Shortlist 2017/G5

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27906 posts

#1 h Jul 10, 2018, 11:08 AM • 13 Y

Y by Amir Hossein, Pluto1708, GammaBetaAlpha, jhu08, Jc426, electrovector, HWenslawski, Adventure10, Mango247, deplasmanyollari, Rounak_iitr, Funcshun840, Retemoeg

Let $ABCC_1B_1A_1$ be a convex hexagon such that $AB=BC$ , and suppose that the line segments $AA_1, BB_1$ , and $CC_1$ have the same perpendicular bisector. Let the diagonals $AC_1$ and $A_1C$ meet at $D$ , and denote by $\omega$ the circle $ABC$ . Let $\omega$ intersect the circle $A_1BC_1$ again at $E \neq B$ . Prove that the lines $BB_1$ and $DE$ intersect on $\omega$ .

This post has been edited 2 times. Last edited by Amir Hossein, Jul 10, 2018, 4:16 PM
Reason: Added another source, thanks to rkm0959.

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MarkBcc168

1594 posts

MarkBcc168

#2 h Jul 10, 2018, 11:10 AM • 7 Y

Y by Amir Hossein, qubatae, jhu08, Adventure10, Mango247, trying_to_solve_br, Funcshun840

Really nice problem which has really short solution.

Solution

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rkm0959

1721 posts

rkm0959

#3 h Jul 10, 2018, 11:11 AM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

This was also south korean TST #2.

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juckter

322 posts

juckter

#4 h Jul 10, 2018, 11:14 AM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

The construction in this problem is really weird, but beside that it felt easy for G5

Let $B'B$ cut $\omega$ again at $X$ and let $XP$ cut $\omega$ again at $E'$ . Let $CP$ and $AP$ cut $AX$ and $X$ at $R$ and $S$ .

Notice that, because $AA'C'C$ is an isosceles trapezoid, the internal angle bisector of angle $APC$ is parallel to $AA'$ and $CC'$ , and thus is also parallel to $BB'$ . Notice that $XB$ is the internal angle bisector of $\angle AXB$ because $B$ is the midpoint of arc $AC$ in $\omega$ . Thus, the internal angle bisectors of angles $APC$ and $AXC$ are parallel. By considering the feet of these bisector onto side $AC$ , it is easy by angle chasing to show that this implies that quadrilateral $ARSC$ is cyclic.

Now, by the Law of Sines in triangle $AEC$

$\frac{AE'}{E'C} = \frac{\sin(\angle ACE')}{\sin(\angle CAE')} = \frac{\sin(\angle AXP)}{\sin(\angle CXP)}$
And from Law of Sines in triangles $APX$ and $CPX$ we have

$\sin(\angle AXP) = \frac{AP \cdot \sin(\angle XAP)}{PX} \qquad \sin(\angle CXP) = \frac{CP \cdot \sin(\angle XCP)}{PX}$
However, we have $\angle XAP = \angle XCP$ because $ARSC$ is cyclic. So the previous equality implies

$\frac{AE'}{E'C} = \frac{\sin(\angle AXP)}{\sin(\angle CXP)} = \frac{AP}{CP}$
This equality implies that the external angle bisectors of $\angle APC$ and $\angle AE'C$ cut line $AC$ at the same point. By simmetry, the external angle bisector of $\angle APC$ cuts $AC$ at the point $Y = AC \cap A'C'$ . This implies that $Y, E', B$ are collinear because $B$ is the midpoint of arc $AEC$ . Finally, by Power of a Point from $Y$ we have

$YE' \cdot YB = YA \cdot YC = YA' \cdot YC'$
Implying that $A', C', B$ and $E'$ are collinear. Thus $E = E'$ and $EP$ and $BB'$ are concurrent at point $X$ on $\omega$ .

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v_Enhance

6870 posts

v_Enhance

#5 h Jul 10, 2018, 12:34 PM • 12 Y

Y by Sskkrr, Amir Hossein, anantmudgal09, AlastorMoody, Wizard_32, Purple_Planet, rcorreaa, v4913, Adventure10, Mango247, Funcshun840, Rounak_iitr

Let $\ell$ denote this perpendicular bisector, and let $V = \overline{AC} \cap \overline{A_1 C_1} \cap \ell$ . Also, let $X = \overline{BB_1} \cap \omega \ne B$ . We contend $E = \overline{BV} \cap \omega \ne B$ ; indeed, this gives $VE \cdot VB = VA \cdot VC = VA_1 \cdot VC_1$ .

Now re-define $D = \overline{XE} \cap \ell$ , so that $\overline{VD} \perp \overline{XB}$ . Then, if we can show that $\ell$ is the external angle bisector of $\angle ADC$ , we would conclude $D = \overline{CA_1} \cap \overline{AC_1}$ and solve the problem.

We present two approaches now.

$[asy] size(12cm); pair X = dir(138); pair A = dir(210); pair C = dir(330); pair M = dir(90); pair B = dir(-90); pair E = dir(235); pair S = extension(M, E, A, C); pair L = extension(E, M, X, B); pair V = extension(E, B, A, C); pair K = extension(X, M, V, B); pair D = extension(S, S+X-L, X, E); draw(X--A--C--cycle, red); draw(unitcircle, red); draw(M--K, grey); draw(X--B, lightcyan); draw(D--S, lightcyan+dashed); draw(M--E, red); draw(K--B, red); draw(V--A, red); draw(X--E, red); pair H_A = foot(A, V, D); pair H_C = foot(C, V, D); pair A_1 = 2*H_A-A; pair C_1 = 2*H_C-C; draw(M--(2*M-X), grey); draw(V--(3*D-2*V), grey); draw(A--A_1, lightcyan); draw(C--C_1, lightcyan); draw(V--C_1, dotted+red); draw(B--(1.5*B-0.5*L), lightcyan); draw(K--(1.8*L-0.8*K), orange+dashed); label("$\infty_2$", 3*D-2*V, dir(D-V)); label("$\infty_2$", 2*M-X, dir(D-V)); label("$\infty_1$", 1.5*B-0.5*L, dir(B-L), lightblue); label("$\infty$", 1.8*L-0.8*K, dir(0), red); pair N = extension(V, D, M, B); draw(M--B, orange); dot("$X$", X, dir(X)); dot("$A$", A, dir(A)); dot("$C$", C, dir(C)); dot("$M$", M, dir(M)); dot("$B$", B, dir(B)); dot("$E$", E, dir(E)); dot("$S$", S, dir(315)); dot("$L$", L, dir(10)); dot("$V$", V, dir(225)); dot("$K$", K, dir(K)); dot("$D$", D, dir(50)); dot("$A_1$", A_1, dir(A_1)); dot("$C_1$", C_1, dir(C_1)); dot("$N$", N, dir(325)); /* TSQ Source: !size(12cm); X = dir 138 A = dir 210 C = dir 330 M = dir 90 B = dir -90 E = dir 235 S = extension M E A C R315 L = extension E M X B R10 V = extension E B A C R225 K = extension X M V B D = extension S S+X-L X E R50 X--A--C--cycle 0.1 lightred / red unitcircle 0.1 yellow / red M--K grey X--B lightcyan D--S lightcyan dashed M--E red K--B red V--A red X--E red H_A := foot A V D H_C := foot C V D A_1 = 2*H_A-A C_1 = 2*H_C-C M--(2*M-X) grey V--(3*D-2*V) grey A--A_1 lightcyan C--C_1 lightcyan V--C_1 dotted red B--(1.5*B-0.5*L) lightcyan K--(1.8*L-0.8*K) orange dashed !label("$\infty_2$", 3*D-2*V, dir(D-V)); !label("$\infty_2$", 2*M-X, dir(D-V)); !label("$\infty_1$", 1.5*B-0.5*L, dir(B-L), lightblue); !label("$\infty$", 1.8*L-0.8*K, dir(0), red); N = extension V D M B R325 M--B orange */ [/asy]$

First completion (projective, Evan Chen) Define $M$ to be the $B$ -antipode of $\omega$ , and let $S = \overline{EM} \cap \overline{AC}$ .

