IMO Shortlist 2011, G4

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1347 posts

#1 h Jul 13, 2012, 11:41 AM • 15 Y

Y by narutomath96, Davi-8191, tenplusten, Amir Hossein, Pondtzx, samrocksnature, RedFlame2112, megarnie, ImSh95, Adventure10, Mango247, Rounak_iitr, Funcshun840, and 2 other users

Let $ABC$ be an acute triangle with circumcircle $\Omega$ . Let $B_0$ be the midpoint of $AC$ and let $C_0$ be the midpoint of $AB$ . Let $D$ be the foot of the altitude from $A$ and let $G$ be the centroid of the triangle $ABC$ . Let $\omega$ be a circle through $B_0$ and $C_0$ that is tangent to the circle $\Omega$ at a point $X\not= A$ . Prove that the points $D,G$ and $X$ are collinear.

Proposed by Ismail Isaev and Mikhail Isaev, Russia

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malcolm

148 posts

malcolm

#2 h Jul 13, 2012, 11:47 AM • 16 Y

Y by zarengold, hvaz, YadisBeles, HiroHamada, myh2910, samrocksnature, mijail, Jalil_Huseynov, RedFlame2112, ImSh95, Zhaom, Adventure10, Rounak_iitr, and 3 other users

Let $\Gamma,O$ be the circumcircle and circumcenter of $\triangle ABC$ , and let $A_0$ be the midpoint of $BC$ . Let $XB_0,XC_0$ meet $\Gamma$ again at $B',C'$ , and let $E=BB' \cap CC'$ . Pascal's Theorem on hexagon $ABB'XC'C$ gives $E \in B_0C_0$ . Since the dilation carrying $\omega$ to $\Gamma$ carries $B_OC_0$ to $B'C'$ , $B'C' \parallel B_0C_0 \parallel BC$ . Then $B'C'BC$ is an isosceles trapezium, so $E$ is the foot of the perpendicular of $A_0$ on $B_0C_0$ . The dilation centered at $G$ taking $\triangle ABC$ to $\triangle A_0B_0C_0$ takes $D$ to $E$ , so $D,G,E$ are collinear. It suffices to show $B'B_0, C'C_0, ED$ are concurrent. Since lines $BB_0, CC_0, ED$ concur at $G$ , $\triangle BCE$ and $\triangle B_0C_0D$ are perspective. Let $U=BE \cap B_0D$ and $V=CE \cap C_0D$ . Desargue's Theorem implies $UV \parallel B_0C_0$ . By Desargue's Theorem again, this implies $\triangle B'C'E$ and $\triangle B_0C_0D$ are perspective, so $B'B_0, C'C_0, ED$ are concurrent as desired.

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daniel73

253 posts

daniel73

#3 h Jul 13, 2012, 11:56 AM • 10 Y

Y by MBGO, Illuzion, alinazarboland, samrocksnature, RedFlame2112, ImSh95, Adventure10, Rounak_iitr, and 2 other users

Alternative solution:

Click to reveal hidden text

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prime04

106 posts

prime04

#4 h Jul 13, 2012, 2:42 PM • 16 Y

Y by pi37, hvaz, MBGO, narutomath96, AnonymousBunny, WiseWolf, RAMUGAUSS, Limerent, samrocksnature, guptaamitu1, RedFlame2112, ImSh95, Adventure10, Mango247, and 2 other users

solution by inversion

This post has been edited 6 times. Last edited by prime04, Jul 17, 2012, 2:53 PM

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paul1703

222 posts

paul1703

#5 h Jul 13, 2012, 4:23 PM • 6 Y

Y by samrocksnature, RedFlame2112, ImSh95, Adventure10, Mango247, and 1 other user

My solution interprets $DX$ as a radical axis, the power of G is easy to calculate (no trigo).

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Goutham

3130 posts

Goutham

#6 h Jul 13, 2012, 6:23 PM • 7 Y

Y by samrocksnature, RedFlame2112, ImSh95, Adventure10, Mango247, and 2 other users

I believe my solution is along similar lines to daniel's solution but I didn't read that fully...

Let $T$ be on $BC$ such that $AT$ tangents cirumcircle of $ABC$ . Now by radical axis theorem on $ABC, AB_0C_0, B_0C_0X$ , their radical centre is midpoint of $AT$ , call it as $S$ . Let $\gamma$ be circle $TAD$ with centre $S$ . Since $SA\perp AO$ where $O$ is cirumcentre of $ABC$ , $\omega$ and $\gamma$ are orthogonal. Also, $SX$ is tangent to $\omega$ and so, $\gamma$ passes through $X$ . Now by Reim's theorem on circles $\gamma, \omega$ with lines $XD$ and $AP$ , we have that if $XD$ intersects $\omega$ at $U$ , then $AU\parallel BC$ . The foot of the altitude on the antimedial triangle from vertex corresponding to $A$ lies on the nine point circle $\omega$ and is also the image of the point $D$ under a homothety with centre $G$ and so, we see that that point is $U$ . This proves the result.

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Zhero

2043 posts

Zhero

#7 h Jul 13, 2012, 7:06 PM • 6 Y

Y by samrocksnature, RedFlame2112, ImSh95, Adventure10, and 2 other users

Let $G'$ and $X'$ be the reflections of $G$ and $X$ across $B_0C_0$ , respectively, let $T = AA \cap B_0 C_0 \cap XX$ be the radical center of $(ABC)$ , $(AB_0C_0)$ , and $(B_0 C_0 X)$ , let $AG'$ meet $B_0 C_0$ at $P$ , let $O$ be the center of $(ABC)$ , let $D'$ be where $AD$ meets $B_0 C_0$ , let $A_1B_1C_1$ be the medial triangle of $\triangle A_0 B_0 C_0$ , let $A'$ be the reflection of $A$ across the perpendicular bisector of $B_0 C_0$ , and let $G_0$ be the centroid of $AB_0 C_0$ .

