Let $ABCDEF$ be a convex hexagon all of whose sides are tangent to a circle $\omega$ with centre $O$. Suppose that the circumcircle of triangle $ACE$ is concentric with $\omega$. Let $J$ be the foot of the perpendicular from $B$ to $CD$. Suppose that the perpendicular from $B$ to $DF$ intersects the line $EO$ at a point $K$. Let $L$ be the foot of the perpendicular from $K$ to $DE$. Prove that $DJ=DL$. Proposed by Japan
Problem
Source: IMO Shortlist 2011, G7
Tags: geometry, circumcircle, IMO Shortlist
13.07.2012 17:43
WakeUp wrote: Let $ABCDEF$ be a convex hexagon all of whose sides are tangent to a circle $\omega$ with centre $O$. Suppose that the circumcircle of triangle $ACE$ is concentric with $\omega$. Let $J$ be the foot of the perpendicular from $B$ to $DF$ intersects the line $EO$ at a point $K$. Let $L$ be the foot of the perpendicular from $K$ to $DE$. Prove that $DJ=DL$. Image not found DJ = DL??
13.07.2012 17:58
Distance from$B$ to $AF$ =dist B to $CD$ Let the projections be $j_1$ and $j_2$ let $l_1$, $l_2$ be on $ED$, $EF$ with the distance we need $x=Cj_1=Aj_2$ perpenendiculars in $l_1$ and $l_2$ meet in $K'$ let $W_1$ be the circle with center $k'$, radius $K'l_1$, $W_2$ the circle with center $B$ radius $BJ$ those have $DF$ as radical axis because $DL$ is tangent to the cricles...etc =>$K'B$ perpendicular on $FD$=> $K=K'$ QED
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14.07.2012 02:10
nquocthuy wrote: DJ = DL?? sorry, edited.
15.07.2012 14:01
because of the concentric circles the powers of $A,C.E$ wrt \omega are equal so by adding the tangents from $D,F,B$ to the tangent from $A$(all to $\omega$) we get $DC=DE,FE=FA,AB=BC$ since $OC=OE, DE=DC$ and $DO=DO $$\triangle DOC \cong \triangle DEO \Rightarrow \angle OCD=\angle OED \Rightarrow \angle FED=\angle BCD$ similarly $\angle BAF=\angle FED= \angle DCB=x$ $FE=FA,OE=OA$ so $FO$ is the perpendicular bisector of $EA$ mow let $CO\cap BJ=Z$ and let $X$ be on $OE$ such that $O-E-X$ and $XE=CZ$ now let $Y$ be symmetric of $Z$ wrt $FO$ $\angle FAY=\angle FEX=180- x/2$ since they are symmetric wrt $FO$ and $EX=AY=CZ$ also $\angle BAY=180- x/2=\angle BCZ$ but also $BA=BC$ so $\triangle BCZ\cong \triangle BAY$ so $\angle JBC=\angle LBA$ where $L=FA\cap BY$ so because $\angle BAF=\angle BCD$ we get $\angle BLF=\angle BJD=90$ so $FY^2-FB^2=AY^2-AB^2$ so $FX^2-FB^2=BC^2-CZ^2=BD^2-DZ^2=BD^2-DX^2$($DZ=DX$ because $OE=OC$ and $OX=OZ$ and so $XZ\parallel EC$ so $OD$ is the perpendicular bisector of $XZ$. from the last fact we have that $BX\perp DF$ but since $X$ is on $EO$ than $X=K$ now $\angle KLE=\angle ZJC$ $\angle JCZ= x/2=\angle LEK$ and $KE=CZ$ so $\triangle CZJ\cong \triangle LEK$ and $LE=JC$ but $DC=DE$ so $DL=DJ$...
19.07.2012 21:00
Note that only $OA=OC$ assumption is necessary, $OC=OE$ is not.
19.06.2013 16:07
Firstly make some observations. $FA=FE, BA=BC, DC=DE$. Let $\angle FAB = \angle BCD = \angle DEF = 2x$. Diagonals of quadrilateral $KFBD$ are perpendicular so $KF^2 + BD^2 = KD^2 + FB^2$. Use cosine rule in $KEF, BCD, KED, FAB$ and the equal lengths ..... so $KE\cos x = BC \cos (180-2x)$. I.e. $LE = CJ$ i.e. $DJ = DL$, done.
