egmo 2018 p5

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microsoft_office_word

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microsoft_office_word

#1 h Apr 12, 2018, 11:06 AM • 7 Y

Y by mkhayech, Mathuzb, Davi-8191, tiendung2006, Mathlover_1, Adventure10, Rounak_iitr

Let $\Gamma$ be the circumcircle of triangle $ABC$ . A circle $\Omega$ is tangent to the line segment $AB$ and is tangent to $\Gamma$ at a point lying on the same side of the line $AB$ as $C$ . The angle bisector of $\angle BCA$ intersects $\Omega$ at two different points $P$ and $Q$ .
Prove that $\angle ABP = \angle QBC$ .

This post has been edited 2 times. Last edited by djmathman, Apr 23, 2018, 1:41 PM

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#2 h Apr 12, 2018, 11:17 AM • 16 Y

Y by microsoft_office_word, Anar24, Ankoganit, Snakes, Wizard_32, AlastorMoody, Dindin, Aryan-23, geometry6, BVKRB-, HoRI_DA_GRe8, ike.chen, Ru83n05, Schur-Schwartz, Adventure10, Mango247

Does EG stand for easy geometry? Let the bisector intersect the circumcircle at $M$ and $AB$ at $D$ . And let Omega intersect the circumcircle at $T$ and $AB$ at $S$ . From homothety follows that $M$ , $S$ and $T$ are colinear. By shooting lemma and PoP $MD\cdot MC=MS\cdot MT=MQ\cdot MP=MB^2$ , thus $MB$ is tangent to the circumcircle of $BQP$ , which implies that $\angle{MBA}+\angle{AMQ}=\angle{MBQ}=\angle{BPM}=\angle{PBM}+\angle{PCB} \ \ \ \blacksquare$

This post has been edited 2 times. Last edited by rmtf1111, Apr 12, 2018, 11:19 AM

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#3 h Apr 12, 2018, 11:31 AM • 21 Y

Y by microsoft_office_word, Anar24, Math-Ninja, cip999, richrow12, Ankoganit, e_plus_pi, tapir1729, anantmudgal09, k.vasilev, v4913, CrazyMathMan, HoRI_DA_GRe8, megarnie, centslordm, HamstPan38825, Mathlover_1, Minkowsi47, Adventure10, Mango247, Rounak_iitr

If we let $M$ denote the midpoint of arc $\widehat{AB}$ then the inversion at $M$ with radius $MA = MB$ fixes $\Omega$ , so it swaps $P$ and $Q$ , thus $\angle MPB = \angle QBM$ .

But $\angle MPB = \tfrac12 \angle C + \angle CBP$ and $\angle QBM = \angle QBA + \tfrac12 \angle C$ , implying desired isogonality.

This post has been edited 1 time. Last edited by v_Enhance, Apr 12, 2018, 3:46 PM
Reason: add a line break

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#4 h Apr 12, 2018, 11:37 AM • 7 Y

Y by v_Enhance, Ankoganit, math_pi_rate, AlastorMoody, guptaamitu1, Adventure10, Mango247

Note that isogonal conjugate of a point on the bisector is a projective map. Since vertex of the bisector swaps with the foot, incenter and $C$ -excenter are fixed; we see that this projective map coincides with inversion $\odot(AIB)$ . As $P, Q$ are inverses, we get $\angle ABP=\angle QBC$ as desired.

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Yaghi

#5 h Apr 12, 2018, 11:38 AM • 4 Y

Y by microsoft_office_word, RopuToran, Adventure10, Mango247

Let I be the incenter of $ABC$ and $M$ be the midpoint of arc $AB$ and let the tangency point of $\Omega$ and $AB,\Gamma$ be $K,L$ respectively,then obviously, $K,L,M$ are collinear.So by POP:
$MK.ML=MP.MQ=MB^2=MI^2$ Which implies that $I$ is the foot of angle bisector of $B$ in triangle $\triangle BPQ$ .so we are done.

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#6 h Apr 12, 2018, 12:44 PM • 11 Y

Y by rmtf1111, Tawan, math_pi_rate, ManuelKahayon, AlastorMoody, Modesti, guptaamitu1, RopuToran, GioOrnikapa, Adventure10, Mango247

Let $\Omega$ touches $AB, \Gamma$ at $K, T$ . Let $TC\cap\Omega = L$ and $M$ be the midpoint of arc $AB$ not containing $C$ . By homothety, $T,K,M$ are colinear and $KL\parallel CM$ .

By Desargues' Involution Theorem on $KTKL$ and line $CM$ , there exists involution swapping $(P,Q), (AB\cap CM, C)$ and $(M, {\infty}_{CM})$ Pencil from B gives involution swapping $(BA, BC), (BM_C, B\infty_{CM}), (BP,BQ)$ but the first two pairs are isogonal w.r.t. $\angle ABC$ so we are done.

EDIT : I have attached the link to my own article on Desargues' Involution Theorem.

This post has been edited 1 time. Last edited by MarkBcc168, Apr 13, 2018, 10:20 AM

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#7 h Apr 12, 2018, 12:53 PM • 2 Y

Y by Adventure10, Mango247

P1 was also geometry. Geoff rule??