It is easy to see that $G'$ is the reflection of $G_0$ across the perpendicular bisector of $B_0 C_0$ . We first claim that $OP \perp B_0 C_0$ . Now, since $D'$ is the foot of the altitude from $A$ to $B_0 C_0$ , $D'A_1 B_1 C_1$ is an isosceles trapezoid. Since a homothety centered at $G_0$ with factor -2 sends $A_1B_1C_1$ to $AB_0C_0$ , and $A'AB_0C_0$ is an isosceles trapezoid, this homothety must send $D'$ to $A'$ as well, so $A'$ , $G_0$ , and $D'$ are collinear. Reflecting these points across the perpendicular bisector of $B_0C_0$ shows that $B_0D' = PC_0$ . Since $A$ and $O$ are antipodes with respect to $(AB_0C_0)$ , $AD' \perp B_0C_0$ , and $B_0D' = PC_0$ , $OP \perp B_0 C_0$ . Since $OA \perp B_0C_0$ , quadrilateral $OPAT$ must be cyclic. Thus, $\angle APT = \angle AOT$ . We thus have

$\begin{align*} \angle G'AT &= \angle B_0 AT + \angle G'AB_0 = \angle B_0 C_0 A + \angle PA B_0 \\ &= \angle C + (180^{\circ} - \angle AB_0P - \angle B_0 PA) \\ &= 180^{\circ} + \angle C - \angle ABC - \angle TPA = 180^{\circ} + \angle C - \angle B - \angle TOA \\ &= 90^{\circ} + \angle C - \angle B + \angle ATO = 90^{\circ} + \angle C - \angle B + \frac{\angle ATX}{2} \\ &= 90^{\circ} + \angle C - \angle B + \frac{\angle ATB_0 + \angle B_0 T X}{2} \\ &= 90^{\circ} + \angle C - \angle B + \frac{\angle ATB_0 + \angle X' T B_0}{2} \\ &= 90^{\circ} + \angle C - \angle B + \frac{\angle ATB_0 + \angle X'AT + \angle ATB_0}{2} \\ &= 90^{\circ} + \angle C - \angle B + \angle ATB_0 + \frac{\angle X'TA}{2} \\ &= 90^{\circ} + \angle C - \angle B + 180^{\circ} - \angle TB_0 A - \angle B_0 AT) + 90^{\circ} - \angle TAX' \\ &= 180^{\circ} - \angle TAX' + \angle C - \angle B + \angle AB_0 C_0 - \angle AC_0 B_0 \\ &= 180^{\circ} - \angle TAX' + \angle C - \angle B + \angle B - \angle C = 180^{\circ} - \angle TAX', \end{align*}$

so $X'$ , $A$ , and $G'$ are collinear. Reflecting these points across $B_0 C_0$ shows that $X$ , $D$ , and $G$ are collinear, as desired.

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v_Enhance

6876 posts

v_Enhance

#8 h Jul 13, 2012, 10:14 PM • 38 Y

Y by mathisfun7, narutomath96, GoJensenOrGoHome, kun1417, Swag00, Achintski_Y_, mickeydomath, Wizard_32, AlastorMoody, amar_04, Toinfinity, Siddharth03, indulged, fjm30, Muaaz.SY, HamstPan38825, samrocksnature, guptaamitu1, nima.sa, GeoKing, Cookierookie, CyclicISLscelesTrapezoid, N1RAV, RedFlame2112, ImSh95, PRMOisTheHardestExam, crazyeyemoody907, SerdarBozdag, lpieleanu, Adventure10, Mango247, amirhsz, MS_asdfgzxcvb, farhad.fritl, and 4 other users

This very short solution is from Evan o'Dorney:

Let $R$ be intersection of $B_0C_0$ and tangents at $A$ and $X$ . Take $E$ s.t. $AEBC$ is an isosceles trapezoid; well-known $D$ , $G$ and $E$ are collinear. $RA = RD = RX \implies \angle AXD = \frac 12 \angle ARD = \angle ARB_0 = \pi - \angle RAE = \angle AXE$ , done.

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r1234

462 posts

r1234

#9 h Jul 15, 2012, 2:39 PM • 7 Y

Y by samrocksnature, RedFlame2112, ImSh95, Adventure10, Mango247, and 2 other users

Here are some facts I want to say about this problems and these will also lead to the solution.....

1. $A_0, B_0, C_0$ are the midpoints of the sides $BC, CA, AB$ . Now we define the points $Y,Z$ such that $\odot YA_0C_0$ us tangent to $\odot ABC$ at $Y$ and similar for $Z$ . Then the lines $AX, BY, CZ$ concur at the isogonal conjugate of the isotomic conjugate of orthocente rof $\triangle ABC$ .

2. Let $D'$ be the isotomic point of $D$ wrt $BC$ . Then $GD$ bisects $AD'$ .

Proof of (1):- Inverting with respect to $A$ with some power, $B_0', C_0'$ becomes the reflection of $A$ wrt $B', C'$ respectively. The point $X'$ is such that the line $B'C'$ is tangent to $\odot X'B_0'C_0'$ at $X'$ Hence $X'$ is the midpoint of the arc $B_0'C_0'$ . So $X'$ lies on the perpendicular bisector of $B_0'C_0'$ . Hence $X'$ is the reflection of the foot of A-altitude of the triangle $\triangle AB'C'$ wrt the midpoint of $B'C'$ . So $AX, BY, CZ$ concur at the isogonal conjugate of isotomic conjugate of the orthocenter of $\triangle ABC$ .

Proof of (2):- Reflect $G$ wrt the midpoint of $BC$ . Let it be $G'$ .Then clearly $G'D'\parallel GD$ . Now $G$ is the midpoint of $AG'$ . So $GD$ bisects $AD'$ .

Main Proof:- Suppose the isogonal ray of $AD'$ cuts $\odot ABC$ at $D_2$ . And $AD'$ cuts the circumcircle at $D_1$ . Then clearly $D_1D_2\parallel BC$ .So the perpendicular bisector of $D_1D_2$ passes through the midpoint of $BC$ which again passes through the midpoint of $AD'$ (easy to see). So $D_2, D, \text{midpoint of} AD'$ are collinear. So $D_1, D, G$ are collinear. Hence proven.

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proglote

958 posts

proglote

#10 h Jul 15, 2012, 11:22 PM • 14 Y

Y by hongduc_cqt, mad, AlastorMoody, HolyMath, samrocksnature, Cookierookie, RedFlame2112, ImSh95, Adventure10, Mango247, Rounak_iitr, and 3 other users

Posted on my blog a while ago, didn't know it was from the shortlist.

http://www.artofproblemsolving.com/Forum/blog.php?u=112643&b=72551

$3. \blacktriangleright$ (ISL 2011 G4) Let $\mathcal{C}$ denote the circle passing through $M_B, M_C$ tangent to the circumcircle $\omega$ of $ABC$ at a point $T_A \neq A.$ Then $T_A, H_A, G$ are collinear.

