IMO 2023 P2

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799786

1052 posts

799786

#1 h Jul 8, 2023, 4:47 AM • 8 Y

Y by Quidditch, PRMOisTheHardestExam, LoloChen, KST2003, trinhquockhanh, Rounak_iitr, Funcshun840, cubres

Let $ABC$ be an acute-angled triangle with $AB < AC$ . Let $\Omega$ be the circumcircle of $ABC$ . Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$ . The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E \neq A$ . The line through $D$ parallel to $BC$ meets line $BE$ at $L$ . Denote the circumcircle of triangle $BDL$ by $\omega$ . Let $\omega$ meet $\Omega$ again at $P \neq B$ . Prove that the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$ .

This post has been edited 5 times. Last edited by Amir Hossein, Aug 8, 2023, 5:43 PM

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khan.academy

633 posts

khan.academy

#3 h Jul 8, 2023, 4:52 AM • 1 Y

Y by LoloChen

Let $ABC$ be an acute-angled triangle with $AB < AC$ . Let $\Omega$ be the circumcircle of $ABC$ . Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$ . The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E \neq A$ . The line through $D$ parallel to $BC$ meets line $BE$ at $L$ . Denote the circumcircle of triangle $BDL$ by $\omega$ . Let $\omega$ meet $\Omega$ again at $P \neq B$ . Prove that the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$ .

@above it is

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LoloChen

475 posts

LoloChen

#4 h Jul 8, 2023, 4:54 AM • 18 Y

Y by 799786, khan.academy, khina, David-Vieta, GeoKing, oolite, ike.chen, Before-the-snow-melted, Aryan-23, sabkx, Dhruv777, Gato_combinatorio, aidan0626, PennyLane_31, AlexCenteno2007, ehuseyinyigit, LLL2019, Sedro

Too easy for P2!

Sketch: construct antipode of A to get rid of L and find concyclics kills it.

https://cdn.artofproblemsolving.com/attachments/c/5/7b46be3b7184784a85e64fdf9db4642e647f9c.png

Construct midpoint of arc $BC$ not containing ${A}$ and $A-$ antipode to be $N$ and $A'$ . Let $AN$ cut $SB$ at $Y$ and $DA$ at $X$ . Let ${AE}$ cut $SA'$ at $T$ .
$\angle L=\angle EBC=\angle BPD$ so $LPDB$ concyclic. Note that $A'S//AN$ and by Reim $PYXB$ concyclic. Since $\angle TEA'=90$ and $SE=SA'$ , $A'S=TS$ so by homothety at ${D}$ , $AY=YX$ . Combining $\angle APX=90$ , $PY=XY$ , $\angle YPD= \angle YXD=\angle PBY$ , so $PY$ is tangent to $\omega$ .

This post has been edited 3 times. Last edited by LoloChen, Jul 8, 2023, 2:56 PM

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tastymath75025

3223 posts

tastymath75025

#9 h Jul 8, 2023, 5:54 AM • 4 Y

Y by khina, centslordm, ehuseyinyigit, jkim0656

Let $A',S'$ be the antipodes to $A,S$ on $(ABC)$ . Defining $X = AS' \cap BS$ , we want to show $PX$ is tangent to $(BPD)$ or $XD\cdot XB = XP^2$ .

First, note by Reim on $(LPDB), (PBEA')$ that since $LD||EA'$ we must have $P,D,A'$ are collinear. Next, note that $\angle DBA = \angle S'AE = \angle XAD$ , so $XA$ is tangent to $(BAD)$ and thus $XD\cdot XB = XA^2$ , so to finish we only need to show $XA=XP$ . The locus of such $X$ is the $A$ -midline of $\triangle AA'P$ , so if we let $Y=AX\cap A'D$ we just need to show $X$ is the midpoint of $AY$ .

But if we let $Q = AE \cap SA'$ then from angle-chasing $QAA'$ is isosceles, so $AS\perp A'S$ implies $S$ is the midpoint of $QA'$ . Taking homothety centered at $D$ therefore shows $X$ is the midpoint of $AY$ , as desired $\blacksquare$

Motivation: The motivation for this solution is to first eliminate $L$ using Reim to get a nicer characterization for $P$ . After this, dealing with $XP$ and $(BPD)$ directly is still not nice, but we bypass this by noticing $XA$ is tangent to $(BAD)$ .

This post has been edited 7 times. Last edited by tastymath75025, Jul 8, 2023, 6:17 AM

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trinhquockhanh

522 posts

trinhquockhanh

#14 h Jul 8, 2023, 6:34 AM • 4 Y

Y by PRMOisTheHardestExam, GeoKing, Rounak_iitr, ehuseyinyigit

$\text{No more cyclic quadrilaterals}$

https://i.ibb.co/yfc8FBc/2023-IMO-P2.png

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Aryan-23

558 posts

Aryan-23

#15 h Jul 8, 2023, 6:40 AM • 1 Y

Y by PRMOisTheHardestExam

Let $A'$ denote the $A$ antipode. We have $\angle BPD = \angle EBC = \angle BCA' =\angle BPA'$ , so $P,D,A'$ are collinear.

Let $AI$ hit $PA'$ at $H$ and $BD$ at $F$ . Let $AE$ and $SA'$ intersect at $J$ . Then Note that, $FH \equiv AI\parallel SA'$ , so that $PFHB$ is cyclic by Reims. Now note that $JA'E$ is a right triangle with $SA'=SE$ (as $A'E \parallel BC$ ), so $SA'=SJ$ . From homothety at $D$ we get $FA=FH$ . Now as $\triangle APH$ is right, we have $FP = FH$ . So we get $\angle FPA'= \angle FHP= \angle SA'P = \angle SBP = \angle DBP$ .

So $FP$ is tangent to $\omega$ , as desired. $\square$ .

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KST2003

173 posts

KST2003

#17 h Jul 8, 2023, 6:50 AM • 2 Y

Y by PRMOisTheHardestExam, Rounak_iitr

Let $M$ be the midpoint of minor arc $BC$ , and let $\overline{PM}$ intersect $\overline{AE}$ at $K$ . Then $\overline{SM} \parallel \overline{DK}$ , so $K$ lies on $\omega$ by Reim's. Now redefine $F$ to be the intersection of $\overline{AM}$ and $\overline{BS}$ . We will show that $\overline{FP}$ is tangent to $\omega$ .

