Zack likes Moving Points

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1207 posts

#1 h Jun 25, 2019, 5:35 PM • 9 Y

Y by anantmudgal09, AFSA, megarnie, qwertyboyfromalotoftime, tiendung2006, Adventure10, Rounak_iitr, Jack_w, farhad.fritl

Let $ABC$ be an acute triangle with orthocenter $H$ and circumcircle $\Gamma$ . A line through $H$ intersects segments $AB$ and $AC$ at $E$ and $F$ , respectively. Let $K$ be the circumcenter of $\triangle AEF$ , and suppose line $AK$ intersects $\Gamma$ again at a point $D$ . Prove that line $HK$ and the line through $D$ perpendicular to $\overline{BC}$ meet on $\Gamma$ .

Gunmay Handa

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

pinetree1

1207 posts

pinetree1

#2 h Jun 25, 2019, 5:47 PM • 22 Y

Y by yayups, Pluto1708, Systematicworker, Wizard_32, lQluVyR, mueller.25, Andrei246, Kanep, srijonrick, Pluto04, AllanTian, aritrads, gambi, Flying-Man, megarnie, tiendung2006, ike.chen, TemetNosce, Kingsbane2139, Adventure10, Mango247, Rounak_iitr

Let $T$ be the point on $\Gamma$ such that $\overline{DT}\perp\overline{BC}$ , and let $Q = \overline{TH}\cap \Gamma$ .
$[asy] defaultpen(fontsize(10pt)); size(250); pair A, B, C, H, E, F, K, D, Q, T, X; A = dir(117); B = dir(210); C = dir(330); H = orthocenter(A, B, C); X = dir(20); E = extension(H, X, A, B); F = extension(H, X, A, C); K = circumcenter(A, E, F); D = IP(Line(A, K, 20), circumcircle(A, B, C), 1); T = IP(Line(D, foot(D, B, C), 20), circumcircle(A, B, C), 1); Q = IP(Line(T, H, 20), circumcircle(A, B, C), 1); draw(A--B--C--cycle, orange); draw(circumcircle(A, B, C), red); draw(E--F, red); draw(circumcircle(A, E, F), heavygreen); draw(A--D, orange); draw(circumcircle(B, E, H), purple); draw(circumcircle(C, F, H), purple); draw(Q--T, magenta); draw(E--K--F, magenta+linewidth(0.9)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$E$", E, dir(180)); dot("$F$", F, dir(35)); dot("$D$", D, dir(D)); dot("$T$", T, dir(T)); dot("$H$", H, dir(290)); dot("$Q$", Q, dir(210)); dot("$K$", K, dir(30)); [/asy]$

Claim: $BEHQ$ and $CFHQ$ cyclic.

Proof. Pure angles:
$\angle BQH = \angle BDT = 90^\circ - \angle DBC = 90^\circ - \angle KAF = \angle AEF.$ A similar angle chase proves $CFHQ$ cyclic. $\blacksquare$

Now $\overline{QHT}$ is the radical axis of $(BEHQ)$ and $(CFHQ)$ . To finish, observe that $\angle KEH = 90^\circ - \angle A = \angle EBH$ , which implies $\overline{KE}$ tangent to $(BEHQ)$ ; similarly, $\overline{KF}$ is tangent to $(CFHQ)$ . Thus, since $KE = KF$ , $K$ lies on the radical axis, as desired.

This post has been edited 1 time. Last edited by pinetree1, Jul 27, 2019, 3:58 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

MarkBcc168

1594 posts

MarkBcc168

#3 h Jun 25, 2019, 6:12 PM • 2 Y

Y by yayups, Adventure10

Let $X$ be the point of $\odot(ABC)$ such that $XD\perp BC$ . First, by angle chasing
$\angle XBC = 90^{\circ} - \angle BXD = 90^{\circ} - \angle EAK = \angle AFE$ so $\triangle XBC\cup O\sim\triangle AFE\cup K$ . Moreover, if $A'$ be the antipode of $A$ w.r.t. $\odot(ABC)$ and $XA'$ meet $BC$ at $P$ . Then
$\angle BXP = 90^{\circ}-\angle C = \angle FAH$ so $\triangle XBC\cup O\cup P\sim\triangle AFE\cup K\cup H$ which means $\angle AHK = \angle OPX$ . Hence we will be done if we prove that $\angle AHX = \angle OPX$ .

Let $M,N$ be midpoints of $BC, XA'$ . Notice that the homothety $\mathcal{H}(A',2)$ takes $\triangle OMN$ to $\triangle AHX$ . Moreover, $O, P, M, N$ are concyclic. Thus $\angle AHX = \angle OMN = \angle OPN$ so we are done.

This post has been edited 2 times. Last edited by MarkBcc168, Apr 19, 2020, 1:19 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

yayups

1614 posts

yayups

#4 h Jun 25, 2019, 6:12 PM • 14 Y

Y by AlastorMoody, tapir1729, Jorvis, amar_04, Aryan-23, Imayormaynotknowcalculus, Kanep, tenebrine, lgkarras, Bongcloud, EpicBird08, Adventure10, Lionking212, aidan0626

I also like moving points

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

jj_ca888

2726 posts

jj_ca888

#5 h Jun 25, 2019, 6:38 PM • 4 Y

Y by amar_04, megarnie, Adventure10, Mango247

wait I like this problem

why do so many of the TSTST problems this year have orthocenters in their geo

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

zacchro

179 posts

zacchro

#6 h Jun 25, 2019, 6:42 PM • 24 Y

Y by AlastorMoody, MarkBcc168, yayups, winnertakeover, WAit_Mng, amar_04, AFSA, Kanep, Gaussian_cyber, tigerzhang, megarnie, tenebrine, David-Vieta, CyclicISLscelesTrapezoid, kamatadu, EpicBird08, nguyenducmanh2705, Adventure10, Mango247, Lionking212, Rounak_iitr, Aelin, aidan0626, Jack_w

Hmm I guess since i've been called out...
Solution with Moving Points

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

TheUltimate123

1740 posts

TheUltimate123

#7 h Jun 25, 2019, 8:37 PM • 5 Y

Y by AlastorMoody, karitoshi, Adventure10, Rounak_iitr, everythingpi3141592

$\measuredangle$ denotes a directed angle modulo $180^\circ$ . Denote by $H_A$ , $H_B$ , $H_C$ the reflections of $H$ over $\overline{BC}$ , $\overline{CA}$ , $\overline{AB}$ , respectively.
$[asy] size(10cm); defaultpen(fontsize(10pt)); pen pri=red; pen sec=orange; pen tri=fuchsia; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair O,A,B,C,H,K,EE,F,D,T,SS,HA,HB,HC; O=(0,0); A=dir(110); B=dir(210); C=dir(330); H=A+B+C; EE=(2A+5B)/7; F=extension(EE,H,A,C); K=circumcenter(A,EE,F); D=2*foot(O,A,K)-A; T=intersectionpoint(H -- (H+(K-H)*100),circle(O,1)); SS=2*foot(O,H,K)-T; HA=2*foot(O,A,H)-A; HB=2*foot(O,B,H)-B; HC=2*foot(O,C,H)-C; filldraw(circle(O,1),fil,pri); filldraw(circumcircle(A,EE,F),sfil,sec); filldraw(circumcircle(D,F,K),sfil,sec); filldraw(circumcircle(B,EE,H),tfil,tri); filldraw(circumcircle(C,F,H),tfil,tri); draw(A--B--C--A,pri); draw(A--D,sec); draw(A--HA,sec); draw(B--HB,sec); draw(C--HC,sec); draw(EE--F,pri); draw(HB--D--HC,tri); draw(SS--T,tri); dot("$A$",A,N); dot("$B$",B,dir(195)); dot("$C$",C,dir(-15)); dot("$H$",H,dir(120)); dot("$K$",K,dir(-40)); dot("$E$",EE,dir(180)); dot("$F$",F,dir(30)); dot("$D$",D,S); dot("$T$",T,T); dot("$S$",SS,SW); dot("$H_A$",HA,NE); dot("$H_B$",HB,HB); dot("$H_C$",HC,HC); [/asy]$

Claim 1. $D$ is the anti-Steiner point of $\overline{EHF}$ , and $DEKF$ is cyclic.

