draw(A--C); draw(B--D); pair E = extension(A, C, B, D); draw(circumcircle(B,O,D),red); path w = circumcircle(B,O,D); path ao = (10*A-9*O)--(10*O-9*A); pair G = OP(ao,w); path co = (10*C-9*O)--(10*O-9*C); pair H = IP(co, w); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$O$", O, dir(50)); dot("$E$", E, dir(E)); dot("$G$", G, dir(G)); dot("$H$", H, dir(H)); draw(A--O--G); draw(H--C--O); draw(H--G); [/asy]$
$\text{Figure 2: A Small Configuration}$
Denote $(BOD)=\omega$ .
Next, let $G=AO\cap\omega, H=CO\cap\omega$ .
Now we note that $\angle BAC=90^{\circ}-\angle DBA = 90^{\circ}-\frac{1}{2}\angle DOA = \angle OAD$ .
Hence $\angle BOC=\angle GOD$ .
Which means that $HG$ is parallel to $BD$ .

Next we consider a larger diagram:

$[asy] size(10cm); pair O = origin; pair A = dir(120); pair B = dir(200); pair C = dir(240); pair D = dir(340); draw(unitcircle); draw(A--B--C--D--cycle); draw(A--C); draw(B--D); pair E = extension(A, C, B, D); draw(circumcircle(B,O,D),red); path w = circumcircle(B,O,D); path ao = (10*A-9*O)--(10*O-9*A); pair G = OP(ao,w); path co = (10*C-9*O)--(10*O-9*C); pair H = IP(co, w); draw(A--O--G); draw(H--C--O); draw(H--G); path eg = (10*E-9*G)--(10*G-9*E); path eh = (10*E-9*H)--(10*H-9*E); pair P = IP(eg,w); pair Q = IP(eh, w); draw(G--P); draw(H--Q); pair foot = foot(O, B, D); path ofoot = (10*O-9*foot)--(10*foot-9*O); pair M = OP(ofoot,w); draw(A--Q--M, orange+dashed); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$O$", O, dir(50)); dot("$E$", E, dir(E)); dot("$G$", G, dir(G)); dot("$H$", H, dir(H)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); dot("$M$", M, dir(M)); [/asy]$
$\text{Figure 3: Extending the Configuration}$
Let $P = GE \cap \omega, Q = HE \cap \omega$ .

Now note that $AE\cdot EC = BE\cdot ED = HE\cdot EQ$ , hence $AQCH$ cyclic.
Also note that $\angle AQH + \angle HQM = \angle ACH + \angle HOM = 180^{\circ}-\angle ACO + \angle HOM = 180^{\circ}$
Hence $A, Q, M$ collinear.

$\angle BEP = \angle HGP = \angle HQP = \angle EQP$ .
Hence $(EQP)$ is tangent to $BD$ at $E$ .

$[asy] size(10cm); pair O = origin; pair A = dir(120); pair B = dir(200); pair C = dir(240); pair D = dir(340); draw(unitcircle); draw(A--B--C--D--cycle); draw(A--C); draw(B--D); pair E = extension(A, C, B, D); draw(circumcircle(B,O,D),red); path w = circumcircle(B,O,D); path ao = (10*A-9*O)--(10*O-9*A); pair G = OP(ao,w); path co = (10*C-9*O)--(10*O-9*C); pair H = IP(co, w); draw(A--O--G); draw(H--C--O); draw(H--G); path eg = (10*E-9*G)--(10*G-9*E); path eh = (10*E-9*H)--(10*H-9*E); pair P = IP(eg,w); pair Q = IP(eh, w); draw(G--P); draw(H--Q); pair foot = foot(O, B, D); path ofoot = (10*O-9*foot)--(10*foot-9*O); pair M = OP(ofoot,w); draw(A--Q--M, orange+dashed); pair I = foot(O, A, C); draw(circumcircle(E, Q, P), green); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$O$", O, dir(50)); dot("$E$", E, dir(E)); dot("$G$", G, dir(G)); dot("$H$", H, dir(H)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(240)); dot("$M$", M, dir(M)); dot("$I$", I, dir(I)); [/asy]$
$\text{Figure 4: The Final Configuration}$
Now let $I$ be the midpoint of $AC$ .
We remark that $\angle AIO = 90^{\circ}=\angle AQO$ .
Note that $\angle EQI = 180^{\circ}-\angle AQI-\angle EQM = 180^{\circ}-\angle AOI - \angle HOM = 90^{\circ}$ .

Similarly, we can show that $\angle EPI = 90^{\circ}$ .
Hence $E, Q, P, I$ are concyclic, with circumcircle tangent to $BD$ at $E$ .

Now invert the diagram with respect to $(O)$ . This results in the first diagram, and we are done.