Since $\angle KBE = \angle LDE = 90^\circ$ , we have
$\measuredangle BFA = \measuredangle SFM = 90^\circ + \measuredangle BMA = 90^\circ + \measuredangle BEA = \measuredangle BKA,$ so $AFKB$ is cyclic. Moreover,
$\measuredangle KBF = \measuredangle KAF = \measuredangle EAM = \measuredangle SMA = \measuredangle SBA = \measuredangle FBA,$ so $FA = FK$ . Finally, $\measuredangle APK = \measuredangle APM = \measuredangle ASM = 90^\circ - \measuredangle SMA = 90^\circ - \measuredangle KAF$ , so $F$ is the circumcenter of $\triangle PAK$ and hence $FP^2 = FA^2 = FD \cdot FB$ . Hence $\overline{FP}$ is tangent to $\omega$ as desired.

This post has been edited 3 times. Last edited by KST2003, Jul 8, 2023, 6:56 AM

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MathLover_ZJ

826 posts

MathLover_ZJ

#19 h Jul 8, 2023, 6:51 AM • 5 Y

Y by David-Vieta, PRMOisTheHardestExam, Amiralizakeri2007, Rounak_iitr, ehuseyinyigit

Without antipodes: Let $\omega$ meet $AE$ at $T\neq D$ .
The circumcenter of $\triangle APT$ is $Q$ .
Connect $PT,BT,TQ,AP$ .
then we have $\angle BPD=\angle BLD=90^{\circ}-\angle BED=\angle APB-90^{\circ},$
so $AP \perp PD$ .
hence $\angle QPT=90^{\circ}-\angle PAT=\angle PDA=\angle PBT,$ which means $PQ$ is tangent to $\omega$ .
by $\angle AQT=360^{\circ}-2\angle APT=180^{\circ}-2\angle DBT=180^{\circ}-\angle ABT$ we know that $A,Q,T,B$ is concyclic.
since $AQ=QT\Rightarrow Q$ lies on $BS$ .
we just need to prove $Q$ lies on the bisector.
$\Leftrightarrow \angle QAS=90^{\circ},$
which is obvious since $\angle AQB=\angle ATB,\angle TBE=90^{\circ}$ .

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a22886

924 posts

a22886

#20 h Jul 8, 2023, 7:15 AM • 5 Y

Y by LoloChen, Aryan-23, David-Vieta, PRMOisTheHardestExam, Alfombra12

We consider the inversion $\Phi$ wrt $D$ that preserves $\Omega$ .

Under $\Phi$ we have $A\leftrightarrow E$ and $B\leftrightarrow S$ . It's natural to find $\Phi(L)=\Phi(BE)\cap\Phi(DL)=\odot ADS\cap DL$ which indicates $\angle AS~\Phi(L)=\angle ADL=90^{\circ}$ , i.e., $\Phi(\omega)=\operatorname{line}S\Phi(L)=SA'$ where $A'$ is the antipodal point of $A$ .
From that, $\Phi(P)=\Phi(\omega)\cap\Phi(\Omega)=A'$ , so $PDA'$ collinear.

Let the antipodal point of $S$ be $T$ . $\angle APD=\angle SBT=90^{\circ}\Longrightarrow Q:=AP\cap BT\in\omega$
But also $AS=ET\Longrightarrow \overset{\LARGE \frown}{AP}+\overset{\LARGE \frown}{ET}=\overset{\LARGE \frown}{PS}\Longrightarrow R:=AE\cap PT\in\omega$
Now we apply Pascal's theorem to $(PPQBDR)$ to get the desired result.

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Parsia--

73 posts

Parsia--

#21 h Jul 8, 2023, 7:20 AM

Y by

Let $A'$ be the antipode of $A$ . We have $\angle BPD = \angle EBC = \angle BCA' =\angle BPA'$ , so $P,D,A'$ are collinear.
Let $AE$ intersect $SA'$ at $I$ and let $AF$ intersect $PA'$ at $H$ .
Since $SE=SA'$ and $\triangle IEA'$ is a right triangle, we get that $SA' = SE = SA$ . Note that $AH \equiv AF || SA'$ so by Thales we get that $\frac{AF}{IS} = \frac{DF}{DS} = \frac{HF}{SA'}$ and so $AF = HF$ . Note that $\triangle APH$ is a right triangle, so $PF = AF = HF$ . Now note that $\angle DAF = \angle FAA' = \angle AA'S = \angle ABS = \angle ABF$ so $\triangle ADF \sim \triangle ABF$ and so $FP^2 = AF^2 = FD.FB$ .
So $FP$ is tangent to $\omega$ .

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giannis2006

45 posts

giannis2006

#22 h Jul 8, 2023, 7:39 AM • 1 Y

Y by Assassino9931

Let $N$ be the midpoint of the other arc $BC$ and $K= AD \cap PN$
Then $\angle BDK=\angle BDE =90-\angle DBC=90-\angle SBC=\angle BSN=\angle A/2=\angle BAN=\angle BPN=\angle BPK$ and so $K$ lies on $\omega$ . Let now $H= AN \cap BS$ . We need to show that $HP$ is tangent to $\omega$ . It is obvious that $AE$ , $SN$ are parallel and so $\angle DAH=\angle DAN=\angle ANS=\angle ABS=\angle ABD$ , so $AH$ is tangent to $ABD$ , i.e. $AH^2=HB*HD$ , so it remains to prove $AH=HP$ . Now we have from the cyclic quadrilaterals and the parallel lines: $\angle AKB=\angle DKB=$ $=180-\angle DLB$ $=180-\angle EBC$ $=180-\angle EAC$ $=180-\angle (90-C)$ $=90+\angle C$ $=90+\angle ANB$ $=\angle HBN+\angle HNB$ $=\angle AHB$ and hence $BKHA$ is cyclic, which gives that $\angle AKH=\angle ABH=\angle ABD=\angle DAH=\angle KAH$ and so $HA=HK$ . Finally: $\angle AHK=\angle AHB+\angle BHK=\angle HBN+\angle HNB+\angle BAK=90+\angle C+\angle BAK=90+\angle C+90-\angle B=2\angle C+\angle A=2(\angle C+\angle A/2)=2(180-(\angle B+\angle A/2))=2(180-\angle ABN)=2(180-\angle APN)= 2(180-\angle APK)$ , from where we conclude that $H$ is the circumcenter of $APK$ and $AH=HP$ , as needed.