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#8 h Apr 12, 2018, 3:45 PM • 1 Y

Y by Adventure10

Another inversion approach:

Let $D$ be the tangency point of $\Omega$ with $AB$ , $M$ be the midpoint of arc $\widehat{AB}$ and $T$ be the tangency point of $\Gamma$ and $\Omega$ . Inverting trough $D$ with radius $DA.DB$ (directed lenghts), $\Gamma$ is fixed. Thus, $\Omega$ becomes the tangent line to $\Gamma$ through $M$ , which is parallel to line $AB$ . Note that such inversion swaps $A$ and $B$ , $M$ and $T$ .

By the inversion, we have that $\angle{ABP} = \angle{DP'A}$ and $\angle{CBQ} = \angle{B} - \angle{DBQ} = \angle{B} - \angle{DQ'A}$ . Through easy angle chasing, we can see that $\angle{DP'A} + \angle{DQ'A} = \angle{B} \iff \angle{MP'A} = \angle{MAQ'}$ , which is equivalent to $MA$ be tangent to $(AP'Q')$ . By Archimedes's Lemma, we have that $MA^2 = MD.MT$ . Moreover, as $P, Q, M$ are collinear, we know that $D, T, P', Q'$ are concyclic. So $MD.MT = MP'.MQ' \Rightarrow MA^2 = MP'.MQ'$ , as we wanted to show. $\blacksquare$

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#9 h Apr 12, 2018, 3:58 PM • 2 Y

Y by Adventure10, Mango247

Let $M$ be the midpoint of arc $AB$ not containing $C$ , and let $CM$ meet $AB$ at $N$ ; let the circle be tangent to $AB$ at $X$ and the circle at $Y$ ; by shooting $MP\cdot MQ = MX\cdot MY = MN\cdot MC = MB^2$ ; hence $(PQB)$ and $(BNC)$ are tangent at $B$ ; let $BP, BQ$ meet $(BNC)$ at $U,V$ , then $UV\parallel NC$ and the result follows.

This post has been edited 1 time. Last edited by Muriatic, Apr 16, 2018, 5:02 AM

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#10 h Apr 12, 2018, 4:49 PM • 3 Y

Y by Jafarly8097, Adventure10, Mango247

Let the $\Omega$ intersect the circumcircle at $T$ and $AB$ at $K$ and let the angle bisector of $\angle BCA$ intersect $\Gamma$ at point $M$ .
By homothety $M,K,T$ are collinear . Because $ATCM$ is cyclic we have $\angle {ATM}=\angle{ACM}$ ,but $\angle {MAB}=\angle{MCA}$ .So we have $\angle {ATM}=\angle{MAB}$ , so $MA$ is tangent to $(KAT)$ . $MA^2=MK*MT$ , but $MX*MY=MQ*MP$ and $MA^2=MB^2$ , so $MB^2=MQ*MP$ .Then we have $MB$ tangent to $(PQB)$ , so $\angle MBQ= \angle MPB$ . $\angle ABQ$ = $\angle MBQ$ - $\angle MBA$ = $\angle MPB$ - $\angle MCB$ = $180$ - $\angle BPC$ - $\angle PCB$ = $\angle PBC$ . So $\angle ABQ$ = $\angle PBC$ which is similar to $\angle ABP = \angle QBC$

This post has been edited 1 time. Last edited by BogdanB, Apr 12, 2018, 4:52 PM
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#11 h Apr 12, 2018, 7:17 PM • 6 Y

Y by microsoft_office_word, BogdanB, AlastorMoody, Aliosman, Adventure10, Mango247

We know $\Omega$ is a $\text{curvilinear incircle}$ of $\triangle ABC.$
Let $M=CP\cap \Gamma,$ $T=\Omega\cap \Gamma,$ and $K=\Omega\cap AB.$
Lemma:
$T,K,M$ are collinear.
Proof:Angle-chasing and homotety
From this lemma we can get $MB^2=MK\cdot MT=MQ\cdot MP,$
then $MB$ is tangent to $(PQB),$
We know $\angle PBC=\angle MPB-\angle MCB=\angle MBQ-\angle MBA=\angle ABQ.$ As desired.

This post has been edited 6 times. Last edited by falantrng, Mar 7, 2019, 6:27 AM

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#13 h Apr 12, 2018, 10:05 PM • 5 Y

