Let $ABCD$ be a cyclic quadrilateral with $AB<CD$. The diagonals intersect at the point $F$ and lines $AD$ and $BC$ intersect at the point $E$. Let $K$ and $L$ be the orthogonal projections of $F$ onto lines $AD$ and $BC$ respectively, and let $M$, $S$ and $T$ be the midpoints of $EF$, $CF$ and $DF$ respectively. Prove that the second intersection point of the circumcircles of triangles $MKT$ and $MLS$ lies on the segment $CD$. (Greece - Silouanos Brazitikos)
Problem
Source: Balkan MO 2016, Problem 2
Tags: geometry, cyclic quadrilateral, Nine Point Circle, Balkan Mathematics Olympiad, Balkan, bir tiyinga qimmat masala
07.05.2016 16:13
We will prove that the second intersection is $X$, the midpoint of $CD$ Note that circle pass through $M,K,T$ is nine point circle of $\triangle{EFD}$ Let $P$ is midpoint of $ED$ Note that $\angle{PMT} =\angle{PDF} =\angle{ADB}=\angle{ACB}$ And since $EC\| PX,XT\|AC$, we get $\angle{ECA}=\angle{PXT}$ So $\angle{PMT} =\angle{PXT}$, so $X\in (MPT)=(MKT)$, this finish the prove
07.05.2016 16:46
Let $P$ and $Q$ the midpoints of $AB$ and $CD$ $\Longrightarrow$ by Newton-Gauss line we get $M$, $P$ and $Q$ are collinear and it's well-know: $ME^2=MP.MQ...(\star)$ and $EF$ cuts $AB$ and $CD$ in $X$ and $Y$ $\Longrightarrow$ combining $(E,F,X,Y)=-1$ and $M$ is midpoint of $EF$ we get $ME^2=MX.MY...(\star\star)$ $\Longrightarrow$ by $(\star)$ and $(\star\star)$ we get $XYQP$ is cyclic. Define $\angle BDA=\alpha$, $\angle BDC=\beta$ and $\angle BEM=\theta$ $\Longrightarrow$ since $ML=MF$ and $LS=SC$ we get $\angle MLS=180-\alpha-\theta$, so from $SQ\parallel BD$ we get $\angle SQC=\beta$ since $XYQP$ is cyclic we get $\beta+\angle SQM=\angle ZQP=\angle BYZ=\alpha+\beta+\theta$ $\Longrightarrow$ $\angle SQM=\alpha+\theta$ hence $MLSQ$ is cyclic simililarly $MKTQ$ is cyclic hence $\odot (MLS)\cap \odot (MKT)\in CD$.
07.05.2016 17:33
Let $G,D$ be the midpoints of $AB$ and $CD$, let $\{X\}=EF\cap AB,\ \{Y\}=EF\cap CD$. It is well known that $XYDG$ is cyclic. Note that $$\widehat{MDT}=\widehat{MDC}-\widehat{DTC}=180^\circ-\widehat{ AXF}-\widehat{FDC}=\widehat{XFA}=180^\circ- \widehat{EFC}=180^\circ-\widehat{MKT}$$so $D\in (MKT)$. Analogously, $D\in (MLS)$, whence the conclusion.