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ericxyzhu

49 posts

ericxyzhu

#23 h Jul 8, 2023, 7:50 AM • 3 Y

Y by nervy, oolite, Funcshun840

This problem is proposed by Tiago Mourão, Nuno Arala, Portugal.

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juckter

322 posts

juckter

#24 h Jul 8, 2023, 7:55 AM • 3 Y

Y by PRMOisTheHardestExam, math90, UI_MathZ_25

Mediocre solution because I'm washed up. Let $T$ be the midpoint of arc $BC$ of $\Omega$ not containing $T$ , let $R$ be the second intersection of $\omega$ with $AE$ , and let $X = AT \cap BS$ .

$[asy] import geometry; unitsize(5cm); defaultpen(fontsize(10pt)); point A = dir(120); point B = dir(210); point C = dir(330); point I = incenter(A, B, C); point O = circumcenter(A, B, C); point T = circumcenter(I, B, C); point S = 2 * O - T; line l = bisector(S, T); point E = 2 * (projection(l) * A) - A; point D = extension(A, E, B, S); point L = intersectionpoint(line(E, B), parallel(D, line(B,C))); point R = 2 * circumcenter(B, D, L) - L; point P = extension(L, S, T, R); point X = extension(A, T, B, S); draw(A--B--C--cycle, lightgray); draw(S--L, gray); draw(P--T, dashed); draw(L--E--A); draw(T--S); draw(A--T, lightgray); draw(B--S); draw(A--P, gray); draw(A--B--R--X--cycle, gray); draw(circumcircle(A, B, C), blue); draw(circumcircle(D, B, L), red); dot(A); dot(B); dot(C); dot(S); dot(T); dot(E); dot(D); dot(L); dot(P); dot(R); dot(X); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, W); label("$E$", E, (0,-1)); label("$L$", L, W); label("$X$", X, (1, 0)); label("$R$", R, (1, 0)); label("$S$", S, N); label("$T$", T, (0,-1)); label("$P$", P, N); [/asy]$

Claim 1. $L$ , $P$ and $S$ are collinear.

Proof. Notice that the tangent to $\Omega$ at $S$ is parallel to $BC$ which is parallel to $DL$ . Since $S, D, B$ are collinear, the converse of Reim's Theorem gives the result.

Claim 2. $T$ , $R$ and $P$ are collinear.

Proof. Since $ST$ is a diameter of $\Omega$ we have $\angle TPS = 90^{\circ}$ . On the other hand

$\angle RPS = 180^{\circ} - \angle RPL = 180^{\circ} - \angle RDL = 180^{\circ} - 90^{\circ}$
And this $\angle RPS = \angle TPS$ , giving the desired collinearity.

To finish, angle chase to find $\angle ABX = \angle XBR = \frac{\angle B - \angle C}{2} = \angle RAX$ , so $AXRB$ is cyclic with $XA = XR$ and

$\angle XRA = \angle XBA = \angle RBX = \angle RBD$
And therefore $XR$ is tangent to $\omega$ . It now suffices to show that $XR = XP$ . By angle chase we have $\measuredangle RXA = 2\measuredangle RPA$ , and since $XA = XR$ this implies that $X$ is the circumcenter of $\triangle APR$ , finishing the problem.

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MarkBcc168

1593 posts

MarkBcc168

#25 h Jul 8, 2023, 8:08 AM • 2 Y

Y by PRMOisTheHardestExam, GeoKing

$[asy] size(9cm); import olympiad; import geometry; pair A = dir(119); pair A_prime = -A; pair E = reflect((0,0), dir(90)) * A_prime; pair S = dir(90); pair D = 0.6*E + 0.4*A; pair B = 2 * foot((0,0), D, S) - S; pair P = 2 * foot((0,0), A_prime, D) - A_prime; pair Q = reflect((0,0), (P+A_prime)/2) * S; pair M = -S; pair X = extension(E,Q,S,D); pair Y = extension(P,Q,A,M); pair L = extension(D+E-A_prime, D, B,E); fill(X--M--Q--cycle, paleblue); fill(A--P--S--cycle, paleblue); draw(unitcircle, linewidth(1)); draw(S--X, red+linewidth(1.2)); draw(P--A_prime, linewidth(0.8)); draw(circle(B,D,P), linewidth(0.8)); draw(P--Q, linewidth(0.8)); draw(X--M--Q--cycle, blue+linewidth(0.8)); draw(A--P--S--cycle, blue+linewidth(0.8)); draw(A--M, linewidth(0.8)); draw(A--E--A_prime, gray+linewidth(0.7)); draw(S--M, gray+linewidth(0.7)); draw(D--L, gray+linewidth(0.7)); draw(M--A_prime); dot("$A$", A, dir(118)); dot("$A'$", A_prime, dir(-62)); dot("$E$", E, dir(-84)); dot("$S$", S, dir(89)); dot("$D$", D, dir(16)); dot("$B$", B, 1.5*dir(-4)); dot("$P$", P, dir(133)); dot("$Q$", Q, dir(17)); dot("$M$", M, dir(-82)); dot("$X$", X, dir(-139)); dot("$Y$", Y, dir(143)); dot("$L$", L, dir(165)); [/asy]$
Let $A'$ be the antipode of $A$ in $\Omega$ . Moreover, let the tangent to $\omega$ at $P$ meet $\Omega$ again at $Q$ . Let $M$ be the other midpoint of arc $BC$ . Let $X=QE\cap MA'$ , and $Y=PQ\cap AM$ . We finish the problem in three steps.

