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Source: 2019 Taiwan TST Round 2

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JustPostTaiwanTST

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JustPostTaiwanTST

#1 h Apr 1, 2020, 6:27 AM • 2 Y

Y by itslumi, Funcshun840

Given a triangle $\triangle{ABC}$ . Denote its incircle and circumcircle by $\omega, \Omega$ , respectively. Assume that $\omega$ tangents the sides $AB, AC$ at $F, E$ , respectively. Then, let the intersections of line $EF$ and $\Omega$ to be $P,Q$ . Let $M$ to be the mid-point of $BC$ . Take a point $R$ on the circumcircle of $\triangle{MPQ}$ , say $\Gamma$ , such that $MR \perp EF$ . Prove that the line $AR$ , $\omega$ and $\Gamma$ intersect at one point.

This post has been edited 1 time. Last edited by JustPostTaiwanTST, Apr 1, 2020, 6:50 AM
Reason: add the definition of $ M $

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#2 h Apr 1, 2020, 8:02 AM • 2 Y

Y by AmirKhusrau, Funcshun840

Just a mix of well known lemmas

. Here's my solution: Suppose that the perpendicular from $D$ to line $EF$ meets line $EF$ and $\omega$ at $Z,S$ respectively. Let $N=AS \cap \omega$ and $X=DN \cap EF$ . Note that $N$ is the $D$ -Queue point of $\triangle DEF$ (since $SENF$ is harmonic), and so $X$ must be the $D$ -Ex point of $\triangle DEF$ . Since $X$ lies on the orthic axis of $\triangle DEF$ , so using the well known fact that the orthic axis is the radical axis of the circumcircle and the tangential circle, we get $XP \cdot XQ=XE \cdot XF=XD \cdot XN \Rightarrow N \in \odot (DPQ)$ Taking $K=EF \cap BC$ , we have $(K,D;E,F)=-1$ , which means that $D,K$ are inverses in the circle with $BC$ as diameter. This means that $KD \cdot KM=KB \cdot KC=KP \cdot KQ \Rightarrow M \in \odot (DPQ)$ Combining the above two observations, we get that $N=\omega \cap \Gamma$ , and so it suffices to prove that $R,N,S$ are collinear (since we have $N \in AS$ ). Since $MR \parallel DS$ , so this follows from $\measuredangle MRN=\measuredangle MDN=\measuredangle BDN=\measuredangle DSN \Rightarrow S \in NR \quad \blacksquare$
NOTE: This configuration is pretty well known, and has appeared numerous times on various contests. Off the top of my head, this example is the foremost one.

This post has been edited 2 times. Last edited by math_pi_rate, Apr 1, 2020, 8:07 AM

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#3 h Apr 1, 2020, 3:34 PM

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#4 h May 14, 2020, 12:43 PM • 8 Y

Y by math_pi_rate, AlastorMoody, niyu, GeoMetrix, sameer_chahar12, Aryan-23, itslumi, sabkx

Surprised that all solutions here involve harmonics or powers. Angle-chase ftw

JustPostTaiwanTST wrote:

Given a triangle $\triangle{ABC}$ . Denote its incircle and circumcircle by $\omega, \Omega$ , respectively. Assume that $\omega$ tangents the sides $AB, AC$ at $F, E$ , respectively. Then, let the intersections of line $EF$ and $\Omega$ to be $P,Q$ . Let $M$ to be the mid-point of $BC$ . Take a point $R$ on the circumcircle of $\triangle{MPQ}$ , say $\Gamma$ , such that $MR \perp EF$ . Prove that the line $AR$ , $\omega$ and $\Gamma$ intersect at one point.

WLOG $AB<AC$ . Let $I, O$ be the centres of $\omega, \Omega$ and $M_A$ be the midpoint of arc $\widehat{BAC}$ of $\Omega$ . Clearly, $D \in \odot(MPQ)$ . The centre $N$ of $\odot(MPQ)$ lies on the perpendicular bisector of $\overline{DM}$ and the line through $O$ parallel to $\overline{AI}$ , hence it is the midpoint of $\overline{IM_A}$ .

