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jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
IMO Genre Predictions
ohiorizzler1434   70
N a minute ago by HoRI_DA_GRe8
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
70 replies
ohiorizzler1434
May 3, 2025
HoRI_DA_GRe8
a minute ago
Two random bijections
alinazarboland   5
N 10 minutes ago by mathematical-forest
Source: Iran MO 3rd round 2023 , D1P2
Does there exist bijections $f,g$ from positive integers to themselves st:
$$g(n)=\frac{f(1)+f(2)+ \cdot \cdot \cdot +f(n)}{n}$$holds for any $n$?
5 replies
alinazarboland
Aug 16, 2023
mathematical-forest
10 minutes ago
p^3-q^3=pq^3-1
parmenides51   3
N an hour ago by Assassino9931
Source: 2016 Belarus TST 1.3
Solve the equation $p^3-q^3=pq^3-1$ in primes $p,q$.
3 replies
+1 w
parmenides51
Nov 4, 2020
Assassino9931
an hour ago
Show that TA=TM
goldeneagle   59
N an hour ago by Siddharthmaybe
Source: Iran TST 2011 - Day 1 - Problem 1
In acute triangle $ABC$ angle $B$ is greater than$C$. Let $M$ is midpoint of $BC$. $D$ and $E$ are the feet of the altitude from $C$ and $B$ respectively. $K$ and $L$ are midpoint of $ME$ and $MD$ respectively. If $KL$ intersect the line through $A$ parallel to $BC$ in $T$, prove that $TA=TM$.
59 replies
1 viewing
goldeneagle
May 10, 2011
Siddharthmaybe
an hour ago
No more topics!
Geometry with parallel lines.
falantrng   33
N Apr 28, 2025 by joshualiu315
Source: RMM 2018,D1 P1
Let $ABCD$ be a cyclic quadrilateral an let $P$ be a point on the side $AB.$ The diagonals $AC$ meets the segments $DP$ at $Q.$ The line through $P$ parallel to $CD$ mmets the extension of the side $CB$ beyond $B$ at $K.$ The line through $Q$ parallel to $BD$ meets the extension of the side $CB$ beyond $B$ at $L.$ Prove that the circumcircles of the triangles $BKP$ and $CLQ$ are tangent .
33 replies
falantrng
Feb 24, 2018
joshualiu315
Apr 28, 2025
Geometry with parallel lines.
G H J
G H BBookmark kLocked kLocked NReply
Source: RMM 2018,D1 P1
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falantrng
252 posts
#1 • 8 Y
Y by microsoft_office_word, itslumi, k12byda5h, mathematicsy, harshmishra, Adventure10, Rounak_iitr, cubres
Let $ABCD$ be a cyclic quadrilateral an let $P$ be a point on the side $AB.$ The diagonals $AC$ meets the segments $DP$ at $Q.$ The line through $P$ parallel to $CD$ mmets the extension of the side $CB$ beyond $B$ at $K.$ The line through $Q$ parallel to $BD$ meets the extension of the side $CB$ beyond $B$ at $L.$ Prove that the circumcircles of the triangles $BKP$ and $CLQ$ are tangent .
This post has been edited 1 time. Last edited by falantrng, Feb 24, 2018, 12:11 PM
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rmtf1111
698 posts
#2 • 5 Y
Y by microsoft_office_word, Euiseu, translate, Adventure10, cubres
Denote by $\omega$ the circumcircle of $ABCD$. Let $\{T\} = DQ \cap \omega$. By converse of Reim's Theorem on the parallel lines $PK \mid \mid CD$ and circle $\omega$ we have that $BDTK$ is cyclic. By converse of Reim's Theorem on the parallel lines $LQ \mid \mid BD$ and circle $\omega$ we have that $CQTL$ is cyclic. Now because $\angle{ACT}=\angle{ABT}$ we have that the lines tangent to the circumcircles of $QCT$ and $BDT$ at $T$ coincide, thus the circumcircles of the triangles $BKP$ and $CLQ$ are tangent at $T$.
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GGPiku
402 posts
#3 • 3 Y
Y by tbn456834678, Adventure10, cubres
Pretty easy problem compared to the ones from the last year. A bit too easy.
Let $DP$ intersect $(ABCD)$ in $D$ and $S$. We can easily observ that $S$ is on both circumcircles of $BKP$ and $CLQ$.
Indeed, $\angle PSB=180-\angle BCB=\angle PKB$, since $PK\parallel CD$, so $P,B,K,S$ are concyclic, and $\angle QSC=\angle DBC=\angle QLC$, so $Q,C,L,S$ are concyclic.
Now, for an easier explanation, if we let $d$ be the tangent in $S$ at $(BKP)$, and $R$ a point on $d$ such that $L,R$ are on opposite sides wrt $SB$, we'll have $\angle RSP=\angle ABS=\angle ACS=\angle QLS$, so $d$ is tangent in $S$ at $(CLQ)$. This concludes the tangency of $BKP$ and $CLQ$.
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WizardMath
2487 posts
#4 • 3 Y
Y by Adventure10, Mango247, cubres
The intersection of $PD$ and $(ABCD)$ is $X$. $\measuredangle PKB = \measuredangle PXB$, so $PKBX, CQLX$ are cyclic. Now $XK, XB$ are isogonal in $XLC$ so we are done.
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BarishNamazov
124 posts
#5 • 4 Y
Y by tenplusten, Adventure10, Mango247, cubres
Too easy.Let $T=DP\cap \odot (ABCD) $.Easy angle-chasing implies that $CLTQ$,$BKTP $ are cyclic.Again chasing some angles $\angle LTK=\angle BAC=\angle CTB$ which yields that $TK,TB $ are isogonals wrt $TLC $ which finishes problem.
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randomusername
1059 posts
#6 • 2 Y
Y by Adventure10, cubres
Letting $O=DP\cap BC$ (assuming it exists), then Power of a Point wrt secants $OD,OC$ shows that $(BKPT)$ and $(CLQT)$ are cyclic, where $T=(ABCD)\cap OD$. (Since the diagram varies continuously as $P$ varies continuously, this proves they are cyclic even if $DP$ and $BC$ are parallel.)