$[asy] import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -3.59, xmax = 13.85, ymin = -4.12, ymax = 4.54; /* image dimensions */ pen ccwwqq = rgb(0.8,0.4,0); pen ttttff = rgb(0.2,0.2,1); pen ttzzqq = rgb(0.2,0.6,0); pen ffcctt = rgb(1,0.8,0.2); pen cccccc = rgb(0.8,0.8,0.8); draw((2.14,3.44)--(0.66,-1.34)--(7.36,-1.36)--cycle, ccwwqq); /* draw figures */ draw((2.14,3.44)--(0.66,-1.34), ccwwqq); draw((0.66,-1.34)--(7.36,-1.36), ccwwqq); draw((7.36,-1.36)--(2.14,3.44), ccwwqq); draw(circle((4.01,0.24), 3.71), ttttff); draw(circle((3.07,0.25), 1.85), ttttff); draw((4.75,1.04)--(4.01,-1.35), ttzzqq); draw((4.01,-1.35)--(1.4,1.05), ttzzqq); draw((2.14,3.44)--(5.91,3.43), red); draw((2.14,3.44)--(2.12,-1.34), ffcctt); draw((4.01,-1.35)--(2.14,3.44), cccccc); draw((1.4,1.05)--(7.36,-1.36), cccccc); draw((4.75,1.04)--(0.66,-1.34), cccccc); draw((5.91,3.43)--(1.34,-2.33), linetype("4 4") + red); draw((4.02,1.04)--(4.01,-1.35)); draw(circle((3.07,-0.67), 2.4), cccccc); draw(circle((3.08,1.84), 1.85), cccccc); draw((2.14,3.44)--(-1.92,1.06)); draw((-1.92,1.06)--(1.34,-2.33)); draw((-1.92,1.06)--(7.64,1.03), ttzzqq); draw((5.91,3.43)--(0.4,1.05), red); draw((5.91,3.43)--(7.64,1.03), red); /* dots and labels */ dot((2.14,3.44),dotstyle); label("$A$", (2.18,3.51), NE * labelscalefactor); dot((0.66,-1.34),dotstyle); label("$B$", (0.54,-1.56), W); dot((7.36,-1.36),dotstyle); label("$C$", (7.43,-1.59), NE * labelscalefactor); dot((4.75,1.04),dotstyle); label("$M_B$", (4.93,0.91), N*1.5); dot((1.4,1.05),dotstyle); label("$M_C$", (1.11,0.96), N); dot((4.01,-1.35),dotstyle); label("$M_A$", (4,-1.54), S); dot((3.39,0.25),dotstyle); label("$G$", (3.18,0.34), NE * labelscalefactor); dot((2.12,-1.34),dotstyle); label("$H_A$", (2.16,-1.57), S*0.25); dot((1.34,-2.33),dotstyle); label("$T_A$", (1.23,-2.07), S*4); dot((5.91,3.43),dotstyle); label("$T'_A$", (5.88,3.19), NE*2); dot((4.01,0.24),dotstyle); label("$O$", (4.06,0.32), NE * labelscalefactor); dot((4.02,1.04),dotstyle); label("$H'_A$", (3.74,1.16), N); dot((-1.92,1.06),dotstyle); label("$P_A$", (-1.98,1.22), NE * labelscalefactor); dot((7.64,1.03),dotstyle); label("$T_2$", (7.71,1.18), NE * labelscalefactor); dot((0.4,1.05),dotstyle); label("$T_1$", (0.17,1.19), W); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$
Proof: Let $\mathcal{C}_2$ denote the circumcircle of $AM_CM_B$ , which is also tangent to $\omega.$ Let $P_A$ be the radical center of $\mathcal{C}, \mathcal{C}_2, \omega$ , and let $T_1, T_2$ be the intersections of $M_BM_C$ and $\omega$ , as above. Then since $P_AA$ and $P_AT_A$ are tangent to $\omega$ , $AT_1T_AT_2$ is a harmonic quadrilateral. Let $T_A^*$ denote the second intersection of $T'_AG$ and $\omega$ , and $H'_A$ denote the image of $H_A$ under the homothety $\mathcal{H}$ which sends $\triangle ABC$ to $\triangle M_AM_BM_C.$ Then $M_AH'_A \perp T_1T_2$ , so $OH'_A \perp T_1T_2$ and $H'_A$ is the midpoint of $T_1T_2.$ We have shown in $(2)$ that $AT'_A \parallel T_1T_2$ , so $T'_A(T_2, T_1; H'_A, A)$ is a harmonic pencil, which implies that $T'_A(T_2, T_1; T_A^*, A)$ is a harmonic pencil, and $AT_1T_A^*T_2$ is a harmonic quadrilateral $\implies T_A^* \equiv T_A$ , as desired.

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Orin

22 posts

Orin

#11 h Jul 24, 2012, 6:50 PM • 9 Y

Y by samrocksnature, RedFlame2112, ImSh95, Adventure10, Mango247, and 4 other users

Let $\odot AB_0C_0=\alpha,\odot XB_0C_0=\omega,M=AD \cap B_0C_0$ .
Let $AD' \bot B_0C_0,D' \in B_0C_0$ , $O$ be the centre of $\Omega$ .
Let $K$ be the intersection of the tangents to $\Omega$ from $A,X$ .
$AK$ is the radical axis of $\Omega,\alpha$ ; $KX$ is the radical axis of $\Omega,\omega$ and $B_0,C_0$ is the radical axis of $\alpha,\omega$ .
So $K \in B_0C_0$ .
$\angle OAK+\angle OXK=90^{\circ}+90^{\circ}=180^{\circ} \rightarrow AKXO$ are cyclic.
A homothety of ratio $-2$ and centre $G$ sends $O$ to $H$ ,the orthocentre of $ABC$ and $\triangle A_0B_0C_0$ to $\triangle ABC$ .
So $O$ is the orthocentre of $\triangle A_0B_0C_0$ .
$\angle OD'K=\angle OAK=90^{\circ} \rightarrow AKXOD'$ is cyclic.
$AB_0C_0 \cong A_0B_0C_0 \rightarrow AM=A_0D'$
$AM \bot B_0C_0,A_0D' \bot B_0C_0 \rightarrow AM \parallel A_0D'$
which implies $AMA_0D'$ is a parallelogram.
Note that a homothety of ratio $-2$ and centre $G$ sends $D'$ to $D$ .
Hence $D,G,D'$ are collinear.
So it suffices to prove that $X,D,D'$ are collinear.
$\angle KD'X=\angle KAX=\angle KXA=\angle KD'A=\angle AD'M$
$=\angle A_0MD'=\angle MA_0D=\angle MD'D$ [The last arguement comes from the fact that $MDA_0D'$ is a rectangle.]
which implies $X,D,D'$ are collinear.