Y by microsoft_office_word, Pluto1708, Adventure10, Mango247, Rounak_iitr

Let $M_C$ be the second intersection of $\Gamma$ and the angle bisector of $\angle{BCA}$ . Invert about $M_C$ with radius $M_CA$ . Then $\Gamma$ and line $AB$ get swapped, while the center of $\Omega$ and the center of its image must lie on a line through $M_C$ . But there is only one circle centered on this line in this position that is tangent to $AB$ and internally tangent to $\Gamma$ , so $\Omega$ is fixed under this inversion. So the power of $M_C$ with respect to $\Omega$ is $M_CA$ . But it is also $M_CP\cdot M_CQ$ , so $\triangle{M_CAP}\sim\triangle{M_CQA}$ . So
$\begin{align*} \measuredangle{PAC}&=\measuredangle{BAC}-\measuredangle{BAP}\\ &=\measuredangle{BAC}-\measuredangle{BAM_C}-\measuredangle{M_CAP}\\ &=\measuredangle{BAC}-\measuredangle{BCM_C}+\measuredangle{APM_C}+\measuredangle{PM_CA}\\ &=\measuredangle{BAC}-\measuredangle{BCM_C}+\measuredangle{APM_C}+\measuredangle{CM_CA}\\ &=\measuredangle{BAC}-\measuredangle{BCM_C}+\measuredangle{APM_C}+\measuredangle{CBA}\\ &=\measuredangle{BCA}+\measuredangle{M_CCB}+\measuredangle{APM_C}\\ &=\measuredangle{M_CCA}+\measuredangle{APM_C}\\ &=\measuredangle{BCM_C}+\measuredangle{APM_C}\\ &=\measuredangle{BAM_C}+\measuredangle{M_CAQ}\\ &=\measuredangle{BAQ}, \end{align*}$ so $AP$ and $AQ$ are isogonal. But $CP$ and $CQ$ are isogonal, so $P$ and $Q$ are isogonal conjugates. Then $BP$ and $BQ$ are isogonal, so $\angle{ABP}=\angle{QBC}$ .

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328 posts

#14 h Apr 13, 2018, 5:54 AM • 2 Y

Y by Adventure10, Mango247

Let $M_C$ denote midpoint of arc $AB$ not containing $C$ .

Its well-known that $\text{Pow}(M_C, \Omega)=M_CB^2$ , so $M_CB^2=M_CP \cdot M_CQ$ , thus by similarity $\angle M_CBP=\angle M_CQB$ .

But, $\angle M_CBP=\angle M_CBA+\angle ABP=\angle ABP+\frac{1}2 \angle C$ and $\angle QBM_C=\angle QBC+\angle QCB=\angle QBC+ \frac{1}2 \angle C$ ; thus $\angle ABP=\angle QBC$ as desired. $\square$

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Loppukilpailija

155 posts

Loppukilpailija

#15 h Apr 13, 2018, 11:53 AM • 2 Y

Y by Adventure10, Mango247

Denote by $T$ the tangent point between the two circles.

By easy (trivial?) angle chasing and similar triangles it's enough to prove $MP \cdot MQ = MB^2$ , where $M$ is the midpoint of arc $\widehat{AB}$ . Now, do inversion around $M$ with radius $MA = MB$ .

After inversion, $A', B', C', T'$ lie on a line, and so do $M, C', P', Q'$ . We also know that the circumcircle of $P'Q'T'$ is tangent to the line $A'B'$ , and the same circle is tangent to the circumcircle of $A'B'M$ . The condition to be proved is basically exactly the same as before: $MQ' \cdot MP' = MB'^2$ . This is a well-known basic configuration: we have a circle $\Omega$ (circumcircle of $A'B'M$ ), a chord of $\Omega$ ( $A'B'$ ), and a circle $\omega$ tangent both to the chord and $\Omega$ (circumcircle of $P'Q'T'$ ). The midpoint of the arc between the ends of the chord ( $M$ ) is on the same line as the tangent points of $\omega$ with $\Omega$ and $A'B'$ . From here it's power of a point and easy similar triangles to prove $MQ' \cdot MP' = MB'^2$ .

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#16 h Apr 15, 2018, 1:35 PM • 2 Y

Y by Adventure10, Mango247

Let $D$ be a point on segment $AB$ such that $CD$ is tangent to $\Omega$ at $Y$ . Let $X$ be the tangency point on $AB$ and $C'=CI \cap AB$ and $M$ be the midpoint of arc $AB$ not containing $C$ .

It is well-known $XIY$ are colinear. Also note $\measuredangle CYI=\measuredangle IXC'$ as they both subtend arc $XY$ hence using sine rule and angle bisector theorem on $\triangle AC'C$ we see:
$\frac{C'X}{CY}=\frac{C'I}{CI}=\frac{AC'}{AC}$ Now by power of a point:
$\frac{C'P}{PC} \cdot \frac{C'Q}{QC}=\left (\frac{C'X}{CY} \right)^2=\left (\frac{AC'}{AC} \right)^2$ But it's well-known this means $P,Q$ are isogonal in $\triangle ACC'$ and hence by symmetry $P,Q$ are isogonal conjugates in $\triangle ABC$ as desired.