Since $\angle BPD = \angle BLD = 90^\circ - \angle C = \angle BPA'$ , we get that $P,D,A'$ are collinear.
Pascal on $QPA'MAE$ gives $PQ\cap AM=Y$ , $PA'\cap AE = D$ , and $A'M\cap EQ = X$ are collinear.
We have and are homothetic center $Y$ . This is just angle chasing. To spell out details, we have
- Reim on $PPQ$ and $BDS$ gives $SQ\parallel PD$ , implying that $A'Q=SP$ , or $\angle PAS=\angle QMX$ ,
- $EM=AS$ , so $\angle APS = MQX$ , and
- clearly, $MX\parallel AS$ .
Therefore, are collinear.

Hence, $S, Y, D, X$ are collinear. $B$ clearly lies on this line, done.

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SQTHUSH

154 posts

SQTHUSH

#26 h Jul 8, 2023, 8:22 AM

Y by

Let $K$ be the mid-point of arc $CB$ which is not containing $A$ ,suppose $AK\cap BS=T$ , $PK\cap AE=E$
Note that $\angle BDE=\dfrac{1}{2}\angle A=\angle BPK$ ,so $P,B,Z,D$ are cyclic
Consider that $\angle ABD=\angle B+\dfrac{1}{2}\angle A-90^{\circ}=90^{\circ}-\dfrac{1}{2}\angle A-\angle C=\angle DBZ=\angle EAK$
Which means that $A,B,Z,T$ are cyclic,and $TA=TC$
Notice $\angle ZAT=\angle TBZ=\angle ABT$ ,so $TA^{2}=TB\cdot TD$
Since $2\angle APK=2\angle B+\angle A=360^{\circ}-\angle ATZ$
Hence $TA=TZ=TP$
So $TP^{2}=TB\cdot TD$
Which means that $TP$ is tangent to $\omega$

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Pyramix

419 posts

Pyramix

#103 h Aug 5, 2024, 4:33 PM

Y by

Let $M$ be the antipode of $S$ in $\Omega$ . Let $E'$ be the reflection of $E$ in $SM$ . Define $J=AS\cap BE'$ .

Claim. $JD\parallel BC$ .
Proof. Applying Pascal's on $AEE'SBC$ , we get that $\underbrace{EE'\cap BC}_{=\infty_{BC}},SE'\cap AC,\underbrace{AE\cap SB}_{=D}$ are collinear. The rest of details are left as an exercise. [My proof involves complex numbers.]

Claim. $L,P,S$ are collinear.
Proof. Apply Pascal's on $SPE'BEA$ to get $SP\cap BE,\underbrace{PE'\cap EA}_{=D},\underbrace{E'B\cap AS}_{=J}$ are collinear. However, since $JD\parallel BC$ , we have $J, L, D$ collinear. So, $L=BE\cap JD$ , which means $L\in SP$ . $\blacksquare$

Note, $(B,M;E,S)\overset{A}{=}(B,K;D,S)\overset{P}{=}(B,KP\cap(BDP);D,L).$ So, if $P'=KP\cap(BDP)$ , we need to show $P=P'$ .
It suffices to show $\frac{\frac{BE}{EM}}{\frac{BS}{SM}}=\frac{\frac{BD}{BL}}{\frac{DP}{LP}}$
Claim. $\frac{DP}{LP}=\frac{BD}{DL}\frac{BS}{SM}\frac{EM}{BE}$
Proof. Note that by Power of Point, $DP=\frac{DB\cdot DS}{DE'}$ . Let $r_2$ be the radius of $(BDL)$ . Then, $LP=2r_2\sin(\angle PDL)$ , while $\angle PDL=\angle DE'E$ as $EE'\parallel LD$ , and since $DE\perp EE'$ , we have $\sin(\angle DE'E)=\frac{DE}{DE'}$ . Hence, $\frac{DP}{LP}=\frac{DB\cdot DS}{2r_2\sin(\angle PDL)DE'}=\frac{DB\cdot DS}{2r_2DE}$ It suffices to show
$\frac{DB\cdot DS}{2r_2DE}=\frac{BD}{DL}\frac{BS}{SM}\frac{EM}{BE}$ $\Longleftrightarrow \frac{DS}{2r_2DE}=\frac{BS\cdot EM}{DL\cdot SM\cdot BE}$ By sine rule, $DL=2r_2\sin(\angle LPD)$ . Note, $\angle LPD=\pi-\angle E'ES=\angle EBS=B+\frac A2=\frac{\pi}2+\frac{B-C}{2}$ . So, $DL=2r_2\cos\left(\frac{B-C}2\right)$ . Similarly, $BS=SM\cos\left(\frac A2\right)$ . Finally, $\frac{EM}{BE}=\frac{\sin\left(\frac{B-C}{2}\right)}{\cos(B)}$ .
Finally, $\frac{DS}{DE}=\frac{DA}{DB}=\frac{\sin\left(\frac{B-C}2\right)}{\cos(B)}$
Plugging in the ratios, we're done.

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SHZhang

109 posts

SHZhang

#104 h Aug 8, 2024, 3:10 PM

Y by

Let $T$ be diametrically opposite $S$ on $\Omega$ , let $M$ be the intersection of $\omega$ with $AE$ other than $D$ , and let $N$ be the intersection of $AP$ with $\omega$ other than $P$ .

Claim: $S$ , $P$ , and $L$ are collinear.
Proof: This follows from $\angle BPL = \angle BDL = \angle SBC = \angle SCB = 180^\circ - \angle BPS.$
Claim: $N$ , $B$ , and $T$ are collinear.
Proof: First note that $\angle SPD = 180^\circ - \angle LPD = \angle LBD = 180^\circ - \angle SBE = \angle STE = \angle AST,$ so $\angle APD = \angle APS + \angle SPD = \angle ATS + \angle AST = 180^\circ - \angle SAT = 90^\circ.$ This gives $\angle DBN = 180^\circ - \angle DPN = \angle APD = 90^\circ$ , and since $\angle DBT = \angle SBT = 90^\circ$ the claim follows.