26.04.2014 05:48
For those who were not innately born with the ability to do the geometry, here goes... So reflect $B$ across $DO$ to the point $B'$. Note that $DJ = DL \Longleftrightarrow B'K \perp DE$ (this is easy with simple angle chase, etc.). Or, another equivalent statement is: Let the perpendicular from $B'$ to $DE$ intersect $EO$ at $K'$. (1) Then we must show that $BK' \perp DF$. So we use complex numbers. Let the inscribed circle be the unit circle. Let the tangency points to $FA, AB, BC, CD, DE, EF$ be $X, U, W, V, Y, Z$. (Sorry about the sucky letter ordering, I used these letters in my work for some reason ) We can simplify our bash by a lot if we WLOG allow $DO$ to be the real axis. Then, we let the complex numbers at $X, U, W, V, Y, Z$ be $x, rx, w, wr, \frac{1}{wr}, \frac{1}{w}$. ($r$ is a (unit) rotation factor, since it is easy to show by simple geometry again that $XU = WV = YZ$.) It's easy to show that $d = \frac{2rw}{r^2w^2+1}$, $b = \frac{2xrw}{xr+w}$, $f = \frac{2x}{xw+1}$, $e = \frac{2}{rw+w}$. The nice thing is that $b' = \overline{b} = \frac{2}{xr+w}$ because $DO$ is the real axis. Now we just need to find $k'$. To find $K'$, we use that $K'B || OY$, so $\frac{k'-b'}{\overline{k'}-\overline{b'}} = \frac{1/rw}{rw} = \frac{1}{r^2w^2}$. Also, $E, O, K'$ are collinear, so $\frac{k'}{\overline{k'}} = \frac{e}{\bar{e}} = \frac{1}{rw^2}$. Solving this system gives $k' = \frac{2(rw-x)}{(xr+w)(rw-w)}$. Now, we just need to show that $\frac{b-k'}{f-d}$ is imaginary. But a direct computation gives that it is equal to $\frac{(r^2w^2+1)(xw+1)}{(xr+w)(rw-w)}$. Now a direct computation shows that the conjugate of the previous term is the negative of it, so we're done.
16.01.2015 00:50
No need for complex bashing Sketch: translate the lengths to lengths in terms of BO and OK, using the fact that the tangents from D to $\omega$ are equal in length. Now, the main problem will be the angle formed by BK. Notice that if the tangents from D and F touch $\omega$ at $X, Y, X_1, Y_1$ then if Z is intersection of the polars of D and F, OZ is parallel to BK, and it is perpendicular to the polar of Z, that passes through $W_1$ and $W_2$ (the other intersections of opposite sides of complete quad $XYX_1Y_1$. After this, using Sine Law many times is enough to finish. If you are curious abut details you can ask
18.01.2015 16:44
$A'B'C'DE'F'O'$: Rotation of $ABCDEFO$ about a point $D$ such that $D, C, E'$ are collinear. $A''B''C''D''E''FO''$: Rotation of $ABCDEFO$ about a point $F$ such that $A, F, E''$ are collinear. $K'$: Intersection of $BJ$ and perpendicular from $B'$ to $DF'$ $M$: Foot of the perpendicular from $B$ to $AF$. $K''$: Intersection of $BM$ and perpendicular from $B''$ to $D'F.$ Since $O'O'' \parallel AC$ and $E'E'' \parallel AC$, and since there is a point $X$ such that rotation of X about a point $D$ is $P'$(with $\angle EDC$), and that rotation of X about a point $F$ is $P''$(with $\angle EFA$), it can suffice to prove that $K'K'' \parallel AC$. Since $B'K' \perp DF'$ and $B''K'' \perp D'F$, it's done. Sorry to Roughly description.