Proof. Let $D'=\overline{EH_C}\cap\overline{FH_B}$ be the anti-Steiner point of $\overline{EHF}$ . First note that
$\begin{align*} \measuredangle KED'&=\measuredangle KEA+\measuredangle AED'=90^\circ-\measuredangle AFE+\measuredangle AEH_C\\ &=90^\circ+\measuredangle EFA+\measuredangle HEA=90^\circ+\measuredangle EFA+\measuredangle FEA, \end{align*}$ which is symmetric, whence $\measuredangle KED'=90^\circ+\measuredangle EFA+\measuredangle FEA=\measuredangle KFD'$ , and $D'EKF$ is cyclic. Now, $\measuredangle AKE=2\measuredangle AFE=2\measuredangle AFH=\measuredangle H_BFH=\measuredangle D'FE=\measuredangle D'KE,$ so $D'\in\overline{AK}$ , as required. $\blacksquare$

Claim 2. Let $S$ be the Miquel point of $EFH_BH_C$ ; $S$ lies on line $HK$ .

Proof. Check that $\frac{SE}{SF}=\frac{EH_C}{FH_B}=\frac{EH}{FH},$ whence $\overline{SH}$ bisects $\angle ESF$ and $\overline{SH}$ passes through $K$ , the midpoint of arc $EF$ on $(DEKF)$ . $\blacksquare$

Claim 3. Quadrilaterals $BEHS$ and $CFHS$ are cyclic.

Proof. Notice that $\measuredangle ESH=\measuredangle ESK=\measuredangle EFK=90^\circ-\measuredangle FAE=\measuredangle ABH=\measuredangle EBH,$ so we are done by symmetry. $\blacksquare$

Finally, let line $HK$ meet $\Gamma$ at $T\ne S$ . Note that $\measuredangle ATH=\measuredangle ATS=\measuredangle ABS=\measuredangle EBS=\measuredangle EHS=\measuredangle FHT,$ so $\overline{AT}\parallel\overline{EF}$ . However, $\overline{DH_A}$ is the reflection of $\overline{EF}$ over $\overline{BC}$ , so $\measuredangle TAH_A=\measuredangle EHA=\measuredangle AH_AD$ , and $AD=TH_A$ . Thus, $ATDH_A$ is an isosceles trapezoid, and $\overline{AH_A}\parallel\overline{DT}$ . This completes the proof. $\square$

This post has been edited 1 time. Last edited by TheUltimate123, Jun 25, 2019, 8:45 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

greenturtle3141

3543 posts

greenturtle3141

#8 h Jun 25, 2019, 9:13 PM • 2 Y

Y by Adventure10, Mango247

What is this moving points technique and where can I read more about it?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

anantmudgal09

1979 posts

anantmudgal09

#9 h Jun 25, 2019, 9:30 PM • 3 Y

Y by Wizard_32, Adventure10, Mango247

pinetree1 wrote:

Let $ABC$ be an acute triangle with orthocenter $H$ and circumcircle $\Gamma$ . A line through $H$ intersects segments $AB$ and $AC$ at $E$ and $F$ , respectively. Let $K$ be the circumcenter of $\triangle AEF$ , and suppose line $AK$ intersects $\Gamma$ again at a point $D$ . Prove that line $HK$ and the line through $D$ perpendicular to $\overline{BC}$ meet on $\Gamma$ .

Gunmay Handa

We first prove the following lemma.

Lemma. In triangle $AEF$ , with circumcenter $K$ point $H$ lies on $\overline{EF}$ , points $B$ and $C$ lie on lines $\overline{AE}$ and $\overline{AF}$ respectively, such that $\overline{BH} \perp \overline{AF}$ and $\overline{CH} \perp \overline{AE}$ . Line $\overline{AK}$ meets $\odot(KEF)$ again at point $D$ . Then $ABCD$ is cyclic and reflection of $D$ in $\overline{BC}$ lies on $\overline{EF}$ .

(Proof) Move $H$ along $\overline{EF}$ and note that $B \mapsto H$ and $H \mapsto C$ are linear maps, SO $B \mapsto C$ is also linear. Suppose $\odot(DAB)$ meets line $\overline{AF}$ at $C'$ . Then we need to show that $C=C'$ . Since, by spiral similarity, $B \mapsto C'$ is linear; we need to check this for two choices of $H$ .

$H=E$ . Then $B=E$ and we need to show that if $\odot(AED)$ meets $\overline{AF}$ at $F'$ , then $\angle AEF'=90^{\circ}$ .

Apply inversion at $A$ of radius $\sqrt{AE \cdot AF}$ followed by reflection in the bisector of angle $EAF$ . Suppose $X \mapsto X^{*}$ under this transformation. Then $E^{*}=F, F^{*}=E$ and $D^{*}$ is the orthocenter of $\triangle AEF$ , so $(F')^{*}=\overline{FD^{*}} \cap \overline{AE}$ hence $\angle A(F')^{*}F=90^{\circ}$ so $\angle AEF'=90^{\circ}$ , and we're done.
$H=F$ . Same proof as above works.

Finally, moving $H$ , since $\triangle DBC$ has fixed shape, so the locus of the reflection of $D$ in $\overline{BC}$ is a line.

For $H=E$ , we need to show that $\angle AED=180^{\circ}-\angle AEF$ since $\angle FEF'=90^{\circ}-\angle AEF$ ; this follows since $\angle AED=\angle AD^{*}F$ and $D^{*}$ is the orthocenter of $\triangle AEF$ .
Similarly, $H=F$ case holds.

The lemma is proved. $\square$

Now we go back to the original problem. Let $L$ be the reflection of $A$ in $K$ and $N=\overline{EF} \cap \overline{AK}$ , then, by our lemma, we have $(AL; ND)=-1$ .

Suppose $P$ lies on $\overline{EF}$ such that $\overline{DP} \perp \overline{BC}$ and $\overline{DP}$ meets $\overline{BC}$ at $S$ and $\Gamma$ again at $Q$ . Reflect $Q$ in $S$ to get $R$ . By the lemma, $S$ is the midpoint of $\overline{DP}$ . Let $S'=\overline{HL} \cap \overline{DP}$ and $Q'=\overline{HK} \cap \overline{DP}$ .

Observe that $-1=(AL; ND) \overset{H}{=} (\infty s', PD)$ , so clearly, $\overline{HL}$ bisects $\overline{DP}$ , so $H, L, S$ are collinear. Finally, since $\overline{AK} \parallel \overline{RH}$ so $-1=(AL; K \infty) \overset{H}{=} (\infty S; Q'R)$ so $\overline{HK}$ passes through $Q$ , as desired.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

rocketscience

466 posts

rocketscience

#10 h Jun 25, 2019, 10:40 PM • 5 Y

Y by AlastorMoody, flokraach, Pluto04, tiendung2006, Adventure10

Define $H_B, H_C$ as the reflections of $H$ over $AC, AB$ , respectively and let $D_A$ be the reflection of $D$ over $BC$ .