Claim: $P$ , $M$ , and $T$ are collinear.
Proof: We have $\angle DMP = \angle DBP = \angle SBP = \angle STP$ ; since $DM \parallel ST$ , this gives $MP \parallel TP$ , so $MP$ must be the same line as $TP$ .

Now Pascal's theorem on $PPNBDM$ gives $A$ , $X$ , and $T$ collinear, as desired.

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fasttrust_12-mn

118 posts

fasttrust_12-mn

#105 h Aug 22, 2024, 2:44 AM

Y by

what a P2!!
let $O$ be the center of the circle $\Omega$ let $X,M$ be the intersection of $AO$ with $BC$ and $\overarc{CB}$ not containing $A$ , let $F$ be the second intersection point $AE$ with $\omega$
we have that $MA'=ME=AP$ since $SM\parallel AE$ so we have $SAEM$ is a isosceles trapezium

$\boxed{\textbf{Claim:} M,K,P \text{are collinear }}$ proof:: $\measuredangle KPB=\measuredangle KDB=\measuredangle BSM =\measuredangle BPM$ thus $M,K,P$ are collinear $\blacksquare$

$\boxed{\textbf{Claim:} B,M,Y \text{are collinear }}$
proof: let $D'$ be the antipode of $D$
we have that $DKD'L$ is a rectangle $\implies \measuredangle D'BK=90^{\circ}$ and since $O$ is the center of $\Omega$ we have that $\measuredangle BSM=90^{\circ}\implies\measuredangle D'BK+\measuredangle BSM=180^{\circ}$ thus $B,M,Y$ are collinear $\blacksquare$

$\boxed{\textbf{Claim:} D',P,A \text{are collinear }}$ Proof:
$\measuredangle KPL+\measuredangle MBS=180^{\circ}$ thus $D',P,A$ are collinear $\blacksquare$

$\boxed{\textbf{Claim:} A,F,M \text{are collinear }}$ Proof: since $FP$ is a tangent we have that by Pascal Theorem that $A,F,M$ are collinear $\blacksquare$

now since $CM=MB\implies \measuredangle CAM=\measuredangle MAB$
hence we are done $\blacksquare$

Attachments:

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EeEeRUT

50 posts

EeEeRUT

#106 h Aug 31, 2024, 9:38 AM

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Let $A', S'$ be antipode of $A$ and $S$ , respectively.
By angle chasing, we get $P, D, A'$ collinear.
Since $A$ and $A'$ are antipodals, we have $PA \perp PA'$ .
Let $O$ be circumcenter of $ABC, BS \cap AS' = X, PA' \cap AS' = Y$
Suppose, we have $AX = XY$ , it implies $XA = XP$ .
By angle chasing, we can obtain $\angle DBS = \angle DAS'$ , which gives $AX^2 = XD \cdot XB = PX^2$ , resulting in $PX$ tangent to $\omega$ .
Thus, it is suffice to show that $XO \parallel DA’$ . We will proceed by cartesian.

Let $A = (a,\sqrt{1-a^2}), B= (k,\sqrt{1 -k^2}), C= (-k,\sqrt{1-k^2}), S= (0,1), S' = (0,-1)$ and $A' = (-a, -\sqrt{1-a^2})$
We have $AE : x = a$ $BS : y = \frac{\sqrt{1-k^2}-1}{-k} x +1$ $AS' : y = \frac{\sqrt{1-a^2} +1}{a} x -1$ So, we have these points $D = (a, \frac{\sqrt{1-a^2}-a-k}{-k})$ $X = ( \frac{2ak}{k\sqrt{1-a^2} + a\sqrt{1-k^2} -a +k}, \frac{k\sqrt{1-a^2} - a\sqrt{1-k^2} +a +k}{k\sqrt{1-a^2} + a\sqrt{1-k^2} -a +k})$ $\text{Slope} XO : \frac{k\sqrt{1-a^2} - a\sqrt{1-k^2} +a +k}{2ak}$ $\text{Slope} DA’ : \frac{-k\sqrt{1-a^2} + a\sqrt{1-k^2} -a -k}{-2ak}$ Since Slope $XO =$ Slope $DA’$ , we get $XO \parallel DA’$ , as desired.

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kamatadu

466 posts

kamatadu

#107 h Sep 24, 2024, 7:10 PM • 1 Y

Y by SilverBlaze_SY

Here we go again. Another day of resolving already solved problems thinking they are unsolved. :stretcher:

Solved with SilverBlaze_SY. Jokes during the gsolve

$[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 are done using bubu-asy.py. This adds the dps, xmin, linewidth, fontsize and directions. https://github.com/Bubu-Droid/dotfiles/blob/master/bubu-scripts/bubu-asy.py */ pair A = (7.17699,14.19874); pair B = (0.24096,-13.95274); pair C = (38.92675,-14.03776); pair S = (19.65434,18.07435); pair X = (19.55821,-25.66198); pair O = (19.60628,-3.79381); pair D = (7.14014,-2.57086); pair K = (32.03556,-21.78637); pair P = (-0.94805,3.67199); pair L = (-9.82506,-2.53358); pair V = (19.58785,-12.17862); pair E = (7.09803,-21.73156); pair G = (10.60853,3.15108); pair M = (40.74894,1.79251); pair Q = (-9.83673,-7.84409); pair R = (7.12847,-7.88137); import graph; size(12cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); real xmin = -5, xmax = 5, ymin = -5, ymax = 5; pen ffxfqq = rgb(1.,0.49803,0.); draw(A--B, linewidth(0.6)); draw(B--C, linewidth(0.6)); draw(C--A, linewidth(0.6)); draw(circle(O, 21.86822), linewidth(0.6)); draw(B--S, linewidth(0.6)); draw(A--X, linewidth(0.6) + linetype("4 4") + blue); draw(A--K, linewidth(0.6) + linetype("4 4") + blue); draw(circle((-1.34829,-5.20748), 8.88849), linewidth(0.6)); draw(E--K, linewidth(0.6)); draw(A--E, linewidth(0.6)); draw(X--S, linewidth(0.6)); draw(L--D, linewidth(0.6)); draw(P--K, linewidth(0.6) + ffxfqq); draw((16.86098,-10.07389)--(15.59234,-10.35588), linewidth(0.6) + ffxfqq); draw((16.86098,-10.07389)--(16.81239,-8.77519), linewidth(0.6) + ffxfqq); draw((15.54375,-9.05718)--(14.27511,-9.33917), linewidth(0.6) + ffxfqq); draw((15.54375,-9.05718)--(15.49515,-7.75849), linewidth(0.6) + ffxfqq); draw(A--Q, linewidth(0.6) + linetype("4 4") + blue); draw(Q--R, linewidth(0.6)); draw(R--G, linewidth(0.6) + red); draw((8.15031,-2.42940)--(9.41983,-2.82985), linewidth(0.6) + red); draw((8.31717,-1.90043)--(9.58668,-2.30088), linewidth(0.6) + red); draw(S--M, linewidth(0.6) + ffxfqq); draw((31.51887,8.91672)--(30.25023,8.63474), linewidth(0.6) + ffxfqq); draw((31.51887,8.91672)--(31.47028,10.21542), linewidth(0.6) + ffxfqq); draw((30.20164,9.93343)--(28.93300,9.65144), linewidth(0.6) + ffxfqq); draw((30.20164,9.93343)--(30.15305,11.23213), linewidth(0.6) + ffxfqq); draw(P--G, linewidth(0.6) + red); draw((4.58316,4.08894)--(4.52321,2.75911), linewidth(0.6) + red); draw((5.13725,4.06396)--(5.07731,2.73414), linewidth(0.6) + red); draw(G--M, linewidth(0.6)); draw(Q--X, linewidth(0.6) + linetype("4 4") + blue); draw(P--X, linewidth(0.6) + linetype("4 4") + blue); draw(L--S, linewidth(0.6) + linetype("4 4") + blue); draw(L--E, linewidth(0.6) + linetype("4 4") + blue); dot("$A$", A, NW); dot("$B$", B, SW); dot("$C$", C, SE); dot("$S$", S, N); dot("$X$", X, dir(270)); dot("$O$", O, NE); dot("$D$", D, NW); dot("$K$", K, SE); dot("$P$", P, NW); dot("$L$", L, NW); dot("$V$", V, NE); dot("$E$", E, SW); dot("$G$", G, NW); dot("$M$", M, NE); dot("$Q$", Q, SW); dot("$R$", R, dir(0)); [/asy]$

Let $X$ be the midpoint of the arc $\widehat{BC}$ not containing $A$ . Let $Q=AP\cap \odot(PLB)$ . Let $R$ denote the second intersection of $AD$ and $\odot(PLB)$ . Let $O$ denote the center of $\odot(ABC)$ . Let $K=PD\cap \odot(ABC)$ . Let $V=DK\cap SX$ . Let $M=$ tangent at $P$ to $\odot(PLB)\cap \odot(ABC)$ .
We also redefine $G$ as $PP\cap RR$ and prove that it lies both on $BD$ and $AX$ .

Claim: $\overline{L-P-S}$ are collinear.
Proof. Note that since $DL \parallel SS$ , by the converse of Reim's, on $\left\{ \odot(PLB),\odot(ABC) \right\}$ with lines $\left\{ PL,DB \right\}$ , we get that $\overline{L-P-S}$ are collinear. $\blacksquare$

Claim: $\overline{A-O-K}$ are collinear.
Proof. Firstly note that since $AE \parallel SX$ , we get that $AEXS$ is an isosceles trapezium. Now, $\measuredangle XKO=\measuredangle OXK=\measuredangle SXK=\measuredangle EXS=\measuredangle XSA = \measuredangle XKA.$ $\blacksquare$

Claim: $\overline{L-B-E}$ are collinear.
Proof. $\measuredangle DBL=\measuredangle DPL=\measuredangle DPS =\measuredangle KPS = \measuredangle KXS = \measuredangle SXE = \measuredangle SBE = \measuredangle DBE$ . $\blacksquare$

Claim: $V$ is the midpoint of $DK$ .
Proof. Note that $DE \parallel VX$ and also that $VX$ is actually the perpendicular bisector of $EK$ . So by using Midpoint Theorem, we get that $V$ is indeed the midpoint of $DK$ . $\blacksquare$

Claim: $SM \parallel PK$ .
Proof. By Reim's on circles $\left\{ \odot(LPBD),\odot(ABC) \right\}$ with lines $\left\{ PP,BD \right\}$ , we get the desired result. $\blacksquare$

Claim: $\overline{P-R-X}$ are collinear.
Proof. $\measuredangle BPR=\measuredangle BDR=\measuredangle SDA =\measuredangle DSO = \measuredangle BXS = \measuredangle BPX$ . $\blacksquare$

Claim: $\overline{B-D-G}$ are collinear.
Proof. Firstly, $(D,B;P,R)\overset{P}{=} (K,B;M,X) \overset{S}{=} (K,D;\infty_{SM},V) = -1.$ Now, since the quadrilateral $DPBR$ is harmonic, we must have that $PP$ , $BD$ and $RR$ are concurrent. $\blacksquare$

Claim: $\overline{Q-B-X}$ are collinear.
Proof. $\measuredangle DBQ=\measuredangle DPQ=\measuredangle DPA \measuredangle KPA = 90^{\circ} = \measuredangle XBS = \measuredangle XBD$ . $\blacksquare$

Claim: $\overline{A-G-X}$ are collinear.
Proof. By Pascal on $PRDBQP$ , we get that $\overline{PR\cap BQ-RD\cap QP-DB\cap PP}$ are collinear, i.e., $\overline{X-A-G}$ are collinear. $\blacksquare$

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cj13609517288

1867 posts

cj13609517288

#108 h Oct 10, 2024, 6:19 PM

Y by

Let $F$ be the minor arc midpoint. Let $X_{ook}$ be a point on the tangent from $S$ to $(ABC)$ on the same of $SF$ as $A$ , then
$\angle PLD=\angle PBD=\angle PBS=\angle X_{ook}SP.$ Then we can show that this implies $L,P,S$ are collinear.