Suppose $\odot(MPQ)$ meets $\omega$ at $S \ne D$ . Since $\overline{IN} \perp \overline{DS}$ , we see that $\overline{AS}$ is the $A$ -mixtilinear cevian (follows from an angle-chase after noticing $\overline{IM_A}$ is a symmedian in $\triangle IBC$ and $\overline{IM}$ is parallel to the $A$ -Nagell cevian). Let $K \ne S$ be the common point of $\overline{AS}$ and $\odot(MPQ)$ . It is enough to prove $\overline{MK} \parallel \overline{AI}$ .

Let $\alpha=\frac{1}{2}\angle MND$ and $\beta=\frac{1}{2}(\angle B-\angle C)$ . Now $\angle KSM=\angle KSD-\angle MSD=90^{\circ}+\beta-\alpha$ so $\angle MNK=2(180^{\circ}-\angle KSM)=180^{\circ}-2\beta+2\alpha$ hence $\angle KMN =\beta-\alpha$ since $N$ lies on the perpendicular bisector of $\overline{MK}$ . Since $\angle NMD=90^{\circ}-\alpha$ we conclude that $\angle KMD=90^{\circ}-\beta=\angle (\overline{AI}, \overline{BC})$ proving the claim.

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635 posts

#5 h May 30, 2020, 12:37 PM

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Let $\odot(AEF)$ intersects $\odot(ABC)$ at $\mathcal {H}\implies\triangle HBF\sim\triangle HCE.$ Also, $\frac {\mathcal {H}B}{\mathcal {H}C}=\frac{BF}{CE}=\frac{BD}{DC}.$ Hence, $\mathcal {H},K,D$ are collinear $\implies KD\cdot K\mathcal{H}=K\mathcal{J}\cdot KA=LB^2.$ $A\mathcal {J}D\mathcal {H}$ is concyclic $(GD;BC)=-1.$ By applying Maclaurin's theorem $GP\cdot GQ=GD\cdot GM.$ Thus, $MDPQ$ is concylic. Applying Radical axis theorem on $\odot(ABC),\odot(MDPQ),\odot(AEF).$ $A\mathcal {H}$ cuts $PQ$ on the radical axis of $\odot(AEF)$ and $\odot(MDPQ).$ If $\odot(MDPQ)$ intersects $(I)$ at $\ell.$ Then $D\ell,A\mathcal {H},PQ$ concur. Let $\odot(A\mathcal {H}D)$ intersects $(I)$ at $\ell'.$ Then $D\ell',A\mathcal {H},PQ$ concur cause applying Radical axis theorem on $\odot(AEF),(I),\odot (AHD).$ So $\ell=\ell'.$ So note by Reim's theorem $MR\parallel A\mathcal {J}.$

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#6 h Jun 9, 2020, 2:30 PM • 4 Y

Y by Aimingformygoal, GeoMetrix, Bumblebee60, takachonn

Taiwan TST 2019 Round 2 wrote:

Given a triangle $\triangle{ABC}$ . Denote its incircle and circumcircle by $\omega, \Omega$ , respectively. Assume that $\omega$ tangents the sides $AB, AC$ at $F, E$ , respectively. Then, let the intersections of line $EF$ and $\Omega$ to be $P,Q$ . Let $M$ to be the mid-point of $BC$ . Take a point $R$ on the circumcircle of $\triangle{MPQ}$ , say $\Gamma$ , such that $MR \perp EF$ . Prove that the line $AR$ , $\omega$ and $\Gamma$ intersect at one point.