Claim: $H_BE$ and $H_CF$ concur at $D$ .
Proof: By Pascal, they concur at a point $D' \in \Gamma$ . Define $A' = AD' \cap \odot(AEF)$ . Observe that
$\begin{align*} \angle AFA' &= \angle EFA' + \angle AFE \\ &=\angle EAA' + (90^{\circ} - \angle FHH_C) \\ &=90^{\circ} + \angle D'AC - \angle D'H_CC \\ &=90^{\circ}, \end{align*}$ so $AA'$ is a diameter and $D=D'$ , as desired. $\Box$

Claim: $D_A \in EF$
Proof: Let $D_B$ be the reflection of $D$ over $AC$ . Then $\angle CED_B = \angle DEC = \angle AEH_B = \angle HEA$ , so $D_B \in EF$ . But it is well-known that the Simson line of $D$ bisects $\overline{DH}$ , i.e. $H, D_A, D_B$ collinear. Hence the claim follows. $\Box$

Define $Y = BC \cap DD_A$ and $X = DD' \cap \Gamma$ , and then $X'$ as the reflection of $X$ over $\overline{BC}$ . Note that $AXD_AH$ is a parallelogram, so $AH = XD_A = DX'$ and $AHX'D$ is also a parallelogram. Now we have
$-1 = (A, A'; D, EF \cap AA') \stackrel{H}{=} (\infty, HA' \cap XD; D, D_A)$ but since $Y$ is the midpoint of $DD_A$ we must have $H, A', Y$ collinear. Finally, we have
$-1=(X,X';Y,\infty) \stackrel{H}{=} (HX \cap AA', \infty; A', A)$ so $HX$ bisects $\overline{AA'}$ , which is what we needed to prove.

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.5) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11.1, xmax = 8.62, ymin = -6.63, ymax = 4.59; /* image dimensions */ pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen wqwqwq = rgb(0.3764705882352941,0.3764705882352941,0.3764705882352941); /* draw figures */ draw(circle((-4.33,0.13020408163265274), 3.110850339950104), linewidth(1) + dbwrru); draw(circle((-5.476318796333001,1.226069786482878), 1.6641554483504366), linewidth(1) + dbwrru); draw((-6.76760594795539,0.17631226765799224)--(-3.8221458329397446,1.0440665902109125), linewidth(1) + wrwrwr); draw((-4.535744853383504,-2.9738350439058583)--(-7.326646840148698,0.965367271071998), linewidth(1) + wrwrwr); draw((-4.535744853383504,-2.9738350439058583)--(-3.454292693656876,3.115254592291552), linewidth(1) + wrwrwr); draw((-3.454292693656876,3.115254592291552)--(-7.2,-1.07), linewidth(1) + wrwrwr); draw((-7.2,-1.07)--(-5.84,2.85), linewidth(1) + wrwrwr); draw((-5.84,2.85)--(-1.46,-1.07), linewidth(1) + wrwrwr); draw((-1.46,-1.07)--(-7.2,-1.07), linewidth(1) + wqwqwq); draw((-5.84,2.85)--(-5.84,0.4495918367346942), linewidth(1) + wrwrwr); draw((-4.535744853383504,3.234243207171163)--(-4.535744853383503,-5.37424320717117), linewidth(1) + wrwrwr); draw((-4.535744853383503,-5.37424320717117)--(-5.84,0.4495918367346942), linewidth(1) + wrwrwr); draw((-7.326646840148698,0.965367271071998)--(-1.46,-1.07), linewidth(1) + wrwrwr); draw((-5.84,2.85)--(-4.535744853383504,-2.9738350439058583), linewidth(1) + wrwrwr); draw((-5.84,0.4495918367346942)--(-4.535744853383504,-1.07), linewidth(1) + linetype("4 4") + wrwrwr); draw((-5.84,0.4495918367346942)--(-4.535744853383504,3.234243207171163), linewidth(1) + linetype("4 4") + wrwrwr); /* dots and labels */ dot((-5.84,2.85),dotstyle); label("$A$", (-5.96,3.05), NE * labelscalefactor); dot((-7.2,-1.07),dotstyle); label("$B$", (-7.7,-1.21), NE * labelscalefactor); dot((-1.46,-1.07),dotstyle); label("$C$", (-1.38,-1.21), NE * labelscalefactor); dot((-7.326646840148698,0.965367271071998),linewidth(3pt) + dotstyle); label("$H_C$", (-7.96,1.19), NE * labelscalefactor); dot((-3.454292693656876,3.115254592291552),linewidth(3pt) + dotstyle); label("$H_B$", (-3.38,3.27), NE * labelscalefactor); dot((-5.84,0.4495918367346942),linewidth(3pt) + dotstyle); label("$H$", (-6.18,0.59), NE * labelscalefactor); dot((-6.76760594795539,0.17631226765799224),dotstyle); label("$F$", (-7.16,-0.05), NE * labelscalefactor); dot((-3.8221458329397446,1.0440665902109125),linewidth(3pt) + dotstyle); label("$E$", (-3.74,1.21), NE * labelscalefactor); dot((-4.535744853383504,-2.9738350439058583),linewidth(3pt) + dotstyle); label("$D$", (-4.46,-2.81), NE * labelscalefactor); dot((-5.112637592666003,-0.39786042703424335),linewidth(3pt) + dotstyle); label("$A'$", (-5.04,-0.23), NE * labelscalefactor); dot((-4.535744853383504,3.234243207171163),linewidth(3pt) + dotstyle); label("$X$", (-4.46,3.39), NE * labelscalefactor); dot((-4.535744853383504,0.8338350439058582),linewidth(3pt) + dotstyle); label("$D_A$", (-5.1,0.88), NE * labelscalefactor); dot((-4.535744853383503,-5.37424320717117),linewidth(3pt) + dotstyle); label("$X'$", (-4.46,-5.21), NE * labelscalefactor); dot((-4.535744853383504,-1.07),linewidth(3pt) + dotstyle); label("$Y$", (-4.46,-0.91), NE * labelscalefactor); dot((-5.476318796333001,1.2260697864828787),linewidth(3pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

This post has been edited 1 time. Last edited by rocketscience, Jun 26, 2019, 1:53 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

MarkBcc168

1594 posts

MarkBcc168

#11 h Jun 26, 2019, 3:47 AM • 7 Y

Y by rmtf1111, yayups, amar_04, Aryan-23, AFSA, Adventure10, Mango247

Here is another moving points solution based on discussion with yayups. We use zacchro's definition of degree of points.

Move point $E$ along $AB$ . Then we observe the following.

Point $F$ has degree $1$ as $E\mapsto F$ is projective.
Let $D'$ be the reflection of $D$ across perpendicular bisector of $BC$ . This means $AD'\perp EF$ so $E\mapsto D'\mapsto D$ are projective maps.
Midpoints $E'$ , $F'$ of $AE, AF$ have degree one.
This means that point $K$ is induced by two pencils from $\infty_{\perp AB}$ and $\infty_{\perp AC}$ i.e. $K$ moves along a hyperbola $\mathcal{H}$ . As $AD$ has degree $1$ , $K$ has degree $2$ .
Let $X$ be the point on $\odot(ABC)$ such that $XD\perp BC$ . Clearly $D\mapsto X$ is projective. In particular, $X$ has degree two.
This means that degree of $H,K,X$ are colinear is $0+2+2=4$ . Therefore we have to check the problem for five cases.

Now, let's check five cases.