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#17 h Apr 20, 2018, 5:21 AM • 2 Y

Y by Adventure10, Mango247

Let $X$ be the midpoint of the arc $AB$ not containing $C$ . Also let $Y$ and $Z$ be the tangency point of $\Omega$ with $AB$ and $\Gamma$ respectively. $D$ is the feet of angle bisector of $\angle ACB$ . $I$ denotes the incenter of $\triangle ABC$

Lemma 1: $X,Y,Z$ collinear.
Proof: Let $O_1$ and $O_2$ denotes the circumcenters of $\Omega$ and $\Gamma$ respectively. We know that, $Z, O_1, O_2$ collinear.
Also, $YO_1 \perp AB$ and $XO_2 \perp AB$ implies $YO_1 || XO_2$ . Now, homothety gives us $X,Y,Z$ collinear. $\square$

Lemma 2: $Y,Z,C,D$ cyclic.
Proof: $\angle YZC = \angle XZC = \angle XAC = \angle A + \frac {\angle C}{2}$
$\angle YDX = \angle A + \frac {\angle C}{2}$ $\square$

Lemma 3: $XB$ is tangent to $\bigodot PQB$
Proof: $\angle XBD = \angle DCB = \frac {\angle C}{2} \Rightarrow XB$ is tangent to $\bigodot BDC \Rightarrow XD.XC=XB^2$
$XP.XQ = XY.XZ = XD.XC = XB^2$ $\square$

Lemma 4: $BI$ bisects $\angle QBP$
Proof: Lemma 3 gives us $XP.XQ=XB^2=XI^2$ $\Rightarrow \frac{XP}{XI}= \frac {XI}{XQ}$ $\Rightarrow \frac{XP-XI}{XI}= \frac {XI-XQ}{XQ}$ $\Rightarrow \frac{PI}{XI}= \frac {QI}{XQ}$ $\Rightarrow \frac{PI}{QI}= \frac {XI}{XQ} = \frac{XP}{XI} = \frac{XP}{XB}$ $\triangle XPB \sim \triangle BPQ \Rightarrow \frac{XP}{XB}=\frac{BP}{BQ}$ $\therefore \frac{PI}{QI} = \frac{BP}{BQ}$ $\square$

Now our problem statement follows immediately.
$Q.E.D$

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Promi

#18 h Apr 20, 2018, 4:00 PM • 5 Y

Y by microsoft_office_word, AlastorMoody, Adventure10, Mango247, Rounak_iitr

.Let $N$ be the intersection point of $(ABC)$ and the angle bisector of $\angle{ACB}$ other than $C$
Suppose $\Omega$ touches $\tau$ at $L$ and $AB$ at $M$
Let $M'=NL \cap AB$ and $XY$ be the common tangent of $\tau$ and $\Omega$
Define $K=XY \cap AB$
$KM=KL$
$\measuredangle KLM' = \measuredangle KLN = \measuredangle KLB + \measuredangle BLN = \measuredangle LAB + \measuredangle BAN = \measuredangle LNB + \measuredangle NBA = \measuredangle M'NB + \measuredangle NBM' = \measuredangle NM'B = \measuredangle LM'B = \measuredangle LM'K$
So, we get $KM' = KL$ which means $MM'=0$
By similarity we get $NB^2=NM.NL=NQ.NP$
So, $\triangle NBQ \sim -\triangle NPB$
$\measuredangle NBQ = \measuredangle BPN$
$\measuredangle NBA+ \measuredangle ABQ = \measuredangle BCN + \measuredangle PBC$
$\measuredangle ABQ = \measuredangle PBC$
$\measuredangle ABP = \measuredangle QBC$

This post has been edited 1 time. Last edited by Promi, Apr 4, 2021, 8:15 AM

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#19 h Jul 27, 2019, 6:16 AM • 2 Y

Y by Adventure10, Mango247

Let $D$ be the second intersection of $\Gamma$ and the bisector $\angle BCA$ .Let $E,F$ be the tangency point
of $\Omega$ with $AB$ and $\Gamma$ respectively.

$Lemma$ 1: $DE \cdot DF=DB^2$

$Proof$ : $\angle DFB= \frac{\angle C}{2}= \angle DBE$ .
So, $\triangle DFB$ is similar to $\triangle DBE$
$\rightarrow DE \cdot DF=DB^2$
$Lemma$ 2: $DP \cdot DQ=DB^2$

$Proof$ : By, POP we get, $P_\Omega(D)=DP \cdot DQ=DE \cdot DF=DB^2$
$Lemma$ 3: $BI$ is the angle bisector of $\angle PBQ$ where $I$ is the incentre of $\triangle ABC$

$Proof$ : From $Lemma$ $3$ we get $DP \cdot DQ=DB^2$ And, $P,Q,D$ collinear. So, $D$ is the centre of the Appolonian Circle of $\triangle BPQ$ .
Let, the Appolonian Circle $(D,DB)$ intersect $PQ$ at $I$ . So, $BI$ is the angle bisector of $\angle PBQ$ .
And from $Fact$ $5$ we get $I$ is the incentre of $\triangle ABC$ .

Now, $\angle PBI=\angle QBI$ $\rightarrow \angle PBI + \angle ABI=\angle QBI + \angle CBI$ $\rightarrow \angle ABP= \angle QBC$ .

Attachments:

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#20 h Feb 10, 2020, 8:53 PM • 2 Y

Y by Adventure10, Mango247

If $M$ is the midpoint of arc $AB$ not containing $C$ , then $P, Q$ are swapped under an inversion at $M$ with radius $MA$ . Thus $\angle ABP = \angle MBP - \angle MBA = \angle MQB - \angle BCQ = \angle CBQ$ as desired.