07.05.2016 18:31
Let $G$ be midpoint of $DC$ and let $P$ and $Q$ be midpoints of $ED$ and $EC$, respectively, thus $\odot (PQG)$ and $\odot (STG)$ are $9$-points circles of $\triangle DEC$ and $\triangle DFC$, respectively. Let $C_1$ and $D_1$ be foots of perpendiculars from $C$ and $D$ to $ED$ and $EC$, respectively. Similarly define $C_2$ and $D_2$ in $\triangle FDC$. Obviously $DC_1C_2D_2D_1C$ is cyclic hexagon with center $G$ and circle $\omega$. Now consider inversion $\Phi$ WRT $\omega$. Obviously $\Phi(\odot (GST))=D_2C_2$ and $\Phi(\odot (PQG))=D_1C_1...\bigstar$, but $$\angle DBA=\angle DCC_2=\angle DD_2C_2 \Longrightarrow C_2D_2||AB...\bigstar \bigstar$$and since $\odot (DC_1D_1C)$ and $\odot (DABC)$ meet at $DC$, by $\textit{Reim}$'s theorem we have $$C_1D_1|| AB \stackrel{\bigstar \bigstar}{\Longrightarrow} C_1D_1|| C_2D_2\stackrel{\bigstar}{\Longrightarrow} \odot (C_1D_1G)\ \& \ \odot (C_2D_2G)$$are tangent at $G$, thus they also concur at $\textit{Poncelet}$ point of points $D$, $E$, $C$ and $F$, thus since circles $\odot (PQG)$ and $\odot (GST)$ are $9$-point circles of $\triangle DEC$ and $\triangle DFC$, respectively, they also passes through $G$. $\blacksquare$
07.05.2016 18:59
Lemma: Let $ABCD$ be a cyclic quadrilateral, and let diagonals $\overline{AC}$ and $\overline{BD}$ intersects at $E.$ Let $P,Q$ denote the foot from $E$ to $\overline{AD},\overline{BC}$ respectively, and let $X,Y$ denote the midpoint of $\overline{AB},\overline{CD}$ respectively. Show that $XY$ is the perpendicular bisector of $PQ.$ (1.) If we let $N$ be the midpoint of $\overline{CD},$ one can find that $\triangle NTK$ is congruent with $\triangle NSL.$ (2.) Some angle-chasing we find that $\angle KTN=180^\circ-\angle DEC.$ (3.) From Newton-Gauss Line and the above lemma we see that $MN$ bisects $\angle KML,$ i.e. $\angle KMN=\angle LMN=\tfrac{1}{2}\angle DEC,$ the result follows.
07.05.2016 19:46
Let $X$, $Y$ be midpoint of $CD, ED$. Observe $(MKYT)$ is nine-point circle of $\triangle EKD$. Since $\angle (YX, XT)=\angle (EC, CF)=\angle (FD, DE)=\angle (FT, TM)=\angle (YM, MT)$, $X\in (MKYT)=(MKT)$. Likewise, $X\in (MLS)$. Done.
07.05.2016 20:01
My solution : Lemma : Let $ABCD$ be a cyclic quadrilateral, $AC$ cuts $BD$ at $P$. $K , l , M $ are the projections of $P$ on $AD, BC, CD$ and $M$ is the midpoint of $CD$ . Then $KLNM$ is cyclic and $NL= NK$ Prove : Let $X,Y$ be the midpoints of $PC$ and $PB$ , $O_1$ is the circumcenter of $(KLM)$ => $XO_1 \perp ML$ , $YO_1 \perp MK$ . $S$ refelcts with $P$ about $O_1$ => $DS \perp KM$ => $\angle SDC = \angle BDA = \angle BCA = \angle SCD$ =>$SN \perp CD$ =>$O_1$ lies on perpendicular bisector of $MN$ =>$KLNM$ is cyclic => $\angle KLN = \angle KMD = 90 - \angle KMP = 90 - \angle KDB = 90 - \angle PCB = 90 - \angle PML = \angle LMC = \angle LKN $ => $NK = NL$ Come back to the problem : Let G be the midpoint of CD and H is the projection of $F$ on CD =>$GK = GL$. And we have $MK= ML$ =>$\triangle MKG = \triangle MLG$ => $MG \perp LK$ => $\angle KGM = 90 - \angle LKG = 90 - \angle KHD = \angle KDF = \angle DKT = \angle KTM$ => $MKTG $ is cyclic Similary => $MLSG$ is also cyclic
07.05.2016 20:25
Let $R$ the midpoint of $CD$.Like $CD>AB$, then $B$ is between $C$ and $E$ ,analogously $A$ is between $D$ and $E$.$TM\perp KF$ and passes through the midpoint of $FK$, then $\angle TMK=\angle TMF=\angle DEF$, plus for angle-chasing $\angle RTK=\angle LFK$, plus tringles $CFL$ and $DFK$ are similar and $2(TK)=DF$ and $2(RT)=CF$, so triangles $RTK$ and $LFK$ are similar, then $\angle TRK=\angle FLK=\angle FED$, so the quadrilateral $RTKM$is cyclic, analogously the quadrilateral $RSLM$ is cyclic, then $R$ is the second point of intersection and obvius that this $CD$.