Claim: We have $\overline{DS} \parallel \overline{XB}$ .

Proof. This is totally projective. Let $K = \overline{XM} \cap \overline{BEV}, \quad L = \overline{EM} \cap \overline{BX}, \quad \infty = \overline{KL} \cap \overline{MM} \cap \overline{BB},$ $\infty_1 = \overline{XB} \cap \overline{AA_1} \cap \overline{CC_1}, \quad \text{and}\quad \infty_2 = \ell \cap \overline{XKM}.$ with the latter three points at infinity. We aim to show $D$ , $S$ , $\infty_1$ are collinear.

Note that $\triangle X\infty_1\infty_2 \quad\text{and}\quad \triangle ESV$ are perspective from a line, since $\overline{X\infty_1} \cap \overline{ES} = L$ , $\overline{\infty_1 \infty_2} \cap \overline{SV} = \infty$ , and $\overline{\infty_2 X} \cap \overline{VE} = K$ , and $\infty \in \overline{KL}$ by Brokard's theorem.

Thus by Desargue's theorem, they are perspective from a line, meaning $\overline{XE}$ , $\overline{\infty_1 S}$ , $\overline{\infty_2 V}$ concur --- at $D$ . $\blacksquare$

Now $-1 = (MB;AC) \overset{E}{=} (SV;AC)$ , and so $\overline{SD}$ and $\overline{SV}$ are the internal and external angle bisectors, as desired.

Remark: The key projective step seems to be essentially the following lemma (once the polar of $\overline{BM}$ is eliminated via Brokard theorem).

Let $ABCD$ be a (complete) quadrilateral. Pick $W \in \overline{AB}$ , $X \in \overline{AD}$ , $Y \in \overline{BC}$ , $Z \in \overline{CD}$ . Then $\overline{WX}$ , $\overline{YZ}$ , $\overline{BD}$ are concurrent, if and only if $\overline{WY}$ , $\overline{XZ}$ , $\overline{AC}$ are concurrent.

The proof is by the same application of Desargue's theorem. For example, if we are given that $\overline{WX}$ , $\overline{YZ}$ , $\overline{BD}$ are concurrent, then $\triangle AWX$ and $\triangle CYZ$ are perspective from a line, hence $\overline{AC}$ , $\overline{WY}$ , $\overline{XZ}$ are concurrent.

Second completion (synthetic, from the official shortlist) Define $M$ to be the $B$ -antipode of $\omega$ , as before. Let $N = \ell \cap \overline{MB}$ . Then $\measuredangle DNB = \measuredangle XMB = \measuredangle XEB$ so $DNEB$ is cyclic. Now, as $VD \cdot VN = VE \cdot VB = VA \cdot VC$ , we see $ADNC$ is cyclic too.

But as $NA = NC$ , it follows that $\ell \equiv \overline{VDN}$ must be the external angle bisector of $\angle ADC$ as desired.

Third solution (synthetic, Ankan Bhattacharya) We present an independent variation on the second solution. Again let lines $AC$ and $A_1C_1$ meet at $V$ (on the common perpendicular bisector $\ell$ ) and let $N$ be the point on $\ell$ with $AN = CN$ .

$[asy] unitsize(50); pair A, B, C, A1, B1, C1, D, E, V, N, Z; A = (-1, 1); B = (0.2, 2.2); C = rotate(120, B) * A; A1 = yscale(-1) * A; B1 = yscale(-1) * B; C1 = yscale(-1) * C; D = extension(A, C1, C, A1); V = extension(A, C, A1, C1); E = 2 * foot(circumcenter(A, B, C), V, B) - B; N = circumcenter(A, C, A1); Z = 2 * foot(circumcenter(A, B, C), B, B1) - B; draw(circumcircle(A, B, C)); draw(circumcircle(A, N, C)); draw(circumcircle(B, D, N)); draw(E--Z, dashed); draw(B--Z); draw(V--B^^V--C^^V--N^^B--N); draw(A--B--C--N--cycle); dot(A^^B^^C^^D^^E^^V^^N^^Z); label("$A$", A, dir(140)); label("$B$", B, dir(100)); label("$C$", C, dir(40)); label("$D$", D, dir(220)); label("$E$", E, dir(130)); label("$V$", V, dir(190)); label("$N$", N, dir(290)); label("$Z$", Z, dir(250)); [/asy]$

We make three observations:

The quadrilateral $ACC_1A_1$ is an isosceles trapezoid and is cyclic, and by radical axis theorem we learn $B$ , $E$ , $V$ are collinear.
Quadrilateral $ABCN$ is also a kite, and thus $A$ and $C$ are reflections about $\overline{BN}$ .
Finally $\ell$ is the external bisector of $\angle ADC$ and so $A$ , $D$ , $N$ , $C$ are concyclic.

Now we may delete points $A_1$ , $B_1$ , $C_1$ ; this is a problem about $\triangle VBN$ . Points $A$ and $C$ belong to its $V$ -altitude and are reflections about $\overline{BN}$ , and the circumcircles $\triangle BAC$ and $\triangle NAC$ meet $\overline{VB}$ , $\overline{VN}$ respectively at $E$ , $D$ . The task is to show that the $B$ -altitude meets $\overline{DE}$ on circle $\omega$ .

Define $Z$ as the second intersection of the $B$ -altitude and $\omega$ , so we show $E$ , $D$ , $Z$ collinear.

$VE \cdot VB = VA \cdot VC = VD \cdot VN$ , so $\overline{DE}$ and $\overline{BN}$ are antiparallel.
As $BA = BC$ the center of $\omega$ lies on $\overline{BN}$ . Thus $\omega$ is tangent to circle with diameter $\overline{BN}$ . The homothety taking $\omega$ to this circle takes $\overline{ZE}$ to the segment joining the $B$ - and $N$ -altitude feet, so $\overline{EZ}$ and $\overline{BN}$ are antiparallel.

Thus $E$ , $D$ , $Z$ are collinear as desired.

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Generic_Username

1088 posts

Generic_Username

#6 h Jul 10, 2018, 6:52 PM • 2 Y

Y by Adventure10, Mango247

Let $T=\overleftrightarrow{A_1C_1}\cap \overleftrightarrow{AC},$ and let $R$ be the center of the circumcircle of the isosceles trapezoid $\odot (A_1C_1CA).$ The following lemma allows us to erase $A_1$ and $C_1$ .

Lemma: $EDRB$ is cyclic.

Proof: Clearly $T,E,B$ are collinear by considering the radical center of $\odot (ABC),\odot (A_1BC_1),\odot(A_1C_1CA).$ Now $2\angle ARC=\angle ARC+\angle A_1RC_1=\angle ADC+\angle A_1DC_1=2\angle ADC,$ so $R\in \odot (ADC).$ By radical axis once more, we see that $DRBE$ is cyclic.