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#21 h Apr 6, 2020, 12:00 PM

Y by

let TANGENCY point of two circles $K$ and TANGENCY point of circle to $AB$ also $S$ obviously $KS$ and $PQ$ concurrent on midpoint of arc $AB$ then compute power of point M . and according this equality $\angle SBM=$ $\angle SKB$ we get $MB^2=MS.MK=MP.MQ$ then with angle chasing we get the solution.

This post has been edited 2 times. Last edited by arzhang2001, Apr 7, 2020, 6:57 PM
Reason: latex

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#22 h May 10, 2021, 8:16 AM

Y by

Let $M$ be the intersection of the angle bisector with circle $\Gamma$ , $K$ be the tangency point of circle $\Omega$ with $AB$ and $T$ be the tangency point of circle $\Omega$ with circle $\Gamma$ . By homothety $M-K-T$ are collinear. $\angle MAB = \angle MTB = \angle MBA$ By converse alternate segment theorem $MB$ is tangent to $(KTB) \implies MB^2 = MK \cdot MT$ and again by POP we have $MP \cdot MQ = MK \cdot MT \implies MB^2 = MP \cdot MQ$ hence $MB$ is tangent to $(BPQ)$ and from here the result follows.

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554183

#23 h Sep 11, 2021, 10:51 AM

Y by

Ah nice
First note that $\overline{C,P,Q,M}$ and $\overline{X,D,M}$ where $M$ is the midpoint of arc $AB$ . We get, $MD \times MX= MB^2 = MQ \times MP$ so $MB$ is tangent to $\odot{BQP}$ which implies the result on angle chase.

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BVKRB-

#24 h Sep 11, 2021, 12:10 PM • 1 Y

Y by Rounak_iitr

Let $M$ be midpoint of $\overarc{AB}$ and let $\omega$ be tangent to $\Gamma$ at $K$
$\text{EGMO Lemma 4.43}$ implies $M-D-K$ are collinear and $M$ obviously lies on the $C$ angle bisector
$[asy]/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -15.82, xmax = 12.98, ymin = -6.83, ymax = 7.65; /* image dimensions */ pen ffvvqq = rgb(1,0.3333333333333333,0); pen fuqqzz = rgb(0.9568627450980393,0,0.6); pen qqffff = rgb(0,1,1); pen qqwuqq = rgb(0,0.39215686274509803,0); draw(arc((2.7730186914271573,-1.7657842667556372),0.6,155.87081438240853,189.67467443386437)--(2.7730186914271573,-1.7657842667556372)--cycle, linewidth(1) + qqwuqq); draw(arc((-4.76,-3.05),0.6,62.29386147827237,96.09772152972822)--(-4.76,-3.05)--cycle, linewidth(1) + qqwuqq); /* draw figures */ draw((-1.98,5.49)--(-4.76,-3.05), linewidth(1)); draw((-4.76,-3.05)--(5.64,-3.05), linewidth(1)); draw((5.64,-3.05)--(-1.98,5.49), linewidth(1)); draw(circle((0.44,-0.020257611241217534), 6.01825048849263), linewidth(1) + ffvvqq); draw(circle((1.070980155961743,1.8759338602372782), 4.019831600039738), linewidth(1) + fuqqzz); draw((-4.76,-3.05)--(-2.7794754967753703,0.7213614896243091), linewidth(1)); draw((-4.76,-3.05)--(2.7730186914271573,-1.7657842667556372), linewidth(1)); draw((-2.7794754967753703,0.7213614896243091)--(2.7730186914271573,-1.7657842667556372), linewidth(1)); draw((2.7730186914271573,-1.7657842667556372)--(5.64,-3.05), linewidth(1)); draw((-5.282674200057174,1.8426269644214222)--(-2.7794754967753703,0.7213614896243091), linewidth(1)); draw((-5.282674200057174,1.8426269644214222)--(-4.76,-3.05), linewidth(1) + qqffff); draw((-5.282674200057174,1.8426269644214222)--(-1.98,5.49), linewidth(1) + qqffff); draw((-5.282674200057174,1.8426269644214222)--(-2.751424150756261,3.1202294073890404), linewidth(1)); draw((-2.751424150756261,3.1202294073890404)--(2.3402005304231963,5.69013439688057), linewidth(1)); draw(circle((-5.282674200057174,1.8426269644214222), 4.920466129584573), linewidth(1) + linetype("2 2") + blue); /* dots and labels */ dot((-1.98,5.49),dotstyle); label("$A$", (-1.9,5.69), NE * labelscalefactor); dot((-4.76,-3.05),dotstyle); label("$B$", (-5.06,-3.43), NE * labelscalefactor); dot((5.64,-3.05),dotstyle); label("$C$", (5.72,-2.85), NE * labelscalefactor); dot((-2.751424150756261,3.1202294073890404),dotstyle); label("$D$", (-2.68,3.33), NE * labelscalefactor); dot((2.3402005304231963,5.69013439688057),linewidth(4pt) + dotstyle); label("$K$", (2.42,5.85), NE * labelscalefactor); dot((2.7730186914271573,-1.7657842667556372),linewidth(4pt) + dotstyle); label("$Q$", (2.86,-1.61), NE * labelscalefactor); dot((-2.7794754967753703,0.7213614896243091),linewidth(4pt) + dotstyle); label("$P$", (-2.7,0.89), NE * labelscalefactor); dot((-5.282674200057174,1.8426269644214222),linewidth(4pt) + dotstyle); label("$M$", (-5.2,2.01), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$
Consider the inversion with radius $MA=MB$
Observe that $(AB \iff \odot(ABC)) \implies (\omega \iff \omega) \implies (P \iff Q) \implies MP \ \cdot \ MQ = MB^2 \implies \angle MQB = \angle MBP$ $\angle MQB = \angle MBP \iff \angle MBA + \angle ABP = \angle QBC + \angle QCB = \angle QBC + \angle QCA \iff \angle ABP=\angle QBC \ \blacksquare$
Remarks