20.09.2016 17:43
Solution with Danielle Wang: We use complex numbers with $\omega$ the unit circle. Denote the tangency points of $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, $\overline{DE}$, $\overline{EF}$, $\overline{FA}$ by $wx$, $y$, $wy$, $z$, $wz$, $x$ in that order, where $w,x,y,z$ are on the unit circle. Let $L'$ be the point on $\overline{DE}$ such that $DJ = DL' > DE$. Let $K'$ be the point on $\overline{OE}$ such that $\angle K'L'E = 90^{\circ}$. We will prove that $\overline{BK'} \perp \overline{DF}$, which implies the desired result. First, $b = \frac{2wxy}{wx+y}$, $d = \frac{2wyz}{wy+z}$, $e = \frac{2wz}{w+1}$. Now, \[ j = \frac{1}{2}\left( b+wy+wy-(wy)^2 \overline b \right) = wy + \frac{1}{2} b - \frac{1}{2} (wy)^2 \overline b \]and then by rotating about $d$ we see \begin{align*} \ell &= d + \frac{z}{wy}(d-j) = \frac{wy+z}{y} d - \frac{z}{wy} j \\ &= 2z - \left( z + \frac{1}{2} \frac{bz}{wy} - \frac{1}{2} \overline b wyz \right) \\ &= z \left( 1 + \frac{wy-x}{wx+y} \right). \end{align*}Now, we use similar triangles $\triangle KLE$ and $\triangle OzE$ (sic!) to obtain that $\frac ke = \frac{\ell - z}{e-z}$, so \begin{align*} k &=\frac{z(wy-x)}{wx+y} \cdot \frac{e}{e-z} \\ &= \frac{z(wy-x)}{wx+y} \cdot \frac{2w}{w-1}. \end{align*}From this we compute \begin{align*} k-b &= \frac{2w}{w-1} \frac{z(wy-x)}{wx+y} - \frac{2wxy}{wx+y} \\ &= \frac{2w}{wx+y} \left( \frac{z(wy-x)}{w-1} - xy \right) \\ &= \frac{2w}{wx+y} \cdot \frac{wyz - xz + (1-w)xy}{w-1}. \end{align*}But also, \begin{align*} d-f &= \frac{2wyz}{wy+z} - \frac{2wzx}{wz+x} \\ &= 2wz \cdot \frac{y(wz+x)-x(wy+z)}{(wy+z)(wz+x)} \\ &= \frac{2wz \left( wyz - xz + (1-w)xy \right)}{(wy+z)(wz+x)}. \end{align*}Taking the quotient is seen to give a pure imaginary number.
06.12.2016 00:36
Let circle $\omega$ touch $DE$ and $DC$ at $M$ and $N$ respectively and $L'$ be a point on ray $DE$ such that $DL'=DJ.$ Let the line perpendicular to $DE$ at $L'$ meet line $EO$ at $K'$. We will show that $BK' \perp DF$ from which the result shall follow. For obvious reasons, $BA=BC, DC=DE, FE=FA$ and $\angle OAF=\angle OAB=\angle OCB=\angle OCD$. Let $T$ be a point on the line $OB$ such that $$\angle FAT=\angle DCT=90^{\circ}.$$From $DM=DN, DE=DC,$ and $DL'=DJ$ we get $$\frac{OK'}{OE}=\frac{ML'}{ME}=\frac{NJ}{NC}=\frac{OB}{OT} \Longrightarrow ET \parallel BK'.$$Finally, $$FE^2-FT^2=FA^2-FT^2=TA^2=TC^2=DE^2-DT^2 \Longrightarrow ET \perp DF$$as desired. The proof is complete. $\square$