Let $X=(BLD)\cap AD$ . Then $\angle LDX=90^{\circ}$ , so $\angle LPX=90^{\circ}$ . Since $\angle SPF=90^{\circ}$ , we get that $X$ lies on $PF$ . Also, $\angle LBX=90^{\circ}$ too, so $\angle EBX=90^{\circ}$ . But $\angle CBE=90^{\circ}-\angle C$ , so $\angle XBC=\angle C$ . So $BX,AC,SF$ concur, say at $Y$ .

Let $Q=AF\cap BS$ . Note that $\angle YAX=90^{\circ}-\angle C=\frac12\angle BYC$ , so $YA=YX$ . Apply Pascal on $FACBSF$ to get that $Q$ , $Y$ , and $\infty_{BC}$ are collinear, which means $QY\parallel BC$ . Since $AX\perp BC$ , we then get that $QA=QX$ also.

Now note that
$(QD)(QB)=\frac{QD}{QS}\cdot (QS)(QB)=\frac{AQ}{AF}\cdot (AQ)(AF)=QA^2=QX^2,$ so $QX$ is tangent to $(BDL)$ . Since we want to prove that $QP$ is tangent to $(BDL)$ , it suffices to show that $(PX;DB)=-1$ . But in fact,
$(PX;DB)\stackrel{E}{=}(PE\cap (BDL),D;X,L),$ so we want to show that $PE\cap (BDL)$ is the reflection of $D$ over $LX$ . This is equivalent to showing that $\angle XPE=\angle XPD$ . Indeed,
$\angle XPE=\angle FPE=\angle FAE=\angle AFS=\angle QFY=\angle QBY=\angle DBX=\angle DPX\;\blacksquare$

This post has been edited 2 times. Last edited by cj13609517288, Oct 10, 2024, 8:22 PM

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L13832

250 posts

L13832

#109 h Oct 11, 2024, 8:39 AM • 1 Y

Y by S_14159

Not 'Too easy for P2', took me 2 attempts and a hint of L-P-S collinear (which was obvious).
Solution
Figure

Remark

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Golden_Verse

5 posts

Golden_Verse

#110 h Oct 31, 2024, 5:09 PM • 1 Y

Y by S_14159

Solution

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bin_sherlo

662 posts

bin_sherlo

#111 h Nov 8, 2024, 5:46 AM • 1 Y

Y by MS_asdfgzxcvb

Let $M$ be the midpoint of arc $BC$ and $MB\cap \omega=N,AB\cap \omega =R,AD\cap \omega=F$ .
Since $\measuredangle FPB=\measuredangle FDB=\frac{\measuredangle A}{2}=\measuredangle MPB$ , we see that $P,F,M$ are collinear. Note that $\measuredangle FBC=\measuredangle C$ because $\measuredangle FBE=90$ .
$\measuredangle FRN=\measuredangle FBM=\measuredangle FBC+\frac{\measuredangle A}{2}=\measuredangle C+\frac{\measuredangle A}{2}=\measuredangle NBR=\measuredangle NFR$ Thus, $NF=NR$ . Pascal on $NPFRBN$ gives $NP\cap RB,M,AM_{\infty}$ are collienar because $RF\parallel AM$ by $\measuredangle FRM=\measuredangle FDB=\frac{\measuredangle A}{2}=\measuredangle MAB$ . Hence $A,P,N$ are collinear. Pascal on $FPPNBD$ implies $M,PP\cap BD,A$ are collinear. So $PP\cap BD$ is on $AM$ or $KP$ is tangent to $\omega$ as desired. $\blacksquare$

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optimusprime154

17 posts

optimusprime154

#112 h Jan 5, 2025, 10:51 AM • 1 Y

Y by SorPEEK

Let $M$ be the midpoint of the arc $BC$ not containing $A$ ,let $A'$ be the antipode of $A$ let $F$ = $BS \cap AA'$ let $T$ = $AM \cap BS$ , let $K$ = $PM \cap AE$
label $\angle ABC = 2b$ , $\angle ACB = 2a$ $\angle PMA = 2c$ i make my first claim: $K$ belongs to the small circle.
this is quite easy to prove: $\angle PLD = \angle PBS = \angle PMA + \angle ABS = 2c + 90 - 2a - 2b$ now, $\angle PAE = 4a + 2b - 90^\circ - 2c$ $\angle MPA = \angle MPB = 180^\circ-2a-b$
and $\angle PKA = 90^\circ + 2c - 2a - b = \angle PBS$ so its proven. next i claim $L,P,S$ are collinear which just follows from $\angle KBE = 90^\circ = \angle KBL = \angle KPL$ and we know $\angle SPA = 90^\circ$ so its also done.
now, its obvious that $\angle DAT = \angle TAF$ and $\angle MAA' = 90$ it follows from a known lemma that $(D, F, T, S) = -1$ project the line $SD$ from $A'$ to the line $AM$
to get $(P_\infty, A, T, N) = -1$ meaning $AT = TN$ . simple angle chasing gives us $\angle PNA = \angle PBS$ now in the right triangle $\triangle NPA$ , $PT$ is the median on the hypotenuse so $\angle TPN = \angle TNP\ = \angle PBS$ which means that $TP$ is a tangent which finishes

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lelouchvigeo

172 posts

lelouchvigeo

#113 h Jan 30, 2025, 9:15 AM • 1 Y

Y by alexanderhamilton124

nice

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ihatemath123

3430 posts

ihatemath123

#114 h Feb 24, 2025, 4:53 AM

Y by

Let $Q$ be the second intersection of $(BLD)$ with line $AE$ , and let $M$ be the midpoint of minor arc $BC$ .

Claim: Quadrilateral $BPDQ$ is harmonic.
Proof: Let $D'$ be the reflection of $D$ across line $LQ$ . Since $\angle LDQ = 90^{\circ}$ , $D'$ lies on $(BLD)$ . In fact, $D'$ lies on line $PE$ , since
$\angle EPQ = \angle EAM = \angle DLQ = \angle DPQ.$ Now, projecting harmonic quadrilateral $LD'QD$ through $E$ back onto $(BLD)$ proves the claim.