Let $X$ be the midpoint of $\widehat{BAC}$ and $X^*$ denote it's antipode WRT $\Delta ABC$ . Let $\odot(DPQ)\cap\odot(I)=A'$ . Let $\overline{PQ}\cap\overline{BC}=\{V\}$ . Then $VP\cdot VQ=VB\cdot VC=VD\cdot VM$ from Ma'laurin. So, $\{D\}\in\odot(MPQ)$ . Then from OGM- (#109) we get that $\overline{AA'}$ is the $A-$ Mixtillinear Cevian of $\Delta ABC$ $(\bigstar)$ . Let $\{I\}$ denote the Incenter of $\Delta ABC$ . So from here (#1,#2) we get that $\overline{X-R-I-T}$ where $\{T\}$ is the $A-$ Mixtillinear Incircle touchpoint with $\odot(ABC)$ and from $(\bigstar)$ we get $\overline{A-A'-T}$ . Let $\overline{AT}\cap\overline{MR}=K$ . Then by Reims $KTXX^*$ is cyclic. If $\overline{MR}\cap\odot(MPQ)=K^*$ . Then $MR\cdot MK^*=RP\cdot RQ=RX\cdot RT\implies K^*\in\odot(XX^*T)\implies K^*\equiv K$ . $\blacksquare$

This post has been edited 3 times. Last edited by amar_04, Jun 9, 2020, 2:34 PM

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#7 h Jun 19, 2020, 10:22 PM • 2 Y

Y by Cindy.tw, v4913

Solution from Twitch Solves ISL:

Let $D$ be the foot from $I$ to $\overline{BC}$ .
Claim: Quadrilateral $MDPQ$ is cyclic.
Proof. Let $X = PQ \cap BC$ . Then $(XD;BC) = -1$ by Ceva/Menelaus with the Gergonne point. Hence, $XD \cdot XM = XB \cdot XC = XP \cdot XQ$ as needed. $\blacksquare$
We let $G$ be the intersection of the $D$ -altitude to $\overline{EF}$ with the incircle. Let line $AG$ meet the incircle again at $H$ .
$[asy] size(12cm); pair A = dir(130); pair B = dir(210); pair C = dir(330); pair I = incenter(A, B, C); pair D = foot(I, B, C); pair E = foot(I, C, A); pair F = foot(I, A, B); pair M = midpoint(B--C); pair Z = orthocenter(D, E, F); pair K = foot(D, E, F); pair N = midpoint(E--F); pair G = -Z+2*K; pair H = -G+2*foot(I, A, G); filldraw(A--B--C--cycle, invisible, blue); filldraw(incircle(A, B, C), invisible, blue); pair X = extension(E, F, B, C); pair S = foot(A, I, -A); pair Y = extension(H, D, E, F); filldraw(unitcircle, invisible, blue); pair P = IP(circumcircle(H, D, M), unitcircle); pair Q = OP(circumcircle(H, D, M), unitcircle); filldraw(circumcircle(H, D, M), invisible, orange); filldraw(circumcircle(A, E, F), invisible, deepgreen); draw(E--X--P, blue); draw(X--B, blue); pair Rp = extension(M, foot(M, P, Q), A, H); draw(H--Rp--M, red); draw(S--I--A, deepgreen); draw(A--Y--D, lightred); draw(D--G, lightblue); dot("$A$", A, dir(A)); dot("$B$", B, dir(240)); dot("$C$", C, dir(300)); dot("$I$", I, dir(I)); dot("$D$", D, dir(D)); dot("$E$", E, dir(350)); dot("$F$", F, dir(F)); dot("$M$", M, dir(M)); dot("$K$", K, dir(K)); dot("$N$", N, dir(N)); dot("$G$", G, dir(G)); dot("$H$", H, dir(H)); dot("$X$", X, dir(X)); dot("$S$", S, dir(S)); dot("$Y$", Y, dir(140)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(160)); dot("$R'$", Rp, dir(Rp)); /* TSQ Source: !size(12cm); A = dir 130 B = dir 210 R240 C = dir 330 R300 I = incenter A B C D = foot I B C E = foot I C A R350 F = foot I A B M = midpoint B--C Z := orthocenter D E F K = foot D E F N = midpoint E--F G = -Z+2*K H = -G+2*foot I A G A--B--C--cycle 0.1 lightcyan / blue incircle A B C 0.1 lightcyan / blue X = extension E F B C S = foot A I -A Y = extension H D E F R140 unitcircle 0.1 palecyan / blue P = IP circumcircle H D M unitcircle Q = OP circumcircle H D M unitcircle R160 circumcircle H D M 0.1 yellow / orange circumcircle A E F 0.1 lightgreen / deepgreen E--X--P blue X--B blue R' = extension M foot M P Q A H H--Rp--M red S--I--A deepgreen A--Y--D lightred D--G lightblue */ [/asy]$