To show tangency, note that by angle chasing $PTK\sim ATC$ and $ATB\sim QTL$. Hence there exist spiral similarities $\phi,\psi$ centered at $T$ with $\phi(PTK)=ATC$ and $\psi(ATB)=QTL$. Then $\phi\circ \psi$ maps $P\to Q$, hence it's a homothety, and circles $(PTK)$ and $(QTL)$ are homothetic (with center $T$), as desired.
This post has been edited 1 time. Last edited by randomusername, Feb 24, 2018, 6:07 PM
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trumpeter
3332 posts
#7 • 2 Y
Y by Adventure10, cubres
Let $E=PD\cap\left(ABCD\right)$. Then \[\measuredangle{PEB}=\measuredangle{DEB}=\measuredangle{DCB}=\measuredangle{PKB},\]so $EPBK$ is cyclic. Similarly, $EQCL$ is cyclic.

Let $\ell_B,\ell_C$ be the tangents to $\left(EPBK\right),\left(EQCL\right)$ at $E$. Then \[\measuredangle{\left(\ell_B,ED\right)}=\measuredangle{EBP}=\measuredangle{EBA}=\measuredangle{ECA}=\measuredangle{ECQ}=\measuredangle{\left(\ell_C,ED\right)},\]so $\ell_B=\ell_C$ and hence $\left(BKP\right)$ and $\left(CLQ\right)$ are tangent.
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djmathman
7938 posts
#8 • 3 Y
Y by Adventure10, Mango247, cubres
asdf took way too long (and Geogebra help) for me to realize that phantom pointing basically kills this problem; time to draw better diagrams I guess :oops:

Let $X = PQD\cap\odot(ABCD)$. Note that \[\angle PXB \equiv\angle DXB = \angle DAB = \angle PKB,\]so $PBKX$ is cyclic. Similarly, $QCLX$ is cyclic. We can thus get rid of $K$ and $L$, since it suffices to show that the circles $\odot(BPX)$ and $\odot(CQX)$ are tangent to each other. But upon letting $O_B$ and $O_C$ be their respective centers, we obtain \[\angle O_BXP = 90^\circ - \angle XBP = 90^\circ - \angle XCQ = O_CXQ,\]so $X$, $O_B$, and $O_C$ are collinear, implying the tangency. $\blacksquare$
This post has been edited 2 times. Last edited by djmathman, Mar 17, 2018, 11:40 PM
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Kala_Para_Na
28 posts
#9 • 3 Y
Y by Adventure10, Mango247, cubres
$PD \cap \odot ABCD = \{ D,X \}$
By Reim's theorem, $X \in \odot BKP$ and $X \in \odot QLC$
Angle chasing yields, $\angle AXC = \angle PXK$ and $\angle AXB = \angle QXL$ which implies $K$ and $B$ are isogonal in $\triangle XLC$
it leads us to our conclusion.
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Kagebaka
3001 posts
#10 • 3 Y
Y by AlastorMoody, Adventure10, cubres
huh no inversion?

Solution

also @post 2 I think you meant $BPTK$ cyclic not $BDTK$ cyclic
This post has been edited 1 time. Last edited by Kagebaka, Jul 7, 2019, 1:18 PM
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IndoMathXdZ
694 posts
#11 • 2 Y
Y by Adventure10, cubres
Nice Problem :) Here is a boring solution.
Denote the intersection of $PQ$ with $(ABCD)$ as $G$, other than $D$.
We claim that the tangency point is $G$.

Claim 01. $G$ lies on both $(BKP)$ and $(CLQ)$. In other words, $BKGP$ and $CLGQ$ are both cyclic.
Proof. Notice that \[ \measuredangle BKP = \measuredangle BCD = \measuredangle BAD = \measuredangle BGD \equiv \measuredangle BGP\]Similarly,
\[ \measuredangle LQG = \measuredangle BDG = \measuredangle BCG \equiv \measuredangle LCG \]
Claim 02. Let $GK$ intersects $(CQL)$ at $H$. Then , $HQ \parallel PK$.
Proof. We'll prove this by phantom point. Notice that
\[ \measuredangle GHQ = \measuredangle GKP = \measuredangle GBP = \measuredangle GBA = \measuredangle GCA = \measuredangle GCQ \]which is what we wanted.
Therefore, $G$ sends $KP$ to $HQ$, which makes $G$ the tangency point of the two circle.
This post has been edited 1 time. Last edited by IndoMathXdZ, Nov 16, 2019, 7:36 AM
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AlastorMoody
2125 posts
#12 • 4 Y
Y by a_simple_guy, Adventure10, Mango247, cubres
Solution (with PUjnk)
This post has been edited 2 times. Last edited by AlastorMoody, Dec 13, 2019, 7:41 PM
Reason: Give credit to poor PUjnk's soul... lol
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amar_04
1916 posts
#13 • 5 Y
Y by mueller.25, GeoMetrix, BinomialMoriarty, Bumblebee60, cubres
Storage.
RMM 2018 Day 1 P1 wrote:
Let $ABCD$ be a cyclic quadrilateral an let $P$ be a point on the side $AB.$ The diagonals $AC$ meets the segments $DP$ at $Q.$ The line through $P$ parallel to $CD$ mmets the extension of the side $CB$ beyond $B$ at $K.$ The line through $Q$ parallel to $BD$ meets the extension of the side $CB$ beyond $B$ at $L.$ Prove that the circumcircles of the triangles $BKP$ and $CLQ$ are tangent .