08.05.2016 00:35
Just remark for the point of intersection is it magic to consider the midpoint as point of intersection ? that 's not because we have that the 3 ninepoint-circles of 3 triangles $PBC,PCA,PAB$ ,where $ABC $ triangle and $P$ a point ,have a common point so for the problem its necessary the midpoint or the foot of $F$ on $DC$
08.05.2016 02:27
The problem was proposed by me, Silouanos Brazitikos
08.05.2016 19:04
Let $X$ be a midpoint of $ED$. Quadrilateral $XKTM$ is cyclic because it belongs to nine point circle of $\Delta EFD$. And let $N$ be a midpoint of $CD$. We shall prove $N \in \odot MKT$ or $N \in \odot MXT$. This is equivalent with $\angle MXN=\angle MTN$ $\iff$ $180^{o}-\angle DXN-\angle EXM=\angle MTF+\angle NTF$ $\iff$ $180^{o}-\angle DEC-\angle EDB=\angle EDF +\angle CFB$ $\iff$ $180^{o}=\angle DEC+\angle EDB+\angle EDF +\angle CFB$ $\iff$ $180^{o}=\angle DEC+2\angle EDB+\angle CFB$ $\iff$ $180^{o}=(\angle DEC+\angle EDB)+\angle EDB+\angle CFB$ $\iff$ $180^{o}=\angle DBC+\angle EDB+\angle CFB$ $\iff$ $180^{o}=\angle FCB+ \angle CFB + \angle CBF$ $\iff$ $180^{o}=180^{o}$ which is true, thus second intersection point of the circumcircles of triangles $MKT$ and $MLS$ lies on the midpoint of segment $CD$
09.05.2016 20:23
My solution : Let $X$ be the midpoint of $DC$ then we have $\measuredangle MLS=\measuredangle MLF+\measuredangle FLS=\measuredangle LFM+\measuredangle CFL=\measuredangle CFE=180^o-\measuredangle EFA$. from other hand we have $\measuredangle MSX=\measuredangle MSA+\measuredangle ASX=\measuredangle BCA+\measuredangle AFD=\measuredangle BDA+\measuredangle AFD=\measuredangle CAE$. see also that $\frac{MS}{EA}=\frac{1}{2}\cdot \frac{EC}{AE}=\frac{1}{2}\cdot \frac{ED}{BE}=\frac{1}{2}\cdot \frac{sin\measuredangle EBD}{sin\measuredangle BDA}=\frac{1}{2}\cdot \frac{sin\measuredangle DAF}{sin\measuredangle FDA}=\frac{1}{2}\cdot \frac{FD}{AF}=\frac{XS}{AF}$. this means that the triangles $MSX$ and $EAF$ are simillar which give us $\measuredangle EFA = \measuredangle SXM$ and thus, the quadrilateral $MLSX$ is Cyclic. Clearly that $MKTX$ is also Cyclic (see the symmetric!) and we are done!
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14.05.2016 14:09
A different solution.
31.05.2016 19:10
The second intersection of those two circles is $K'$ the midpoint of $CD$ Proof: Let $X$ be the midpoint of $EC$ By the nine point theorem in triangle $FEC$ we get that $XLSM$ is cyclic Now we have: $MX||AC$ and $MS||AC$ so $XMS=ECA$ And $XK'||ED$ and $K'S||BD$ so $XK'S=ADB$ And it's easy to see $ADB=ACB$ So $MXSK'$ is cyclic too and so $MXLSK'$ is cyclic In the same way we get $MKTK'$ is cyclic So done
20.06.2016 18:13
Also here : http://forumgeom.fau.edu/FG2012volume12/FG201212.pdf .