Claim: Quadrilateral $HDPQ$ is cyclic.
Proof. [Proof $HDPQ$ cyclic] For this proof, we need to introduce several points:

Let $K$ be the foot from $D$ to $\overline{EF}$ .
Let $N$ be the midpoint of $\overline{EF}$ .
Let $S$ be the inverse of $K$ with respect to the incircle, which is known to satisfy $\angle AIS = 90^{\circ}$ .
Let $Y = \overline{HD} \cap \overline{EF}$ .

We have $-1 = (EF;GH) \overset{D}{=} (EF;KY)$ . Since $\overline{SK}$ bisects $\angle FSE$ , this implies $Y$ also lies on line $AS$ . We can then calculate $YH \cdot YD = YF \cdot YE = YK \cdot YN = YS \cdot YA = YP \cdot YQ$ which implies the concyclic condition. $\blacksquare$
We let $R'$ be a point on line $\overline{AGH}$ such that $\overline{R'M} \perp \overline{PQ}$ . Then $\measuredangle HR'M = \measuredangle HGD = \measuredangle HDB = \measuredangle HDM$ so $R' = R$ and the problem is solved (the concurrence point in the problem is $H$ ).

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MP8148

#9 h Sep 13, 2020, 9:50 PM

Y by

why are there 2 threads lol
$[asy] size(10cm); defaultpen(fontsize(10pt)); pair A = dir(130), B = dir(210), C = dir(330), I = incenter(A,B,C), D = foot(I,B,C), E = foot(I,C,A), F = foot(I,A,B), M = (B+C)/2, P = intersectionpoint(unitcircle,E--E+dir(E--F)*100), Q = intersectionpoint(unitcircle,F--F+dir(F--E)*100), R = intersectionpoint(foot(M,E,F)--foot(M,E,F)+dir(M--foot(M,E,F))*100,circumcircle(P,Q,M)), S = intersectionpoints(incircle(A,B,C),circumcircle(P,Q,M))[0], T = extension(I,foot(I,D,S),A,S), K = extension(A,S,D,foot(D,E,F)), L = dir(90), N = (I+L)/2; draw(A--B--C--A); draw(unitcircle); draw(incircle(A,B,C)); draw(circumcircle(M,P,Q), blue); draw(P--Q^^M--R^^D--K^^I--A--L, heavyred); draw(R--T--L^^T--D, orange); draw(I--D^^M--L, magenta); dot("$A$", A, dir(140)); dot("$B$", B, dir(210)); dot("$C$", C, dir(330)); dot("$D$", D, dir(270)); dot("$E$", E, dir(80)); dot("$F$", F, dir(135)); dot("$P$", P, dir(200)); dot("$Q$", Q, dir(20)); dot("$R$", R, dir(110)); dot("$S$", S, dir(210)); dot("$M$", M, dir(270)); dot("$T$", T, dir(225)); dot("$K$", K, dir(140)); dot("$L$", L, dir(90)); dot("$I$", I, dir(330)); dot("$N$", N, dir(140)); [/asy]$
Claim: $D \in \Gamma$ , where $D$ is the tangency point of $\omega$ on $\overline{BC}$ .

Proof. Define a function $f$ such that $f(X) = XB/XC$ for any point $X$ . By Ratio Lemma $f(P)f(Q) = f(\overline{PQ} \, \cap \, \overline{BC}) = f(\overline{EF} \, \cap \, \overline{BC}) = f(D) = f(D)f(M)$ and we may conclude.

Claim: Let $T$ be the $A$ -mixtilinear intouch point. If $\overline{AT}$ meet $\omega$ at $K$ and $S$ where $AKST$ are in that order, then $S \in \Gamma$ as well.