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sjaelee

485 posts

sjaelee

#12 h Sep 8, 2012, 4:44 AM • 7 Y

Y by samrocksnature, RedFlame2112, ImSh95, Rushery_10, Adventure10, Rounak_iitr, and 1 other user

Lemma: If $DG$ intersects the circumcircle at $E$ , then $ABCE$ is an isosceles trapezoid.

Proof

Now we note that the tangent from $X$ , $C_0B_0$ , and the tangent from $A$ are concurrent at the radical center $M$ of circles $AB_0C_0,B_0C_0X,ABC$ . Then we note that $MX=MA$ from PoP and $MD=MA$ since $C_0B_0$ is the perpendicular bisector of $AD$ . Then $M$ is the center of circle $ADX$ and $2\angle AXE=\angle DMA=2\angle AXD$ , thus $XD$ and $XE$ are equivalent lines and $X,D,E$ are collinear, implying $X,G,D$ are collinear.

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MBGO

315 posts

MBGO

#13 h Feb 26, 2013, 6:28 PM • 6 Y

Y by samrocksnature, A64298347, RedFlame2112, ImSh95, Adventure10, Mango247

new elementary proof :

Let the circumcircles of $XC_0B$ and $XB_0C$ meet at point $K$ different from $X$ ,because circle $\omega$ and circumcircle of $ABC$ are tangent,it follows that $K$ is on $B_0C_0$ . Let $XB_0$ and $XC_0$ intersect the circumcircle of $ABC$ again at $M$ , $N$ . We have $B,K,M$ and $C,K,N$ collinear, all else we should prove is $K,G,X$ collinear, equivalent to proving that $G$ lies on Radical axes of circumcircles of $XB_0C$ and $XC_0B$ . Let $XB_0C$ and $XC_0B$ meet lines $BB_0$ and $CC_0$ again at $B_1$ , $C_1$ respectively,hence we should prove that $BB_1C_1C$ is a cyclic quadrilateral,which is obvious as we have $\angle BB_1C = \angle CC_1B = 180 - \angle CBM = 180 - \angle BCN$ .

and it's done

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leader

339 posts

leader

#14 h May 4, 2013, 5:23 PM • 5 Y

Y by samrocksnature, RedFlame2112, ImSh95, Adventure10, Mango247

another way $E\in BC$ $CE=BD$ $R\in \odot ABC$ $\angle BAR=\angle CAE(R,A$ are on opposite sides of $BC$ ) $Y$ -midpoint of $AE$ .Since $ARC\sim ABE$ , $ARB\sim ACE$ and $Y$ is on perp bisector of $ED$ (and $BC$ ) then if $RB_{0},RC_{0}$ meet $\odot ABC$ in $B',C'$ then $\angle C'RB=\angle YCB=\angle YBC=\angle B'RC$ so $B'C'||BC||B_{0}C_{0}$ and $\odot RB_{0}C_{0}$ and $\odot ABC$ touch so $X=R$ but if $AE$ meets $\odot ABC$ again at $P$ then $XB=CP$ $BD=CE$ $\angle XBD=\angle PCE$ so $PCE\cong XBD$ and $\angle XDB=\angle PEC=\angle AED$ but $G$ is centroid of $ADE$ so $\angle AED=\angle GDE$ and $G,D,X$ are collinear.

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mathocean97

606 posts

mathocean97

#15 h May 23, 2013, 1:52 PM • 6 Y

Y by butter67, samrocksnature, RedFlame2112, ImSh95, Adventure10, Mango247

A less ingenious solution.

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Cusofay

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Cusofay

#121 h Feb 3, 2024, 8:12 PM

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For storage

This post has been edited 1 time. Last edited by Cusofay, Feb 3, 2024, 8:13 PM
Reason: typo

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shendrew7

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shendrew7

#122 h Feb 12, 2024, 8:01 PM

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Denote the point $Y$ as the intersection of ray $DG$ with $(ABC)$ . Nine-point circle homothety at $G$ tells us this is the antipode of the reflection of the orthocenter $H'$ over $BC$ . Now it suffices to show that $X$ , $D$ , and $Y$ are collinear.

Consider the radical center $S$ of $(AB_0C_0)$ , $(XB_0C_0)$ , and $(ABC)$ , or the intersection of $B_0C_0$ and the tangents at $A$ and $X$ . Noting that $SX = SA = SD$ , we have
$\measuredangle AXD = \measuredangle ASC_0 = \measuredangle OAD = \measuredangle AH'O = \measuredangle ACY = \measuredangle AXY. \quad \blacksquare$

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thdnder

194 posts

thdnder

#123 h Mar 1, 2024, 8:35 AM

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WLOG, assume $AC > AB$ , then $\angle ABC > \angle ACB$ .

Since the radical axes of $(AB_0C_0), \Omega, \omega$ are concurrent, so the tangent of $\Omega$ at $A$ , the tangent of $\Omega$ at $X$ , $B_0C_0$ are concurrent and call this point $R$ . Then $RA = RX$ and since $D$ is the reflection of $A$ wrt $B_0C_0$ , hence $RA = RD$ . Therefore, $R$ is the center of $(ADX)$ , so $\angle AXD = \angle ARC_0 = \angle ABC - \angle ACB$ . Now assume ray $DG$ meets $\Omega$ at $Y$ . Then by homothety, we get $AY \parallel BC$ , which means $\angle AXY = \angle ABC - \angle ACB$ , hence $X, D, Y$ are collinear, thus $D, G, X$ are collinear. Hence we're done. $\blacksquare$

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HamstPan38825

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HamstPan38825

#124 h Mar 2, 2024, 6:28 PM

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Let $E = \overline{XX} \cap \overline{AA} \cap \overline{B_0C_0}$ by radical axis, and let $K$ be on $(ABC)$ with $\overline{AK} \parallel \overline{BC}$ ; negative homothety at $G$ implies $D, G, K$ collinear. Set $K' = \overline{DX} \cap (ABC)$ ; we wish to show $K'=K$ . Let $N$ be on $(ABC)$ so that $\overline{XN} \parallel \overline{BC}$ .