29.09.2016 19:28
It can also be solved using this problem from Iran's national olympiad (2nd round): http://artofproblemsolving.com/community/c6h1086158p4802510
17.11.2016 16:09
Please everyone. Read this carefully and answer my question. How did this come to your mind that their intersection must be midpoint?
17.11.2016 17:34
Murad.Aghazade wrote: Please everyone. Read this carefully and answer my question. How did this come to your mind that their intersection must be midpoint? Just by careful drawing
09.01.2017 16:39
İs there any solution without defining any point and using similarity?
15.01.2017 05:07
Murad.Aghazade wrote: Please everyone. Read this carefully and answer my question. How did this come to your mind that their intersection must be midpoint? I was trying to see how to use the angles $\angle TND$ ($N$ being the point of intersection).and $\angle SNC$
19.12.2017 16:31
Solution: Let $U$ be the midpoint of $CD$, $V $ and $W $ those of $CE $ and $DE$ respectively. Claim: $U $ is the Poncelet point of $EDCF $. Proof of the claim: Let $l$ be the tangent to $\odot (STU) $ at $U $. Observe that $\measuredangle(l,US)=\measuredangle ACD $. Since, $US\parallel BD $ and $UV\parallel ED$, so, $\measuredangle VUS=\measuredangle ADB=\measuredangle ACB $. Thus, $\measuredangle(l,UV)=\measuredangle BCD=\measuredangle UWV $. Hence, $l $ is also tangent to $\odot (UVW) $ at $U\implies \odot (STU) $ and $\odot (UVW) $ are tangent to each other at $U\implies U $ is the Poncelet point of $EDCF $. Main problem: The result is now obvious as $\odot (MKT) $ and $\odot (MLS)$ are the nine-point circles of $\Delta EFD $ and $\Delta EFC $ respectively.
19.12.2017 16:45
Consider triangle $EDC$. Note that isogonal conjugate of perpendicular bisector of $CD$ is a hyperbola centered at $U$, the midpoint of $CD$. Since $F$ lies on this hyperbola, $U$ is the Poncelet point of $CDEF$. But $\odot(MKT), \odot(MLS)$ are nine points circle of $\Delta DEF, \Delta EFC$ hence it pass through $U\in CD$ as desired.
30.12.2020 01:27
Eray wrote: Let $ABCD$ be a cyclic quadrilateral with $AB<CD$. The diagonals intersect at the point $F$ and lines $AD$ and $BC$ intersect at the point $E$. Let $K$ and $L$ be the orthogonal projections of $F$ onto lines $AD$ and $BC$ respectively, and let $M$, $S$ and $T$ be the midpoints of $EF$, $CF$ and $DF$ respectively. Prove that the second intersection point of the circumcircles of triangles $MKT$ and $MLS$ lies on the segment $CD$. (Greece - Silouanos Brazitikos) Sorry for reviving this post but nobody has mentioned this so far. Firstly, this is an incredibly rich configuration, I have the same solution as the people above using the nine point circle and the fact that the two circles intersect at the midpoint of $DC$. I just wanted to mention the fact that the circumcircles of $\triangle MKT$ and $\triangle MLS$ have equal radii which follows from the fact that $\angle KTM$ = $\angle MSL$ and $MK$ = $ML$ as both $K$ and $L$ lie on the circle with diameter $EF$ which has center $M$.
20.04.2021 17:16
Let $G$ be the midpoint of $CD$, we prove that $G$ is indeed the intersection of the circles. Let $Y$ be the midpoint of $EC$ and $X$ be the midpoint of $ED$, note that $(MLS)$ is the nine point circle of $\triangle EFC$ so $Y \in (MLS)$, similarly $X \in MKT$.
yields $\angle TGX = \angle TMX$, so $G \in (MXKT)$, similarly $\angle SGY = \angle SMY \implies G \in (MXKT)$ and we are done!