Let $DP\cap\odot(ABC)=X$. Then $\angle KPX=\angle CDX=\angle XBK\implies X,K,B,P$ are concyclic. And $\angle XQL=\angle XDB=\angle XCL\implies L,C,Q,X$ are concyclic, Now $$\angle LXK=\angle LXQ-\angle KXP=(180^\circ-\angle BCA)-(180^\circ-\angle ABL)=\angle ABL-\angle BCA=\angle BAC=\angle BXC$$So, $\{XK,XB\}$ are isogonal WRT $\triangle LXC\implies\odot(BKP)$ and $\odot(CLQ)$ are tangent to each other at $X$. $\blacksquare$
This post has been edited 4 times. Last edited by amar_04, Mar 14, 2020, 9:16 PM
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itslumi
284 posts
#14 • 1 Y
Y by cubres
1)Let $(CLQ)n(ABCD)=X$ ,prove that $(BPXK)$-cyclic and $(BPXK)$ tangent to $(CLQ)$.

2)Prove that $X-Q-D$ collnear
$\angle CLQ=\angle CXQ$
but
$\angle CXD=\angle CAD$,which implies that $\angle CXD=\angle CXQ$,which implies the desired collinearity

3)Prove that $(BKXP)-cyclic$

$\angle DCY=\angle DAB=\angle BKP$
and
its obvious that $\angle BXP=\angle BAD$,which implies the desired claim

4)Let $\ell$ be a line that passes through $X$ and tangent to $(CQL)$.Prove that $\ell$ is tangent to $(KBPX)$ at $X$.
$\angle QLX=\angle QX=\angle QCX=\angle ACX=\angle ABX=\angle PKX$
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554183
484 posts
#15 • 1 Y
Y by cubres
:D
Let $PQ \cap \odot{ABCD}=E$. I claim that $E$ is the point of tangency.
First I will prove that $EQCL$ is cyclic
$$\angle{QLC}=\angle{DBC}=\angle{DEC}$$Similarly,
$$\angle{BKP}=180-\angle{DCB}=180-(180-\angle{DEB})=\angle{DEB}$$Hence $EPBK$ is cyclic.
To finish, we present the following claim :
Claim : $\angle{PBE}=\angle{QLE}$
Proof. $$\angle{QLE}=\angle{QCE}=\angle{ABE}=\angle{PBE}$$Now, draw a tangent to $\odot{BKP}$ at $E$. Let $G$ be an arbitrary point on the tangent inside $\odot{ABC}$. We see that
$$\angle{GEQ}=\angle{GEP}=\angle{EBP}=\angle{ELQ}$$So we are done $\blacksquare$
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BVKRB-
322 posts
#16 • 1 Y
Y by cubres
Nice diagram = Problem done! :D (And this time it's on paper!)

Let $\odot(ABC) \cap \odot(LQC) = X$
We claim that $X$ is the desired tangency point

Claim: $D-Q-P-F$ are collinear
Proof

Hence One of $\odot(CLQ) \cap \odot(BKP)=X$

Now draw the line that is tangent to $\odot(LQC)$ at $T$ and name it $\ell$ and let a point on $\ell$ on the same side of $F$ as $L$ be $Y$
We know that $$\angle YFL= \angle  XCL = \angle XQL = \angle  XDB$$and after some easy angle chasing we get $\angle  XPK = \angle  XDC = \angle  YXL + \angle  CXB$ Therefore it suffices to show that $XK,XB$ are isogonal in $\triangle XLC$
This is true after some angle chasing which I am too lazy to write, just assume variables and calculate each angle, which finally gives the desired conclusion $\blacksquare$
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HoRI_DA_GRe8
599 posts
#17 • 1 Y
Y by cubres
solution
This post has been edited 1 time. Last edited by HoRI_DA_GRe8, Nov 22, 2021, 6:24 PM
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REYNA_MAIN
41 posts
#18 • 2 Y
Y by BVKRB-, cubres
Shortage
Just orsing BVKRB
HELP?
Attachments:
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UI_MathZ_25
116 posts
#19 • 1 Y
Y by cubres
The line $DP$ meets the circumcircle of triangle $CLQ$ at $R$. We get

$\angle RDB = \angle RQL = \angle RCL = \angle RCB \Rightarrow$ $R$ lie on $\odot(ABCD)$.

Since $PK \parallel CD$, by the Reim's Theorem we get that $RBKP$ is cyclic.