Notice that as the nine-point circle is congruent to $(AB_0C_0)$ , $E$ is the center of $(AXD)$ . Then $\angle XNK' = \angle EXD = 90^\circ - \angle XAD = \angle NXA$ hence $AK'NX$ is an isosceles trapezoid and $K=K'$ .

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Pyramix

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Pyramix

#125 h Mar 4, 2024, 6:35 PM

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Nice, cute problem

.
Here's my solution. All angles are directed.

Note that taking homothety at $A$ with factor 1/2, we get the circumcircle becomes $(AB_0C_0)$ , which is tangent to the circumcircle. Let $Y$ be the radical centers of the circles $(B_0C_0X),(ABC),(AB_0C_0)$ . Then, $Y,B_0,C_0$ are collinear and $YA=YX$ . Let $T$ be the point on $AY$ such that $AT=2AY$ . From homothety at $A$ with factor 2, we get that $T,B,C$ are collinear.
Consider the circle with center $Y$ and radius $\overline{YA}$ . Clearly, $T$ lies on this circle. Moreover, $\triangle TDA$ is a right-angle triangle, so $YA=YT=YD$ . Hence $D$ also lies on this circle.

Let $\angle XYO=\theta\Longrightarrow\angle AOX=180^\circ-2\theta\Longrightarrow\angle ACX=90^\circ-\theta$ and $\angle ADX=180^\circ-\theta$ . Hence, $\angle CXD$ and $\angle DAC$ are complementary. Hence, $\angle CXD=\angle ACD=\angle ACB$ . Let $E$ be the second intersection of $XD$ with circumcircle $(ABC)$ . Then, $\angle ACB=\angle CXE$ . So, $AB=CE$ . It follows that $AE\parallel BC$ .

Let $G'$ be the intersection of $AA_0$ and $DX$ , where $A_0$ is the mid-point of $\overline{BC}$ . We know that $E$ is the reflection of $A$ in the line $A_0O$ . Since $A_0O$ and $AD$ are both perpendicular to $AE$ , it follows that $AE=2DA_0$ . Moreover, $\angle DG'A_0=\angle EG'A$ and $\angle AEG'=\angle G'DA_0$ . Hence, $\triangle AEG'\sim-\triangle A_0DG'$ . So, $\frac{AG'}{G'A_0}=\frac{AE}{DA_0}=2$ . Hence, $G'$ is the centroid. So, $G'=G$ , as required. $\blacksquare$

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Aiden-1089

278 posts

Aiden-1089

#126 h Jul 12, 2024, 2:25 AM

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Note that $\Omega$ and $(AB_0C_0)$ are tangent at $A$ by homotehty. Using radax on $\Omega, \omega, (AB_0C_0)$ , we see that the tangent to $A, X$ at $\Omega$ and line $B_0C_0$ are concurrent. Call this point $S$ .
Let $T$ be the intersection between the tangent to $\Omega$ at $A$ and line $BC$ . Then $S$ is the midpoint of $AT$ .
Let $A'$ be the reflection of $A$ across the perpendicular bisector of $BC$ , recall that $D,G,A'$ are collinear.
Let $Y$ be the antipode of $A$ in $\Omega$ , $E$ be the point on $BC$ such that $BD=CE$ . Note that $A',E,Y$ lie on a line perpendicular to $BC$ . Also let $A_1$ be the point such that $A'$ is the midpoint of $AA_1$ .

Take $\sqrt{bc}$ -inversion.
Recall that $(A',T)$ invert to each other, so $(A_1,S)$ invert to each other. Also, $(D,Y)$ invert to each other.
Note that $(AA_1E)$ is tangent to $BC$ at $E$ . After inversion, this means that $SE'$ is tangent to $(ABC)$ , so $(X,E)$ invert to each other.
Note that $\measuredangle TEY = 90^\circ = \measuredangle TAY$ , so $A,E,T,Y$ are concyclic. Inverting, $X,D,A'$ are collinear.
Hence $D,G,X$ are collinear.

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Eka01

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Eka01

#127 h Aug 4, 2024, 4:05 PM • 1 Y

Y by Sammy27

It is well known that $\overline{D-G-A_1}$ are collinear where $A_1$ is the point on $(ABC)$ such that $AA_1 || BC$ .
Now we do $\sqrt{\frac{bc}{2}}$ inversion for all points(ignoring $G$ ) and then apply homothety with center $A$ and factor $2$ for all points except $B_0$ and $C_0$ (So we almost do a $\sqrt{bc}$ inversion but not exactly.)

Now the new problem is :

Quote:

In $\Delta ABC$ , $X$ is the point on the line $BC$ such that $AX$ is tangent to $(ABC)$ at $A$ , $A'$ is the $A$ antipode, $Y$ is on the $A$ midline such that the midline is tangent to $(BCY)$ at $Y$ . $AY$ meets $BC$ at $Z$ . Show that $A,X,A' , Z$ are concyclic.

Note that by trivial angle chasing, $Y$ lies on the perpendicular bisector of $BC$ , so $\overline{Y-O-A_0}$ are collinear where $O$ is the circumcenter of $(ABC)$ and $A_0$ is midpoint of $BC$ . Also notice that since $YO \perp B_0C_0$ $\implies$ due to homothety, $ZA_0 \perp BC$ and $\angle A'AX=90°$ due to tangency, hence $A,X,A',Z$ are concyclic. Now inverting back we get the desired result.

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cj13609517288

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cj13609517288

#128 h Dec 9, 2024, 7:18 PM

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$\sqrt{bc/2}$ invert; then $(BCX')$ is tangent to $B_0C_0$ at $X'$ . Therefore $X'B=X'C$ , so if $M$ is the midpoint of $BC$ , then $X'$ is the foot of the perpendicular from $M$ to $B_0C_0$ . Thus $AX'$ and the $A$ -altitude are isotomic. Therefore,
$AX'=(?:S_B:S_C)$ $AX=(?:\frac{b^2}{S_B}:\frac{c^2}{S_C})$ $AX=(a^2t:b^2S_C:c^2S_B).$
Plug this into the circumcircle formula to get
$a^2b^2c^2 S_BS_C+a^2b^2c^2 S_Bt+a^2b^2c^2 S_Ct=0$ $S_BS_C+a^2t=0$ $a^2t=-S_BS_C.$ Thus
$\begin{align*} X =& (-S_BS_C:b^2S_C:c^2S_B) \\ D =& (0:S_C:S_B) \\ G =& (1:1:1). \end{align*}$ Now we want to prove the following equation:
$0 = \begin{vmatrix} -S_BS_C & b^2S_C & c^2S_B \\ 0 & S_C & S_B \\ 1 & 1 & 1 \end{vmatrix}$ $\Leftrightarrow 0 = \begin{vmatrix} -S_BS_C & (S_B+b^2)S_C & (S_C+c^2)S_B \\ 0 & S_C & S_B \\ 1 & 0 & 0 \end{vmatrix}$ $\Leftrightarrow 0 = \begin{vmatrix} (S_B+S_A+S_C)S_C & (S_C+S_A+S_B)S_B \\ S_C & S_B \end{vmatrix}$ which is true. $\blacksquare$