Now, let $X=\overleftrightarrow{DE}\cap \odot (ABC).$ Note that it suffices to show $\overline{BX}\perp \overline{TDR}$ from the perpendicular bisector conditions. To finish, let $B'$ be the antipode of $B$ in $\odot (ABC),$ and let $\overline {BX}\cap \overline{TDR}=U.$ Then $\angle (\overline{XB},\overline{DR})=\angle RUB=180^{\circ}-\angle DRB-\angle UBR=\angle XEB-\angle XBR=180-\angle XB'B-\angle XBR=90^{\circ},$ finishing the problem.

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math_pi_rate

1218 posts

math_pi_rate

#7 h Jul 11, 2018, 2:19 PM • 2 Y

Y by Adventure10, Mango247

Nice problem! Quite a rich configuration(although, surprisingly, quite easy for a G5).

My solution: In the upcoming solution, WLOG I assume $AA_1>CC_1$ .

Let $\ell$ denote the common perpendicular bisector of $AA_1, BB_1, CC_1$ . Note that $\triangle A_1B_1C_1$ is the reflecton of $\triangle ABC$ about $\ell$ .

Let $BB_1 \cap \omega = P, BB_1 \cap AC = T, AC \cap A_1C_1 = X, BX \cap \omega = E', \odot (DEX) \cap AC = F$ .
Due to symmetry, $D$ and $X$ lie on $\ell$ .

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.5) + fontsize(15); defaultpen(dps); /* default pen style */ pen dotstyle = lightcyan; /* point style */ real xmin = -12.091253394599342, xmax = 13.089144910773957, ymin = -6.60064145094755, ymax = 4.9708087626666355; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-1.0195668114733758,3.3754094807231514)--(-4.633242293465882,1.9658197537048119), linewidth(2) + wrwrwr); draw((1.681337782237601,0.591400130282612)--(-1.0195668114733758,3.3754094807231514), linewidth(2) + wrwrwr); draw((xmin, 0.16282894736842105*xmin-1.0835197368421052)--(xmax, 0.16282894736842105*xmax-1.0835197368421052), linewidth(2) + wrwrwr); /* line */ draw((2.1258477372440456,-2.1385195933933305)--(0.44768124451565317,-5.635568479290281), linewidth(2) + wrwrwr); draw((1.681337782237601,0.591400130282612)--(2.1258477372440456,-2.1385195933933305), linewidth(2) + wrwrwr); draw((-4.633242293465882,1.9658197537048119)--(-3.4265104455575948,-5.445220281933963), linewidth(2) + wrwrwr); draw((-1.0195668114733758,3.3754094807231514)--(0.44768124451565317,-5.635568479290281), linewidth(2) + wrwrwr); draw((-3.4265104455575948,-5.445220281933963)--(0.44768124451565317,-5.635568479290281), linewidth(2) + wrwrwr); draw((-4.633242293465882,1.9658197537048119)--(2.1258477372440456,-2.1385195933933305), linewidth(2) + wrwrwr); draw((1.681337782237601,0.591400130282612)--(-3.4265104455575948,-5.445220281933963), linewidth(2) + wrwrwr); draw(circle((-1.765148213852266,-0.05006072360510188), 3.505672224882686), linewidth(2) + wrwrwr); draw((0.02863768282365131,-3.0620484236464707)--(0.7721296488121127,2.369021498219366), linewidth(2) + wrwrwr); draw((-4.633242293465882,1.9658197537048119)--(5.363852980862329,-0.21012920212932454), linewidth(2) + wrwrwr); draw((-3.4265104455575948,-5.445220281933963)--(5.363852980862329,-0.21012920212932454), linewidth(2) + wrwrwr); draw((-1.0195668114733758,3.3754094807231514)--(5.363852980862329,-0.21012920212932454), linewidth(2) + wrwrwr); draw(circle((2.1337507479366074,-0.5838045523523681), 3.251644768807941), linewidth(2) + wrwrwr); draw((0.7721296488121127,2.369021498219366)--(1.681337782237601,0.591400130282612), linewidth(2) + wrwrwr); draw((-0.01853058372847902,0.9613902818697114)--(0.7721296488121127,2.369021498219366), linewidth(2) + wrwrwr); draw((-0.01853058372847902,0.9613902818697114)--(0.3063184592594781,-1.033642224561368), linewidth(2) + wrwrwr); /* dots and labels */ dot((-1.0195668114733758,3.3754094807231514),dotstyle); label("B", (-0.9426800990905536,3.5676262616801977), NE * labelscalefactor); dot((-4.633242293465882,1.9658197537048119),dotstyle); label("A", (-5.171449280145612,2.2413304730765784), NE * labelscalefactor); dot((1.681337782237601,0.591400130282612),dotstyle); label("$C$", (1.7867981904995291,0.7804829378030267), NE * labelscalefactor); dot((-3.4265104455575948,-5.445220281933963),dotstyle); label("$A_1$", (-4.133478662977551,-5.754887614736546), NE * labelscalefactor); dot((2.1258477372440456,-2.1385195933933305),dotstyle); label("$C_1$", (2.1712317524136253,-1.6222268241600517), NE * labelscalefactor); dot((0.44768124451565317,-5.635568479290281),dotstyle); label("$B_1$", (0.787270929522879,-5.812552649023661), NE * labelscalefactor); dot((0.3063184592594781,-1.033642224561368),linewidth(4pt) + dotstyle); label("$D$", (0.5181674361830118,-1.7760002489256888), NE * labelscalefactor); dot((0.7721296488121127,2.369021498219366),linewidth(4pt) + dotstyle); label("$E$", (0.6527191828529454,2.625764034990671), NE * labelscalefactor); dot((0.02863768282365131,-3.0620484236464707),linewidth(4pt) + dotstyle); label("$P$", (-0.8273500305163248,-3.0254093251464895), NE * labelscalefactor); dot((5.363852980862329,-0.21012920212932454),linewidth(4pt) + dotstyle); label("$X$", (5.6695771658319005,-0.6995862755662295), NE * labelscalefactor); dot((-0.6488331877428701,1.0985807612469138),linewidth(4pt) + dotstyle); label("$T$", (-1.307891982908945,1.5301283835355073), NE * labelscalefactor); dot((-0.01853058372847902,0.9613902818697114),linewidth(4pt) + dotstyle); label("$F$", (-0.32758640002799977,0.5306011225588666), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Now, $XB \cdot XE' = XA \cdot XC = XA_1 \cdot XC_1 \Rightarrow A_1BC_1E'$ is cyclic $\Rightarrow E' = E$

Also, $\angle XEC = \angle BAC = \angle BCA \Rightarrow \angle ACE = \angle XEC + \angle CXE = \angle BCA + \angle CXE$

$\Rightarrow \angle BCE = \angle CXE \Rightarrow BC$ is tangent to $\odot (CXE) \Rightarrow BC^2 = BE \cdot BX$

But, As $B$ is the midpoint of $\overarc {AC}$ , by the Shooting Lemma, $BC^2 = BT \cdot BP$

$\Rightarrow BT \cdot BP = BE \cdot BX \Rightarrow PTEX$ is cyclic.