Finally, let $X$ be the intersection of the tangent to $\odot(CLQ)$ at $R$ with the line $CD$. Thus

$\angle XRQ = \angle RCQ = \angle RCA = \angle RBA = \angle RBP \Rightarrow$ $XR$ is tangent to$\odot (RBP)$.

Therefore, the circumcircles of the triangles $BKP$ and $CLQ$ are tangent to the line $XR$ at $R$ $\blacksquare$
Attachments:
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Reason: Figure
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Mahdi_Mashayekhi
695 posts
#20 • 1 Y
Y by cubres
Let $DQ$ meet $ABCD$ at $S$.
$\angle QLC = \angle DBC = \angle QSC \implies SLCQ$ is cyclic. $\angle BKP = \angle 180 - \angle BCD = \angle BSP \implies BKSP$ is cyclic. Let $L_1$ be line tangent to $SLCQ$ at $S$ we have $\angle SCQ = \angle SCA = \angle SBA = \angle SBP \implies L_1$ is tangent to $BKSP$ as well so our circles are tangent at $S$.
we're Done.
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SatisfiedMagma
461 posts
#21 • 2 Y
Y by Rounak_iitr, cubres
Let $E= DQ \cap \odot(ABCD) \ne D$. We will prove that $E$ is the common point of tangency.

[asy]
            import graph; size(12cm);
            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 = 1.9, xmax = 11.2, ymin = -2.5862804185721906, ymax = 5.031020679728067;  /* image dimensions */
            pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882);
            /* draw figures */
            draw(circle((5.852809635608784,1.367535756966066), 3.5453838571445635), linewidth(0.8) + blue);
            draw((3.741438563859456,4.215668284058398)--(9.,3.), linewidth(0.8) + green);
            draw((9.,3.)--(6.68,-2.08), linewidth(0.8) + green);
            draw((6.68,-2.08)--(2.373492429373072,0.6862880245710273), linewidth(0.8) + green);
            draw((3.741438563859456,4.215668284058398)--(2.373492429373072,0.6862880245710273), linewidth(0.8) + green);
            draw((9.,3.)--(2.373492429373072,0.6862880245710273), linewidth(0.8) + green);
            draw(circle((8.003294213009466,3.084537618648431), 1.0002844769300214), linewidth(0.8) + red);
            draw((9.,3.)--(9.306489730729668,3.6711068241839295), linewidth(0.8) + green);
            draw((9.306489730729668,3.6711068241839295)--(4.739858471858444,2.07662144581455), linewidth(0.8) + green);
            draw((8.719877477199834,2.3866282690410148)--(7.070615206872309,3.4460329938583647), linewidth(0.8) + green);
            draw((4.739858471858444,2.07662144581455)--(8.141511444765205,4.075226788687628), linewidth(0.8) + green);
            draw((8.141511444765205,4.075226788687628)--(9.,3.), linewidth(0.8) + green);
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Claim: $E \in \odot(PBK)$ and $E \in \odot(CQL)$.

Proof: For $PBKE$ we have
\[ \measuredangle DEB = \measuredangle DCB = \measuredangle PKB. \]For $CQLE$ we have
\[ \measuredangle ECL = \measuredangle ECB = \measuredangle EDB = \measuredangle EQL. \]This shows the claim. $\square$