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alex_g

#9 h Jul 22, 2018, 2:38 PM • 2 Y

Y by Adventure10, Mango247

Consider the notations in the attached figure.
$F$ is defined as the second intersection of circles $\odot(AOCP)$ and $\odot(PXY)$ .
Notice that $F$ is the center of spiral similarity sending segment $XY$ to segment $CA$ , and moreover as triangles $PXY$ and $OAC$ are isosceles and similar, it follows that this spiral similarity centered at $F$ also sends $P$ to $O$ . Also, $\angle PFO = 90^{\circ}$ .
$OP \perp CA$ , hence $OP \parallel BD$ . Thus $OP$ is a common tangent of circles $\odot(PXY)$ and $\odot(OBD)$ .
Then the midpoint of this segment lies on the radical axis of the aforementioned circles. Let $L$ be the midpoint of $JE$ .
We want to prove that $L$ lies on the radical axis of circles $\odot(PXY)$ and $\odot(OBD)$ .
Now, Desargues' Involution Theorem yields the existence of some involution on line $BD$ , with pairs of points $(B,D)$ , $(X,Y)$ and $(E,E)$ .
However, by looking at our initial spiral similarity we get that :
$\frac{YJ}{JX} = \frac{AE}{EC} = \frac{DE}{EB}$ .
This means that in the involution on line $BD$ we got earlier, $(J,J)$ is also a pair. As every involution is also an inversion, there is an inversion centered at $L$ , with radius $LE^2 = LJ^2$ , swapping $B$ with $D$ and $X$ with $Y$ .
Thus : $LB \cdot LD = LX \cdot LY$ , so $L$ also lies on the radical axis of circles $\odot(PXY)$ and $\odot(OBD)$ .
So, $LK$ is the radical axis of $\odot(PXY)$ and $\odot(OBD)$ , $K$ being the midpoint of $PO$ . Now, $LK$ is also a median in triangle $FPO$ , so it contains $F$ .
Because $KP = KF$ , $KF$ is a tangent to $\odot(PXY)$ and since $KF$ is the radical axis of $\odot(PXY)$ and $\odot(OBD)$ , we get that these two circles are tangent at point $F$ and the conclusion follows.

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Reason: Typo

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#10 h Jul 24, 2018, 12:12 PM • 2 Y

Y by Adventure10, Mango247

A different solution: Let $BD \cap AA = X, BD \cap CC = Y, AA \cap CC = Z, AO \cap \odot (ABCD) = A'$ .

Note that $OZ \perp AC \Rightarrow OZ \parallel BD \Rightarrow OZ$ is tangent to $\odot (BOD)$ . Also, $\angle ZYX = \angle YZO = \angle AZO = \angle ZXY \Rightarrow ZX = ZY$ .

Now, $\triangle AA'C \sim \triangle ZOA \Rightarrow \frac{AA'}{AC} = \frac{OZ}{ZA} \Rightarrow \frac{2AO}{AC} = \frac{OZ}{ZA}$

And, $\triangle OAC \sim \triangle ZXY \Rightarrow \frac{ZX}{XY} = \frac{AO}{AC} = \frac{OZ}{2ZA} \Rightarrow 2ZA \cdot ZX = XY \cdot OZ$

Also, $XA+CY = (ZX+ZA)+(ZC-ZY) = 2ZA \Rightarrow (XA+YC) \cdot ZX = XY \cdot OZ \Rightarrow XA \cdot ZY+YC \cdot ZX = XY \cdot OZ \Rightarrow ZY \cdot \sqrt{XB \cdot XD} + ZX \cdot \sqrt{YB \cdot YD} - ZO \cdot XY = 0$

Thus, Using the aforementioned equality, and by the converse of Casey's Theorem on point circles $\odot (X), \odot (Y), \odot (Z)$ and circle $\odot (BOD)$ , we get that $\odot (XYZ)$ is tangent to $\odot (BOD)$ .

This post has been edited 2 times. Last edited by math_pi_rate, Jul 26, 2018, 4:42 PM

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#11 h Dec 26, 2018, 1:55 PM • 1 Y

Y by Adventure10

$[asy] size(20cm); pair O = origin; pair A = dir(120); pair B = dir(200); pair C = dir(240); pair D = dir(340); draw(unitcircle, cyan); draw(A--B--C--D--cycle,magenta); draw(A--C); draw(B--D); pair E = extension(A, C, B, D); draw(circumcircle(B,O,D), heavygreen); path w = circumcircle(B,O,D); path ao = (10*A-9*O)--(10*O-9*A); pair G = OP(ao,w); path co = (10*C-9*O)--(10*O-9*C); pair H = IP(co, w); draw(A--O--G); draw(H--C--O); draw(H--G); path eg = (10*E-9*G)--(10*G-9*E); path eh = (10*E-9*H)--(10*H-9*E); pair P = IP(eg,w); pair Q = IP(eh, w); draw(G--P); draw(H--Q); pair foot = foot(O, B, D); path ofoot = (10*O-9*foot)--(10*foot-9*O); pair M = OP(ofoot,w); pair I = foot(O, A, C); draw(circumcircle(A,C,H), red); draw(circumcircle(A, I, Q), orange); draw(circumcircle(E, Q, P), purple); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$O$", O, dir(50)); dot("$E$", E, dir(E)); dot("$G$", G, dir(G)); dot("$H$", H, dir(H)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(240)); dot("$M$", M, dir(M)); dot("$I$", I, dir(I)); [/asy]$

INVERTED FIGURE (after inversion at O with circle (ABCD))

Here’s a sketch.

Angle chasing gives $GH // AB$
Also by PoP, $AQCH$ is cyclic.
Angle chasing gives $A,Q,M$ to be collinear.
Also $A,I,Q,O$ is cyclic.
Angle chasing gives $(EQP)$ to be tangential to BD.
Also $E,Q,P,I$ are concyclic.

In the original figure, after inversion,

$X \Rightarrow I$
$Y \Rightarrow P$
$(BOD) \Rightarrow BD$
$Z \Rightarrow Q$ .

So, we are done.

This post has been edited 11 times. Last edited by MathInfinite, Mar 19, 2019, 4:41 AM

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#13 h Feb 11, 2022, 11:14 AM • 1 Y

Y by DroneChaudhary

In this solution we will convert this to an incircle problem.