When $E=B$ , we get that $F=H_b$ is the foot of altitude from $B$ . Thus $K$ is midpoint of $AB$ , $D=B$ and $X$ is the antipode of $C$ . The result is then obvious.
When $F=C$ , we get a similar conclusion.
When $E=A$ , we get $F=K=D=A$ . Thus $X=AH\cap\odot(ABC)$ and the result is obvious.
When $BCEF$ is concyclic, we get $D=AH\cap\odot(ABC)$ therefore $X=A$ . The result is obvious as $K$ lie on $A$ -altitude.
When $EF\parallel BC$ , point $D$ becomes $A$ -antipode and $X$ is located so that $AXBC$ is isosceles trapezoid. Thus if $M,N$ be the midpoints of $BC,EF$ , then it suffices to show that
$\frac{AH}{AA'} = \frac{KN}{HN}$ but this is obvious as $KN : HN = OM : H_aM = AH : AA'$ . ( $H_a$ is foot from $A$ -altitude).

This post has been edited 1 time. Last edited by MarkBcc168, Jun 26, 2019, 3:47 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

LeVietAn

375 posts

LeVietAn

#12 h Jun 26, 2019, 7:10 AM • 2 Y

Y by AlastorMoody, Adventure10

When we have $D_A\in EF$ at #10, we can completes the proof from https://artofproblemsolving.com/community/c6h1429293p8050432

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

jbaca

225 posts

jbaca

#13 h Jun 28, 2019, 6:07 AM • 2 Y

Y by Adventure10, Mango247

Solution. Redefine $D$ as $\overline{AK}\cap \Gamma,\ D\neq A$ and let $P$ and $Q$ the intersection points of $HK$ with $\Gamma$ , so that $Q$ lies on the shorter arc $BC$ . Clearly
$\angle KFH=\angle HEK=90^\circ-\angle A=\angle FCH=\angle HBE$ so $KF$ is tangent to $(CFH)$ and $KE$ is tangent to $(BEH)$ . Since $KE^2=KF^2$ , we conclude that $K$ lies on the radical axis of these two circles. Define $Q'$ as its second common point. We get $\angle CQ'B=\angle HQ'B+\angle CQ'H=\angle HEA+\angle AFH=180^\circ-\angle A$ so $Q'$ lies on $\Gamma$ , thus implying that $Q=Q'$ . Then
$KA^2=KE^2=KH\cdot KQ$ hence, $\angle QPD=\angle QAD=\angle KHA$ , so $PD\parallel AH$ and thus $PD\perp BC$ , so our definition of $D$ coincides with the original one. $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

math_pi_rate

1218 posts

math_pi_rate

#14 h Jun 30, 2019, 1:39 PM • 2 Y

Y by Adventure10, Mango247

Here's my solution: WLOG assume that $AB<AC$ and $AE>AF$ (all other cases can be similarly handled). Let $I$ be the antipode of $A$ in $\odot (AEF),$ and suppose $CH$ meets $\Gamma$ again at $X (\neq C).$ Then $EI \parallel HX$ and $EH=EX$ gives that $\angle EXH=\angle EHX=\angle IEH=\angle IAF=\angle DAC=\angle DXH \Rightarrow E \in DX$ Defining $Y=BH \cap \Gamma,$ we similarly get that $F \in DY.$ Then $\angle EDA=\angle XCA=90^{\circ}-\angle BAC=\angle YBA=\angle FDA \Rightarrow DA \text{ bisects } \angle EDF$ Using Fact 5, this gives that $I$ is the incenter of $\triangle DEF$ , while $A$ is its $D$ -excenter. Let $T$ be the point on $EF$ such that $DT \perp BC.$ Also let $DT \cap \Gamma=P.$ Then $180^{\circ}-\angle PAH=\angle APD=\angle AYD=\angle AYF=\angle AHF \Rightarrow AP \parallel EF$ This means that $APTH$ is a parallelogram, or equivalently that $PH$ bisects segment $AT.$ So for showing that $K \in PH$ (i.e. $PH$ bisects $AI$ ), we just need to prove that $KH \parallel IT.$ Taking $AD \cap EF=Z$ , and since $\angle HAZ=\angle TDZ$ , we have $\triangle HAZ \sim \triangle TDZ \Rightarrow \frac{HA}{TD}=\frac{AZ}{DZ}=\frac{KI}{DI}=\frac{KA}{DI}$ where the second equality is well known (wrt $\triangle DEF$ ). Again from $\angle HAK=\angle TDI$ , we get $\triangle HAK \sim \triangle TDI \Rightarrow \angle AKH=\angle DIT \Rightarrow KH \parallel IT \quad \blacksquare$

This post has been edited 2 times. Last edited by math_pi_rate, Jul 9, 2020, 6:26 AM
Reason: Removed unnecessary stuff

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

mathman3880

423 posts

mathman3880

#15 h Jul 2, 2019, 2:55 AM • 6 Y

Y by Ultimate_Drinker, Einstein314, AlastorMoody, Wizard_32, Adventure10, Rounak_iitr

I haven't seen a solution that utilizes the nice Miquel point in the problem yet, so here's one that does.

$[asy] size(10cm); defaultpen(fontsize(10pt)); pen ct=chartreuse; pen gr=green; pen cy=cyan; pen sg = springgreen; pair A, B, C, H, D, E, F, K, T, N, S, h, P, X, Q; A = dir(110); B = dir(210); C = dir(330); H = orthocenter(A, B, C); X = dir(10); E = extension(H, X, A, B); F = extension(H, X, A, C); K = circumcenter(A, E, F); h = 2*(foot(H, B, C)) - H; D = IP(Line(A, K, 20), circumcircle(A, B, C), 1); N = extension(A, K, E, F); S = IP(circumcircle(A, B, C), circumcircle(A, E, F), 1); P = extension(E, F, B, C); T = extension(P, S, D, foot(D, B, C)); Q = IP(Line(T, K, 20), circumcircle(A, B, C), 1); draw(A--B--C--A, ct); draw(circumcircle(A, B, C), sg); draw(P--F, ct); draw(A--D, gr); draw(P--T, gr); draw(P--D, gr); draw(circumcircle(A, E, F), sg); draw(circumcircle(A, H, N), sg); draw(Q--T, ct); draw(D--T, ct); draw(P--B, ct); draw(A--T, ct); draw(A--h, ct); draw(circumcircle(S, H, h), dashed+cy); draw(circumcircle(E, B, S), dashed + cy); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$K$", K, dir(K)); dot("$P$", P, dir(P)); dot("$D$", D, dir(D)); dot("$T$", T, dir(T)); dot("$N$", N, dir(250)); dot("$Q$", Q, dir(Q)); dot("$H'$", h, dir(h)); dot("$S$", S, dir(172)); dot("$H$", H, dir(105)); [/asy]$

Let $T$ be the point on $\odot (ABC)$ such that $AT \parallel EF$ and let lines $EF$ and $BC$ meet at $P$ . Line $TP$ intersects $\odot (ABC)$ again at $S$ , lines $AK$ and $EF$ meet at $N$ , and let $H'$ be the reflection of $H$ about $BC$ .

Claim 1: $DH'$ and $EF$ are reflections of each other about $BC$ .

Let $H_B$ and $H_C$ be the reflections of $H$ about $AC$ and $AB$ respectively. Then $H_BF$ and $H_CE$ meet at a point $D'$ satisfying $2\measuredangle BAC = \measuredangle H_BDH_C = \measuredangle FDE,$ so $D'$ lies on $\odot (ABC)$ as well as $\odot (EKF)$ . Then since $\measuredangle H_BDH_C = 2\measuredangle H_BDA = 2\measuredangle FDK$ it follows that $D'$ lies on $AK$ and thus $D = D'$ . Then $D$ is just the Anti-Steiner point of line $EF$ , so $EF$ and $DH'$ are reflections in line $BC$ . $\blacksquare$

Claim 2: S is the Miquel point of both $HH'DN$ and $BEFC$ .