Proof. Let $L$ be the midpoint of arc $BAC$ so that $T \in \overline{LI}$ . From the fact that $\angle DTI = \angle ATI$ (well-known), we know that $\overline{IL}$ is the perpendicular bisector of $\overline{DS}$ . It follows that the midpoint $N$ of $\overline{IL}$ is the circumcenter of $\triangle DMS$ . But since $\overline{AL} \parallel \overline{PQ}$ and $\angle LAI = 90^\circ$ , we have $NP = NQ$ , so $N$ is also the center of $\Gamma$ which implies the conclusion.

Claim: Redefine $R = \overline{AS} \cap \Gamma$ ; then $\overline{MR} \perp \overline{PQ}$ .

Proof. By Reim's $\overline{DK} \parallel \overline{RM}$ , but it is well-known that $\overline{DK} \perp \overline{EF}$ , so we are done.

$\blacksquare$

This post has been edited 1 time. Last edited by MP8148, Sep 13, 2020, 9:52 PM

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gambi

#11 h May 9, 2022, 7:05 PM

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Let $L$ midpoint of arc $BC$ , $N$ midpoint of arc $BAC$ in $\Omega$ . Let $M$ midpoint of $BC$ , and $D=\omega\cap BC$ . Let $G=AI\cap EF$ . Let $H$ be the projection of $D$ on $EF$ . Let $K=NI \cap EF$ , and $S=ND\cap \Omega$ . Let $T=NI\cap \Omega$ and $X=AT\cap \omega$ . Let $V=NI\cap BC$ . Let $Y\in \omega$ such that $DY\perp EF$ . Let $Z=AY\cap BC$ .

$-1=(B,C;N,L)\stackrel{(S)}=(B,C;D,LS\cap BC),$ so $LS, EF, BC$ concur at a certain point $W$ .
$WD\cdot WM=WS\cdot WL=WP\cdot WQ \Longrightarrow D\in \Gamma$ Also,
$\frac{NK}{NI}=\frac{AG}{AI}=\frac{NM}{NL}, \Longrightarrow MK\parallel LI \Longrightarrow M-K-R \thickspace \text{collinear.}$ Now,
$TK\cdot KN=PK\cdot KQ=MK\cdot KR \Longrightarrow TMNR \thickspace \text{cyclic.}$ Now,
$\measuredangle NTR=\measuredangle NMR=\measuredangle NLA=\measuredangle NTA \Longrightarrow A-T-R \thickspace \text{collinear.}$ Also, by $\triangle YFHE\sim IBDC$ we get
$\frac{FH}{HE}=\frac{BD}{DC} \Longrightarrow AFHE\sim NBDC \Longrightarrow \measuredangle FAH=\measuredangle BND \Longrightarrow A-H-S \thickspace \text{collinear.}$ By radical axis on $(LTI), (BIC), \Gamma$ we get that $U=LT\cap BC$ satisfies $UI\perp LI$ , so $(\overline{UI})=(UTDI)$ tangent to $LI$ ,
$\measuredangle DSV=\measuredangle DTV=\measuredangle DTI=\measuredangle DIL=\measuredangle NLA=\measuredangle NSA=\measuredangle DSA \Longrightarrow A-S-V \thickspace \text{collinear.}$ where we have used $DVTS$ cyclic, which is trivial using inversion $NB=NC$ . Therefore $A-H-V-S$ collinear.

Claim. $A-Y-T$ collinear.
Proof.
Let $X'=AY\cap \omega$ .
Taking polars wrt $\omega$ , Brocard on $DEFY$ gives $DE\cap FY\in \mathcal{P}(H)$ .
Pascal on $DX'YFEE$ gives $DX'\cap FE - A - DE\cap FY$ collinear, and $A\in \mathcal{P}(H)$ by La Hire.
Hence $DX'\cap FE\in \mathcal{P}(H)\Longrightarrow H\in \mathcal{P}(DX'\cap FE)$ .
And clearly by La Hire, $A\in \mathcal{P}(DX'\cap FE)$ .
Since $A-H-V$ collinear, then
$V\in \mathcal{P}(DX'\cap FE) \Longrightarrow DX'\cap FE\in\mathcal{P}(V) \Longrightarrow VX'=VD \Longrightarrow VI \thickspace \text{is the perpendicular bisector line of} \thickspace \overline{XD'}.$ And since $T,V\in IN$ , then $T\in VI$ , so
$\measuredangle VTX'=\measuredangle DTV=\measuredangle DSV=\measuredangle NSA=\measuredangle NTA=\measuredangle VTA$ Therefore $A-X'-T$ collinear, and so $X\equiv X'$ , which proves the Claim.