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HamstPan38825

8857 posts

HamstPan38825

#129 h Jan 3, 2025, 4:33 AM

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Let $E = \overline{XX} \cap \overline{MN} \cap \overline{AA}$ be the radical center of $(AMN)$ , $(ABC)$ , and $(B_0C_0X)$ . Let $X'$ be on $\overline{MN}$ so that $\overline{AD}$ and $\overline{AX'}$ are isotomic. Then:

$D, G, X'$ are collinear by homothety at $G$ with ratio $-\frac 12$ ;
$(AX'OXE)$ is cyclic as $\overline{OX'} \perp \overline{MN}$ , and $\sqrt{\frac{bc}2}$ inversion swaps $X$ and $X'$ , so $X', D, X$ collinear.

Thus $D, G, X$ are also collinear.

Remark: I need to stop resolving geometry problems.

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Ihatecombin

58 posts

Ihatecombin

#130 h Feb 21, 2025, 3:28 PM

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We use phantom points, let $X'$ be the intersection of the ray $GD$ with $(ABC)$ . We shall show that $(B_0C_0X')$ is tangent to the circumcircle.
Let $A_0$ be the midpoint of $BC$ , notice that $C_0D = C_0A = A_0B_0$ , thus $C_0DA_0B_0$ is cyclic.
We perform a homothety of scale $-2$ to obtain that $D$ , $G$ and $Q$ (the reflection of $A$ over $BC$ ) are colinear.

We invert $\sqrt{bc}$ , this sends $Q$ to the intersection of the tangent at $A$ with $BC$ . It also sends $D$ to the $A$ antipode.
Finally we perform a homothety of scale $\frac{1}{2}$ . This causes the original $C_0$ to switch with $C$ and $B_0$ to switch with $B$ , and $Q$ goes to the intersection of the tangent at $A$ with segment $B_0C_0$ (call it $Q'$ ),
it also sends $D$ to the circumcenter. Thus $X = QD \cap (ABC)$ is sent to $X' = (AQ'O) \cap B_0C_0$ , we simply need to show that $(X'BC)$ is tangent to $B_0C_0$ . A diagram of this is shown below
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -17.004127728201293, xmax = 24.399765993405712, ymin = -7.426237579785017, ymax = 15.08341516718577; /* image dimensions */ pen ffxfqq = rgb(1,0.4980392156862745,0); /* draw figures */ draw(circle((5,5), 5), linewidth(0.4)); draw((3.45460988321868,9.755183423061052)--(0.6972183465910291,2.4532235977441674), linewidth(0.4)); draw((0.6972183465910291,2.4532235977441674)--(9.302781653408971,2.4532235977441674), linewidth(0.4)); draw((9.302781653408971,2.4532235977441674)--(3.45460988321868,9.755183423061052), linewidth(0.4)); draw((-7.779497912657037,6.104203510402609)--(6.3786957683138255,6.104203510402609), linewidth(0.4)); draw((-7.779497912657037,6.104203510402609)--(3.45460988321868,9.755183423061052), linewidth(0.4)); draw(circle((-1.3897489563285186,5.5521017552013054), 6.413556585311909), linewidth(0.4) + red); draw((0.6972183465910291,2.4532235977441674)--(5,6.104203510402609), linewidth(0.4)); draw((5,6.104203510402609)--(9.302781653408971,2.4532235977441674), linewidth(0.4)); draw(circle((5,1.7432394622107437), 4.360964048191866), linewidth(0.4) + ffxfqq); /* dots and labels */ dot((5,5),linewidth(3pt) + dotstyle); label("$O$", (5.1093155094751745,5.153414340414279), NE * labelscalefactor); dot((3.45460988321868,9.755183423061052),linewidth(3pt) + dotstyle); label("$A$", (3.54924056661558,9.907928451985667), NE * labelscalefactor); dot((0.6972183465910291,2.4532235977441674),linewidth(3pt) + dotstyle); label("$B$", (0.8005370958629621,2.6028156243108787), NE * labelscalefactor); dot((9.302781653408971,2.4532235977441674),linewidth(3pt) + dotstyle); label("$C$", (9.393330828756282,2.6028156243108787), NE * labelscalefactor); dot((2.0759141149048546,6.104203510402609),linewidth(3pt) + dotstyle); label("$C_{0}$", (2.1872703784048233,6.242990490982722), NE * labelscalefactor); dot((6.3786957683138255,6.104203510402609),linewidth(3pt) + dotstyle); label("$B_{0}$", (6.471285697685931,6.242990490982722), NE * labelscalefactor); dot((-7.779497912657037,6.104203510402609),linewidth(3pt) + dotstyle); label("$Q'$", (-7.668441165374834,6.242990490982722), NE * labelscalefactor); dot((5,6.104203510402609),linewidth(3pt) + dotstyle); label("$X'$", (5.1093155094751745,6.242990490982722), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
However this is obvious since
$90 = \angle OAQ' = \angle OX'Q'$ Hence $X'$ lies on the perpendicular bisector of $BC$ , meaning $\triangle X'BC$ is isosceles, taking advantage of the fact that $C_0B_0 \parallel BC$ we have
$\angle C_0X'B = \angle X'BC = \angle X'CB$ so we are done.

This post has been edited 2 times. Last edited by Ihatecombin, Feb 21, 2025, 3:31 PM
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Pekiban

9 posts

Pekiban

#131 h Feb 28, 2025, 9:24 AM

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New solution, relying mostly on harmonic bundles:

Rays $AG, BG, CG$ meet $(ABC)$ for the second time at $A_1, B_1, C_1$ , respectively. Let $A_0$ be the midpoint of $BC$ . Tangents to $(ABC)$ at $A$ and $X$ meet at $T$ . As in all other solutions, let $A'$ be the reflection of $A$ across perpendicular bisector of $BC$ and as it's well known that $D, G, A'$ are collinear, we will just show that $G, A', X$ are collinear.