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590 posts

#25 h Oct 19, 2021, 8:58 AM

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microsoft_office_word wrote:

Let $\Gamma$ be the circumcircle of triangle $ABC$ . A circle $\Omega$ is tangent to the line segment $AB$ and is tangent to $\Gamma$ at a point lying on the same side of the line $AB$ as $C$ . The angle bisector of $\angle BCA$ intersects $\Omega$ at two different points $P$ and $Q$ .
Prove that $\angle ABP = \angle QBC$ .

Inversion makes me feel guilty some times,
Sol

This post has been edited 4 times. Last edited by HoRI_DA_GRe8, Jan 18, 2022, 10:30 AM

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#26 h Oct 19, 2021, 12:26 PM

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BRUH , a P5?
Let $M$ be the mid point of $\widehat{AB}$ .Let the tangency point with $AB$ and $\Gamma$ be $K$ and $T$ , respc. Then $M-K-T$ and $C-P-Q-M$ .Also, $MP.MQ=MK.MT=MB^2=MA^2 \implies$ $MB$ is tangent to the circumcircle of $BPQ$ .Thus $\angle MPB=\angle MBQ \implies \angle CBP=\angle ABQ$ $\blacksquare$

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559 posts

#27 h Oct 19, 2021, 8:22 PM

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Fix $ABC$ and vary $P$ on bisector of angle $C$ . Let $I,$ $J,$ $Q'$ be incenter, $C-\text{excenter}$ and isogonal conjugate of $P$ in $\triangle ABC.$
Clearly $P\mapsto Q'$ is an involution fixing $I,$ $J,$ so it's nothing but inversion wrt $\odot (AIBJ),$ and hence $Q=Q'$ as desired.

This post has been edited 1 time. Last edited by JAnatolGT_00, Oct 20, 2021, 2:53 PM

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DakuMangalSingh

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DakuMangalSingh

#28 h Oct 25, 2021, 6:33 PM • 1 Y

Y by Rounak_iitr

Let $M$ is the midpoint of minor arc $AC$ . $BQ\cap \Gamma$ , $MN\cap AB=X$ . $\Omega$ is tangent to $\Gamma$ at $D$ and $AB$ at $E$ .
By shooting lemma, $MN.MX=MD.ME=MQ.MP\implies$ points $X,P,Q,N$ are concyclic.
So, $\angle MBX =\angle MCA=\angle BCM =\angle BMN=\angle MPX\implies$ points $X,P,B,M$ are concyclic.
So, $\angle QBC=\angle XMP=\angle XBP=\angle ABP$ .

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1162 posts

#29 h Dec 13, 2021, 5:31 AM

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Suppose $\Omega$ touches $AB$ and $\Gamma$ at $K, T$ respectively, and let the $C$ -bisector meet $\Gamma$ again at $M$ . Because $\Omega$ is a Curvilinear Incircle wrt $\Gamma$ , we know $M, K, T$ are collinear.

Now, the Shooting Lemma implies $MB^2 = MK \cdot MT = Pow_{\Omega}(M) = MP \cdot MQ$ so $MB$ is tangent to $(BPQ)$ . This yields $\angle ABP = \angle MBP - \angle MBA = \angle BQP - \angle MAB = \angle BQP - \angle MCB = \angle QBC$ as desired. $\blacksquare$

Remark: Let $X$ lie on $BC$ such that $KX \parallel PQ$ . Then, Steiner's Ratio Lemma implies $CX$ is tangent to $(PQX)$ .

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#30 h Dec 20, 2021, 3:40 PM • 1 Y

Y by Rounak_iitr

Shortage

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#31 h Jun 19, 2022, 6:04 PM

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Let $\Omega$ be tangent to $ABC$ and $AB$ at $R$ and $S$ and Let $T$ be midpoint of arc $AB$ not containing $C$ . It's well Known that $R,S,T$ are collinear. $TP.TQ = TS.TR = TB^2 \implies \angle TQB = \angle PBT$ so $\angle CBQ = \angle TQB - \angle TCB = \angle PBT - \angle ABT = \angle PBA$ .