Now, $XE$ is the external angle bisector of $\triangle AEC \Rightarrow \frac{XC}{XA} = \frac{EC}{EA}$

Also, $XD$ is the external angle bisector of $\triangle ADC \Rightarrow \frac{DC}{DA} = \frac{XC}{XA} = \frac{EC}{EA}$

$\Rightarrow \odot (DEX)$ is the $E$ -Appolonius Circle of $\triangle EAC \Rightarrow FX$ is a diameter of $\odot (DEX)$

$\Rightarrow \angle FDX = 90^{\circ} \Rightarrow FD \perp \ell \Rightarrow FD \parallel TP$

$\Rightarrow \angle XFD = \angle XTP \Rightarrow \angle XED = \angle XEP \Rightarrow E, D, P$ are collinear.

$\blacksquare$

This post has been edited 4 times. Last edited by math_pi_rate, Jul 13, 2018, 3:33 PM

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anantmudgal09

1979 posts

anantmudgal09

#8 h Jul 12, 2018, 9:05 AM • 3 Y

Y by PNT, Adventure10, Mango247

fastlikearabbit wrote:

Let $ABCC_1B_1A_1$ be a convex hexagon such that $AB=BC$ , and suppose that the line segments $AA_1, BB_1$ , and $CC_1$ have the same perpendicular bisector. Let the diagonals $AC_1$ and $A_1C$ meet at $D$ , and denote by $\omega$ the circle $ABC$ . Let $\omega$ intersect the circle $A_1BC_1$ again at $E \neq B$ . Prove that the lines $BB_1$ and $DE$ intersect on $\omega$ .

Let $P=\overline{BE} \cap \overline{AC}$ and $\ell$ be the perpendicular bisector of $\overline{BB_1}$ . By radical axis theorem on $\odot(AA_1CC_1), \odot(ABC), \odot(A_1BC_1)$ we conclude that $P \in \overline{A_1C_1} \implies P \in \ell$ . Note that $\ell=\overline{PD}$ as $AA_1C_1C$ is an isosceles trapezoid. Let $U=\overline{AA_1} \cap \overline{PD}$ and $V=\overline{BB_1} \cap \overline{PD}$ ; then $(\overline{UV}; \overline{DP})=-1$ .

Project on line $\overline{AC}$ ; if $T \in \overline{AC}$ with $\angle PDT=90^{\circ}$ then $(\overline{AC}; \overline{TP})=-1$ . Now $\overline{EP}$ bisects angle $AEC$ externally, so $\overline{ET}$ bisects angle $AEC$ internally. Let $\overline{BB'}$ be a diameter of $\odot(ABC)$ ; then $E,T,B'$ are collinear. Note also that $P,D,T,E$ are concyclic. So $\measuredangle DET=\measuredangle DPT$ .

Now let $X=\overline{BB_1} \cap \odot(ABC)$ with $X \ne B$ . Then $\overline{BX} \perp \ell$ and $\overline{BB_1} \perp \overline{AC}$ so $\measuredangle XEB'=\measuredangle XBB'=\measuredangle DPT$ and we're done. $\blacksquare$

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PROF65

2016 posts

PROF65

#9 h Jul 12, 2018, 9:57 PM • 2 Y

Y by Adventure10, Mango247

i have the same proof as @ above a part from i ve proved that $PETD$ is cyclic whereas he has takes it for obvious which i don't estimate so then if necessary i will post the proof at least the fore-mentionned part .
RH HAS

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math_pi_rate

1218 posts

math_pi_rate

#10 h Jul 13, 2018, 12:07 PM • 2 Y

Y by Adventure10, Mango247

PROF65 wrote:

i have the same proof as @ above a part from i ve proved that $PETD$ is cyclic whereas he has takes it for obvious which i don't estimate so then if necessary i will post the proof at least the fore-mentionned part .
RH HAS

I have shown that $PETD$ is cyclic in my proof in post#7(In my solution $P$ has been replaced by $X$ and $T$ has been replaced by $F$ )

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PROF65

2016 posts

PROF65

#11 h Jul 13, 2018, 3:10 PM • 1 Y

Y by Adventure10

math_pi_rate wrote:

...Also, $XD$ is the external angle bisector of $\triangle A\color{red}{D}$ $C$
$\blacksquare$

you haven't proved this result which is not simple!

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math_pi_rate

1218 posts

math_pi_rate

#12 h Jul 13, 2018, 3:37 PM • 2 Y

Y by Adventure10, Mango247

PROF65 wrote:

math_pi_rate wrote:

...Also, $XD$ is the external angle bisector of $\triangle A\color{red}{D}$ $C$
$\blacksquare$

you haven't proved this result which is not simple!

I don't get it. Isn't it obvious.

$XD$ is the perpendicular bisector of $CC_1$ in $\triangle CDC_1$ , which is isosceles, and so $XD$ is the internal angle bisector of $\triangle CDC_1$ , i.e. $XD$ is the external angle bisector of $\triangle ADC$ .

This post has been edited 1 time. Last edited by math_pi_rate, Jul 13, 2018, 3:37 PM

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jbaca

225 posts

jbaca

#13 h Jul 13, 2018, 3:38 PM • 3 Y

Y by Adventure10, Mango247, Rounak_iitr

Since $\overline{AA_1},\ \overline{BB_1}$ and $\overline{CC_1}$ have the same perpendicular bisector, quadrilaterals $AA_1B_1B,\ AA_1C_1C$ and $BB_1C_1C$ must be isosceles trapezoids. Let $F=\overline{BB_1}\cap \omega,\ P=\overline{A_1C}\cap \omega,\ Q=\overline{AC_1}\cap \omega,\ P\neq C,\ Q\neq A$ . It suffices to show that $EF,\ PC$ and $AQ$ concur at $D$ . By Ceva's theorem on circumference $\omega$ , it suffices to prove that,
$\begin{eqnarray} \dfrac{EA}{AP}\dfrac{PF}{FQ}\dfrac{QC}{CE}=\frac{EA}{CE}\dfrac{PF}{QF}\dfrac{QC}{AP}=1 \end{eqnarray}$ By the radical axes theorem applied to $\omega,\ (A_1BC_1),\ (AA_1C_1C)$ we conclude that $BE,\ AC,\ A_1C_1$ concur at a point, say $K$ . Since $B$ is the midpoint of $\widehat{ABC}$ , $KB$ is the external bisector of $\angle AEC$ , so,
$\begin{eqnarray} \dfrac{EA}{CE}=\dfrac{KA}{KC}=\dfrac{AA_1}{CC_1} \end{eqnarray}$ On the other hand, notice that $\angle A_1AP=\angle A_1C_1-\angle PAQ=\angle A_1CC_1-\angle PCQ=\angle QCC_1$ and we have $\angle PA_1A=\angle QC_1C$ , so $\bigtriangleup A_1AP\sim \bigtriangleup C_1CQ$ , thus,
$\begin{eqnarray} \dfrac{CQ}{AP}=\dfrac{CC_1}{AA_1} \end{eqnarray}$ Let $J=\overline{A_1C}\cap \overline{BB_1},\ L=\overline{AC_1}\cap \overline{BB_1}$ . Since $BB_1\parallel AA_1\parallel CC_1$ , we get $\angle FLQ=\angle AC_1C=\angle AA_1C=\angle CJB$
then $\dfrac{1}{2}\left(\widehat{QF}+\widehat{AB}\right)=\dfrac{1}{2}\left(\widehat{BC}+\widehat{PF}\right)$ whence $\widehat{QF}=\widehat{PF}$ , so $PF=CF$ . Substituting this fact together with (2) and (3) in (1) we conclude the required equality.