16.04.2017 21:42
Here's a length bash. Define $f(P)=DP^2-FP^2$ for a general point $P$ in the plane. Then note that $f(P)$ is linear. Let $O$ be the center of $(ACE)$. Also, write $\theta=\angle A=\angle C=\angle E>\pi/2$ and $b=BA=BC, d=DC=DE, f=FE=FA$ and $R$ the radius of the larger circle. First, note that $f(K)=f(B)$ since $KB\perp DF$. Then by the Law of Cosines, $BD^2=b^2+d^2-2bd\cos\theta$ and $FD^2=b^2+f^2-2bf\cos\theta$. Thus $f(K)=(d-f)(d+f-2b\cos\theta)$. Trivially, $f(E)=d^2-f^2=(d-f)(d+f)$. Now, write $t$ as the tangent length from $E$ to the inscribed circle; then $t=R\cos (\theta/2)$. If $ED$ is tangent to the smaller circle at $T$, then $OD^2-OE^2=DT^2-TE^2=(DT+TE)(DT-TE)=d\cdot (d-2TE)=d(d-2t)$. Similarly, $OF^2-OE^2=f(f-2t)$. Thus $f(O)=d(d-2t)-f(f-2t)=(d-f)(d+f-2t)$. Now, note that since $f$ is linear, $\frac{f(O)-f(E)}{f(E)-f(K)}=\frac{OE}{EK}=\frac{t}{-b\cos\theta}=\frac{R\cos(\theta/2)}{-b\cos\theta}$. Since $OE=R$, it follows $EK=\frac{-b\cos\theta}{\cos(\theta/2)}$. But since $\angle LEK=\theta/2$ we have $EL=-b\cos\theta=CJ$, and since $DE=DC$ we have $DJ=DL$ as desired. $\square$
08.12.2018 12:46
Another completely elementary solution. The concentric condition gives lengths of tangents from $A, C, E$ to the incircle are equal. This gives $DC=DE, FE=FA, AB=BC$ and $\angle BAF=\angle FED=\angle DCB$. Thus $\angle BAE = \angle DEA$ and $\angle BCE = \angle FEC$ so there exists points $P, Q$ on $DE, DF$ such that quadrilaterals $BPEA$ and $BQEC$ are isosceles trapezoid. This gives $EP=AB=BC=EQ$. Let $K'$ be the orthocenter of $\triangle EPQ$. We claim that $K=K'$. Obviously $EK'$ bisects $\angle FED$ thus $E, K', O$ are colinear. Now we proceed to show that $BK'\perp DF$. First, note that $EO\perp PQ$ thus $$\angle EPQ = 90^{\circ}-\angle(BP, OE) = 90^{\circ}-\angle AEO = \angle ACE$$. Similarly we get $\angle EQP = \angle AEC$ so $\triangle ACE\sim\triangle EPQ$. Now extend $PQ$ to meet $\odot(BPK')$ and $\odot(BQK')$ at $X, Y$. Since \begin{align*} \angle BK'X &= \angle BPQ = \angle ACE = \angle FOE \\ \angle XBK' &= \angle QPK' =\angle OEF \end{align*}, we get $\triangle BK'X\sim\triangle EOF$. Similarly $\triangle BK'Y\sim\triangle EOD$ thus $BK'XY\sim EOFD$ or $\angle(BK', XY) = \angle (EO, FD)$. But $XY\perp EO$ hence we get $BK'\perp DF$ as claimed. To finish, notice that $\triangle QEL\cong\triangle BCJ$ so $CJ=EL\implies DJ=DL$.