Let the tangents to $(O)$ at $A$ and $C$ meet at $Z$
Let the tangents to $(O)$ at $D$ and $C$ meet at $X$
Let the tangents to $(O)$ at $A$ and $D$ meet at $Y$
Let $BD$ meet $ZC$ and $ZA$ at $P$ and $Q$
Let $BD$ intersect $AC$ at $M$
Let $(AZCO)$ intersect $(XYZ)$ at $N$
We will prove that $N$ is the desired point of tangency

By the sharky-devil theorem $O, M, N$ collinear and $DN$ is angle bisector of $\angle XNY$
$\angle MNC=\angle ONC=\angle OZC=\angle MPC$
$\implies MPNC$ cyclic
$\angle ZNP=\angle ZNC-\angle PNC=180-\angle ZAC-\angle PMC=90-\angle ZAC=\angle AZO=\angle ZQP$
$\implies ZPNQ$ cyclic
$MO\cdot MN=MA\cdot MC=MB\cdot MD$
$\implies DOBN$ cyclic
$\angle OCM=\angle CNM$
$\implies OM\cdot ON=OC^2=OD^2$
$\implies \Delta ODM \sim \Delta ODN$
$\implies \angle OMD=\angle ODN$
$\angle ZQN+\angle ODN=\angle NPC+ \angle OMD=\angle NMC+\angle PMN=90=\angle ZNO$
$\implies (ZPNQ)$ and $(DOBN)$ are tangent as desired

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1184 posts

L567

#14 h Mar 22, 2022, 4:51 AM

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Solved with Aayuu

Let the $A,C$ tangents meet $BD$ at $Y,Z$ and intersect each other at $X$ . Let $AC \cap BD = P$ and finally, let $M$ be the miquel point of quadrilateral $ABCD$ which is well known to lie on $(BOD)$ and $(AOC)$ and on line $OP$ .

See that $\angle AMP = \angle AMO = \angle ACO = 90 - \angle ADC = 90 - \angle PAY = \angle AYP$ so $M \in (APY)$ and since it lies on $(AXC)$ too, it must be the miquel point of $PYXC$ and so lies on $(XYZ)$ .

To show tangency, it suffices to show $90 = \angle XMO = \angle XZM + \angle MDO$ . But $\angle XZM + \angle MDO = \angle MYA + \angle MDO = 90 - \angle OPD + \angle MDO = 90$ since $OP.OM = OD^2$ , so we are done. $\blacksquare$

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#15 h Jun 10, 2022, 10:03 AM

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We perform a normal inversion with respect to $(ABCD)$ with center $O$ . Let $E=(AOC)\cap (BOD)$ and $P^*$ is the midpoint of $AC$ . That inversion swaps $\{AC, (AOC)\}, \{PC, (P^*OC)\}$ . The points $Q, E, C, Q^*, E^*$ are concyclic by the obvious reason. Notice $(P^*OC)$ is tangent to $PC$ . Now
$\begin{align*} \angle P^*Q^*O+\angle OQ^*E^*&=\angle P^*CO+\angle QCE^*=\angle PCO=90^\circ \end{align*}$ Means $\angle P^*Q^*E^*$ is right angle and $P^*E^*$ is the diameter of $\triangle P^*Q^*E^*$ . Similarly $\angle P^*R^*E^*=90^\circ$ . So $P^*Q^*E^*R^*$ is cyclic. The line $BD$ is tangent to the cyclic quadrilateral $P^*Q^*E^*R^*$ at $E^*$ . So the cyclic quadrilateral $PQER$ is tangent to $(BOD)$ at $E$ .

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#16 h Aug 15, 2022, 7:17 AM

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Let $AA\cap CC=P, AA\cap BD=Q, CC\cap BD=R$ and $p,q,r$ be the power of $P,Q,R$ to $(BOC)$ respectively . Now $BD$ is radical axis of $(ABCD),(BOC)$ so $q=QA,r=RC$ . Also, $OP$ is tangent to $(BOC)$ so $p=OP$ . Notice that $PQ/QR=OP/(2PA)$ and $PQ=PR,$
$PR\cdot q+PQ\cdot r=PR\cdot QA+PQ\cdot RC=PQ\cdot (QA+RC)=2PQ\cdot PA=QR\cdot OP=QR\cdot p.$ Thus from the converse of Casey's theorem, $(PQR)$ and $(BOC)$ are tangent to each other.

