External Homothety center of BOC and Nine point circle

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External Homothety center of BOC and Nine point circle

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Source: STEMS 2021/CAT B/P1

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#1 h Jan 24, 2021, 3:19 PM • 5 Y

Y by Atpar, Severus, shalomrav, microsoft_office_word, GeoKing

An acute angled triangle $\mathcal{T}$ is inscribed in circle $\Omega$ .Denote by $\Gamma$ the nine-point circle of $\mathcal{T}$ .A circle $\omega$ passes through two of the vertices of $\mathcal{T}$ , and centre of $\Omega$ .Prove that the common external tangents of $\Gamma$ and $\omega$ meet on the external bisector of the angle at third vertex of $\mathcal{T}$ .

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#2 h Jan 25, 2021, 12:49 AM • 1 Y

Y by Atpar

Pluto1708 wrote:

An acute angled triangle $\mathcal{T}$ is inscribed in circle $\Omega$ .Denote by $\Gamma$ the nine-point circle of $\mathcal{T}$ .A circle $\omega$ passes through two of the vertices of $\mathcal{T}$ , and centre of $\Omega$ .Prove that the common external tangents of $\Gamma$ and $\omega$ meet on the external bisector of the angle at third vertex of $\mathcal{T}$ .

I will give a very brief outline of my solution as it is basically lenght chasing.
Let $T$ denote the center of homothety taking $\Omega$ to $\omega$ .
We get that the ratio of homothety is $m = \frac{1}{2 cos A}$

Let $L'$ be center of arc $BC$ of $\Omega$ containing $A$ .

Then we calculate $TL' = \dfrac{2Rcos A}{1-2 cos A}$

Now let $H$ be orthocenter .
Then note that first homothety with center $H$ and ratio $2$ and then with center $T$ and ratio $m$ takes $\Gamma$ to $\omega$ .
Let center of this overall homothety be $S$ , then $S$ is the intersection of the direct common tangents of $\Gamma$ and $\omega$ .
Again lenght chasing gives $\frac{HS}{ST} = 1-2 cos A = \frac{AH}{TL'}$
And also we have $AH \parallel TL'$
Therefore $A,M,L'$ are collinear.
Hence proved.

This post has been edited 1 time. Last edited by BOBTHEGR8, Jan 25, 2021, 12:52 AM

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#3 h Jan 25, 2021, 2:51 AM • 2 Y

Y by Atpar, BOBTHEGR8

Let $H$ be orthocentre of $\triangle ABC$ and let $O$ be circumcentre of $\triangle ABC$ . Let midpoint of arc $BAC$ of $\Omega$ be $L$ .

Claim 1: $A$ is exsimilicentre of $\Gamma$ and $\odot (BHC)$ .
Proof: Homothety with centre $A$ and ratio $2$ takes $\Gamma$ to $\odot (BHC)$ . Hence $A$ is exsimilicentre of $\Gamma$ and $\odot (BHC)$ .

Claim 2: $L$ is exsimilicentre of $\odot (BHC)$ and $\omega$ .
Proof: Let $M$ be midpoint of arc $BC$ of $\Omega$ not containing $A$ . Let $M'$ be it's reflection in $BC$ . Then $M'$ lies on $\odot (BHC)$ . Let $LB$ and $LC$ meet $\omega$ again at $B'$ and $C'$ . Then we will show that $\exists$ homothety with centre $L$ taking $\triangle B'OC'$ to $\triangle BM'C$ . All we need to show is that $BM'\parallel B'O$ and $CM'\parallel C'O$ . But $\angle LBM'= 90^{\circ}-\angle BAC=\angle BCO=\angle BB'O=\angle LB'O$ . So $BM'\parallel B'O$ . Similarly $CM'\parallel C'O$ .
Now note that the above homothety maps $\omega$ to $\odot (BHC)$ . Hence $L$ is exsimilicentre of $\omega$ and $\odot (BHC)$ .

Hence by Monge's thm, we obtain that exsimilicentre of $\omega$ and $\Gamma$ lies on $AL$ . Thus the common external tangents of $\Gamma$ and $\omega$ meet on the external bisector of the $\angle BAC$ as desired.
Q.E.D.

This post has been edited 3 times. Last edited by Euler365, Jul 3, 2021, 6:59 AM

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#4 h Jan 25, 2021, 8:46 AM • 1 Y