Claim: $T$ lies on $B_0C_0$
Proof: Note that pairwise radical axes of $(ABC), (AB_0C_0), (XB_0C_0)$ concur. This implies the result as those radical axes are precisely tangent to $(ABC)$ at $A$ , tangent to $(ABC)$ at $X$ and line $B_0C_0$ .

Claim: $T$ lies on $B_1C_1$
Proof: Apply Pascal's theorem to degenerate hexagon $AABB_1C_1C$ . It implies that intersection of tangent to $(ABC)$ at $A$ and $B_1C_1$ , $B_0$ and $C_0$ are collinear. As tangent to $(ABC)$ at $A$ and $B_0C_0$ meet at $T$ , this implies the claim.

Note that above claim implies that $(A, X; B_1, C_1) = -1$

Claim: $(A_1, A'; B, C) = -1$
Proof: We have that:

$\measuredangle BA'A_0 = \measuredangle A_0AC = \measuredangle A_1AC = \measuredangle A_1A'C$ Which implies that $A'A_1$ is a symmedian of $A'BC$ and hence the claim.

Final claim: $X, A', G$ are collinear
Proof: $XG$ meets $(ABC$ ) for the second time at $A''$ . Note that $-1 = (A, X; B_1, C_1) \overset{G}{=} (A_1, A'', B, C)$ . As $(A_1, A', B, C) = -1$ , we conclude that $A' \equiv A''$ , so we are done.

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14.9512cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(12); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.3842, xmax = 30.9938, ymin = -10.4096, ymax = 7.3004; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); draw((7.4538,3.6264)--(4.58,-5.92)--(17.3318,-5.9436)--cycle, linewidth(2.) + zzttqq); /* draw figures */ draw((7.4538,3.6264)--(4.58,-5.92), linewidth(2.) + zzttqq); draw((4.58,-5.92)--(17.3318,-5.9436), linewidth(2.) + zzttqq); draw((17.3318,-5.9436)--(7.4538,3.6264), linewidth(2.) + zzttqq); draw(circle((10.962000621479165,-2.63544809415139), 7.177618905652082), linewidth(2.)); draw(circle((9.198669363255293,-4.492286594963629), 4.616918543577667), linewidth(2.)); draw((6.019399775473924,-7.840148857100443)--(-1.042647039206303,-1.133734761357199), linewidth(2.)); draw((-1.042647039206303,-1.133734761357199)--(16.80641217125435,1.5312132541739771), linewidth(2.)); draw((7.4538,3.6264)--(-1.042647039206303,-1.133734761357199), linewidth(2.)); draw((12.3928,-1.1586)--(-1.042647039206303,-1.133734761357199), linewidth(2.)); draw((17.3318,-5.9436)--(4.155399948837515,-0.35764203612791956), linewidth(2.)); draw((12.329793680426208,-9.681536037306122)--(7.4538,3.6264), linewidth(2.)); draw((4.58,-5.92)--(16.80641217125435,1.5312132541739771), linewidth(2.)); draw((12.329793680426208,-9.681536037306122)--(17.3318,-5.9436), linewidth(2.)); draw((12.329793680426208,-9.681536037306122)--(4.58,-5.92), linewidth(2.)); draw((4.58,-5.92)--(14.493354975263156,3.61337176105207), linewidth(2.)); draw((14.493354975263156,3.61337176105207)--(17.3318,-5.9436), linewidth(2.)); draw((10.9559,-5.9318)--(14.493354975263156,3.61337176105207), linewidth(2.)); draw((12.329793680426208,-9.681536037306122)--(14.493354975263156,3.61337176105207), linewidth(2.)); draw(circle((9.207900310739584,0.4954759529243046), 3.588809452826042), linewidth(2.)); /* dots and labels */ dot((7.4538,3.6264),dotstyle); label("$A$", (7.5418,3.8464), NE * labelscalefactor); dot((4.58,-5.92),dotstyle); label("$B$", (4.6598,-5.7016), NE * labelscalefactor); dot((17.3318,-5.9436),dotstyle); label("$C$", (17.4198,-5.7236), NE * labelscalefactor); dot((6.0169,-1.1468),linewidth(4.pt) + dotstyle); label("$C_0$", (6.1118,-0.9716), NE * labelscalefactor); dot((12.3928,-1.1586),linewidth(4.pt) + dotstyle); label("$B_0$", (12.4698,-0.9936), NE * labelscalefactor); dot((10.9559,-5.9318),linewidth(4.pt) + dotstyle); label("$A_0$", (11.0398,-5.7456), NE * labelscalefactor); dot((9.788533333333334,-2.745733333333334),linewidth(4.pt) + dotstyle); label("$G$", (9.8738,-2.5776), NE * labelscalefactor); dot((4.155399948837515,-0.35764203612791956),linewidth(4.pt) + dotstyle); label("$C_1$", (4.2418,-0.1796), NE * labelscalefactor); dot((12.329793680426208,-9.681536037306122),linewidth(4.pt) + dotstyle); label("$A_1$", (12.4258,-9.5076), NE * labelscalefactor); dot((16.80641217125435,1.5312132541739771),linewidth(4.pt) + dotstyle); label("$B_1$", (16.8918,1.7124), NE * labelscalefactor); dot((14.493354975263156,3.61337176105207),linewidth(4.pt) + dotstyle); label("$A'$", (14.5818,3.7804), NE * labelscalefactor); dot((-1.042647039206303,-1.133734761357199),linewidth(4.pt) + dotstyle); label("$T$", (-0.9502,-0.9496), NE * labelscalefactor); dot((6.019399775473924,-7.840148857100443),linewidth(4.pt) + dotstyle); label("$X$", (6.1118,-7.6596), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$
Remark: Slight generalization of problem is to have $G$ be an arbitrary point on $A$ -median of $ABC$ and $B_0, C_0$ be on $AC/AB$ such that $B_0, C_0, G$ are collinear and that $B_0C_0$ is parallel to $BC$ . My solution still works under these circumstances.

This post has been edited 1 time. Last edited by Pekiban, Feb 28, 2025, 9:26 AM
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balllightning37

389 posts

balllightning37

#132 h Feb 28, 2025, 2:05 PM

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Solved without scannose.