15.03.2019 23:10
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We will show that $KB\perp DF$ using complex numbers. Let $X_1,X_2,Y_1,Y_2,Z_1,Z_2$ be the tangency points as shown in the diagram. We'll set $(X_1X_2Y_1Y_2Z_1Z_2)$ to be the unit circle. We see that $\widehat{X_1X_2}=\widehat{Y_1Y_2}=\widehat{Z_1Z_2}$, so there exist $x,y,z,r$ on the unit circle so that $X_1=x$, $X_2=rx$, $Y_1=y$, $Y_2=ry$, $Z_1=z$, and $Z_2=rz$. We may now compute \[a=\frac{2rx^2}{x(r+1)}=\frac{2rx}{r+1}\text{ and }\bar{a}=\frac{2}{rx+x}=\frac{2}{(r+1)x}.\]We have similar formulas for $c,e$. We also see that \[b=\frac{2rxy}{rx+y}\text{ and }\bar{b}=\frac{2}{rx+y},\]with cyclic variations for $d,f$. Now, $J$ is the foot from $B$ to the tangent to the unit circle at $Y_2$, so \begin{align*} j &= \frac{1}{2}(b+2y_2-y_2^2\bar{b}) \\ &=\frac{1}{2}\left(\frac{2rxy}{rx+y}+2ry-\frac{r^2y^2\cdot 2}{rx+y}\right) \\ &= \frac{rxy}{rx+y}+\frac{ry(rx+y)}{rx+y}-\frac{r^2y^2}{rx+y} \\ &= \frac{yr}{rx+y}(x+rx+y-ry) \\ &= yr\left(1+\frac{x-ry}{rx+y}\right). \end{align*}To compute $\ell$, we'll rotate $J$ about $O$ by $z_1/y_2$, then reflect about $z_1$, so \begin{align*} \ell &= 2z_1-\frac{z_1}{y_2}j \\ &= z_1\left(2-\frac{j}{ry}\right) \\ &= z\left(1+\frac{ry-x}{rx+y}\right). \end{align*}We see that $KO/EO=LZ_1/EZ_1$, so \begin{align*} k/e &= \frac{\ell-z_1}{e-z_1} \\ &= \frac{z\cdot\frac{ry-x}{rx+y}}{\frac{2rz}{r+1}-z} \\ &=\frac{r+1}{r-1}\cdot\frac{ry-x}{rx+y}, \end{align*}so \[k=\frac{2rz}{r+1}\cdot\frac{r+1}{r-1}\cdot\frac{ry-x}{rx+y}=\frac{2rz}{r-1}\cdot\frac{ry-x}{rx+y}.\]We now compute \begin{align*} k-b &= \frac{2rz}{r-1}\cdot\frac{ry-x}{rx+y}-\frac{2rxy}{rx+y} \\ &= \frac{2r}{rx+y}\left(\frac{z(yr-x)}{r-1}-xy\right) \\ &= \frac{\frac{2r}{r-1}}{rx+y}(rxy-xz-rxy-xy), \end{align*}and \begin{align*} d-f &= \frac{2ryz}{ry+z}-\frac{2rzx}{rz+x} \\ &= 2rz\left(\frac{y}{ry+z}-\frac{x}{rz+x}\right) \\ &= \frac{2rz}{(ry+z)(rz+x)}(rzy+xy-rxy-xz). \end{align*}Thus, \[\alpha:=\frac{k-b}{d-f}=\frac{\frac{2r}{r-1}}{rx+y}\cdot\frac{(ry+z)(rz+x)}{2rz}=\frac{1}{z(r-1)}\cdot\frac{(ry+z)(rz+x)}{rx+y}.\]We wish to show that $\alpha\in i\mathbb{R}$. We see that \[\alpha/\bar{\alpha}=\frac{1}{z^2(-r)}\cdot\frac{ryz\cdot rzx}{rxy}=-1,\]so $\alpha\in i\mathbb{R}$, so $KB\perp DF$, as desired.
24.03.2020 21:36
mOViNg PoiNTS sol: Fix $\omega$, $C$, and $E$. We phantom point to reduce to: Vary $A$ on the circle concentric with $\omega$ through $C$ and $E$, define $B$ and $F$ to fit. Let the line through $B$ perpendicular to $CD$ meet $CO$ at $T$, and reflect $T$ over $DO$ to $T'$. Prove that rotating $\infty_{BT'}$ by 90 degrees maps it to $\infty_{DF}$. We proceed with moving points. Move $A$ projectively on the circle centered at $O$ through $C,E$. Tether $B$ to line $CB$, $F$ to line $EF$ (which are fixed since the circle is fixed), $T$ to $CO$, $T'$ to $EO$, and all 3 points at infinity to the line at infinity. $A \rightarrow B$ is a projective map because $A \rightarrow \frac{A+C}{2} \rightarrow B$ is a perspectivity centered at $O$. Similarly, all the points move projetively. We get that $B,F,T,T'$ have degree $1$ each, so $\infty_{DF}$ has degree $1$, $\infty_{BT'}$ has degree $2$, its rotation by 90 degrees has degree $2$ as well. Now, we just need to check $1+2+1=4$ cases. $A=C$, $CDA$ collinear, $EA=EC$, and $A=E$ are easy.