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5000 posts

#32 h Aug 8, 2022, 6:04 PM

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Let $\Omega$ touch $\overline{AB}$ at $K$ and $\Gamma$ at $T$ , and let $M$ be the arc midpoint of $\widehat{AB}$ not containing $C$ . It is then well-known that $M,K,T$ are collinear and $MK \cdot MT=MB^2$ , hence by PoP we find $MB^2=MP\cdot MQ$ , i.e. $\overline{MB}$ is tangent to $(BPQ)$ . Then, WLOG let $C,Q,P,M$ lie on $\overline{CM}$ in that order, so
$\begin{align*} \angle ABP&=\angle QBC &\iff\\ \angle MBP-\angle MBA&=\angle MBQ-\angle MBC &\iff\\ \angle MQB-\frac{\angle C}{2}&=\angle MBQ-\left(\angle B+\frac{\angle C}{2}\right) &\iff\\ \angle B+\angle C&=\angle MQB+\angle MBQ &\iff\\ \angle A&=\angle QMB=\angle CMB, \end{align*}$ which is evident. $\blacksquare$

This post has been edited 1 time. Last edited by IAmTheHazard, Aug 8, 2022, 6:05 PM

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#33 h Nov 5, 2022, 2:46 PM • 1 Y

Y by jelena_ivanchic

Solved with mxlcv and jelena_ivanchic!

Let $M$ be the midpoint of arc $AB$ not containing $C$ . The angle bisector of $C$ passes through $M$ . It's well known that $MA^2=MB^2$ is power of $M$ wrt $\Omega$ .
Hence $MA^2=MP\cdot MQ \implies MA$ tangent $(APQ) \implies \angle BAP=\angle MAP -\angle MAB=\angle AQP-\angle MCB=\angle QAC$ .
As $\angle BAP=\angle CAQ$ and $\angle ACP=\angle BCQ$ , hence $P, Q$ are isogonal conjugate wrt $\triangle ABC$ . Hence $\angle ABP = \angle QBC$ , and we are done!

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8857 posts

#34 h Feb 2, 2023, 5:02 PM

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Let $T$ be the tangency point, and $M$ the arc midpoint of minor arc $\widehat{AB}$ . It is well-known that $M, E, T$ are collinear and also $MP \cdot MQ = ME \cdot MT = MB^2.$ As a result, $\triangle BQM \sim \triangle PBM$ . Now angle chase: $\measuredangle ABQ = \measuredangle ABM + \measuredangle MBQ = \measuredangle BPM + \measuredangle MCB = \measuredangle PBC,$ which suffices.

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#35 h Feb 20, 2023, 11:47 AM

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Easy. Solved on the plane

. Let $M$ be the midpoint of $\overarc{AB}$ . $L$ be intersection with $\Omega$ $F$ be circles' intersection . So we have $M,L,F$ collinear from shooting lemma. We have $MB^2=ML.MF=MP.MQ$ from PoP. So from similarities if we chasing angle we get desired result.

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465 posts

#36 h Jul 12, 2023, 6:48 PM • 2 Y

Y by HoripodoKrishno, Eka01

I literally forgot that there existed something called Shooting Lemma. :yaw: :rotfl:

:rotfl:

$[asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions (otherwise throws error due to the missing xmin) are done using bubu-asy.py. This adds back the dps, real xmin coordinates, changes the linewidth, the fontsize and adds the directions to the labellings. */ pair A = (-37.57730,75.56340); pair B = (-100.84251,-43.18470); pair C = (39.83866,-44.14172); pair D = (-83.79684,-11.19018); pair M = (-101.25884,33.26400); pair T = (-56.51806,-80.63562); pair P = (-60.13324,10.70260); pair Q = (7.98004,-26.66416); import graph; size(10cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); draw(arc(B,23.68739,-0.38976,8.63225)--B--cycle, linewidth(0.75) + blue); draw(arc(B,23.68739,52.93070,61.95272)--B--cycle, linewidth(0.75) + blue); draw(A--B, linewidth(0.5)); draw(B--C, linewidth(0.5)); draw(C--A, linewidth(0.5)); draw(circle((-30.23601,-4.57470), 80.47365), linewidth(0.5)); draw(circle((-40.49370,-34.26071), 49.06538), linewidth(0.5)); draw(M--T, linewidth(0.5) + blue); draw(M--C, linewidth(0.5)); draw(B--M, linewidth(0.5) + blue); draw(B--P, linewidth(0.5) + red); draw(B--Q, linewidth(0.5) + red); dot("$A$", A, N); dot("$B$", B, W); dot("$C$", C, dir(0)); dot("$D$", D, W); dot("$M$", M, NW); dot("$T$", T, SW); dot("$P$", P, N); dot("$Q$", Q, NE); [/asy]$

Let $T$ denote the tangency point of $\Omega$ and $\Gamma$ and $D$ denote the tangency point of $\Omega$ and $AB$ .

Now from Shooting Lemma, we have that $\overline{M-D-T}$ are collinear and that $MD\cdot MT=MB^2$ . This further gives that $MB^2=MD\cdot MT=MP\cdot MQ$ which means that $\measuredangle MBP=\measuredangle BQP$ .