By Claim 1, it suffices to check that $S$ lies on both $\odot (PHH')$ and $\odot (PEB)$ . For the former, by construction, we have $\measuredangle HPS = \measuredangle ATS = \measuredangle AH'S$ so $S$ lies on $\odot (PHH')$ . For the latter, simply note that $\measuredangle EPA = \measuredangle HPA = \measuredangle HH'S = \measuredangle ABS$ and we're done. $\blacksquare$

Claim 3: $\odot (AHN)$ and $\odot (ABC)$ are inverses about $\odot (AEF)$ .

First, note that Claim 2 implies that $S$ lies on both $\odot (AEF)$ and $\odot (AHN)$ . Let $\infty_{HH_C}$ be the point at infinity along $HH_C$ ; then $-1 = (H, H_C; HH_C \, \cap \, AB , \infty_{HH_C}) \overset{E}{=} (N, D; A, AK \, \cap \, \odot(AEF))$ so by a well-known property of harmonic divisions it follows that $N$ and $D$ are inverses about $\odot (AEF)$ . Finally, note that $A$ and $S$ are fixed upon inversion about $\odot(AEF)$ , so the result easily follows. Let the inverse of $H$ about $\odot(AEF)$ be $Q \in \odot(ABC)$ . $\blacksquare$ .

To finish up, let $T'$ be the second intersection of line $QH$ with $\odot(ABC)$ . Then $HN$ and $QD$ are antiparallel in $\angle HKN$ by inversion, as are $QD$ and $AT'$ , so it follows that $T' = T$ . Finally, since the reflection of $H'D$ about $BC$ is parallel to $AT$ , it follows that $ATDH'$ is an isosceles trapezoid and thus $TD \perp BC$ . So the desired point of intersection is precisely $T$ .

Remark: Interestingly enough, the first three claims alone are enough to instantly solve 2005 ISL G4.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Markas

105 posts

Markas

#70 h May 5, 2024, 12:29 PM

Y by

Denote $\angle HAE = \alpha$ , $\angle HAK = \beta$ and $\angle KEF = \angle KFE = \gamma$ $\Rightarrow$ $\angle KAE = \angle AEK = \alpha + \beta$ and $\angle KAF = \angle KFA = 90 - \alpha - \beta - \gamma$ . Now $\angle HBE = 180 - 90 - \angle BAC = 90 - (90 - \gamma) = \gamma$ $\Rightarrow$ $\angle KEF = \angle EBH = \gamma$ $\Rightarrow$ KE is tangent to (BEH). Similarly $\angle HCF = 180 - 90 - \angle BAC = 90 - (90 - \gamma) = \gamma$ $\Rightarrow$ $\angle KFE = \angle FCH = \gamma$ $\Rightarrow$ KF is tangent to (CFH).

Now let $(BEH) = \omega_1$ and $(CFH) = \omega_2$ . PoP of K w.r.t $\omega_1$ is equal to $KE^2$ . PoP of K w.r.t $\omega_2$ is equal to $KF^2$ , but KE = KF $\Rightarrow$ $KE^2 = KF^2$ $\Rightarrow$ $K \in \rho (\omega_1, \omega_2)$ $\Rightarrow$ $H \in \omega_1$ , $H \in \omega_2$ $\Rightarrow$ $H \in \rho (\omega_1, \omega_2)$ $\Rightarrow$ $KH \equiv \rho (\omega_1, \omega_2)$ .

Denote the second point of intersection of $\omega_1$ and $\omega_2$ be Y. Now $\angle BYH = \angle AEH = \alpha + \beta + \gamma$ and $\angle HYC = \angle HFA = 90 - \alpha - \beta$ $\Rightarrow$ $\angle BYC = \angle BYH + \angle HYC = \alpha + \beta + \gamma + 90 - \alpha - \beta = 90 + \gamma$ $\Rightarrow$ $\angle BAC + \angle BYC = 90 - \gamma + 90 + \gamma = 180^{\circ}$ $\Rightarrow$ BACY is cyclic $\Rightarrow$ $Y \in \Gamma$ .

We know that $HY \equiv \rho (\omega_1, \omega_2) \equiv KH$ $\Rightarrow$ K, H, Y lie on one line. Let $HK \cap \Gamma = P$ . Now it is left to show that $PD \perp BC$ . Let $PD \cap BC = Z$ $\Rightarrow$ we now want to show that $\angle PZB = 90^{\circ}$ . We have that $\angle PZB = 180 - \angle BPZ - \angle ZBP = 180 - \angle BPD - \angle CBP = 180 - \angle BAD - \angle CYP = 180 - (\alpha + \beta) - \angle HFA = 180 - (\alpha + \beta) - (90 - \alpha - \beta) = 180 - \alpha - \beta - 90 + \alpha + \beta = 90^{\circ}$ $\Rightarrow$ $\angle PZB = 90^{\circ}$ which is exactly what we wanted to prove. We are ready.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Martin2001

132 posts

Martin2001

#71 h Jun 11, 2024, 2:03 PM

Y by

Move $E$ along $\overline{AB}$ with degree $1.$ Then $E \mapsto{H} F$ by perspectivity. Note that the midpoint of of $AE$ and $AF$ move with degree $1,$ and projecting from a point at infinity for both of them, the moving line that is the perpendicular bisector of $AE,AF$ has degree $1+1-1=1$ by Zack's lemma. Thus, $K$ moves with $1+1-0=2.$ Let $X$ be where the intersection of the line through $D$ that is perpendicular to $\overline{BC}$ intersects $(ABC).$ Note that $K\mapsto{A} D \mapsto{\infty_{\overline{AH}}} X.$ Thus, $X$ moves with degree $2$ as well. Therefore, as $\text{deg} H+ \text{deg}K+\text{deg}X=4,$ we simply need to check $4+1=5$ cases. These are

$E=A.$
$F=C.$
$EF \| BC$ because $\triangle AKH \sim \triangle AOH_A \sim \triangle DKG.$
$EFBC$ concyclic(Isogonal conjugate of orthocenter is circumcenter).
$E=B$ (Reflection of orthocenter). $\blacksquare$

This post has been edited 3 times. Last edited by Martin2001, Jun 11, 2024, 2:07 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Markas