Finally, $\triangle ZDY$ and $\triangle ZMR$ are clearly homothetic from $Z$ because $DY\parallel MR$ , so
$\frac{ZM}{ZR}=\frac{ZD}{ZY}=\frac{ZD}{\frac{ZD^2}{ZX}}=\frac{ZX}{ZD} \Longrightarrow ZM\cdot ZD=ZX\cdot ZR,$ which means $Z\in \Gamma$ , as desired.

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#12 h May 9, 2022, 9:42 PM

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Cute problem, took like 20 mins (yay, strict oly training back).
Ok so let $I$ the incenter of $\triangle ABC$ , let $D$ the intouch point of $\omega$ with $BC$ , let the perpendicular from $D$ to $EF$ meet $\omega$ again at $G$ and hit $EF$ at $K$ , let $EF$ meet $BC$ at $S$ , let $J$ the $D$ -queque point on $\triangle DEF$ , let $DJ$ meet $EF$ at $U$ and let $(AEF)$ meet $\Omega$ again at $V$
Claim 1: $PJDMQ$ is cyclic
Proof: Clearly $-1=(S, D; E, F)$ so by McLaurins theorem and PoP
$SD \cdot SM=SB \cdot SC=SP \cdot SQ \implies PDMQ \; \text{cyclic}$ Now since its well known that $VK$ bisects $\angle EVF$ and $-1=(U, K; E, F)$ projecting
$-1=(A, I; E, F) \overset{V}{=} (AV \cap EF, K; E, F) \implies AV \cap EF \cap DJ=U$ And now by PoP
$UP \cdot UQ=UA \cdot UV=UE \cdot UF=UD \cdot UJ \implies PJDQ \; \text{cyclic}$ Hence $PJMDQ$ is cyclic.
Finishing: Using Claim 1 let $JA \cap \Gamma =R'$ , since its known that $J,G,A$ are colinear by Reim's theorem on $\omega, (PJMDQ)$ we get $GD \parallel MR'$ which means that $R=R'$ hence $AR,\omega, \Gamma$ share $J$ as a common point.
Thus, we are done

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Project_Donkey_into_M4

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Project_Donkey_into_M4

#15 h May 25, 2022, 8:14 AM

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In the diagram attached below I have swapped $R$ and $N$

JustPostTaiwanTST wrote:

Given a triangle $\triangle{ABC}$ . Denote its incircle and circumcircle by $\omega, \Omega$ , respectively. Assume that $\omega$ tangents the sides $AB, AC$ at $F, E$ , respectively. Then, let the intersections of line $EF$ and $\Omega$ to be $P,Q$ . Let $M$ to be the mid-point of $BC$ . Take a point $R$ on the circumcircle of $\triangle{MPQ}$ , say $\Gamma$ , such that $MR \perp EF$ . Prove that the line $AR$ , $\omega$ and $\Gamma$ intersect at one point.

Say $G \in \omega$ such that $GENF$ is harmonic where $N \in \omega,DN \perp EF$ .By the harmonic condition we have $A-G-N$ .Say $A_{SD}$ be the $\text{A-Sharkydevil}$ point.So by Incircle Inversion $I,A_{SD},DN\cap EF=J$ are collinear.Now say $EF \cap BC=X$
Claim 1: $D \in \Omega$
Proof : Since $AD,BE,CF$ concurr,by ceva-menelaus we have, $(XD;BC)=-1$ .So by midpoint lengths,
$XD.XM=XB.XC=XP.XQ \implies D \in \Omega \text{ }\blacksquare$ Claim 2: $G \in \Omega$
Proof : Say $AA_{SD}\cap EF=H$ and $DG\cap EF=H'$ ,we will show that $H\equiv H'$ and then prove the claim.Consider the following Cross-ratio chase
$(H,J;E,F)\stackrel{A_{SD}}=(A,I;E,F)=-1=(G,N;E,F)\stackrel{D}=(H',J;E,F)$ This proves $H\equiv H'$ .Now by Power of Point,
$HD.HG=HE.HF=HA_{SD}.HA=HP.HQ \implies G\in \Omega \text{ } \blacksquare$ Finally say $GR \cap MR=R'$
Claim 3: $N' \equiv N$ and thus $A-G-R$
Proof : Since $DR || MN$ , $\angle GND=\angle GR'M$ .Thus,
$\begin{align*}\angle GR'M+\angle GDM &=\angle GND+\angle GDN+\angle MDN \\ &=\angle GND+\angle GDN+\angle DGN \\ &=180 \end{align*}$ Which means $R'\in \Omega \implies R'\equiv R,R-A-G$ ,we get $AR,\Gamma,\omega$ meet at the same point $G$ $\blacksquare$