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PROF65

2016 posts

PROF65

#14 h Jul 13, 2018, 5:10 PM • 2 Y

Y by Adventure10, Mango247

math_pi_rate wrote:

PROF65 wrote:

math_pi_rate wrote:

...Also, $XD$ is the external angle bisector of $\triangle A\color{red}{D}$ $C$
$\blacksquare$

you haven't proved this result which is not simple!

I don't get it. Isn't it obvious.

$XD$ is the perpendicular bisector of $CC_1$ in $\triangle CDC_1$ , which is isosceles, and so $XD$ is the internal angle bisector of $\triangle CDC_1$ , i.e. $XD$ is the external angle bisector of $\triangle ADC$ .

since you neither have $(A,C;F,X)=-1$ nor $XD\perp DF$ you can't conclude it

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math_pi_rate

1218 posts

math_pi_rate

#15 h Jul 13, 2018, 5:19 PM • 2 Y

Y by Adventure10, Mango247

PROF65 wrote:

since you neither have $(A,C;F,X)=-1$ nor $XD\perp DF$ you can't conclude it

I guess you didn't notice but I have taken $F$ as the point where $\odot (DEX)$ meets $AC$ , and so $F$ is the foot of internal angle bisector of $\angle AEC$ , as it lies on the $E$ -Appolonius circle of $\triangle AEC$ . Thus $FX$ is a diameter of $\odot (DEX)$ , and so $FD \perp \ell$ .

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Mahdi_Mashayekhi

689 posts

Mahdi_Mashayekhi

#64 h May 17, 2022, 10:09 AM • 3 Y

Y by Mango247, Mango247, Mango247

Case $1 : AA_1 = CC_1$ .
Note that in this case both centers of $A_1BC_1$ and $ABC$ lie on $BB_1$ so circles are tangent to each other at $B$ so there exists no $E$ .
Case $2 : AA_1 \neq CC_1$ .
WLOG assume $AA_1 < CC_1$ . Let $S$ be intersection of $AC$ and $A_1C_1$ and perpendicular bisector of $AA_1$ . Note that by Radical Axis on $ABC$ and $A_1BC_1$ and $AA_1C_1C$ we have $S,E,B$ are collinear.
Let $O$ be center of $AA_1C_1C$ .
Claim $: ADOC$ is cyclic.
Proof $:$ Note that $\angle AOC = 2\angle AC_1C = \angle DC_1C + \angle DCC_1 = \angle ADC$ .
Let $BB_1$ meet $\omega$ at $K$ and Let $KE$ meet perpendicular bisector of $AA_1$ at $D'$ and Let $BO$ meet $\omega$ at $T$ .
Claim $: BED'O$ is cyclic.
Proof $:$ Note that $\angle SED' = \angle SEK = \angle KTB$ and note that $KT \perp KB \perp SD'$ so $\angle KTB = \angle D'OB$ .
Claim $: AD'OC$ is cyclic.
Proof $:$ Note that $SD'.SO = SE.SB = SA.SC$ .
Now Note that $AD'DOC$ meets $SD'$ at $3$ points so either $D'$ is $O$ or $D$ . Note that $D'$ can't be $O$ cause $\angle KOE < \angle KOB < B_1OB < 180$ so $D'$ is $D$ and we're Done.

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JAnatolGT_00

559 posts

JAnatolGT_00

#65 h Jun 15, 2022, 9:59 AM

Y by

Let $AC_1,BB_1,CA_1$ meet $\omega$ again at $F,G,H$ respectively. By radical axis for $\omega, \odot (AB_1C_1),\odot (ACC_1A_1)$ lines $BE,AC,A_1C_1$ concur at $S.$ Observe that $\measuredangle CFC_1=-\measuredangle AHA_1,\measuredangle CC_1F=-\measuredangle AA_1H\implies CC_1F\stackrel{-}{\sim} AA_1H\implies |FG|=|GH|.$ Furthermore $S$ is the foot of $E-\text{external bisector}$ in $AEC,$ therefore $\frac{|AE|}{|EC|}\cdot \frac{|CF|}{|FG|}\cdot \frac{|GH|}{|HA|}=\frac{|AS|}{|SE|} \cdot \frac{|CC_1|}{|AA_1|}=1,$ which yields that $AC_1,CA_1,EG$ concur at $D$ as desired.

This post has been edited 1 time. Last edited by JAnatolGT_00, Jun 15, 2022, 10:00 AM

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VicKmath7

1386 posts

VicKmath7

#66 h May 11, 2023, 3:45 PM

Y by

Solution

This post has been edited 5 times. Last edited by VicKmath7, Jan 15, 2024, 12:47 PM

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KST2003

173 posts

KST2003

#67 h May 21, 2023, 10:59 AM

Y by

Let $F = \overline{BB'} \cap \overline{DE}$ , $S = \overline{AC} \cap \overline{A_1C_1}$ , $\Omega$ be the circumcircle of $ACC_1A_1$ , and $O$ be its center. Then by radical axes, $\overline{BE}$ passes through $S$ . Moreover, since $ADOC$ is also cyclice, we have $SD \cdot SO = SA \cdot SC = SE \cdot SB$ , so $BEDO$ is cyclic, and
$\measuredangle EFB = \measuredangle FES + \measuredangle EBF = \measuredangle SOB + \measuredangle SBB' = 90^\circ + \measuredangle OSC + 90^\circ + \measuredangle BSO = \measuredangle BSC.$ Finally, $\triangle BCS \sim \triangle BEC$ , so $\measuredangle BSC = -\measuredangle BCE = \measuredangle ECB$ and so $F$ lies on $\omega$ .

This post has been edited 1 time. Last edited by KST2003, May 21, 2023, 11:03 AM

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asdf334

7586 posts

asdf334

#68 h Nov 29, 2023, 1:20 AM • 1 Y

Y by Ywgh1

so i failed to solve 2015 G7 and then 2016 G4 (both were extremely easy by like one construction), and i went to sleep basically depressed lmao. well we're coming back with a vengeance

i hope im starting to get the hang of this

Let $BB_1\cap AC=P$ and let $BB_1\cap \omega$ be $L$ . Also define $\omega_1=(A_1BC_1)$ . Finally, let $AC\cap A_1C_1=F$ . Wait, we also define $N\in AC$ such that $\angle NDF=90^{\circ}$ . In particular, $N$ is the harmonic conjugate of $F$ w.r.t $AC$ .
Since
$\text{Pow}_F(\omega)=FA\cdot FC=FA_1\cdot FC_1=\text{Pow}_F(\omega_1)$ it follows that $F\in EB$ , or that $E=BF\cap \omega$ .
Let $J$ be the midpoint of $AC$ . Verify $FA\cdot FC=FN\cdot FJ=FE\cdot FB$ hence $EBNJ$ cyclic, hence $E\in (NF)$ .

By Shooting Lemma $BE\cdot BF=BP\cdot BL=BA^2=BC^2$ , hence $FEPL$ cyclic. We have $FEND$ cyclic, and $ND\parallel PL$ . Hence by Reim's $L\in ED$ . Done!