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jelena_ivanchic

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jelena_ivanchic

#17 h Nov 1, 2022, 2:04 AM

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Oo very cute but hard problem.. Thanks to L567( for spoiling the tangency point).
$[asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(12.3cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.8, xmax = 17.5, ymin = -11.9, ymax = 3.2; /* image dimensions */ pen wrwrwr = rgb(0.4,0.4,0.4); draw((2.1,-2.8)--(3.6,1.3)--(10.2,-2.9)--cycle, linewidth(0.4) + wrwrwr); draw((2.1,-2.8)--(3.5,-5.2)--(10.2,-2.9)--cycle, linewidth(0.4) + wrwrwr); /* draw figures */ draw((2.1,-2.8)--(3.6,1.3), linewidth(0.4) + wrwrwr); draw((3.6,1.3)--(10.2,-2.9), linewidth(0.4) + wrwrwr); draw((10.2,-2.9)--(2.1,-2.8), linewidth(0.4) + wrwrwr); draw(circle((6.2,-2), 4.1), linewidth(0.4) + wrwrwr); draw((2.1,-2.8)--(3.5,-5.2), linewidth(0.4) + wrwrwr); draw((3.5,-5.2)--(10.2,-2.9), linewidth(0.4) + wrwrwr); draw((10.2,-2.9)--(2.1,-2.8), linewidth(0.4) + wrwrwr); draw(circle((-0.4,-3), 1.1), linewidth(0.4) + wrwrwr); draw(circle((6.1,-11.7), 9.7), linewidth(0.4) + wrwrwr); draw((0.2,-3.9)--(6.2,-2), linewidth(0.4) + wrwrwr); draw((0.2,-3.9)--(3.5,-5.2), linewidth(0.4) + wrwrwr); draw((3.6,1.3)--(3.5,-5.2), linewidth(0.4) + wrwrwr); draw(circle((2.9,-2), 3.3), linewidth(0.4) + wrwrwr); draw(circle((4.5,-6.8), 5.1), linewidth(0.4) + wrwrwr); draw((0.2,-3.9)--(2.9,-2), linewidth(0.4) + wrwrwr); draw((2.9,-2)--(6.2,-2), linewidth(0.4) + wrwrwr); draw((6.1,-11.7)--(6.2,-2), linewidth(0.4) + wrwrwr); draw((-1.5,-2.8)--(2.1,-2.8), linewidth(0.4) + wrwrwr); draw((-1.5,-2.8)--(3.6,1.3), linewidth(0.4) + wrwrwr); draw((-0.4,-1.9)--(3.5,-5.2), linewidth(0.4) + wrwrwr); /* dots and labels */ dot((2.1,-2.8),linewidth(5pt) + dotstyle); label("$A$", (2.2,-2.7), NE * labelscalefactor); dot((3.6,1.3),linewidth(5pt) + dotstyle); label("$B$", (3.6,1.4), NE * labelscalefactor); dot((10.2,-2.9),linewidth(5pt) + dotstyle); label("$C$", (10.3,-2.7), NE * labelscalefactor); dot((3.5,-5.2),linewidth(4pt) + dotstyle); label("$D$", (3.6,-5.1), NE * labelscalefactor); dot((6.2,-2),linewidth(4pt) + dotstyle); label("$O$", (6.2,-1.8), NE * labelscalefactor); dot((-0.4,-1.9),linewidth(4pt) + dotstyle); label("$E$", (-0.4,-1.8), NE * labelscalefactor); dot((-1.5,-2.8),linewidth(4pt) + dotstyle); label("$F$", (-1.5,-2.7), NE * labelscalefactor); dot((0.7,-2.8),linewidth(4pt) + dotstyle); label("$G$", (0.7,-2.7), NE * labelscalefactor); dot((0.2,-3.9),linewidth(4pt) + dotstyle); label("$M$", (0.3,-3.8), NE * labelscalefactor); dot((3.5,-2.8),linewidth(4pt) + dotstyle); label("$P$", (3.6,-2.7), NE * labelscalefactor); dot((-0.4,-3),linewidth(4pt) + dotstyle); label("$O_1$", (-0.4,-2.9), NE * labelscalefactor); dot((6.1,-11.7),linewidth(4pt) + dotstyle); label("$O_3$", (6.1,-11.6), NE * labelscalefactor); dot((2.9,-2),linewidth(4pt) + dotstyle); label("$O_2$", (2.9,-1.8), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$
We introduce the miquel point, $M$ for quadrilateral $ABCD$ . Define $O_2$ as the center of $BOD$ and $O_3$ as the center of $COA$ .

Claim: $O_2MO_3O$ cyclic quadrilateral
Proof: Note that $O_2O_3$ is the perpendicular bisector of $OM$ . Note that $OO_2||AC,O_3O||BD, \text{ as } BD\perp AC\implies O_2O\perp OO_3\implies O_2MO_3O\text{ cyclic}$
Now, we define $M'=(EFG)\cap (BODE)$ .

Claim: $M'=M$ i.e $O-P-M',P=BD\cap AC.$
Proof: Note that $M'$ is the spiral center taking $E\rightarrow B, F\rightarrow B, G\rightarrow D.$ But note that $\angle EM'O=90\implies \angle FM'B=90=\angle GMD$ $\implies GM'DP, FM'PB\text{ cyclic}$ $\implies \angle PM'D=\angle PGD=\angle EFG=\angle EM'G=\angle OM'D\implies O-P-M'$ $\implies M'=M.$
Define $O_1$ as the center of $(EFG).$ Then note that $\angle O_1MO_2=90.$ But $\angle O_2MO_3=90\implies O_1-M-O_3.$ So $M$ is the tangency point.

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1481 posts

CT17

#18 h Aug 9, 2023, 3:02 AM

Y by

Let $X = AA\cap BD$ , $Y = CC\cap BD$ , $Z = AA\cap CC$ , $E = AC\cap BD$ . We claim the tangency point is the miquel point $M$ of $ABCD$ .

First, we have

$\angle MAX = \angle MAB - \angle BAX = \angle MDC - \angle BDA = 180^\circ - \angle MCD - \angle CBD = \angle YCD - \angle MCD = \angle MCY$
and

$\frac{AX}{CY} = \frac{AE}{CE} = \frac{BA}{BC}\cdot\frac{DA}{DC} = \frac{MA}{MC}$
so $\triangle MAX\sim\triangle MCY$ and hence $M$ is the spiral center mapping $AX\to CY$ . In particular, $M$ lies on $(XYZ)$ . Now since $M$ lies on the line through $AB\cap CD$ and $AD\cap BC$ by pascal, we have $\angle OMZ = 90^\circ$ . To prove the tangency, it suffices to show

$\angle MDO + \angle MYZ = \angle OMZ = 90^\circ.$
We have

$\angle MYZ = 180^\circ - \angle MEC = \angle AEM$
and if $W = BB\cap DD$ ,

$\angle MDO = \angle MWO = 90^\circ - \angle WOM = 90^\circ - \angle AEM$
as desired.