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Eka01

#16 h Sep 7, 2024, 9:21 AM

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We first prove that $D$ and $H$ lie on $(MPQ)$ where the incircle touches $BC$ at $D$ and $H$ is the $D$ queue point of $\Delta DEF$ .
Let $EF \cap BC= J$ then it is well known that $(JD;BC)=-1$ so by the harmonic midpoint lemma, $JB.JC=JD.JM=JP.JQ$ so $D$ lies on $(MPQ)$ .
Now let $S$ be the $A$ sharky devil point and $AS \cap EF=K$ . It is well known that $K$ is the $D$ expoint in $\Delta DEF$ so it is collinear with $D$ and $H$ . Now by power of point- $KP.KQ=KS.KA=KF.KE=KH.KD$ so $D,H \in (MPQ)$ .
Now let $D$ altitude cuts $EF$ and $\omega$ at $G$ and $T$ respectively. It is well known that $\overline{I-G-S}$ and $\overline{A-T-H}$ so it suffices to prove that $R,T,H$ are collinear which follows by some angle chasing.

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#17 h Dec 12, 2024, 10:58 AM

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$EF\cap BC=T$ Let $L,N$ be the midpoints of arcs $BC,BAC$ respectively and let $S$ be $A-$ sharky devil point. Let $D$ be the tangency point of incircle with $BC$ . Since $T,S,N$ and $S,D,L$ are collinear, we have $TP.TQ=TS.TN=TD.TM$ thus, $D,M,P,Q$ are concyclic. Invert around the incircle.
New Problem Statement: $ABC$ is a triangle with circumcenter $O$ . Nine point circle and $(BOC)$ meet at $P,Q$ and $M$ is the intersection of $(APQ)$ and the circle with diameter $AO$ . $(APQ)$ intersects $(ABC)$ at $V$ and $(VON)$ meets $(APQ)$ at $W$ , prove that $(MOW)$ is tangent to the perpendicular bisector of $BC$ .
Perform $\sqrt{\frac{bc}{2}}$ inversion and reflect over the angle bisector of $\measuredangle CAB$ .
New Problem Statement: $ABC$ is a triangle with circumcenter $O$ and its nine point circle meets $(BOC)$ at $P,Q$ . Let $K,L$ be the midpoints of $AC,AB$ respectively and $PQ$ intersects $BC,KL$ at $M,V$ . $S$ is $A-$ dumpty point and $(DSV)$ meets $PQ$ at $W$ . Prove that $(DAS)$ and $(DWM)$ are tangent to each other.
Since the radical axises of $(AO),(BOC)$ and nine point circle are concurrent, we conclude that $SO,PQ,KL$ meet at $V$ which is equavilent to the collinearity of $S,O,V$ . Note that $\measuredangle ASV=\measuredangle ASO=90$ . Let $l$ be the tangent to $(DAS)$ at $D$ . If $E$ is the midpoint of $AD$ , then $A,E,S,V$ are concyclic.
$\measuredangle (WD,l)=\measuredangle (SD,l)+\measuredangle WDS=\measuredangle SAD+\measuredangle WVS=\measuredangle SVE+\measuredangle WVS=\measuredangle WVK=\measuredangle WMD$ As desired. $\blacksquare$
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