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akasht

83 posts

akasht

#69 h Dec 21, 2023, 7:53 AM • 1 Y

Y by sabkx

Cool problem! First, observe that $AA_1C_1C$ is an isosceles trapezium, and $AA_1 // BB_1 // CC_1$ . Let $X= BB_1 \cap \omega$ , $F = AC \cap A_1C_1$ , and let $T$ be the point on $AC$ such that $DT // AA_1$ . From radical axis on $(ACC_1A_1)$ , $(ABC)$ and $(A_1BC_1)$ , we have that $B,E,F$ are collinear.

Claim: $DTEF$ is cyclic

Proof: Note that since $B,E,F$ are collinear, $F$ is the foot of the external angle bisector of $E$ onto $AC$ . Thus, $\frac{TA}{TC} = \frac{DA}{DC_1} = \frac{AA_1}{CC_1} = \frac{FA}{FC}$ which means that $(A,C;T,F)=-1$ . Since $EF$ is an external angle bisector, $ET$ must be the internal angle bisector, so $ET \perp EF$ . Furthermore, $DF$ is the perpendicular bisector of $AA_1$ , so $DT // AA_1 \implies DT \perp DF$ . Thus, $DTEF$ is cyclic with diameter $TF$ .

Hence, $\begin{align*} \angle XEF = \angle BAX = 180^{\circ} - \angle ABX - \angle AXB &= \angle BAA_1 - \angle ACB \\ &=\angle BAC + \angle CAA_1 - \angle ACB \\ &= \angle CAA_1 \\ &= \angle DTF \\ &= \angle DEF \end{align*}$ implying $D,X,E$ are collinear as desired.

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awesomeming327.

1682 posts

awesomeming327.

#70 h Dec 25, 2023, 3:16 PM

Y by

https://cdn.aops.com/images/0/7/8/078b969a3aaa12ff54fd02d2f4f677fecaa4d8be.png

Note that $ABEC$ , $A'BEC'$ , and $AA'C'C$ are all cyclic. Thus, $AC$ , $A'C'$ , $BE$ concur at a point, $P$ on the perpendicular bisector. Let $D'$ be the point on the perpendicular bisector such that $ACD'Y$ is cyclic. Let the angle bisector of $\angle ABC$ , which is also the diameter of $\omega$ from $B$ , intersect $\omega$ at $X$ and the perpendicular bisector at $Y$ . We have
$\measuredangle CD'P=\measuredangle CD'Y=\measuredangle CAY=\measuredangle YCA=\measuredangle YDA$ By symmetry, $\measuredangle CD'C=\measuredangle A'D'A$ so $D'=D$ . Thus, we have $PD\cdot PY=PA\cdot PC=PE\cdot PB$ , so $BEDY$ is cyclic. Indeed, let $BB'$ intersect $\omega$ at $F$ then $BF\perp FX$ because $BX$ is a diameter, and $BF\perp DY$ . Thus, $DY\parallel FX$ .
$\measuredangle BED=\measuredangle BYD=\measuredangle BXF=\measuredangle BEF$ so $F$ is on $ED$ as desired.

This post has been edited 1 time. Last edited by awesomeming327., Dec 25, 2023, 3:17 PM

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starchan

1602 posts

starchan

#71 h Dec 25, 2023, 4:37 PM • 2 Y

Y by L567, mxlcv

decent problem
solution

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GrantStar

815 posts

GrantStar

#72 h Apr 19, 2024, 3:33 PM

Y by

Let $D'=AC\cap A_1C_1$ . By radical axis, $B,E,D'$ are collinear. Moreover, $E$ lies on the inverse of the perpendicular bisector of $AC$ under inversion at $D'$ fixing $(ACA_1C_1)$ . Call this inverse $\Omega$ .
Let $D_1$ be the point on $\Omega$ with $D'D_1\parallel BB_1$ . Also, let $\omega$ and $\Omega$ intersect at $K \neq E$ . By Reim, $T,K,D_1$ are collinear.

Claim: $TT \parallel A_1C_1D'$ .
Proof. Let $X=CT\cap A_1C_1$ , as $\angle XC_1B_1 = \angle BAC=\angle BTC$ we find $TXC_1B_1$ is cyclic. Then, $\measuredangle (TT,CT)=\measuredangle TBC=\measuredangle C_1B_1T=\measuredangle C_1XT$ and we conclude the claim. $\blacksquare$

Claim: $DD_1 \parallel A_1C_1D'$
Proof. Invert around $D'$ and $D$ is sent to $O$ , the circumcenter of $ACA_1C_1$ . It suffices to show some circle tangency, or and angle equality: $\angle C_1D'O$ is equal to the angle between the perpendicular bisector of $AC$ and line $BB_1$ . I This is not hard to show since if $\angle CC_1D'=x$ then both are $90^{\circ} - x$ after some light angle chase by dropping feet from $O$ to $AC$ and $A_1C_1$ $\blacksquare$

Thus $DD_1 \parallel TT$ are we finish by Reim.

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pokmui9909

185 posts

pokmui9909

#73 h Apr 27, 2024, 12:39 PM

Y by

Let $K = BB_1 \cap \omega$ , redefine $E = DK \cap \omega$ . We'll prove that $A_1, B, E, C_1$ are concyclic. Let $Z = AC \cap A_1C_1$ , and $F \in AC$ so that $BK \parallel FD$ .
$[asy] pair A = (-5.2,4.52); pair B = (-7,0.64); pair C = (-3.50172,-1.82099); pair P = (1.36,4.52); pair Q = (3.16,0.64); pair R = (-0.33827,-1.82099); pair K = (-0.55640,0.64); pair E = (-6.03000,-0.95759); pair D = (-1.92,0.24200); pair Z = (-1.92,-7.72683); pair F = (-4.05424,0.24200); import graph; size(12cm); pen ttqqtt = rgb(0.2,0,0.2); pen zzccff = rgb(0.6,0.8,1); pen ffttww = rgb(1,0.2,0.4); pen ffwwzz = rgb(1,0.4,0.6); pen ttffcc = rgb(0.2,1,0.8); pen zzwwff = rgb(0.6,0.4,1); pen qqwuqq = rgb(0,0.39215,0); pen ffzzqq = rgb(1,0.6,0); pen ttttff = rgb(0.2,0.2,1); pen ttzzqq = rgb(0.2,0.6,0); draw(circle((-3.77820,1.50287), 3.33534), linewidth(0.7) + zzccff); draw(circle((-2.64064,2.19355), 4.62790), linewidth(0.7) + linetype("4 4") + ffttww); draw(K--E, linewidth(0.7) + ffwwzz); draw(A--B, linewidth(0.7) + ttffcc); draw(B--C, linewidth(0.7) + ttffcc); draw(C--R, linewidth(0.7) + ttffcc); draw(R--Q, linewidth(0.7) + ttffcc); draw(Q--P, linewidth(0.7) + ttffcc); draw(P--A, linewidth(0.7) + ttffcc); draw(A--K, linewidth(0.7) + zzwwff); draw(K--C, linewidth(0.7) + zzwwff); draw(B--Q, linewidth(0.7) + qqwuqq); draw(A--Z, linewidth(0.7) + ffzzqq); draw(A--D, linewidth(0.7) + ttttff); draw(D--C, linewidth(0.7) + ttttff); draw(F--D, linewidth(0.7) + ttzzqq); dot("$A$", A, dir((-10, 10))); dot("$B$", B, dir((-10, -10))); dot("$C$", C, dir((-5, -10))); dot("$A_1$", P, dir((10, 10))); dot("$B_1$", Q, dir((10, -10))); dot("$C_1$", R, dir((5, -10))); dot("$K$", K, dir((10, -10))); dot("$E$", E, dir((-10,-10))); dot("$D$", D, dir((5, -10))); dot("$Z$", Z, dir((10, 10))); dot("$F$", F, dir((-10, 0))); [/asy]$
As $\angle CDZ = \angle C_1DZ$ , we have $\angle ADF = \angle CDF$ , so $(A, C; F, Z) = -1$ . Now, we get $\angle KAD = \angle KCD$ since $\measuredangle KAD = \measuredangle FDA - \measuredangle BKA = \measuredangle CDF - \measuredangle CKB = \measuredangle DCK.$ This implies the radius of $(KAD)$ and $(KCD)$ are equal. By the law of sines, $\frac{AE}{EC} = \frac{\sin \angle AKD}{\sin \angle CKD} = \frac{AD}{DC} = \frac{AF}{FC}$ , and so $EF$ bisects $\angle AEC$ . This means $BE \perp EF, ZE \perp EF$ . (Remark that $AB = BC$ , $(A, C; F, Z) = -1$ .) Therefore, $ZB \cdot ZE = ZA \cdot ZC = ZA_1 \cdot ZC_1$ , we have $A_1, B, E, C_1$ are concyclic. $\blacksquare$