Now to finish, we have,
$\begin{align*} \measuredangle ABP&=\measuredangle MBP-\measuredangle MBA\\ &=\measuredangle BQP-\measuredangle MCA\\ &=\measuredangle BQP-\measuredangle BCM\\ &=\measuredangle BQC+\measuredangle QCB\\ &=\measuredangle QBC .\end{align*}$
This gives us what we wanted and we are done.

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57 posts

#37 h Oct 11, 2023, 7:27 AM

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Let $S$ be the midpoint of $\overarc{AB}$ , $T$ be the tangency point of both circles and $K$ be the touch point of $AB$ to $\Omega$ .It's a well known lemma that $T,K,S$ are collinear.Let $\angle SCB=\alpha$ and
$\angle SBP=\beta \implies \angle PBC=\beta- \alpha$ .
So we need $\angle ABQ=\beta-\alpha$ .
$\angle SBA=\angle BTK$ $\implies$ $BS^2=SK\cdot ST$ and $SK\cdot ST=SQ\cdot SP$ .
This finishes the problem.

This post has been edited 2 times. Last edited by rafigamath, Oct 11, 2023, 7:28 AM
Reason: .

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174 posts

#38 h Jan 17, 2024, 7:49 AM

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Shooting lemma does it

$MB^2=MP\cdot MQ$ .
Therefore we have , $\triangle BQM \sim \triangle PBM$ .
Now by angle chasing we are done
Remarks

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116 posts

Iveela

#39 h Mar 27, 2024, 6:08 AM • 1 Y

Y by Rounak_iitr

This took an embarassingly long time.

Relabel $A, C$ so that they switch places. Suppose $\Omega$ is tangent to $BC$ and $\Gamma$ at $D$ and $T$ respectively. Let the angle bisector of $\angle BAC$ intersect $\Gamma$ at $M$ . Note that $\angle MTC = \angle CBM = \angle BCM$ which implies $MC^2=MD \cdot MT$ . Since $TDQP$ is cyclic, this implies $MP \cdot MQ = MD \cdot MT = MC^2$ . Therefore, $\angle CAM + \angle ACP = \angle MPC = \angle MCQ = \angle BCQ + \angle BCM$ $\implies$ $\angle ACP = \angle BCQ$ , as desired. $\blacksquare$

Attachments:

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ismayilzadei1387

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ismayilzadei1387

#40 h Mar 28, 2024, 4:57 PM

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Let $\omega$ tangents $\gamma$ at $T$ and $BC$ at $D$
Let $M=TD \cap \gamma$
From 5 stars lemma $MB$ tangents to $(BQP)$ and we are done

This post has been edited 2 times. Last edited by ismayilzadei1387, Mar 31, 2024, 2:23 PM

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Eka01

#41 h Aug 27, 2024, 2:40 PM • 1 Y

Y by Sammy27

Lol the flagship shooting/archimedes lemma problem.

Let the circles be tangent at $X$ , let the circle be tangent to $BC$ at $D$ and let the midpoint of arc $AB$ not containing $C$ be $M$ . Also I take $P$ to be nearer to $M$ than $Q$ .(The same words can probably be copy pasted for both the cases but just in case.)
Then by archimedes' lemma, $\overline{X-D-M}$ is collinear and $MD.MX=MP.MQ=MA^2=MB^2$ .
This implies that $MB$ is tangent to $(PBQ)$ so $\angle MBP= \angle MQB$ .
Note that $\angle MBP= \frac{C}{2} + \angle ABP$ and $\angle MQB= \frac{C}{2}+ \angle QCB$ so the desired result follows.

This post has been edited 1 time. Last edited by Eka01, Aug 27, 2024, 2:42 PM

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Avron

#42 h Mar 7, 2025, 6:19 PM

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Let $\Omega$ be tangent to $\Gamma$ at $T$ and to $AB$ at $K$ , and $M$ be the midpoint of arc $BC$ . It is well known that $M,K,T$ are collinear and $MB^2=MK\cdot MT$ . Then also $MQ\cdot MP=MK\cdot MT=MB^2$ so $\angle{MBQ}=\angle{MPB}$ and:
$\angle{ABQ}=\angle{MBQ}-\angle{ABM}=\angle{MPB}-\angle{MCB}=180-(\angle{BPC}+\angle{MCB})=\angle{CBP}$ so $\angle{ABP}=\angle{ABQ}+\angle{QBP}=\angle{CBP}+\angle{QBP}=\angle{CBQ}$ and we're done.

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#43 h Mar 10, 2025, 6:01 AM

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Another solution
Let $\Omega$ is tangent to $AB$ and $\Gamma$ , at $X$ , $Y$ respectively, and let $M$ be the midpoint of arc $AB$ not including $C$ , $I_c$ be the $C-excenter$ . From Archimede we get that:
$MI^2=MB^2=MX\cdot MY=MP\cdot MQ$
So we get that $(P,Q;I,I_c)=-1$ and from $\angle{I_cBI}=90$ we get that $BI$ bisector of $\angle QBP$ and we are done¡

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