09.12.2020 13:53
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Notice that $Pow(C,\omega)-Pow(E,\omega)=CO^2-EO^2=0$, therefore, $CD=DE$, and similarly $EF=FA,CB=BA$. Therefore, $B_1E=BA$, hence $OF$ is the axis of symmetry of quadrilateral $B_1BAE$, this implies $BB_1\|AE$, similarly $BB_2\|CE$. CLAIM. The perpendicular $l_1$ from $B_1$ to $EF$, the perpendicular from $B$ to $DF$ and the perpendicular $l_2$ from $B_2$ to $DE$ are concurrent. Proof. By Carnot's theorem it suffices to show that the perpendicular from $F$ to $B_1B$, the perpendicular from $D$ to $BB_2$ and the perpendicular from $E$ to $DF$ are concurrent. However, since $BB_1\|AE$, and $BB_2\|CE$ they all pass through $O$. $\blacksquare$. Now nnotice that $OE$ bisects $\angle DEF$, hence $l_1$ and $l_2$ are symmetric with respect to $OE$, hence $l_1\cap l_2\cap OE$, this implies $l_1\cap l_2\cap OE\cap BK=K$ as desired. Now we have $\angle DEO=\angle OCE=\angle OCA=\angle OEF$, hence $$EL=EB_2\cos(180^{\circ}-\angle FED)=BC\cos(180^{\circ}-\angle BCD)=CJ$$Combining with $CD=DE$ we have $DJ=DL$ as desired.
21.01.2021 16:49
WakeUp wrote: Let $ABCDEF$ be a convex hexagon all of whose sides are tangent to a circle $\omega$ with centre $O$. Suppose that the circumcircle of triangle $ACE$ is concentric with $\omega$. Let $J$ be the foot of the perpendicular from $B$ to $CD$. Suppose that the perpendicular from $B$ to $DF$ intersects the line $EO$ at a point $K$. Let $L$ be the foot of the perpendicular from $K$ to $DE$. Prove that $DJ=DL$. Proposed by Japan $80 \%$ of the difficulty of this problem lies on the diagram construction and the weird statement... Fortunately, we could bash this out. Claim 01. $\angle FAB = \angle FED = \angle BCD = 2\theta$. Proof. Let $\omega$ tangent to $AB$ and $AF$ at $A'$ and $F'$ respectively. Define $B', C', D', E'$ similarly (in the cyclic order). Now, notice that $OF'AA'$, $OB'CC'$, $OD'EE'$ are all congruent since $OA = OC = OE$ and $OA' = OB' = OC' = OD' = OE' = OF'$ and the $\angle OF'A = \angle OA'A = \angle OB'C = \angle OC'C = \angle OD'E = \angle OE'E = 90^{\circ}$. Therefore, these are all congruent, forcing $\angle FAB = \angle FED = \angle BCD$. Claim 02. $AB = BC$, $CD = DE$ and $EF = FA$ Proof. This follows from the previous lemma. The congruent condition forces $FA = FF' + F'A = FE' + EE' = FE$, and others follow similarly. Claim 03. $EL = CJ$. Proof. We are ready to bash. Notice that $OE$ bisects $\angle DEF$. Therefore, since $KB \perp FD$, we have \begin{align*} BF^2 - BD^2 &= KF^2 - KD^2 \\ (BA^2 + AF^2 - 2 \cdot BA \cdot AF \cdot \cos 2 \theta) - (BC^2 + CD^2 - 2 \cdot BC \cdot CD \cdot \cos 2 \theta) &= (KE^2 + EF^2 - 2 \cdot KE \cdot EF \cdot \cos (180^{\circ} - \theta) ) - (KE^2 + DE^2 - 2 \cdot KE \cdot DE \cdot \cos (180^{\circ} - \theta) ) \\ 2 \cdot AB \cdot (DE - EF) \cdot \cos 2\theta &= 2 \cdot KE \cdot (DE - EF) \cdot \cos (180^{\circ} - \theta) \\ AB \cdot \cos 2 \theta &= KE \cdot \cos (180^{\circ} - \theta) \\ BC \cdot \cos \angle BCD &= KE \cdot \cos \angle KEL \\ CJ &= EL \end{align*}(Note: throughout the bash, we only care about the magnitude of the value and not the sign.) Thus, we conclude that \[ DJ = DE + EJ = CD + CL = DL \]done.