105 posts

Markas

#72 h Sep 22, 2024, 10:08 PM

Y by

Let $DO \cap \Gamma = D'$ , so now we want to show that D', H, K lie on one line. Since $AH \parallel DD'$ and $\angle HAK = \angle KDD'$ , to prove that D', H, K lie on one line it is enough to show that $\triangle AKH \sim \triangle DKD'$ $\Leftrightarrow$ $\frac{AK}{DK} = \frac{AH}{DD'}$ . Let $\angle EAH = \alpha$ , $\angle HAK = \beta$ , $\angle FEK = \gamma$ . Now we know that $AH = 2R\cos \angle BAC = 2R\cos (90 - \gamma) = 2R\sin \gamma$ . From $\triangle DD'C$ , we have that $DD' = 2R\sin \angle D'CD = 2R\sin (2\alpha + 2\beta + \gamma)$ $\Rightarrow$ $\frac{AH}{DD'} = \frac{2R\sin \gamma}{2R\sin (2\alpha + 2\beta + \gamma)} = \frac{\sin \gamma}{\sin (2\alpha + 2\beta + \gamma)}$ . By law of sines on $\triangle AEH$ , we get $\frac{AE}{\sin \angle AHE} = \frac{AH}{\sin \angle AEH}$ $\Rightarrow$ $\frac{AE}{\sin (180 - 2\alpha - \beta - \gamma)} = \frac{AH}{\sin (\alpha + \beta + \gamma)}$ $\Rightarrow$ $AE = \frac{AH \cdot \sin (2\alpha + \beta + \gamma)}{\sin (\alpha + \beta + \gamma)} = \frac{2R\sin \gamma \cdot \sin (2\alpha + \beta + \gamma)}{\sin (\alpha + \beta + \gamma)}$ . Let P be the midpoint of AE $\Rightarrow$ $AP = \frac{R\sin \gamma \cdot \sin (2\alpha + \beta + \gamma)}{\sin (\alpha + \beta + \gamma)}$ . For $\triangle APK$ , we have that $\frac{AP}{AK} = \cos \angle PAK = \cos (\alpha + \beta)$ $\Rightarrow$ $AK = \frac{AP}{\cos (\alpha + \beta)}$ $\Rightarrow$ $AK = \frac{R\sin \gamma \cdot \sin (2\alpha + \beta + \gamma)}{\sin (\alpha + \beta + \gamma) \cdot \cos (\alpha + \beta)}$ . Now from $\triangle ADC$ , we have that $AD = 2R\sin \angle ACD = 2R \sin (2\alpha + \beta + \gamma)$ $\Rightarrow$ $DK = AD - AK = \frac{2R\sin (2\alpha + \beta + \gamma) \cdot \sin (\alpha + \beta + \gamma) \cdot \cos (\alpha + \beta) - R\sin \gamma \cdot \sin (2\alpha + \beta + \gamma)}{\sin (\alpha + \beta + \gamma) \cdot \cos (\alpha + \beta)} = \frac{R\sin (2\alpha + \beta + \gamma) \cdot (2\sin (\alpha + \beta + \gamma) \cdot \cos (\alpha + \beta) - \sin \gamma)}{\sin (\alpha + \beta + \gamma) \cdot \cos (\alpha + \beta)}$ $\Rightarrow$ $\frac{AK}{DK} = \frac{R\sin \gamma \cdot \sin (2\alpha + \beta + \gamma)}{R\sin (2\alpha + \beta + \gamma) \cdot (2\sin (\alpha + \beta + \gamma) \cdot \cos (\alpha + \beta) - \sin \gamma)} = \frac{\sin \gamma}{2\sin (\alpha + \beta + \gamma) \cdot \cos (\alpha + \beta) - \sin \gamma}$ . We want to show that $\frac{AK}{DK} = \frac{AH}{DD'}$ $\Leftrightarrow$ $\frac{\sin \gamma}{2\sin (\alpha + \beta + \gamma) \cdot \cos (\alpha + \beta) - \sin \gamma} = \frac{\sin \gamma}{\sin (2\alpha + 2\beta + \gamma)}$ $\Leftrightarrow$ $2\sin (\alpha + \beta + \gamma) \cdot \cos (\alpha + \beta) - \sin \gamma = \sin (2\alpha + 2\beta + \gamma)$ , which is obvious by trig formula $\Rightarrow$ $\frac{AK}{DK} = \frac{AH}{DD'}$ , $\triangle AKH \sim \triangle DKD'$ $\Rightarrow$ D', H and K lie on one line and we are ready.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

ellayang2015

22 posts

ellayang2015

#73 h Sep 22, 2024, 11:05 PM

Y by

I also like moving points.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

L13832

262 posts

L13832

#74 h Sep 25, 2024, 7:49 AM • 1 Y

Y by MihaiT

Cool Problem! My first good problem of TSTST I did a few months ago

Solution:
Let $L$ be such that $DL\perp BC$ and $LH \cap \odot(ABC)=J$ .
So we need to show that $(HEB)\cap(HEC)=J$ and $K\in LHJ$ . We angle chase to get
$\measuredangle CJH =\measuredangle CDL=90^{\circ}- \measuredangle DCB= 90^{\circ}-\measuredangle KAE= \measuredangle AFE=\measuredangle HFC\implies \odot(CFHJ)$ Similarly we get $\odot(BEHJ)$ .
Now to prove $K\in LHJ$ , note that this can be done if $KE$ and $KF$ are tangents this is because by Radical axis and Power of Point we will have $\text{Pow}_{\odot(BEHJ)}(K)=\text{Pow}_{\odot(CFHJ)}(K)$ and we get $K\in LHJ$ , so by angle chasing we have $\measuredangle KEH = 90 - \measuredangle BAC = \measuredangle EBH$ and we are done!

This post has been edited 1 time. Last edited by L13832, Sep 25, 2024, 7:50 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

eulerleonhardfan

50 posts

eulerleonhardfan

#75 h Sep 30, 2024, 10:55 AM

Y by

Really nice problem, although trying to not get circular arguments was kinda tricky.

Let $D' \in (ABC)$ be the Anti-Steiner point of $\overline{EF}$ w.r.t $(ABC)$ , $G = (AEF) \cap \overline{AK}$ and $P$ be the intersection of $(ABC)$ and the line through $D$ perpendicular to $\overline{BC}$ .

Notice that the reflection of $\overline{EF}$ over $\overline{AB}, \overline{AC}$ passes through $D'$ , so they are the external bisectors of $\angle D'EF$ and $\angle D'FE$ respectively, thus $A$ is the $D'$ -excenter of $\triangle D'EF$ . Then, we have $G$ is the incenter of $\triangle D'EF$ , and since $EK = FK$ , we have $\overline{D', G, K}$ . Hence, we have $\overline{A, K, G, D'}$ , so $D = D'$ .

Let $Q = \overline{HK} \cap (DEF)$ . Then $\angle EGH = \angle EFK = 90 - \angle A = \angle EBH$ so $BQHE$ is cyclic. Similarly, $CQHF$ is cyclic. After that, because $\angle BQC = \angle BQH + \angle CQH = \angle AEF + \angle AFE = 180 - \angle A$ so $Q \in (ABC)$ .

Let $H'$ denote the reflection of $H$ across $BC$ , which is also the intersection of $\overline{AH}$ and $(ABC)$ . Since $D$ is the Anti-Steiner point of $\triangle ABC$ , the reflection of $\overline{EF}$ across $\overline{BC}$ is $\overline{H'D}$ . Note that $\overline{AH'} // \overline{DP}$ , so $\angle APD = \angle H'DP = \angle(\overline{EF}, \overline{DP})$ which means that $\overline{AP} // \overline{EF}$ . From which we get $\angle BQH = \angle AEF = 180 - \angle PAE$ Let $\overline{QHK} \cap (ABC) = P'$ . This means that $\angle PAE = 180 - \angle BQH = \angle P'AE$ , so $P = P'$ , i.e. $\overline{Q, H, K, P}$ .

Therefore, $\overline{H, K, P}$ , as desired.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Eka01

204 posts

Eka01

#76 h Oct 16, 2024, 6:00 PM

Y by

Let the perpendicular from $D$ to $BC$ meet $\Gamma$ at $S$ and let $SH$ meet $\Gamma$ again at $L$ . We prove that $K$ lies on $\overline{S-H-L}$ . Now note that by some angle chasing, we get that $BEHL$ , $CFHL$ are cyclic and that $KE$ and $KF$ respectively are tangent to these circles so $K$ must lie on the radical axis which is $LH$ and thus we are done.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Mathandski

738 posts

Mathandski

#77 h Oct 17, 2024, 9:57 PM

Y by

$\text{[math]}$

Attachments:

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Saucepan_man02

1311 posts

Saucepan_man02

#78 h Dec 10, 2024, 2:31 AM

Y by

Define point $T$ on $(ABC)$ such that: $DT \perp BC$ . Let $TH$ intersect $(ABC)$ at $S$ .