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8857 posts

#19 h Aug 13, 2023, 9:26 PM • 1 Y

Y by axolotlx7

Very clean and beautiful problem statement, a great hard 1/4 or standard 2/5.

First, we look at images of points under inversion about $(ABCD)$ . Let $X = \overline{AA} \cap \overline{BD}$ , $Y = \overline{CC} \cap \overline{BD}$ , and $P = \overline{AA} \cap \overline{CC}$ . Additionally, denote by $E, F, G$ the images of $P, X, Y$ under inversion about $(ABCD)$ . Furthermore, set $M = (AOC) \cap (BOD)$ . Furthermore let $T = \overline{AC} \cap \overline{BD}$ be the image of $M$ .

Claim. $MATFX$ is cyclic (and so is its symmetric variant).

Proof. Note that $(MAT)$ is orthogonal to $(ABCD)$ , so it suffices to show $X$ lies on it. But $\angle ATX = 90^\circ$ and $\overline{OA}$ is tangent to $(MAT)$ by orthogonality, so $\overline{AX}$ is a diameter and the result follows. $\blacksquare$

Now, I claim that $(EFG)$ is tangent to $\overline{BD}$ at $T$ , which will imply the original problem. It suffices to show $\angle TFE = \angle TGE = 90^\circ$ . To do so, note $\angle TFE = \angle TFO + \angle EFO = \angle CAP + \angle EAO = 90^\circ.$ The other angle follows similarly.

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1729 posts

OronSH

#20 h Oct 19, 2023, 7:39 PM

Y by

Invert first about circle $O,$ then about the circle with diameter $AC.$ Denote by $C',D',O'$ the images of $C,D,O,$ and $M$ the midpoint of $AC.$ Let $X$ be the center of circle $AB'CD',$ and let $E,F$ be the intersections of circle $B'D'O'$ with lines $AO'$ and $CO'.$ First, since $M$ is on the polar of $O'$ with respect to circle $X,$ we have that $M,O'$ are images under an inversion at circle $X,$ so circles $B'D'M$ and $B'D'O'$ are images of each other. Thus, since $XM$ is tangent to circle $B'D'M$ we have that $XO'$ is tangent to circle $B'D'O'.$ Since $XO'$ is an angle bisector of lines $AO'$ and $CO',$ we get $EO'=FO'.$ From here, we may let $P$ be the midpoint of $EF.$ A boring trig bash tells us that $PO'$ has the same length as the diameter of circle $B'D'M,$ so $EF$ is tangent to $B'D'M$ at the antipode of $M,$ and inverting back gives us our desired result.

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794 posts

#21 h Dec 25, 2023, 4:46 AM • 1 Y

Y by Math_legendno12

Define the points $R = BD \cap AA$ , $S = BD \cap CC$ , and $T = AA \cap CC$ . Inverting about circle $O$ , we note

$T$ maps to the midpoint of $AC$ , which we denote as $M$ .
$R$ maps to $(AOM) \cap (BOD)$ , which we denote as $P$ .
$S$ maps to $(COM) \cap (BOD)$ , which we denote as $Q$ .

Let $T = AC \cap BD$ . It is clear that $APXR$ and $CQXS$ are cyclic from the right angles. Thus
$\measuredangle MPX = \measuredangle MPR + \measuredangle RPX = \measuredangle MAO + \measuredangle RAX = \measuredangle RAO = 90,$
and analogously for $Q$ , so $MPXQ$ is cyclic, with its circumcircle tangent to $BD$ . Inverting back, we get the desired. $\blacksquare$

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cursed_tangent1434

609 posts

cursed_tangent1434

#22 h Jun 3, 2024, 1:58 AM

Y by

Solved with kingu. Really nice problem which simply dies to inversion. Let $G$ be the intersection of the tangents to $(ABCD)$ at $A$ and $C$ . Let $F$ and $H$ be the intersections of $BD$ and the tangents to $(ABCD)$ at $A$ and $C$ respectively. We define $X= (AOC) \cap \overline{OE}$ . Our goal is to show that this is the desired tangency point.

Consider the inversion $\Omega$ centered at $O$ with radius $OA$ . It is clear that $(ABCD)$ is fixed under this inversion. Further, $G$ maps to the midpoint of $AC$ - $M$ . Now, we do the proof in three parts.

Claim : $X$ lies on $(BOD)$ .
Proof : Note that since $AOCG$ is clearly cyclic, $X$ maps to $\overline{AC} \cap \overline{OE}$ which is simply $E$ . But, since $E$ lies on $\overline{BD}$ , it is clear that $X$ lies on $(BOD)$ as desired.

Claim : $X$ lies on $(FGH)$ .
Proof : Let $H^*$ and $F^*$ be the inverses of $H$ and $F$ respectively under $\Omega$ . It is clear that $H^* = (BOD) \cap (MOC)$ and $F^* = (BOD) \cap (MOA)$ . Now, note that since $O-H-H^*$ under inversion,
$\measuredangle CH^* H = \measuredangle CH^* O = \measuredangle CMO = \measuredangle = 90^\circ = \measuredangle CEH$ Thus, $H^* HCE$ must be cyclic. Now, this allows us to see that,
$\measuredangle MH^* E = \measuredangle MH^* C + \measuredangle CH^* E = \measuredangle MOC = \measuredangle CHE = \measuredangle MOC + 90 + \measuredangle HCE = 90^\circ$ Similarly, we can show that $\measuredangle EF^* M = 90^\circ$ . Thus, $MF^* E H^*$ is cyclic, which implies the claim.