28.01.2022 15:33
When I checked this problem, it reminded me the Newton-Gauss line..... So I googled <<Newton-Gauss line>> and this is the link from Wikipedia: https://en.wikipedia.org/wiki/Newton%E2%80%93Gauss_line When I went on applications to cyclic quadrilaterals things became really interesting
Very nice problem though... Congratulations to Silouanos who proposed the problem.
05.02.2022 22:11
Since $ABCD$ is cyclic, the lines $E\infty_{BD}$ and $E\infty_{AC}$ are isogonal in $\angle DEC$
05.02.2022 22:32
Isogonal line lemma solution: We use the following well known lemma. Lemma (Isogonal line lemma): Supose that the lines $\{AP, AQ\}$ and $\{AS, AT\}$ are isogonal with respect to a certain angle. Then if $X=PS\cap QT$ and $Y=PT\cap QS$, $\{AX, AY\}$ are also isogonal pairs. Proof: Well known. Apply polar duality at $E$ to get that it is equivalent to the inscribed angle theorem. $\blacksquare$ With this lemma in mind, we can now approach the problem. I claim the concurrency point is $N$, the midpoint of $CD$. The pith of the problem is the following Claim: Let $F'$ be the reflexion of $F$ over $N$. Then $\{AF, AF'\}$ is an isogonal pair with respect to $\Theta:=\angle DEC$ Proof: Since $ABCD$ is cyclic, the lines $E\infty_{BD}$ and $E\infty_{AC}$ are isogonal in $\Theta$. Now, by isogonal line lemma on $\{ED, EC\}$ and $\{E\infty_{BD}, E\infty_{AC}\}$, we get $D\infty_{DB}\cap C\infty_{AC}=F$ and $D\infty{AC}\cap C\infty{BD}=F'$ are isogonal. $\blacksquare$ Now we just need to prove that $MKSN$ is cyclic. Indeed, let $P$ be the midpoint of $EC$, such that $(MPKS)$ is the nine point circle of $\triangle ECF$. Then $$\angle NMS=\angle (NM, MS)=\angle (EF, EC)=\angle F'EC=\angle DEF=\angle NPS$$ so $MPKSN$ is cyclic. Then end $\blacksquare$
19.02.2022 16:43
we will prove their intersection point is midpoint $N$ of $CD$. we will use a well-known lemma : Lemma : Let $ABCD$ be a cyclic quadrilateral. The diagonals intersect at the point $F$. Let $K$, $L$ and $P$ be orthogonal projections of $F$ onto $AD$, $BC$ and $CD$. Let $N$ be midpoint of $CD$ then $KLNP$ is cyclic. Claim1 : $NM$ is perpendicular bisector of $KL$. Proof: $\angle KLN = \angle KPD = \angle KFD = \angle LPC = \angle LPN = \angle LKN$ so $NK = NL$. $\angle EKF = \angle ELF = 90$ so $KM = AM = LM$. Now we have $\angle KNM = \angle KNL/2 = \angle KPL/2 = \angle KDF = \angle KTM$ so $MKTN$ is cyclic. with same approach we prove $MLSN$ is cyclic. we're Done.
14.03.2022 06:27
Let $P$ be the midpoint of $CD$. We claim that $P$ is the second intersection of $(MKT)$ and $(MLS)$. Let $P', K', L'$ be the reflections of $F$ across $P, K, L$ respectively. Taking a homothety centered at $F$ with ratio $2$, we need to show that $EL'CP'$ and $EK'DP'$ are cyclic. Since $\angle EDF = \angle ADB = \angle ACB = \angle ECF$ and $DFCP'$ is a parallelogram, then by the Parallelogram Isogonality Lemma, lines $EF$ and $EP'$ are isogonal wrt $\angle DEC$. Then $\angle P'EL' + \angle P'CL' = \angle P'EL + \angle LEL' + \angle P'CF + \angle FCL' = \angle DEF + \angle FEL + (180^{\circ} - \angle DFC) + 2\angle FCB = 180^{\circ} + \angle DEC + \angle ECF + \angle FDE - \angle DFC = 180^{\circ}$, hence, $EL'CP'$ is cyclic. Similarly, we can show that $EK'DP'$ is cyclic, and we are done.