Let $F$ be the intersection of the parallel line at $A$ to $BC$ with $(ABC)$ . We first show that $F$ , $G$ , $D$ are collinear.

Let $A_0$ be the midpoint of $BC$ . The homothety at $G$ with ratio $-\frac12$ takes $A$ to $A_0$ and $(ABC)$ to the nine-point circle of $ABC$ . Thus, $F$ goes to the intersection of the parallel line at $A_0$ to $BC$ (which is just $BC$ ) with the nine-point circle, and that point is $D$ . Thus, $F$ , $G$ , and $D$ are collinear.

Now, let $H$ be the intersection of the tangent at $A$ to $(ABC)$ with $B_0C_0$ . Define $X$ as the intersection of $FG$ with $(ABC)$ . If $A_1$ is the intersection of $AD$ with $(ABC)$ , we get $\angle AHD=2\angle AHB_0=2(90^{\circ}-\angle HAD)=2(90^{\circ}-\angle ACA_1)=2(90^{\circ}-(\angle C+\angle BCA_1))=2(90^{\circ}-(\angle C+90^{\circ}-\angle B))=2(\angle B-\angle C).$ We also have $\angle AXD=\angle AXF=\angle ACF=\angle BCF-\angle C=\angle B-\angle C.$ Thus, $\angle AHD=2\angle AXD$ , and $H$ is the circumcenter of $ADX$ .

This means that $HA=HX$ , but since $HA$ is tangent to $(ABC)$ , $HX$ is also tangent to $(ABC)$ .

Lastly, notice that $(AC_0B_0)$ is tangent to $(ABC)$ at $A$ by homothety. Therefore, $HX^2=HA^2=HC_0\cdot HB_0,$ which implies that $(XB_0C_0)$ is tangent to $HX$ . Since $HX$ is a tangent to $(ABC)$ at $X$ , $(XB_0C_0)$ is tangent to $(ABC)$ . Then, $(XB_0C_0)$ must be $\omega$ , so we are done because we already know that $X$ is collinear with $F$ , $G$ , and $D$ .

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Retemoeg

58 posts

Retemoeg

#133 h Mar 1, 2025, 1:40 PM

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We redefine $X$ as the intersection of $DG$ with $(ABC)$ such that $X$ lies on arc $BC$ not containing $A$ , and we'll instead prove that $(B_0C_0X)$ is tangent to $(ABC)$ .
This certainly made the problem a lot easier. Let $XC_0$ and $XB_0$ intersect $(ABC)$ again at $C_1$ and $B_1$ . Our condition translates to $B_1C_1 \parallel B_0C_0 \parallel BC$ . So we'll show that $BC_1B_1C$ is an isoceles trapezoid, or arcs $BC_1$ and $CB_1$ are equal, meaning $\angle BXC_0 = \angle CXB_0$ . Let's prove this.

Let $DG$ intersect $C_0B_0$ at $J$ , and $(ABC)$ at $X$ . $AD$ intersects $B_0C_0$ at $T$ , construct isoceles trapezoid $ABCN$ . Denote $M$ the midpoint of $BC$ . As $\cfrac{NA}{DM} = 2, AN \parallel DN$ and $A, G, M$ are collinear with $\cfrac{GA}{GM} = 2$ , we deduce that $N, G, D$ are collinear. Thus, $\angle BXJ = \angle BXN = \angle BCN = \angle ABC = 180^{\circ} - \angle BC_0J$ , implying that $BXJC_0$ is cyclic. Similarly, $CXJB_0$ is cyclic. Now, if we let $X, Y$ be orthogonal projections from $B, C$ to $B_0C_0$ , along with the observation that $J$ and $T$ are equidistant from midpoint of segment $B_0C_0$ , one can deduce:
$JX = JC_0 + XC_0 = B_0T + TC_0 = YB_0 + JB_0 = JY$ Henceforth triangles $BJX$ and $CJY$ are congruent, implying $\angle BJC_0 = CJB_0$ . We should be done now:
$\angle BXC_0 = \angle BJC_0 = \angle CJB_0 = \angle CXB_0$ And this concludes the problem.

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lian_the_noob12

173 posts

lian_the_noob12

#134 h Mar 8, 2025, 5:22 PM

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Let $X'$ be the reflection of $A$ over the perpendicular bisector of $BC$ and $M$ be the midpoint of $BC$
From $\textbf{Azerbaijan 2022 Junior National Olympiad}$ $D,G,X'$ are collinear
Denote by $X$ the other intersection of $DG$ with the circle. $MZ \perp B_{0}C_{0}$ , $B_{0}C_{0}$ intersect the circle at $P,Q$
$AD=2MZ \parallel MZ \implies \triangle AGD \sim \triangle MGZ \implies D,G,Z$ collinear
Also $OZ$ is the perpendicular bisector of $PQ$
$-1=(PQ;Z\infty)\overset{X'}{=} (AX,PQ)$ $\implies PQ,$ tangent at $A,X$ are concurrent at $R$
$RX^2=RP \cdot RQ=RA^2=RC_{0} \cdot RB_{0} \blacksquare$

This post has been edited 1 time. Last edited by lian_the_noob12, Mar 8, 2025, 11:57 PM

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Davdav1232

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Davdav1232

#135 h Wednesday at 8:05 PM

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Let $B_0C_0$ intersect the circumcircle of triangle $ABC$ at points $U$ and $V$ . By the Radical Axis Theorem applied to the circles $AB_0C_0$ , $XB_0C_0$ , and $ABC$ , we conclude that there exists a point $Y$ on $B_0C_0$ such that $AY$ and $AX$ are tangent to the circumcircle. This implies that the quadruple $(A, X; U, V)$ is harmonic.

Let line $GD$ intersect $B_0C_0$ at point $M$ , and intersect the circumcircle again at $A'$ , such that $AA'BC$ is a trapezoid (this follows by a homothety with ratio $-2$ from $G$ to $D$ ).

The point $M$ is the midpoint of segment $UV$ , because a homothety centered at $G$ with ratio $-\frac{1}{2}$ maps line $BC$ to line $B_0C_0$ . From known Euler line ratios, this homothety also maps point $D$ to a point on the perpendicular bisector of $BC$ , and thus maps $D$ to the midpoint of $UV$ .

Since $XA$ is a symmedian in triangle $UXV$ , the line $XA'$ must be a median. Therefore, points $X$ , $M$ , and $A'$ are collinear, which implies $X$ , $D$ , and $G$ collinear and we are done.

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