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Source: IMO Shortlist 2011, G7

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1347 posts

WakeUp

#1 h Jul 13, 2012, 11:52 AM • 3 Y

Y by Adventure10 and 2 other users

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

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27 posts

#2 h Jul 13, 2012, 2:43 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

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??

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222 posts

#3 h Jul 13, 2012, 2:58 PM • 6 Y

Y by CyclicISLscelesTrapezoid, Adventure10, Mango247, and 3 other users

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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This post has been edited 7 times. Last edited by paul1703, Jul 15, 2012, 9:13 AM

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1347 posts

WakeUp

#4 h Jul 13, 2012, 11:10 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

nquocthuy wrote:

DJ = DL??

sorry, edited.

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339 posts

leader

#5 h Jul 15, 2012, 11:01 AM • 5 Y

Y by karitoshi, Adventure10, Mango247, and 2 other users

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 $DL=DJ$ ...

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180 posts

#6 h Jul 19, 2012, 6:00 PM • 3 Y

Y by Adventure10, uwu_, and 1 other user

Note that only $OA=OC$ assumption is necessary, $OC=OE$ is not.

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109 posts

#7 h Jun 19, 2013, 1:07 PM • 4 Y

Y by karitoshi, Adventure10, Mango247, and 1 other user

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.

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606 posts

#8 h Apr 26, 2014, 2:48 AM • 3 Y

Y by Adventure10 and 2 other users

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.

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366 posts

#9 h Jan 15, 2015, 10:50 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

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

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#10 h Jan 18, 2015, 2:44 PM • 2 Y

Y by Adventure10 and 1 other user

$A'B'C'DE'F'O'$ : 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.

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6870 posts

#11 h Sep 20, 2016, 2:43 PM • 3 Y

Y by v4913, Adventure10, Mango247

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.

This post has been edited 2 times. Last edited by v_Enhance, Sep 20, 2016, 2:44 PM
Reason: I done goof

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1979 posts

#12 h Dec 5, 2016, 9:36 PM • 4 Y

Y by BobaFett101, Systematicworker, PNT, Adventure10

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. $\square$

This post has been edited 2 times. Last edited by anantmudgal09, Jan 8, 2018, 9:07 PM

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2130 posts

DrMath

#14 h Apr 16, 2017, 6:42 PM • 2 Y

Y by Adventure10, Mango247

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. $\square$

This post has been edited 2 times. Last edited by DrMath, Apr 16, 2017, 6:49 PM

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1594 posts

#15 h Dec 8, 2018, 9:46 AM • 3 Y

Y by Adventure10, Mango247, Om245

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 $CJ=EL\implies DJ=DL$ .

This post has been edited 1 time. Last edited by MarkBcc168, Dec 8, 2018, 9:47 AM