16.03.2022 02:37
Let $P$ be the midpoint of segment $CD$. We claim that both $(MKT)$ and $(MLS)$ pass through $P$, and hence their second point of intersection lies on $CD$. We prove that $MLSP$ is cyclic, and a similar argument shows $PTKM$ is cyclic. Claim 1: $AEBF \sim SMTP$. Proof: First, $\triangle AEB \sim \triangle CED$ as $AB$ is anti-parallel to $CD$ wrt $\triangle ECD$. Then, $\triangle CED \sim \triangle SMT$ by a homothety centre $F$ scale factor $1/2$ (let this be $\Theta$ from now on). This means $\triangle AEB \sim \triangle SMT$, with opposite similarity. Next, $\triangle FAB \sim \triangle FDC$ by two applications of same segment theorem and then $AA$ similarity. $\triangle FDC \sim \triangle FTS$ by $\Theta$. It's well-known that the lines joining the midpoints of edges of a triangle splits it into four congruent triangles, and so $\triangle FTS \sim \triangle PST$, as $T, S, P$ are these midpoints for $\triangle DFC$. Thus, $\triangle FAB \sim \triangle PST$ oppositely. Now, $AEBF \sim SMTP$ since: quadrilateral $AEBF$ is a concatenation of $\triangle AEB$ and $\triangle FAB$, and $SMTP$ is a concatenation of $\triangle SMT$ and $\triangle PST$, the edge where concatenation occurs correspond in both quadrilaterals, similarity is oppositely oriented for all triangles. $\square$. Claim 2: $\triangle FMS \cong \triangle LMS$. Proof: By $\Theta$, $EC \to MS$ means $EC || MS$. But then because $FL \perp BC$, $FL \perp MS$, and $\Theta$ maps $L \to X$ on $MS$ where $FX \perp MS$. However, $XF = XL$ means $MS$ is actually the perpendicular bisector of $FL$, and so $MF = ML$, $SF = SL$, $MS = MS$, and $\triangle FMS \cong \triangle LMS$. $\square$. \begin{align*} \angle PML + \angle LSP &= \angle PMS + \angle SML + \angle LSM + \angle MSP \\ &= (\angle SML + \angle LSM) + (\angle PMS + \angle MSP) \\ &= (\angle SMF + \angle FSM) + (180 - \angle SPM) \\ &= \angle SMF + \angle FSM + (180 - \angle AFE) \\ &= \angle SMF + \angle FSM + \angle SFM = 180. \end{align*} $\therefore$ by converse of opposite angles in a cyclic quad, $MLSP$ cyclic. $\blacksquare$.
01.07.2022 05:49
I was wondering whether we can approach this problem via projective transformation, we can fix a projection that takes ABCD to a Quad where AD=BC and ABCD is still cyclic Then we can easily prove that M,K,T,G (where G is defined to be the midpoint of side CD) is cyclic as well as M,L,S,G. Now since Projection preserves collinearity, we are done. (I am not entirely sure about this solution and would appreciate it someone corrects it)
15.08.2022 23:47
Let the midpoints of $CD, DE, EC$ be $N, X, Y$ respectively. The existence of the Nine-Point Circle implies $MXKT$ and $MYLS$ are cyclic. Now, we define $D_1$ as the reflection of $D$ in $M$. Observe that $$\measuredangle ED_1F = \measuredangle FDE = \measuredangle BDA = \measuredangle BCA = \measuredangle ECF$$so $D_1EFC$ is cyclic. Thus, taking a homothety at $D$ with scale factor $\frac{1}{2}$ implies $MXTN$ is cyclic. Similarly, we deduce $N \in (MYS)$, which suffices. $\blacksquare$ Remark: It's pretty reasonable to guess that the midpoint of $CD$ is the concurrency point, as there is a lot of symmetry in this configuration.