Now, note that circle $(MF^* E H^*)$ is tangent to $\overline{BD}$ at $E$ since $ME \perp BD$ and the center of $(MF^* E H^*)$ lies on $ME$ since we showed $\measuredangle EF^* M = 90^\circ$ . Thus, we must have that the circumcircles of $BOD$ and $\Delta$ are tangent as desired.

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Ywgh1

#23 h Aug 23, 2024, 6:33 AM

Y by

Tuymaada 2018

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10.9064240992493cm); 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 = -247.90374708817487, xmax = 231.0026770110744, ymin = -60.907577424272496, ymax = 213.95772936508536; /* image dimensions */ pen ffcqcb = rgb(1.,0.7529411764705882,0.796078431372549); pen zzffff = rgb(0.6,1.,1.); pen afeeee = rgb(0.6862745098039216,0.9333333333333333,0.9333333333333333); pen ffttww = rgb(1.,0.2,0.4); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen eqeqeq = rgb(0.8784313725490196,0.8784313725490196,0.8784313725490196); pen wwzzff = rgb(0.4,0.6,1.); /* draw figures */ draw(circle((-2.520434060887578,2.5734323754399524), 53.57768597454855), linewidth(1.) + ffcqcb); draw((-52.20454218520992,22.624812235139977)--(42.450367652169426,31.69722724616801), linewidth(1.) + ffcqcb); draw(circle((51.53274545315879,124.50296489253384), 65.43989162003422), linewidth(1.) + zzffff); draw(circle((48.841827430532,7.496365089185839), 51.597646962017876), linewidth(1.) + zzffff); draw((-2.520434060887578,2.5734323754399524)--(50.028156533291515,59.080372254943896), linewidth(1.) + afeeee); draw((-4.877087266520244,27.161019740653998)--(-1.405537984204764,19.223233421990987), linewidth(1.) + ffttww); draw((22.812867954764343,29.8150264460206)--(-1.405537984204764,19.223233421990987), linewidth(1.) + ffttww); draw((-4.877087266520244,27.161019740653998)--(10.73469725235309,42.283773753821364), linewidth(1.) + ffttww); draw((10.73469725235309,42.283773753821364)--(22.812867954764343,29.8150264460206), linewidth(1.) + ffttww); draw(circle((-8.064523620234793,60.416387832445906), 58.108040967178), linewidth(1.) + dtsfsf); draw(circle((30.820139267697137,49.655762419115284), 21.39558827966574), linewidth(1.) + eqeqeq); draw((-52.20454218520992,22.624812235139977)--(8.972773200180354,174.21240239757438), linewidth(1.) + wwzzff); draw((8.972773200180354,174.21240239757438)--(42.450367652169426,31.69722724616801), linewidth(1.) + wwzzff); draw((8.972773200180354,174.21240239757438)--(29.53841590307158,-40.35439744829668), linewidth(1.) + ffcqcb); draw(circle((8.967890344122047,28.488023093337336), 13.908427047557144), linewidth(1.) + dotted + ffttww); draw((-13.608613179582,118.25934328945185)--(42.450367652169426,31.69722724616801), linewidth(1.) + wwzzff); draw((-2.520434060887578,2.5734323754399524)--(-13.608613179582,118.25934328945185), linewidth(1.) + eqeqeq); /* dots and labels */ dot((-2.520434060887578,2.5734323754399524),dotstyle); label("$O$", (-0.8622290228931107,6.9656041602161745), NE * labelscalefactor); dot((-52.20454218520992,22.624812235139977),dotstyle); label("$C$", (-50.607790681089774,26.779514312209763), NE * labelscalefactor); dot((42.450367652169426,31.69722724616801),dotstyle); label("$A$", (44.246034514624206,36.054110553568464), NE * labelscalefactor); dot((22.812867954764343,29.8150264460206),dotstyle); label("$E$", (24.432124362630617,33.946247771441485), NE * labelscalefactor); dot((20.80062641772244,50.8092756099098),linewidth(4.pt) + dotstyle); label("$D$", (22.32426158050364,54.18173047986047), NE * labelscalefactor); dot((29.53841590307158,-40.35439744829668),linewidth(4.pt) + dotstyle); label("$B$", (31.177285265436947,-36.877941708024956), NE * labelscalefactor); dot((-13.608613179582,118.25934328945185),linewidth(4.pt) + dotstyle); label("$H$", (-11.823115489953393,121.63333950792375), NE * labelscalefactor); dot((19.18991088322482,67.61429759206256),linewidth(4.pt) + dotstyle); label("$F$", (21.059543911227454,71.0446327368763), NE * labelscalefactor); dot((8.972773200180354,174.21240239757438),linewidth(4.pt) + dotstyle); label("$G$", (10.520230000592568,177.70248951250136), NE * labelscalefactor); dot((50.028156533291515,59.080372254943896),linewidth(4.pt) + dotstyle); label("$T$", (51.834340530281324,62.61318160836838), NE * labelscalefactor); dot((-4.877087266520244,27.161019740653998),linewidth(4.pt) + dotstyle); label("$K$", (-3.3916643614454833,30.57366732003832), NE * labelscalefactor); dot((10.73469725235309,42.283773753821364),linewidth(4.pt) + dotstyle); label("$L$", (12.628092782719545,45.75027935135256), NE * labelscalefactor); dot((-1.405537984204764,19.223233421990987),linewidth(4.pt) + dotstyle); label("$M$", (0.40248864638307585,22.563788747955808), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Let $H=AA \cap CC$ , $G= CC \cap BD$ and $F= AA \cap BD$ . Let $E$ be the intersection of diagonals of $ABCD$ .
We begin by inverting around $(ABCD)$ , let $L$ and $M$ be the inverses of $F$ and $G$ , we know that $L$ and $M$ lie on $(BOD)$ . Let $K$ be the inverse of $H$ , then $K$ lies on $AC$ , finally let $T$ be the inverse of $E$ . We show that $T$ is the desired tangency point. We can easily get that $FLEAT$ is cyclic, hence by easy angle chase we show that $\angle KLE = KME=90$ , hence $(KLME)$ is cyclic and tangent to $BD$ , inverting back and we are done.