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1614 posts

yayups

#16 h Mar 15, 2019, 8:10 PM • 3 Y

Y by Systematicworker, Adventure10, Mango247

$[asy] unitsize(0.3inches); /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.298226713745567, xmax = 7.495330967836562, ymin = -8.945386667743351, ymax = 8.0563696926648; /* image dimensions */ /* draw figures */ draw(circle((-1.7696032982363763,1.4316711242853086), 4.821139936586497), linewidth(2)); draw(circle((-1.7696032982363763,1.4316711242853086), 4.008253781330843), linewidth(2)); draw((-1.8054453437151061,6.252677827861352)--(-5.7078314615401435,3.602208064796849), linewidth(2)); draw((-5.7078314615401435,3.602208064796849)--(-5.8647600837464084,-1.1125539424570945), linewidth(2)); draw((-5.8647600837464084,-1.1125539424570945)--(-1.8887596833218707,-2.9153283955934555), linewidth(2)); draw((-1.8887596833218707,-2.9153283955934555)--(2.180028431104426,-1.33307108625325), linewidth(2)); draw((2.180028431104426,-1.33307108625325)--(2.285470450293811,3.562164799030479), linewidth(2)); draw((2.285470450293811,3.562164799030479)--(-1.8054453437151061,6.252677827861352), linewidth(2)); draw((-1.7696032982363763,1.4316711242853086)--(2.180028431104426,-1.33307108625325), linewidth(2)); draw((2.180028431104426,-1.33307108625325)--(4.839653908335962,-3.1948089203153236), linewidth(2)); draw((-1.8887596833218707,-2.9153283955934555)--(2.285470450293811,3.562164799030479), linewidth(2)); draw((4.839653908335962,-3.1948089203153236)--(-5.7078314615401435,3.602208064796849), linewidth(2)); draw((-5.8647600837464084,-1.1125539424570945)--(-7.5077897086321,-0.36758122234019663), linewidth(2)); draw((-7.5077897086321,-0.36758122234019663)--(-5.7078314615401435,3.602208064796849), linewidth(2)); draw((3.8614013526643376,-0.6792241618011571)--(2.180028431104426,-1.33307108625325), linewidth(2)); draw((4.839653908335962,-3.1948089203153236)--(3.8614013526643376,-0.6792241618011571), linewidth(2)); /* dots and labels */ dot((-1.7696032982363763,1.4316711242853086),dotstyle); label("$O$", (-1.7113396812036203,1.5944475335428832), NE * labelscalefactor); dot((-1.8054453437151061,6.252677827861352),dotstyle); label("$A$", (-1.742708235374115,6.409520598713827), NE * labelscalefactor); dot((-5.8647600837464084,-1.1125539424570945),dotstyle); label("$C$", (-6.479359915118809,-1.3541965584836226), NE * labelscalefactor); dot((2.180028431104426,-1.33307108625325),dotstyle); label("$E$", (2.3979409151311804,-1.495355052250849), NE * labelscalefactor); dot((-5.7078314615401435,3.602208064796849),linewidth(4pt) + dotstyle); label("$B$", (-5.6480932296007005,3.7275092171365256), W * 2*labelscalefactor); dot((-1.8887596833218707,-2.9153283955934555),linewidth(4pt) + dotstyle); label("$D$", (-1.8211296208003516,-2.7971500503263806), S * 4*labelscalefactor); dot((2.285470450293811,3.562164799030479),linewidth(4pt) + dotstyle); label("$F$", (2.350888083875438,3.6804563858807837), SW * labelscalefactor*5); dot((-7.5077897086321,-0.36758122234019663),linewidth(4pt) + dotstyle); label("$J$", (-7.702733527768101,-0.17787577709006977), NE * labelscalefactor); dot((4.839653908335962,-3.1948089203153236),linewidth(4pt) + dotstyle); label("$K$", (4.907425248770753,-3.0637827607755863), NE * labelscalefactor); dot((3.8614013526643376,-0.6792241618011571),linewidth(4pt) + dotstyle); label("$L$", (3.9193157924001714,-0.5542984271360066), NE * labelscalefactor); dot((-4.021652394997616,4.747447540349859),linewidth(4pt) + dotstyle); label("$X_2$", (-3.9541913043939885,4.872461444359584), N * labelscalefactor); dot((-5.775638634947014,1.565010003006739),linewidth(4pt) + dotstyle); label("$Y_1$", (-5.710830337941689,1.6885531960543674), NE * labelscalefactor); dot((-3.424806791732395,-2.2188630517869523),linewidth(4pt) + dotstyle); label("$Y_2$", (-3.35818877515459,-2.091357581490249), NE * labelscalefactor); dot((-0.3168659236451125,-2.304055377483616),linewidth(4pt) + dotstyle); label("$Z_1$", (-0.25270191227561856,-2.1854632440017334), NE * labelscalefactor); dot((2.2377209725810996,1.345354474268119),linewidth(4pt) + dotstyle); label("$Z_2$", (2.303835252619696,1.4689733168609043), NE * labelscalefactor); dot((0.43289693138426233,4.780565137847703),linewidth(4pt) + dotstyle); label("$X_1$", (0.5001433878162532,4.903829998530079), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Redefine $K$ to be on $OE$ such that the foot from $K$ to $DE$ (we'll call it $L$ because I'm lazy) satisfies $DL=DJ$ . 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.

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mira74

#17 h Mar 24, 2020, 6:36 PM

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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.

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#18 h Dec 9, 2020, 10:53 AM

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Let $B_1$ , $B_2$ be points on the ray $ED,EF$ respectively such that $EB_1=EB_2=BC$ .
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.

This post has been edited 1 time. Last edited by mathaddiction, Dec 9, 2020, 11:12 AM

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#19 h Jan 21, 2021, 1:49 PM • 1 Y

Y by KaiDaMagical336

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.

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1385 posts

#20 h May 11, 2023, 7:56 PM

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length bash

This post has been edited 2 times. Last edited by VicKmath7, May 11, 2023, 8:04 PM

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PNT

#21 h May 30, 2023, 6:01 PM

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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.

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OronSH

#22 h Aug 1, 2024, 10:51 AM

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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.

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#23 h Mar 18, 2025, 4:05 PM

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The incircle condition gives us $AB = BC$ and variants, as well as $\angle BCO = \angle DCO$ and variants. Let point $P$ be such that $\angle PAF = \angle PCD = 90^{\circ}$ . By symmetry, $P$ lies on the perpendicular bisector of $\overline{AC}$ .

Claim: We have $\overline{PE} \perp \overline{DF}$ .
Proof: We use "synthetic" methods. Click to reveal hidden text

Since $P$ lies on the perpendicular bisector of $\overline{AC}$ , there exists a homothety at $O$ that sends $P$ to $B$ . This homothety sends $E$ to $K$ and $C$ to the intersection $K'$ of lines $BJ$ and $OC$ . So, $\overline{KK'} \parallel \overline{EC}$ . Then, since $J$ is the foot from $K'$ to $\overline{DC}$ and $L$ is the foot from $K$ to $\overline{DE}$ , $DJ = DL$ follows from symmetry.

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