23.10.2023 23:12
Let the midpoint of $CD$ be $N$. We try to show that N is on both circumcircles of MKT and MLS. By Simple Lemma, in triangle $MTS$, $\angle TMN= \angle FMS$. Let $\angle BEM=y$ and $\angle AEM=x$. By $MS || EC$ , $\angle FMS= \angle FEC=y$. So, $\angle NMS=x$. Let $\angle TKF=\alpha$ $\Longrightarrow$ $\angle FSM=90-\alpha$. After some simple angle chasing, we get that $\angle MKT=90-x+\alpha$. So we must find that $\angle MNT=90+x-\alpha$. let $AC$ intersects $MN$ at $P$. So $\angle NPS=\angle PSM+\angle PMS=90+x-\alpha$. By $AC || TN$, $\angle SPN=\angle TNM$ So, $MKTN$ is cyclic. With the same algorithm, we also find that $MLSN$ is cylic. We are done!
16.03.2024 21:38
Really nice! Let $P$ and $Q$ denote the reflections of $C$ with respect to $L$ and $M$ respectively. Then we have $ \angle FPC=\angle FCB=\angle FDE$ which implies that $DEPF$ is cyclic, and Since $CFQE$ is a parallelogram, we get that $\angle FPC= \angle FCE=\angle FQE$ thus $QEPF$ is cyclic. So we get that $DFPEQ$ is a cyclic pentagon which implies that $QPFD$ is cyclic. Next Applying homothety centered at $C$ with ratio $\frac{-1}{2}$ To points $D,F,P,Q$ implies: $NSLM$ is cyclic. Similarly $MKTN$ is cyclic and we are done.
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15.04.2024 12:05
Eray wrote: Let $ABCD$ be a cyclic quadrilateral with $AB<CD$. The diagonals intersect at the point $F$ and lines $AD$ and $BC$ intersect at the point $E$. Let $K$ and $L$ be the orthogonal projections of $F$ onto lines $AD$ and $BC$ respectively, and let $M$, $S$ and $T$ be the midpoints of $EF$, $CF$ and $DF$ respectively. Prove that the second intersection point of the circumcircles of triangles $MKT$ and $MLS$ lies on the segment $CD$. (Greece - Silouanos Brazitikos) Let $X,Y,N$ be the midpoints of $DE,CE,DC$ the by Euler we have that:$K,X,M,T$ lie on a ciecle and then:$<TNX=<ACE=<ACB=<ADB=<XMT$ so $N$ belongs to $(KXMT)$ similary $N$ belongs to $(MYLS)$
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12.08.2024 16:42
The problem was really beautiful! Let <BCA = <BDA = a, <ACD = b, <BDC=c, as <FKE = <FLE = 90°, so quad. FKEL is cyclic ; so <LFK = 180°- <DEC = <ECD + <EDC = 2a+b+c T is the midpoint of DF and let N be the midpoint of CD so TN=FC/2=FS, <NTK = <FTK + <FTN = 2a+b+c = <LFK. in tri. FKD we can see that FT=KT=DT ,tri. LFC and tri. FKD are similar because (LF/FC) = (FK/FD) , (LF/FS) = (FK/FT) then we will get (LF/FK) = (TN/KT) and <NTK = <LFK . We also have M as the midpoint of FE so FM = EM =KM , then <TNK = <FLK = <FEK = <MKE in tri FDE, we have TM || ED so, <TMK = <MKE = <TNK, so quad. MKTN is cyclic, similarly we can show that quad. MLSN is cyclic Finally, N is the 2nd intersect point of the circumcircle of tri. MKT and tri. MLS which lies on CD.
13.10.2024 23:04
kinda messy solution