This post has been edited 1 time. Last edited by Ywgh1, Aug 23, 2024, 6:39 AM

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754 posts

#24 h Sep 20, 2024, 7:52 PM

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$\text{[math]}$

Attachments:

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1750 posts

#25 h Sep 22, 2024, 6:00 PM

Y by

Let $\overline{AA} \cap \overline{CC} = T, \overline{AA} \cap BD = X,$ and $\overline{CC} \cap BD = Y.$ Also let $AC$ and $BD$ intersect at $K.$

Now consider inverting with respect to $(ABCD)$ . This sends $T$ to the midpoint $M$ of $AC.$ We will show that it sends the circumcircle of $\Delta$ to the circle with diameter $MK,$ which finishes the problem since $(BOD)$ inverts to $BD.$ It suffices to prove that $X$ is sent to a point on this circle, because similar logic will hold for $Y.$

The inverse of $\overline{AA}$ is the circle with diameter $OA.$ Additionally, if $V = \overline{BB} \cap \overline{DD},$ then $BD$ is sent to the circle with diameter $OY.$ Therefore, $X$ is sent to the intersection of these two circles, which is the foot of the altitude from $O$ to $AV.$ Taking a homothety of scale $2$ at $A,$ the problem reduces to showing that if the reflection of $A$ across $BC$ is $A'$ and $AV$ intersects $(ABCD)$ at point $W,$ then $\angle CWA' = 90^\circ.$ However, it is well-known that $W$ is the Humpty point of $\triangle BA'D,$ which finishes.

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#26 h Sep 22, 2024, 7:29 PM

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Let $AC \cap BD = E$ , $AA \cap BD = P$ , $CC \cap BD = Q$ , $AA \cap CC = R$ . Also let $X$ be the Miquel point of the lines $AA,CC,AC,BD$ .

We claim that $X$ is also the Miquel point of $ABCD$ .
Trivially $X$ lies on $(ACR)$ , and since $\measuredangle OAR = 90^\circ = \measuredangle OCR$ we also have $O$ on $(ACR)$ , so $(ACROX)$ are concyclic.
Note that $RP=RQ$ since $RA=RC$ and $AC \perp BD$ . Since $(AEPX)$ and $(CEQX)$ are concyclic, $\measuredangle AXE = \measuredangle APE = \measuredangle RPQ = \measuredangle PQR = \measuredangle EQC = \measuredangle EXC$ , so $XE$ is an angle bisector of $\angle AXC$ . $XE$ must pass through an arc midpoint of $AC$ on $(ACROX)$ , which is either $R$ or $O$ . But by definition, $X,E,R$ cannot be collinear, so $XE$ passes through $O$ .
The inverted image of $X$ wrt $(ABCD)$ must thus lie on lines $AC$ and $OE$ , so it must be $E$ . This implies that $X$ is the Miquel point of $ABCD$ .

Next we claim that $(BOD)$ and $(XPQR)$ are tangent at $X$ .
Firstly, since $E$ lies on $BD$ , inversion about $(ABCD)$ shows that $X$ lies on $(BOD)$ .
It suffices to show that $\measuredangle OXR = \measuredangle OBX + \measuredangle XPR$ . Clearly $\measuredangle OXR = 90^\circ$ .
Again by inversion about $(ABCD)$ we see that $\measuredangle OBX = - \measuredangle OEB$ , then $\measuredangle OBX + \measuredangle XPR = \measuredangle BEO + \measuredangle XPA = \measuredangle PEX + \measuredangle XEA = \measuredangle PEA = 90^\circ$ , so we are done. $\square$

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3446 posts

#27 h Jan 1, 2025, 6:21 PM • 1 Y

Y by OronSH

We claim that the tangency point is the second intersection of $(AOC)$ and $(BOD)$ (incidentally the Miquel point, though this is not used). Let this point be $E$ . Let $T$ be where the tangents of $(ABCD)$ at $A$ and $C$ meet; let $X$ be where the tangent to $(ABCD)$ at $A$ meets line $BD$ ; let $Y$ be where the tangent to $(ABCD)$ at $C$ meets line $BD$ .

Invert the problem at $O$ and refer to points' images by their original names (except for $O$ ). Note that point $X$ is now the $A$ -dumpty point of $\triangle ABD$ , and point $Y$ is now the $C$ -dumpty point of $\triangle CDB$ . To show that $(TXY)$ is tangent to $\overline{BD}$ at $E$ , it suffices to show that $TE$ is a diameter of $(TXY)$ .

Claim: We have $\angle TYE = 90^{\circ}$ .
Proof: Let $E'$ , $T'$ and $Y'$ be the images of their respective points under a homothety of factor $2$ centered at $C$ . Then, $E'$ is the reflection of $C$ across $\overline{BD}$ ; $T'$ is the orthocenter of $\triangle C'BD$ ; $Y'$ is the harmonic conjugate of $C$ in $\triangle CDB$ , so it is the $C'$ -humpty point in $\triangle C'BD$ . Then, we obviously have $\angle T'Y'E' = 90^{\circ}$ so we're done.