Claim: $BEHS$ , $CFHS$ are cyclic.
Proof: Notice that: $\angle HSB = \angle TDB = 90^\circ - \angle DBC = \angle 90^\circ - \angle DAC = 90^\circ - \angle KAF = \angle AEH$ which implies $EHSB$ is cyclic. Similarly, $CFHS$ is cyclic and we are done.

Claim: $KE, KF$ are tangent to $(BEH), (CFH)$ respectively.
Proof: Notice that: $KE=KF$ and $\angle KEH = \angle HBE = 90^\circ - A$ which concludes the $KE$ to be tangent to $(BEH)$ . Similarly, $KF$ is tangent to $(CFH)$ and the claim is proved.

Note that: $HS$ is the radical axis of $(BEH), (CFH)$ and $T, K \in HK$ (due to the above two claims). We are done.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

ihatemath123

3441 posts

ihatemath123

#79 h Dec 30, 2024, 10:27 PM

Y by

Let $D'$ as the point on $(ABC)$ such that $\overline{DD'} \perp \overline{BC}$ ; let $H'$ be the reflection of $H$ across $\overline{BC}$ . Excessive angle chasing gives us that lines $EF$ and $H'D$ are reflections across $\overline{BC}$ . So, if $X$ is the intersection of lines $EF$ and $DD'$ , it follows that $XHH'D$ is an isosceles trapezoid. Moreover, $AD'XH$ is a parallelogram.

Claim: $D$ lies on the circumcircle of $EKF$ .
Proof: Firstly, $BEXD$ is cyclic since
$\angle EBD + \angle DXE = \angle AH'D + \angle HXD = 180^{\circ}.$ So, we have
$\begin{align*}\angle FED &= \angle XED = \angle XBD = 2 \angle CBD \\ &= 2 \angle CAK = 180^{\circ} - \angle AKF = \angle FKD. ~ \blacksquare \end{align*}$
Let $I$ be the incenter of $\triangle DEF$ , which coincides with the second intersection of line $AD$ with $(AEF)$ . Let $G$ be the intersection of line $AD$ with line $EF$ . Since $A$ is the $D$ -excenter of $\triangle DEF$ , we have by $\sqrt{ef}$ inversion that
$\begin{align*} DA \cdot DI = DG \cdot DK & \implies \tfrac{DK}{IK} = \tfrac{DA}{GA} \\ & \implies \tfrac{DK}{AK} = \tfrac{DG + GA}{GA} \\ & \qquad \qquad ~ ~ = \tfrac{DX + AH}{AH} \\ & \qquad \qquad ~ ~ = \tfrac{DD'}{AH},\end{align*}$ which is sufficient to imply the collinearity.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Haris1

69 posts

Haris1

#80 h Mar 3, 2025, 7:53 PM

Y by

Rijul saini wrote:

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13.990475469712791cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.392836903540378, xmax = 18.588114035885205, ymin = -3.9502545418864212, ymax = 8.778407142868678; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); /* draw figures */ draw((xmin, -3.3656937511256912*xmin-7.113540727919715)--(xmax, -3.3656937511256912*xmax-7.113540727919715), linewidth(0.8) + wrwrwr); /* line */ draw(circle((-3.1689781644455857,2.5210015993932653), 2.987996768535966), linewidth(0.8) + sexdts); draw(circle((1.4404547101423844,2.485815088900227), 5.966112587259111), linewidth(0.8) + rvwvcq); draw(circle((0.8751152227051443,5.41984234724763), 4.60956717228993), linewidth(0.8) + wrwrwr); draw((-3.7343176518828236,5.455028857740668)--(-5.62,-0.39), linewidth(0.8) + wvvxds); draw((-5.62,-0.39)--(2.24,-0.45), linewidth(0.8) + wvvxds); draw((2.24,-0.45)--(-3.7343176518828236,5.455028857740668), linewidth(0.8) + wvvxds); draw((-4.896586860217193,1.8523557617332833)--(5.141457092485179,7.165246760561282), linewidth(0.8) + rvwvcq); draw((-5.62,-0.39)--(-1.765841507070283,3.5093826570856295), linewidth(0.8) + wrwrwr); draw((-4.896586860217193,1.8523557617332833)--(2.24,-0.45), linewidth(0.8) + wrwrwr); draw((-3.7343176518828236,5.455028857740668)--(-3.7790434683792795,-0.4040531032948146), linewidth(0.8) + wrwrwr); /* dots and labels */ dot((-3.7343176518828236,5.455028857740668),dotstyle); label("$A$", (-4.490319846120439,5.462692087505555), NE * labelscalefactor); dot((-5.62,-0.39),dotstyle); label("$B$", (-6.550804404963088,-0.8187458784879161), NE * labelscalefactor); dot((2.24,-0.45),dotstyle); label("$C$", (2.5621579349633765,-0.8924284352737634), NE * labelscalefactor); dot((-3.764606694809693,1.4871642343209155),linewidth(4.pt) + dotstyle); label("$H$", (-3.590335556654888,0.5970005979441104), NE * labelscalefactor); dot((-3.7790434683792795,-0.4040531032948146),linewidth(4.pt) + dotstyle); label("$P$", (-4.232430897369973,-1.1555871568808397), NE * labelscalefactor); dot((-1.2735184164308264,-2.827267751794997),linewidth(4.pt) + dotstyle); label("$K'$", (-2.2403748329910101,-3.3213257780456405), NE * labelscalefactor); dot((-2.9672994199669978,2.873480387582097),linewidth(4.pt) + dotstyle); label("$D'$", (-3.0903512671893368,3.212747074376137), NE * labelscalefactor); dot((-4.896586860217193,1.8523557617332833),linewidth(4.pt) + dotstyle); label("$R$", (-5.721875641122304,1.875921506517665), NE * labelscalefactor); dot((-1.765841507070283,3.5093826570856295),linewidth(4.pt) + dotstyle); label("$Q$", (-1.6930097194193152,3.6574694462523), NE * labelscalefactor); dot((-2.603638677008338,-0.41302565895413484),linewidth(4.pt) + dotstyle); label("$X$", (-2.577200400849485,-1.2108490744702251), NE * labelscalefactor); dot((5.141457092485179,7.165246760561282),linewidth(4.pt) + dotstyle); label("$T'$", (5.159468061664501,7.378438563937582), NE * labelscalefactor); dot((-4.510216689312586,2.056852037390861),linewidth(4.pt) + dotstyle); label("$S'$", (-4.371899206923977,2.3969848874096586), NE * labelscalefactor); dot((-5.342382262673688,0.4705286503918257),linewidth(4.pt) + dotstyle); label("$E'$", (-6.319232899426774,0.15754799892455892), NE * labelscalefactor); dot((-0.228555834301656,1.9899260782361348),linewidth(4.pt) + dotstyle); label("$F'$", (0.0016890866551767842,1.9627706401778147), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Let $AP, BQ, CR$ be the altitudes of $\displaystyle \triangle ABC$ .

Invert at $A$ with radius $\sqrt{AR \cdot AB}$ . Then $B \mapsto R$ , $C \mapsto Q$ , $H \mapsto P$ . A line through $H$ becomes a circle through $A,P$ , so if this circle intersects $BC$ at $X$ then the inverses of $E,F$ which we denote as $E', F'$ are the foot of perpendiculars from $X$ to $AB, AC$ .