Claim: If line $BD$ meets the bisector of $\angle A$ at $T$ , then $\overline{TQ}$ is tangent to $(BLD)$ .
Proof: We have that $ATQB$ is cyclic, since
$\angle AQB = \angle DLE = \angle CBE = \angle ATB.$ (The last equality comes from a hefty angle chase.) Now, letting $X$ be the intersection of lines $LQ$ and $BD$ , we have
$\begin{align*}\angle LQT &= 180^{\circ} - \angle BTQ - \angle LXB = 180^{\circ} - \angle BAE - \left( 180^{\circ} - \angle QLB - \angle DBL\right) \\ & = \angle QLB + \angle DBL - \angle BAE = \angle MPB + 180^{\circ} - \angle SBE - \angle BAE \\ &= \angle MPB + \angle SME - \angle BAE = 90^{\circ}.\end{align*}$ (The last equality comes from a hefty angle chase.) Since $\overline{LQ}$ is a diameter in $(BLD)$ , this implies the tangency.

So, line $TP$ is tangent to $(BLD)$ as well, which is what we wanted to show.

remark

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Ilikeminecraft

298 posts

Ilikeminecraft

#115 h Mar 11, 2025, 12:47 AM

Y by

Note that $L, P, S$ are collinear since $\angle BPL = \angle LDB = \angle DBC = 180 - \angle BPS.$

Let $N$ be the $S$ antipode in $(ABC).$ Let the tangent to $(BPD)$ at $P$ meet $(ABC)$ again at $Q.$

Observe that $SQ \parallel PD$ since $\angle SQP = \angle PBS = \angle DPQ.$ Thus, $\angle PSQ = \angle LPD = \angle DBE = 180 - \angle SQE,$ which implies $PS\parallel EQ,$ or $PSQE$ is isosceles trapezoid. Hence, $\angle PAX = \angle PAD + \angle DAX = \angle SNQ + \angle ANS = \angle ANQ.$ Thus, $PA\parallel NQ.$ Thus, $PAD, QNS$ are homothetic, which implies the result.

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Frd_19_Hsnzde

10 posts

Frd_19_Hsnzde

#117 h Monday at 11:31 PM

Y by

A little bit easy for P2? :maybe:

Let $M$ be the midpoint of arc $BC$ not containing $A$ , this means $SM$ is diameter of $\Omega$ , $AM$ is angle bisector $\angle BAC$ of and let $Q$ be the intersection of the lines $AM$ and $BS$ , let $T$ be the intersection of $AE$ and $MP$ .Soo we will prove that $QP$ tangent to $(BDL)$ and this means we will prove that $QP^2 = QD\cdot QB$ .

$\angle ABS = \angle AME = \angle AMS$ this implies $\triangle DAQ\sim \triangle ABQ$ therefore $QA^2 = QB\cdot QD$ .Soo the proof switches to prove that $QP = QA$ .

According to the Reim's Theorem $BDLT$ i.e. $BDLPT$ is cyclic.Since $ABKD$ is cyclic, $\angle EBT = 90$ , $\angle TBL = 90$ .

$\angle MBQ + \angle BMQ = \angle AQB$ and $\angle MBQ = 90$ from these infos it is easy to see that $\angle AQB = \angle ATB$ this implies $ABTQ$ is cyclic quadrilateral.

$\angle ATQ = \angle ABQ = \angle TBQ = \angle TAQ$ this means $QA = QT$ and the proof switches to prove that $Q$ is the circumcenter of $(APT)$ because when it is true $QA = QP = QT$ .

$\angle TBD = \angle TPD = \angle TAQ = \angle ATQ = x$ and $\angle AQT = 180 - 2x$ .

$\textbf{Claim:}$ $AP \perp DP$ .

$\textbf{Proof:}$ We will prove that $90 - \angle ADP = \angle DAP$ .

$90 - \angle ADP = \angle LDP = \angle LBP = 180 - \angle PBE = \angle EAP = \angle DAP$ . $\square$ (In this $\textbf{Claim}$ ' s $\textbf{Proof}$ i take a hint about $\textbf{Proof}$ ' s angle-chasing :blush:

) .

Finally, we have $\angle APT = 90+x$ , which sufficiently proves $Q$ is the center of $(APT)$ .

Soo we are done. $\blacksquare$ (this problem is taking my time so long.I proved this problem in about 2 hours 30 minutes.

basdim bi tik :oops_sign:

) .

This post has been edited 1 time. Last edited by Frd_19_Hsnzde, Yesterday at 12:12 AM

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kaede_Arcadia

16 posts

kaede_Arcadia

#118 h Yesterday at 1:36 AM

Y by

Generalization: Given a $\triangle ABC$ with the midpoint $U$ of the arc $BAC$ and an arbitrary point $P$ on $\odot (ABC)$ . Let $V$ be the point on $\odot (ABC)$ such that $AP \parallel UV$ and let $D = AP \cap BU$ . The line through $D$ parallel to $BC$ meets $PB$ at $S$ . Let $T$ be the second intersection of $\odot (BDS)$ with $\odot (ABC)$ . Then the line tangent to $\odot (BDS)$ at $T$ meets $BU$ on $AV$ .

Prove: Let $M$ be the antipode of $U$ on $\odot (ABC)$ and let $Q$ be the point on $\odot (ABC)$ such that $PQ \parallel BC$ . Let $E = BU \cap AV, F = TD \cap AM$ and $Q'$ be the reflection of $Q$ across $AU$ . Obviously, we know $Q',A,P$ are collinear and $T,D,Q$ are collinear, so by the $\angle VAM = \angle UQ'Q$ and $AF \parallel QQ'$ , we see that $AEF$ and $Q'UQ$ are homothetic. Therefore by the simple angle-chasing, we see that $ET=EA=EF$ . on the other hand, it follows from Reim's theorem that $T,B,E,F$ are concyclic. Thus we see that $\angle ETD = \angle EFT = \angle EBT$ , as desired.

This post has been edited 3 times. Last edited by kaede_Arcadia, Yesterday at 11:40 PM

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