This post has been edited 1 time. Last edited by ihatemath123, Jan 1, 2025, 9:29 PM

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Trasher_Cheeser12321

13 posts

Trasher_Cheeser12321

#28 h Feb 13, 2025, 9:51 AM

Y by

Nice problem with inversion.
$[asy] size(8.5cm); pair A = dir(115); pair C = dir(245); pair B = dir(160); pair D = dir(20); pair E = extension(A, C, B, D); pair O = origin; pair X = extension(A, A+dir(90)*A, C, C+dir(90)*C); pair Z = extension(B, D, C, X); pair Y = extension(A, X, D, Z); pair[] x = intersectionpoints(circumcircle(B, O, D), O--O+(E-O)*50); pair P = x[0]; draw(A--B--C--D--cycle); draw(A--X^^C--Z^^Z--D^^A--C); draw(unitcircle); draw(circumcircle(B, O, D)); draw(circumcircle(A, O, C)); draw(circumcircle(X, Y, Z)); dot("$A$", A, dir(80)); dot("$B$", B, dir(222)); dot("$C$", C, dir(270)); dot("$D$", D, dir(-10)); dot("$E$", E, dir(45)); dot("$O$", O, dir(-45)); dot("$X$", X, dir(235)); dot("$Y$", Y, dir(110)); dot("$Z$", Z, dir(200)); dot("$P$", P, dir(55)); [/asy]$

In the diagram above, define $P$ as the point where $E$ gets mapped to after inverting around $(ABCD)$ . Our goal is to ultimately prove that $P$ is the point of tangency.

Claim. Quadrilateral $APYE$ is cyclic.

Proof. Since $E$ lies on $\overline{AC}$ , $P$ is the second intersection of $\overline{OE}$ and $(AOC)$ . A simple angle chase tells us that
$\angle APE = \angle APO = \angle AXO = \angle AYE$ with the last equality holding since $\overline{YE}\parallel\overline{XO}$ due to both of them being perpendicular to $\overline{AC}$ . $\blacksquare$

Therefore, because $\angle AEY = 90^\circ$ , we know that $\angle APY = 90^\circ$ as well. Now, we can examine the image after we invert about $(ABCD)$ .

First, it's important to note that $X$ gets sent to the midpoint of $\overline{AC}$ . Next, $Y=\overline{BD}\cap \overline{AX}$ implies that $Y^*=(BOD) \cap (AOX^*)$ . Similarly, we can draw $Z^*$ as the intersection of $(BOD)$ and $(COX^*)$ .
$[asy] size(8.5cm); pair A = dir(115); pair C = dir(245); pair B = dir(160); pair D = dir(20); pair E = extension(A, C, B, D); pair O = origin; pair X = 0.5(A+C); pair Z = intersectionpoints(circumcircle(O, X, C), circumcircle(B, O, D))[1]; pair Y = intersectionpoints(circumcircle(O, X, A), circumcircle(B, O, D))[0]; draw(A--B--C--D--cycle); draw(unitcircle); draw(arc(circumcenter(B, O, D), intersectionpoints(circumcircle(B, O, D), circumcenter(B, O, D)--circumcenter(B, O, D)+50*dir(210))[0], intersectionpoints(circumcircle(B, O, D), circumcenter(B, O, D)--circumcenter(B, O, D)+50*dir(330))[0])); draw(circumcircle(X, Y, Z)); draw(circumcircle(X, O, A)); draw(circumcircle(X, O, C)); draw(A--C^^B--D); dot("$A$", A, dir(A)); dot("$B$", B, dir(207.5)); dot("$C$", C, dir(C)); dot("$D$", D, dir(-10)); dot("$E$", E, dir(45)); dot("$O$", O, dir(-90)); dot("$X^*$", X, 1.4dir(302)); dot("$Z^*$", Z, 2dir(35)); dot("$Y^*$", Y, dir(245)); [/asy]$
Here, we want to show that $\overline{BD}$ is tangent to $(X^*Y^*Z^*)$ at point $E$ . We already know that $\overline{X^*E}\perp\overline{BD}$ , so it suffices to show that $\overline{X^*E}$ is the diameter of $(X^*Y^*Z^*)$ . To do this, we will try to prove that $\angle X^*Y^*E = 90^\circ$ using that
$\angle X^*Y^*E = \angle X^*Y^*O + \angle EY^*O$ Two angle chases give us
$\begin{align*} \angle X^*Y^*O = \angle YXO = \angle AXO = \angle APO \qquad \text{and}\qquad \angle EY^*O = \angle YPO \end{align*}$ Adding them up, we conclude that
$\angle X^*Y^*E = \angle APO+\angle YPO = \angle APY$ which we have shown earlier is indeed equal to $90^\circ$ . $\blacksquare$

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#29 h Feb 20, 2025, 4:39 AM

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Let $T$ be the pole of $AC.$ Let $K = AT\cap BD, L = TC\cap BD.$ Invert about $O.$ Let $E$ map to $P.$
Claim: $APKE$ is cyclic.
Proof: First, note that $APTO$ is cyclic since $AET^*$ is collinear. Hence, $\angle APE = \angle APO =\angle ATO = \angle AKE$ which finishes the claim.
Now, by angle chase, $\angle EKT^* = \angle EK^*O = \angle OK^*T^* = \angle OPK + \angle OTK = \angle EAK + \angle OTK = 90.$
By inversion, we also have that $(PKTL)$ maps to $EK^*T^*L^*,$ so that is cyclic. However, since $BD\parallel OT,$ we also have that $(EK^*T^*L^*)$ is tangent to $BD.$ This finishes, as $(BOD)$ maps to $BD$ and $(PKTL)$ maps to $(EK^*T^*L^*)$ under inversion.

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