To finish it off, it suffices to show that $\measuredangle EBP = \measuredangle ECQ$ by considering a tangent to either $\odot(PBKE)$ or $\odot(CQLE)$ at $E$. This part is obviously true as
\[\measuredangle EBP = \measuredangle EBA = \measuredangle ECA = \measuredangle ECQ.\]So, we are done. $\blacksquare$
This post has been edited 1 time. Last edited by SatisfiedMagma, May 8, 2022, 6:07 PM
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Mogmog8
1080 posts
#22 • 2 Y
Y by centslordm, cubres
Let $T=(ABC)\cap\overline{PQ}.$ Note $$\measuredangle TPK=\measuredangle PDC=\measuredangle TBC=\measuredangle TBK$$and $$\measuredangle CKQ=\measuredangle CBD=\measuredangle CTD$$so $T$ lies on $(BKP)$ and $(CLQ).$ Let $\ell$ be the line tangent to $(CLQ)$ at $T$; we claim $\ell$ is tangent to $(BKP)$ at $T.$ Indeed, $$\measuredangle (\overline{DT},\ell)=\measuredangle QCT=\measuredangle ACT=\measuredangle ABT.$$$\square$
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IAmTheHazard
5003 posts
#23 • 1 Y
Y by cubres
Neat. Let $X=\overline{DP} \cap (ABCD)$. Reim's implies that $BKPX$ and $CLQX$ are cyclic. Now, to show that the two circles are tangent at $X$, it is sufficient to prove that $\measuredangle XBP=\measuredangle XCQ$ reason. This follows by
$$\measuredangle XBP=\measuredangle XBA=\measuredangle XCA=\measuredangle XCQ.~\blacksquare$$
This post has been edited 1 time. Last edited by IAmTheHazard, Aug 12, 2022, 2:06 PM
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HamstPan38825
8867 posts
#24 • 1 Y
Y by cubres
Let $E = \overline{DP} \cap (ABC)$. I claim that $E$ lies on both $(BKP)$ and $(CLQ)$. This is because $$\measuredangle PKB = \measuredangle DCK = \measuredangle DEB$$and similarly $\measuredangle QLC = \measuredangle DBC = \measuredangle DEC$. Thus it suffices to show that the tangent at $E$ to $(BKP)$ is also tangent to $(CLQ)$, which is equivalent to $\measuredangle EBP = \measuredangle ECQ$. This is evident.
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AwesomeYRY
579 posts
#25 • 1 Y
Y by cubres
Let $X = DP\cap \omega$. Then, note that

Claim 1: $X\in (PBK)$
Proof: Angle chase with the following directed angles:
\[\angle PKB = \angle DCB = \angle DAB = \angle DXB = \angle PXB\]
Claim 2: $X\in (QLC)$.
Proof: Angle chase:
\[\angle XQL = \angle XDB = \angle XCB = \angle XCL\]
Thus, $X = (PKB)\cap (QLC)$. Let $\ell_1$ be the tangent to $(PKB)$ at $X$ and $\ell_2$ be the tangent to $(QLC)$ at $X$. Then, we have
\[\angle (\ell_1, XP) = \angle XBP = \angle XBA = \angle XCA = \angle XCQ = \angle (\ell_2, XQ)\]Since $X,P,Q$ are collinear, this means that $\ell_1$ and $\ell_2$ are the same line, which means that $(PKB)$ and $(QLC)$ share a tangent line with the same orientation at $X$, and therefore the two circles are tangent. $\blacksquare$.
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SQTHUSH
154 posts
#26 • 1 Y
Y by cubres
Let $T= \odot(PKB)\cap \odot(ABCD), S=\odot(QLC)\cap \odot(ABCD$
Since $PK//DC\Rightarrow \angle PKB=\angle PTB=\angle DAB$
So $T,P,D$ is collinear
Similarly, $S,Q,D$ is collinear
It shows that$T=S$
Finally,suppose $l_{1},l_{2}$ through point $T$,and tangent to $\odot(PKB)$,$\odot(QLC)$ respectively
$\measuredangle (l_{1},TK) = \angle TDC= \angle TDB+ \angle BDC= \angle LTK+ \angle TCL=\measuredangle (l_{2},TK)\Rightarrow l_{1}=l_{2}$
Which meas that $\odot(PKB)$and $\odot(QLC)$ are tangent.
This post has been edited 2 times. Last edited by SQTHUSH, Mar 20, 2023, 12:28 PM
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SHZhang
109 posts
#28 • 1 Y
Y by cubres
Let $E = (ABCD) \cap DP$. Then \[\angle EPK = \angle EDC = 180^\circ - \angle EBC = \angle EBK,\]so $(EPBK)$ is cyclic. Similarly \[\angle EQL = \angle EDB = \angle ECB = \angle ECL\]gives $(EQCL)$ cyclic.