The circle $(AEF)$ becomes the line $E'F'$ , so $K$ maps to the reflection of $A$ across $K$ . The line $KH$ maps to the circle $(APK')$ . Also, the circle $(ABC)$ maps to the line $QR$ . Thus, $D$ maps to $D' = AK' \cap QR$ .

Suppose $HK$ intersects $(ABC)$ at $S,T$ . Then the intersection of $(APK')$ with $QR$ are the inverses $S',T'$ of $S,T$ . We want to show that $AH \parallel DT$ , so that translates to proving that $AP$ tangent to $(AD'T')$ , which reduces to showing that $AS'PK'$ is an isosceles trapezium.

Now, simply note that the reflections of $P$ across $AB$ and $AC$ lie on $QR$ , so by Simson Line for $P$ and $\displaystyle \triangle AE'F'$ , the reflection of $P$ across $E'F'$ also lies on $QR$ . On the other hand $E'F'$ passes through the center of $(ADK')$ , thus this reflection also lies on $(APK')$ , hence must be $S'$ . Thus $AS'PK'$ is an isosceles trapezium. $\square$

My solution is exactly like this , i was surprised how the problem dies after $\sqrt AHAD$ inversion, it immediately becomes a reflection problem which is trivial, truly a good inversion , reason why other inversions like bc or ad don't work are because they either send H too far or makes the problem work on nine point circle which isnt that good.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

YaoAOPS

1509 posts

YaoAOPS

#81 h Mar 11, 2025, 4:47 AM

Y by

Keep moving those points baby

Let the line through $T$ perpendicular to $BC$ intersect the circle again at $X$ .
Move $E$ with degree $1$ so $\deg F = 1$ and $K$ moves with degree $2$ on some conic $\mathcal{C}$ . Since when $E = F = A$ , $K = A$ , $\deg AK = 1, \deg D = 2, \deg X = 2$ and thus the conclusion has degree $4$ .
We now check $5$ cases. For $E = B$ , we get that $K$ is the midpoint of $AB$ and thus $X$ is the $C$ -antipode, and it is well known this lies on $HK$ . $F = C$ is a similar case.
When $E = \infty_{AB}$ , $K = \infty_{\perp AC}$ , and thus we get $D$ as the $C$ -antipode so $X$ becomes $B$ and $B, H, K$ are collinear. $F = \infty_{AC}$ is similar.
Finally, when $K = \mathcal{C} \cap AH$ , $D$ , then the result follows trivially.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Ilikeminecraft

337 posts

Ilikeminecraft

#82 h Mar 16, 2025, 2:44 AM

Y by

Note $\angle KEF = 90-\angle BAC = \angle HBE,$ so $KE$ is tangnet to $(BHE).$ Similarly, $KF$ is tangent to $(CHF).$
Thus, the radax of $(BEH), (HFC)$ is obviously $KH.$
Miquel Theorem on $AEF$ wrt $BHC$ implies $(BEH), (CHF), (ABC)$ are concurrent.
Let this concurrency be $M.$
Let $X$ be on $(ABC)$ so that $XD\perp BC.$
We get $\angle BMX = \angle BDX = 90 - \angle DBC = 90 - \angle DAC = 90 - \angle KAF = \angle AEF = \angle HMB$ which implies $B,K,X$ is collinear.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

cj13609517288

1887 posts

cj13609517288

#83 h Mar 28, 2025, 1:24 AM

Y by

TSTST 2019/5 Solution

Animate $E$ projectively along $AB$ , then $F$ moves projectively around $AC$ . Then the midpoints of $AE$ and $AF$ also move projectively, so
$K=\left(\frac{A+E}{2}\right)\infty_{\perp AB}\cap \left(\frac{A+F}{2}\right)\infty_{\perp AC}$ has degree $2$ .

It is well known that a degree- $n$ point lies on a locus of degree at most $n$ . So in this case, $K$ must lie on some conic $\mathcal C$ . But in particular, if $E=A$ , then $F=A$ and $K=A$ as well. So $A$ lies on $\mathcal C$ as well.

Let $D'$ be the intersection of the line through $D$ perpendicular to $BC$ and $\Gamma$ . Note that this is also the reflection of $D$ over the diameter of $\Gamma$ parallel to $BC$ . So since the map
$K\to D\to D'\qquad\text{by}\qquad\mathcal{C}\stackrel{A}{\to}\Gamma\stackrel{\text{refl.}}{\to}\Gamma$ is projective, $D'$ also moves with degree $2$ . Thus the degree that $H,K,D'$ are collinear is degree at most $4$ . So it suffices to check $5$ cases.

Now if $E=B$ , then $K$ is the midpoint of $AB$ , then $D=B$ and $D'$ is the $C$ -antipode, so the problem becomes the "reflecting the orthocenter" lemma. If $E=CH\cap AB$ , then $K$ is the midpoint of $AC$ , which similarly works.

If $E=\infty_{AB}$ , then $K=\infty_{\perp AC}$ . Then $D$ is the $C$ -antipode, so $D'=B$ , so $K$ indeed lies on $HD'$ . If $E$ is the point such that $F=\infty_{AC}$ , then $K=\infty_{\perp AB}$ , which similarly works.

Thus it suffices to check one more case. Let's check when $EH\parallel BC$ . Let $P$ be the foot from $A$ to $BC$ , then
$\frac{AH}{AP}=\frac{AK}{AO}$ by homothety. Now let's use complex:
$\begin{align*} k&=\frac{(h-a)(-a)}{p-a}+a \\ &=-\frac{2a^2(b+c)}{-a^2+ab+ac-bc}+a \\ &=a+\frac{2a^2(b+c)}{a^2-ab-ac+bc} \\ &=\frac{(a+b)(a+c)}{a(a-b)(a-c)}. \end{align*}$ Since $D$ is the $A$ -antipode, $d'$ is just $\frac{bc}{a}$ . Then
$a+b+c-\frac{bc}{a}=\frac{a^2+ab+ac-bc}{a}$ and
$k-a-b-c=\frac{a^3+a^2b+a^2c+abc-(a^3-ab^2-abc-ac^2+b^2c+bc^2)}{(a-b)(a-c)}=\frac{(b+c)(a^2+ab+ac-bc)}{(a-b)(a-c)}.$ (This was a lot more obvious on paper with CDN, but I can't type CDN here.)

Dividing, we want to show that
$\frac{a(b+c)}{(a-b)(a-c)}$ is real. Indeed, it is equal to its conjugate, as desired. $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

endless_abyss

40 posts

endless_abyss

#84 h Apr 8, 2025, 9:01 PM

Y by

This problem is a delight in so many ways! I solved with hints on OTIS web

Notation -
Let $\angle E H B = \theta$
Claim - $K E$ , $K F$ are tangent to $B E H$ and $H F C$
Alternate segment theorem as-

$\angle K E H = \angle E B H = 90 - a$
$\angle K F H = \angle H F C = 90 - a$

Claim - The circumcircles of $E B H$ , $H F C$ and $A B C$ concur

We use phantom points!
Let $X$ be the point of intersection of the circumcircles $B E H$ and $A B C$ , so

$\angle H X B = 180 - \angle H E B = 90 + \theta - a$
and since
$\angle B X C = 180 - a$
$\angle H X C$ is forced to be $\angle 90 - \theta$ !

comments and a confession of undeniable stupidity.

The rest is just angle chase, just notice that -
( Let $L$ denote the intersection of $H K$ and $B C$ and let $\alpha$ denote $\angle X B C$ )
$\angle H L C = 90 + a - \alpha -\theta$
and
$\angle H X D = \theta - a + \alpha$

and triangle sum proves that $X D$ is perpendicular to $B C$ as required.
$\square$

:starwars:

Attachments:

Z K Y