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jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
Interesting inequalities
sqing   2
N 7 minutes ago by sqing
Source: Own
Let $ a,b>0  $ . Prove that
$$ \frac{a^2+b^2}{ab+1}+ \frac{4}{ (\sqrt{a}+\sqrt{b})^2} \geq 2$$$$ \frac{a^2+b^2}{ab+1}+ \frac{3}{a+\sqrt{ab}+b} \geq 2$$$$  \frac{a^3+b^3}{ab+1}+ \frac{4}{(a+b)^2}  \geq 2$$$$  \frac{a^3+b^3}{ab+1}+ \frac{3}{a^2+ab+b^2}  \geq 2$$$$\frac{a^2+b^2}{ab+2}+ \frac{1}{2\sqrt{ab}}  \geq \frac{2+3\sqrt{2}-2\sqrt{2(\sqrt{2}-1)}}{4} $$
2 replies
sqing
Today at 4:34 AM
sqing
7 minutes ago
Looks like power mean, but it is not
Nuran2010   5
N 8 minutes ago by pooh123
Source: Azerbaijan Al-Khwarizmi IJMO TST 2025
For $a,b,c$ positive real numbers satisfying $a^2+b^2+c^2 \geq 3$,show that:

$\sqrt[3]{\frac{a^3+b^3+c^3}{3}}+\frac{a+b+c}{9} \geq \frac{4}{3}$.
5 replies
Nuran2010
Yesterday at 11:51 AM
pooh123
8 minutes ago
Interesting inequalities
sqing   9
N 8 minutes ago by sqing
Source: Own
Let $ a,b,c\geq 0 , (a+k )(b+c)=k+1.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{2k-3+2\sqrt{k+1}}{3k-1}$$Where $ k\geq \frac{2}{3}.$
Let $ a,b,c\geq 0 , (a+1)(b+c)=2.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq 2\sqrt{2}-1$$Let $ a,b,c\geq 0 , (a+3)(b+c)=4.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{7}{4}$$Let $ a,b,c\geq 0 , (3a+2)(b+c)= 5.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{2(2\sqrt{15}-5)}{3}$$
9 replies
1 viewing
sqing
May 10, 2025
sqing
8 minutes ago
Expressing polynomial as product of two polynomials
Sadigly   4
N 23 minutes ago by TBazar
Source: Azerbaijan Senior NMO 2021
Define $P(x)=((x-a_1)(x-a_2)...(x-a_n))^2 +1$, where $a_1,a_2...,a_n\in\mathbb{Z}$ and $n\in\mathbb{N^+}$. Prove that $P(x)$ couldn't be expressed as product of two non-constant polynomials with integer coefficients.
4 replies
Sadigly
Yesterday at 9:10 PM
TBazar
23 minutes ago
No more topics!
IMO 2023 P2
799786   92
N May 2, 2025 by L13832
Source: IMO 2023 P2
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 $\angle BAC$.
92 replies
799786
Jul 8, 2023
L13832
May 2, 2025
IMO 2023 P2
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO 2023 P2
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EeEeRUT
72 posts
#106
Y by
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.
This post has been edited 1 time. Last edited by EeEeRUT, Dec 15, 2024, 7:51 AM
Reason: E
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kamatadu
480 posts
#107 • 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
1916 posts
#108
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
268 posts
#109 • 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
#110 • 1 Y
Y by S_14159
Solution
This post has been edited 3 times. Last edited by Golden_Verse, Oct 31, 2024, 5:13 PM
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bin_sherlo
727 posts
#111 • 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
21 posts
#112 • 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
182 posts
#113 • 2 Y
Y by alexanderhamilton124, L13832
nice
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ihatemath123
3446 posts
#114
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
631 posts
#115
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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
20 posts
#117 • 1 Y
Y by Nuran2010
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. :o basdim bi tik :oops_sign: ) .
This post has been edited 1 time. Last edited by Frd_19_Hsnzde, Mar 18, 2025, 12:12 AM
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kaede_Arcadia
23 posts
#118
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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 $AV$.

Proof: 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 4 times. Last edited by kaede_Arcadia, Mar 23, 2025, 11:03 AM
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wu2481632
4239 posts
#119
Y by
Only realized after finishing the problem that I'd solved it almost two years ago : P but this time found a different solution

Note that $\angle EAC = 90 - \angle C$ and $\angle SBC = 90 - \frac{1}{2}\angle A$, so $\angle DBE = 180 - \angle C - \frac{1}{2}\angle A$ and thus $\angle DBL = \angle C + \frac{1}{2}\angle A$.

Then $\angle LPB = \angle LDB = \angle SBC = 90 - \frac{1}{2}\angle A$ and $\angle LPD= 180 - \angle C - \frac{1}{2}\angle A$, so $\angle BPD = 90 - \angle C$.

Next, let $X$ be the antipode of $A$ and let $M_A$ be the midpoint of arc $BC$ not containing $A$. As $\angle LPD = \angle SBE$ and $\angle SPX = \angle SXE$, it follows that $L, P, S$ are collinear. Let $AD$ meet $(BDL)$ again at $Y$; evidently $\angle LPY = 90$, but as $\angle SPM_A = 90$ it follows that $P, Y, M_A$ are collinear.

Finally, let $AP$ meet $(BDP)$ again at $Z$. Note that as $\angle APD = 90$, $DZ$ is a diameter of $(BDP)$ and $DLZY$ is a rectangle. Thus $\angle ZBD = \angle SBM_A = 90$ and so $Z, B, M_A$ are collinear.

We finish by Pascal's theorem on hexagon $PPZBDY$.
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ND_
52 posts
#120 • 1 Y
Y by L13832
Let tangent to $\omega$ at $P \cap \Omega = H$

Claim 1: $S, P, L$ are collinear
Proof: $\measuredangle BPL = \measuredangle BDL = \measuredangle CBD = \measuredangle CBS = \measuredangle BCS = \measuredangle BPS$

Claim 2: $SH \parallel PD$
Proof: $\measuredangle DPH = \measuredangle DBP = \measuredangle SBP = \measuredangle SHP $

Claim 3: $PS \parallel EH$
Proof: $\measuredangle SPH = \measuredangle LPH = \measuredangle LBP = \measuredangle EBP = \measuredangle EHP$

Claim 4: $PA \parallel M_AH$
Proof: $\measuredangle PAM_A =\measuredangle PAD +\measuredangle DAM_A = \measuredangle HM_AS + \measuredangle SM_AA  =\measuredangle HM_AA$

Hence, using Claim 2, 4 and $AD \parallel M_AS$, we have that the 3 lines are concurrent.
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L13832
268 posts
#121
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L13832 wrote:
Not 'Too easy for P2', took me 2 attempts and a hint of L-P-S collinear (which was obvious).
Solution
Figure

Remark
probably a new sol:

After getting $\angle APD=90^{\circ}$ we have $\angle M_AAF=\angle AM_AS=\angle ABS$ so $AM_A$ is tangent to $\odot(ABD)$ and we get that $XA^2=XB\cdot XD$, so the problem is equivalent to proving $XA=XP$.

Since $AD\parallel SM_A$ and $\angle ADP=90^{\circ}$, we must have $\overline{N-X-O}$ where $N$ is the midpoint of $AD$ and since $OA=OP$ and $NA=NP$ we have $XA=XP$ and we are done!!!
This post has been edited 1 time. Last edited by L13832, May 2, 2025, 11:37 AM
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