Now invert at $E$; in the inverted diagram, we have $ABCD$ collinear, $PEQD$ collinear, $(ABEP)$ cyclic, and $(ACQE)$ cyclic. Then $\angle ACQ = 180^\circ - \angle AEQ = \angle AEP = \angle ABP$, so $BP \parallel CQ$. Inverting back gives $(EPB) = (BKP)$ tangent to $(ECQ) = (CLQ)$, as desired.
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thdnder
198 posts
#29 • 1 Y
Y by cubres
Let $DP$ meets $(ABCD)$ at $T$. Then Reim implies $BKPT$ and $CLQT$ are cyclic. Now by homothety centered $T$, it suffices to show that $\angle TBP = \angle TCQ$, which follows from $\angle TBP = \angle TBA = \angle TCA = \angle TCQ$. $\blacksquare$
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Mathandski
771 posts
#30 • 2 Y
Y by ehuseyinyigit, cubres
All directed angles

Rating (MOHs): 0
Attachments:
This post has been edited 1 time. Last edited by Mathandski, Aug 23, 2024, 9:30 PM
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Saucepan_man02
1350 posts
#31 • 1 Y
Y by cubres
Angle-Chase:

Let $X=DQ \cap (ABC)$. Then: $$\angle XQL = \angle XDB = \angle XCB = \angle XCL \implies X \in (CQL)$$$$\angle PXB = \angle DXB = 180^\circ - \angle DCB = 180^\circ - \angle BKP \implies X \in (BKP).$$Let $T$ be a point on $CD$ such that $TX$ is tangent. Then: $$\angle TXQ=\angle TXP = \angle XBP = \angle XBA = \angle XCA = \angle XCQ$$which implies $XT$ is also tangent to $(CQL)$ and we are done.
This post has been edited 1 time. Last edited by Saucepan_man02, Nov 18, 2024, 6:28 AM
Reason: EDIT
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math004
23 posts
#32 • 1 Y
Y by cubres
Let $X=(DP)\cap (ABCD)$ other than $D.$ $(PK)\parallel CD$ so by Reim's theorem, we have $XPBK$ is cyclic. Similarily, $(QL) \parallel (BD)$ implies, by Reim's theorem, that $(XQCL)$ is cyclic. Thus, it suffices to prove that $\angle XBP=\angle XCQ.$ Indeed,
\[\angle XBP= \angle XBA=\angle XCA=\angle XCQ.\]

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[/asy]
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Ilikeminecraft
658 posts
#33 • 1 Y
Y by cubres
Let $E = DP \cap (ABCD)$ that isn't $D.$ By the given conditions, $\angle EBK = \angle EDC = \angle EPK$ and $\angle QEC = \angle DBC = \angle QLC,$ which tells us $BKEP, CLEQ$ are both concyclic. To finish, note that $EBAC$ is concyclic, so $\angle EBP = \angle ECA.$
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endless_abyss
51 posts
#34 • 1 Y
Y by cubres
The parallel condition is practically begging us to angle chase.

Claim - The tangency point is none other than the intersection of $P Q$ and the circumcircle of $A B C D$

Note that -
Let $T$ denote the intersection of $P Q$ and the circumcircle of $A B C D$
$\angle B A D = \angle B K P = \angle B T D$
and
$\angle D T C = \angle M L C$

$\square$

:starwars:
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joshualiu315
2534 posts
#36
Y by
Let $\overline{DP}$ intersect $(ABCD)$ again at $E$. We claim that $E$ is the desired tangency point. First note that

\[\angle EPK = \angle EDC = \angle EBK,\]
and

\[\angle EQL = \angle EDB = \angle ECB = \angle ECL.\]
This means that $E$ lies on both $(PKB)$ and $(QLC)$, so it suffices to show that $\angle EBP = \angle ECQ$, but this is just a simple angle chase:

\[\angle EBP = \angle EBA = \angle ECA = \angle ECQ. \blacksquare\]
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