This post has been edited 3 times. Last edited by pokmui9909, Apr 27, 2024, 12:41 PM

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OronSH

1728 posts

OronSH

#74 h Jul 12, 2024, 2:49 AM

Y by

Letting $F=AC\cap A_1C_1$ we get by Radical Axis theorem that $F,B,E$ collinear. Next, by observing that $FD$ is the external bisector of $\angle ADC,$ we can delete points $A_1,B_1,C_1.$ Our new problem is: (with some points renamed)

In $\triangle ABC$ let $D$ be the intersection of the external $\angle A$ bisector with $BC.$ Let $E$ be a point with $BE=CE$ and let $DE$ meet $(BCE)$ at $F.$ If $AF$ meets $(BCE)$ at $X,$ show that $EX\perp AD.$

Letting $(BCE)$ meet $AB$ and $AC$ at points $P,Q$ we get that the conclusion is equivalent to $X$ being the midpoint of minor arc $PQ$ by some angle chasing. Now notice that $E$ is an arc midpoint, so $DE$ is the external bisector of $\angle BFC.$ Thus this implies $AB\cdot CF=AC\cdot BF.$ Finally, invert at $F.$ Our new problem is: (with some points renamed)

In $\triangle ABC$ let $P$ satisfy $BP=CP$ and let $X=AP\cap BC.$ Let $(ABP),(ACP)$ meet $BC$ again at $D,E$ respectively. Show that $AP$ bisects $\angle DAE.$

Now this is because $\angle DAP=\angle DBP=\angle CBP=\angle BCP=\angle ECP=\angle EAP$ so we are done.

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fearsum_fyz

48 posts

fearsum_fyz

#75 h Jul 12, 2024, 6:05 PM

Y by

Let lines $BB_1$ and $DE$ intersect at $K$ . We would like to show that $A, B, C, E, K$ are concyclic.
A little bit of observation yields the following claim, which helps us make use of the weird circle conditions:

Claim 1: $BE, AC, A_1C_1$ and the common perpendicular bisector of $AA_1, BB_1, \text{and } CC_1$ concur.
Proof. Radical Axis Concurrence Theorem on $(ABEC), (AA_1C_1C), \text{and } (A_1BEC_1)$ .

Let us call this point of concurrence $F$ . An obvious observation after constructing $F$ is the following claim:
Let $G$ be the intersection of $BB_1$ and $AC$ .

Claim 2: $G, E, F, K$ are concyclic.

Now note that $\Delta{BCE} \sim \Delta{BFC}$ , so the claim would imply $\measuredangle{BKE} = \measuredangle{GKE} = \measuredangle{GFE} = \measuredangle{CFB} = \measuredangle{BCE}$ as desired. Hence it would suffice to prove Claim 2.

We have to find a way to use the $AB = BC$ condition, so a natural next step is to add in the perpendicular bisector of $AC$ .
Alternatively, we have that $FB \cdot FE = FC \cdot FA = FC' \cdot FA' = \text{Pow}(F)$ , so it is also natural to think about the point on the common perpendicular bisector such that $FD \cdot F \text{point} = \text{Pow}(F)$ . Either way, we are led to the following claim:
Let $I$ be the intersection of the perpendicular bisector of $AC$ with the common perpendicular bisector of $AA_1, BB_1, \text{and } CC_1$ .

Claim 3: $A, C, D, I$ are concyclic.
Proof. Trivial

This also implies that $B, I, D, E$ are concyclic (by power of a point). Some angle chasing now yields Claim 2:
Let $M$ be the midpoint of $AC$ and $J$ be the intersection of $BB_1$ and the common perpendicular bisector of $AA_1, BB_1, \text{and } CC_1$ , then we have that $G, M, I, J$ are concyclic, implying that $\measuredangle{KGF} = \measuredangle{BGA} = \measuredangle{BGM} = \measuredangle{JIM} = \measuredangle{DIB} = \measuredangle{DEF} = \measuredangle{KEF}$ implying that $G, E, F, K$ are concyclic as desired, which finishes the problem.

Attachments:

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Reason: improved writing

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abeot

123 posts

abeot

#76 h Jul 27, 2024, 2:20 PM • 1 Y

Y by centslordm

$[asy] size(12cm); import geometry; pair A = dir(0); pair B = dir(105); pair C = dir(210); pen dps = linewidth(0.7); defaultpen(dps); line l = line( (0, -0.8), (1, -0.8)); pair A_1 = reflect(l) * A; pair C_1 = reflect(l) * C; pair B_1 = reflect(l) * B; point D = intersectionpoint(line(A, C_1), line(A_1, C)); point X = intersectionpoints(line(B, B_1), circle(A, B, C))[0]; point T = intersectionpoint(line(A, C), line(A_1, C_1)); point E = intersectionpoint(line(B, T), line(D, X)); point F = intersectionpoint(line(B, X), line(A, C)); point M = midpoint(segment(A, C)); point O = intersectionpoint(line(B, M), line(T, D)); point N = intersectionpoint(line(B, X), line(T, D)); draw(circle(A, B, C)); draw(circle(A_1, B, C_1)); draw(A--B--C--cycle); draw(T--B, red); draw(T--O, red); draw(T--A, red); draw(circle(B,E,D), cyan+dashed); draw(E--X, blue+dashed); draw(B--O); draw(B--X); draw(circle(A,C,O), green+dashed); draw(circle(F,M,N), red+dashed); draw(circle(A,C,C_1)); draw(T--A_1); draw(A--A_1); draw(C--C_1); dot("$A$", A, dir(140)); dot("$B$", B, dir(100)); dot("$C$", C, dir(40)); dot("$D$", D, dir(220)); dot("$E$", E, dir(130)); dot("$T$", T, dir(190)); dot("$O$", O, dir(290)); dot("$X$", X, dir(250)); dot("$F$", F, dir(140)); dot("$M$", M, dir(45)); dot("$N$", N, dir(45)); dot("$A_1$", A_1, dir(315)); dot("$C_1$", C_1, dir(180)); [/asy]$
First, notice that by Radical Axis on $(ABC)$ , $(A_1BC_1)$ and $(ACA_1C_1)$ , then we find that $BE$ , $AC$ and $A_1C_1$ concur. Now, define $X$ as the intersction of $BB_1$ with $(ABC)$ such that $X \neq B$ , and define $F$ as the intersection of $BX$ with $AC$ . Notice that by Shooting Lemma, we have $BE \cdot BT = BC^2 = BF \cdot BX$ Hence $\boxed{TEFX \text{ is cyclic}}$ . Let $M$ be the midpoint of $AC$ . Let line $TD$ (which is the perpendicular bisector of $CC_1$ and $AA_1$ ) intersect $BM$ at $O$ , so then $O$ is the center of $(ACC_1A_1)$ . Notice that $\angle AOC = 2\angle AC_1C = 2(90 - \angle TDC) = 180 - \angle TDC - \angle ADO = \angle ADC$ Hence we find that $ACDO$ is cyclic; but then by radical axis, $TN \cdot TO = TC \cdot TA = TE \cdot TB \implies \boxed{BEDO \text{ is cyclic}}$ Now, the rest is just angle chasing. Define $N$ to be the intersection of $BX$ with $TO$ . Then $\angle BNO = 90 = \angle BMT$ which implies $MNOF$ cyclic. Then $\measuredangle TED = -\measuredangle TOB = \measuredangle MON = \measuredangle MFN = \measuredangle TFX = \measuredangle TEX$ as desired. $\blacksquare$

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Saucepan_man02

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Saucepan_man02

#77 h Nov 15, 2024, 6:12 AM

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EeEeRUT

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EeEeRUT

#78 h Mar 5, 2025, 8:09 AM

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Let us analyze the problem first.
Denote $\ell$ a common perpendicular bisector and let $\ell_1$ be perpendicular bisector of $AC$
We want to find a way to remove $A_1B_1C_1$
The line $BB_1$ can be replaced with a line through $B$ perpendicular to $\ell$
Let $G$ be spiral center sending $AC$ to $A_1C_1$
One can easily see that $G$ lies on $\ell$ and $\ell_1$ . Also, $GDAC$ is cyclic.
So, we could redefine $G$ and $D$ as $\ell_1$ intersects $\ell$ and second intersection of $(ACG)$ with $\ell$ .
The point $E$ is simply $AC$ intersect $\ell$ ( By radical axis).
So, we have a new problem as the following
Let $ABC$ be a triangle with orthocenter $H$ . An arbitrary point $D$ is chosen on $AH$ . Let $D’$ be reflection of $D$ over $BC$ . The circumcircle of $\triangle BDD’$ intersects $BA$ and $BH$ at $E$ and $F$ , respectively. The circumcircle of $\triangle CDD’$ intersects $CA$ and $CH$ at $U$ and $V$ , respectively. Prove that $E,F,U,V$ are collinear.

Proof : Denote feet of altitudes from $B$ , $C$ as $X$ and $Y$ . It is known that $BCXY$ is cyclic centered on $BC$ .
Notice that $\triangle BDD’$ and $CDD’$ is symmetry across $BC$ , thus their circumcenters lie on $BC$ .
Hence, the circumcircle of $BDD’$ and $CDD’$ are tangent to $BCXY$ at $B$ and $C$ , respectively.
Thus, $EF \parallel XY \parallel UV$
By radical axis, $BCEU$ is cyclic. By Reim, $EU \parallel XY$ .
Thus, $EF, UV, EU$ are the same line. That is they are collinear.

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.418140984792641, xmax = 1.6711159988323634, ymin = -1.6466677230142428, ymax = 7.582726487821194; /* image dimensions */ /* draw figures */ draw((-5.6840849814603915,5.662281596605231)--(-7.808728709890522,1.788239905648463), linewidth(1.2)); draw((-7.808728709890522,1.788239905648463)--(-2.048609546746767,1.5556523895332042), linewidth(1.2)); draw((-2.048609546746767,1.5556523895332042)--(-5.6840849814603915,5.662281596605231), linewidth(1.2)); draw(circle((-6.632429459508671,1.7407421867461974), 1.177257813628698), linewidth(0.8)); draw(circle((-4.046505704634745,1.6363253220990237), 1.9995242383507932), linewidth(0.8)); draw((-7.304793099109443,2.707108109103057)--(-6.536935120791933,2.914120553268642), linewidth(1.2)); draw((-5.188487842110301,3.27765831008138)--(-3.884312476712269,3.629260478236424), linewidth(1.2)); draw((-5.188487842110301,3.27765831008138)--(-6.536935120791933,2.914120553268642), linewidth(1.2)); draw((-6.574889120301237,4.038003452383763)--(-2.048609546746767,1.5556523895332042), linewidth(1.2)); draw((-7.808728709890522,1.788239905648463)--(-4.694860099606157,4.5448542160224825), linewidth(1.2)); draw((-5.6840849814603915,5.662281596605231)--(-5.878957481438487,0.8361892927633063), linewidth(1.2)); /* dots and labels */ dot((-5.6840849814603915,5.662281596605231),linewidth(3pt) + dotstyle); label("$A$", (-5.62849578610112,5.7508315762765845), NE * labelscalefactor); dot((-7.808728709890522,1.788239905648463),linewidth(3pt) + dotstyle); label("$B$", (-7.7540532407177665,1.8772827938501966), NE * labelscalefactor); dot((-2.048609546746767,1.5556523895332042),linewidth(3pt) + dotstyle); label("$C$", (-1.9926738242568562,1.6395559732680718), NE * labelscalefactor); dot((-5.767550179070379,3.595233922513285),linewidth(3pt) + dotstyle); label("$H$", (-5.712399369835988,3.68120984414985), NE * labelscalefactor); dot((-5.808479625710513,2.5816004572851785),linewidth(3pt) + dotstyle); label("$D$", (-5.754351161703421,2.660382908708961), NE * labelscalefactor); dot((-5.878957481438487,0.8361892927633063),linewidth(3pt) + dotstyle); label("$D'$", (-5.8242708148158115,0.9263755115216972), NE * labelscalefactor); dot((-7.304793099109443,2.707108109103057),linewidth(3pt) + dotstyle); label("$E$", (-7.250631738308561,2.7862382843112625), NE * labelscalefactor); dot((-6.536935120791933,2.914120553268642),linewidth(3pt) + dotstyle); label("$F$", (-6.481515554072274,2.9959972436484312), NE * labelscalefactor); dot((-5.188487842110301,3.27765831008138),linewidth(3pt) + dotstyle); label("$V$", (-5.139058214314392,3.359579439832858), NE * labelscalefactor); dot((-3.884312476712269,3.629260478236424),linewidth(3pt) + dotstyle); label("$U$", (-3.824568735801466,3.709177705394806), NE * labelscalefactor); dot((-4.694860099606157,4.5448542160224825),linewidth(3pt) + dotstyle); label("$X$", (-4.635636711905186,4.63211712647835), NE * labelscalefactor); dot((-6.574889120301237,4.038003452383763),linewidth(3pt) + dotstyle); label("$Y$", (-6.523467345939708,4.128695624069144), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

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