11.05.2023 22:56
30.05.2023 21:01
anantmudgal09 wrote: Let circle $\omega$ touch $DE$ and $DC$ at $M$ and $N$ respectively and $L'$ be a point on ray $DE$ such that $DL'=DJ.$ Let the line perpendicular to $DE$ at $L'$ meet line $EO$ at $K'$. We will show that $BK' \perp DF$ from which the result shall follow. For obvious reasons, $BA=BC, DC=DE, FE=FA$ and $\angle OAF=\angle OAB=\angle OCB=\angle OCD$. Let $T$ be a point on the line $OB$ such that $$\angle FAT=\angle DCT=90^{\circ}.$$From $DM=DN, DE=DC,$ and $DL'=DJ$ we get $$\frac{OK'}{OE}=\frac{ML'}{ME}=\frac{NJ}{NC}=\frac{OB}{OT} \Longrightarrow ET \parallel BK'.$$Finally, $$FE^2-FT^2=FA^2-FT^2=TA^2=TC^2=DE^2-DT^2 \Longrightarrow ET \perp DF$$as desired. The proof is complete. $\square$ Good solution. Although we don’t need the phantom points.
01.08.2024 13:51
First notice $AB=BC,CD=DE,EF=FA$ by symmetry. Now let $X,Y$ be the reflections of $B$ over $OD,OF.$ Then $X,Y$ lie on $EF,ED$ respectively by symmetry. Additionally let $P,Q,R,S$ be the tangency points of $\omega$ with $EF,ED,AF,CD$ respectively, and let $AF$ meet $CD$ at $T.$ Let $PR$ meet $QS$ at $N,$ let the perpendiculars at $A$ and $C$ to $AF$ and $CD$ meet at $H,$ and finally, redefine $K$ to be the orthocenter of $XEY.$ First, notice $K$ lies on $OE,$ since $OX=OB=OY$ and $EX=CB=AB=EY$ so $OE$ is the perpendicular bisector of $XY$ and $KE$ is perpendicular to $XY$ by definition. Next we have $AE\perp OF\perp PR,$ so $AE\parallel RN$ and similarly $CE\parallel SN.$ Then $AC$ and $RS$ are both perpendicular to the bisector of $\angle ATC,$ so $\triangle ACE$ and $\triangle RSN$ are homothetic with center $T.$ Additionally this homothety takes $H$ to $O,$ so $HE\parallel ON.$ Additionally $N$ is the polar of $DF$ with respect to $\omega$ so $HE\perp DF.$ Now we claim that $BK\parallel HE$ holds. To do this, rotate $E$ around $O$ at fixed angular velocity. Then $X$ and $Y$ rotate at the same angular velocity: this is because we have $ABYE$ is an isosceles trapezoid so $\triangle ABO\cong\triangle EYO\cong\triangle EXO$ so $O$ is the spiral center of $EX$ and $AB$ and similarly for $Y.$ Thus this implies $K$ rotates at the same angular velocity as well. In particular $OK/OE$ is fixed, so $BK$ being parallel to $HE$ is fixed as well. Now it suffices to show $BK\parallel HE$ for some $E$ as it rotates. Now the key is to choose $E$ such that point $F$ goes to infinity along line $AT.$ Then the perpendicular bisector of $AE,$ which is $OF,$ becomes parallel to $AT,$ so $AE\perp AT$ and $A,H,E$ collinear. Next we have $YK\perp EX\parallel AT\perp AE$ and $BY\perp OF\perp AE,$ so $B,Y,K$ are collinear along a line parallel to $HE.$ In particular $BK\parallel HE$ as desired. Thus $BK\parallel HE$ always, and $BK\perp DF.$ Since $K$ also lies on $OE$ we see $K$ is the same point defined in the problem. Thus $J$ is also the foot from $X$ to $DE.$ Now finally $BL$ and $XJ$ are reflections over $OD,$ so $DJ=DL$ which was what we wanted.