Y by Atpar

Proposed by Anant Mudgal

Official Solution:
Let $A$ be the third vertex and $O$ be the centre of $\Omega$ . Suppose $BC$ is horizontal.
Let $M_1$ (resp. $M_2$ ) be the top (resp. bottom) point of $\Omega$ , and let $N_1$ (resp. $N_2$ ) be the top (resp. bottom) point of $\Gamma$ . Suppose tangents to $\Omega$ at $B$ and $C$ meet at $T$ . Note that $O,T$ are the top and bottom points of $\omega$ .
$[asy] import cse5; import geometry; import olympiad; size(9cm); pair A=dir(130), B=dir(215), C=dir(325),O=(0,0),M1=dir(90),M2=dir(270),H,G,N1,N2,D,T,P,N3; H=A+B+C; G=H/3; N1=H+0.5*(M1-H); N2=H+0.5*(M2-H); D=(B+C)/2; T=2*circumcenter(B,O,C)-O; N3=extension(N1,N2,A,M1); P=extension(T,N2,O,N1); draw(tangent(P,(O+T)/2,(abs(T)/2),2)--P--tangent(P,(O+T)/2,(abs(T)/2),1),magenta+dotted); draw(circumcircle(B,O,C)^^circumcircle(N1,N2,D)^^circumcircle(A,B,C),green); draw(MP("A",A,N)--MP("B",B,SW)--MP("D",D,SE)--MP("C",C,SE)--A,blue); draw(A--MP("H",H,S)--MP("O",O,SE),heavycyan); draw(MP("M_1",M1,N)--MP("M_2",M2,SE)--MP("T",T,SE)--B^^T--C,heavycyan); draw(M2--MP("N_2",N2,SW)--H--MP("N_1",N1,W)--M1^^N2--MP("N_3",N3,N),heavycyan); draw(T--MP("P",P,W)--O^^P--M1,heavycyan); dot(A^^B^^C^^M1^^M2^^H^^O^^N1^^N2^^D^^T^^P^^N3); [/asy]$

Suppose the two external common tangents of $\Gamma$ and $\Omega$ meet at $P$ . Clearly, it is the centre of positive homothety mapping $\omega$ to $\Gamma$ , so $P=ON_1\cap TN_2$ . Let $N_3=N_1N_2 \cap AM_1$ .

The orthocentre $H$ of $\triangle ABC$ is the centre of homothety with ratio $+2$ sending $\Gamma$ to $\Omega$ , so $H, N_1, M_1$ and $H, N_2, M_2$ are collinear, since $N_i$ map to $M_i$ under the dilation for $i=1,2$ .

The orthocentre $H$ of $\triangle ABC$ is the centre of homothety with ratio $+2$ sending $\Gamma$ to $\Omega$ . Since it sends $N_1$ to $M_1$ , $N_1$ is the midpoint of $HM_1$ . Similarly $N_2$ is the midpoint of $HM_2$ .

Notice that $N_1N_2$ is a diameter of $\Gamma$ hence equals $OM_2$ . Since $N_1N_3 \parallel AH$ and $N_1$ bisects $HM_1$ , we see that $N_1N_3=\tfrac{1}{2}AH=OD$ . Note that $\triangle OBD \sim \triangle OTB$ so $OM_2^2=OD\cdot OT\implies \frac{OM_2}{OT}=\frac{OD}{OM_2}=\frac{N_1N_3}{N_1N_2}.$ Thus $N_2N_1N_3$ is homothetic to $TOM_1$ . So $TN_2$ , $ON_1$ , $M_2N_3$ concur, proving the claim.

This post has been edited 2 times. Last edited by kapilpavase, Jan 25, 2021, 9:07 AM

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#5 h Jan 26, 2021, 1:47 PM • 9 Y

Y by Pluto1708, kapilpavase, Atpar, GeoMetrix, RudraRockstar, centslordm, IAmTheHazard, Prabh2005, mathleticguyyy

Pluto1708 wrote:

An acute-angled triangle $\mathcal{T}$ is inscribed in circle $\Omega$ . Denote by $\Gamma$ the nine-point circle of $\mathcal{T}$ . A circle $\omega$ passes through two of the vertices of $\mathcal{T}$ , and the centre of $\Omega$ . Prove that the common external tangents of $\Gamma$ and $\omega$ meet on the external bisector of the angle at the third vertex of $\mathcal{T}$ .

Mine. A fun anecdote about its conception: before IMO 2020, we decided to come up with an easy geometry problem with as less named points in the statement as possible, and came up with a zero-point geometry problem.

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#6 h Jan 27, 2021, 7:48 AM • 2 Y

Y by Mango247, Mango247

Nice problem!
Solution: Let $H$ be the orthocenter of $ABC$ . Let $\gamma$ be $\odot BHC$ . Let $M$ be the midpt of arc $BAC$ in $\Omega$ . Its easy to see that $A$ is the exsimilicenter of $\Gamma$ and $\gamma$ . We claim that $M$ is the exsimilicenter of $\omega$ and $\gamma$ , after which we will be done by de monge. A one liner is to note that if $H'$ is orthocenter of $MBC$ then $\sqrt{bc}$ inversion in $MBC$ with center $M$ swaps $\odot BOC=\omega$ and $\odot BH'C=\gamma$ , simply bcoz $H'$ and $O$ are isogonal conjugates in $MBC$ .

This post has been edited 3 times. Last edited by kapilpavase, Jan 27, 2021, 7:51 AM

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#7 h Jan 27, 2021, 5:09 PM

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$[asy] size(8cm); defaultpen(fontsize(10pt)+linewidth(0.4)); dotfactor *= 1.5; pair A = dir(130), B = dir(220), C = dir(320), O = origin, H = orthocenter(A,B,C), L = dir(90), N = (H+O)/2, K = circumcenter(B,O,C), P = extension(A,L,K,N), T1 = intersectionpoints(circle(P,sqrt(abs(P-K)*abs(P-K)-abs(K-O)*abs(K-O))),circumcircle(B,O,C))[0], T2 = intersectionpoints(circle(P,sqrt(abs(P-K)*abs(P-K)-abs(K-O)*abs(K-O))),circumcircle(B,O,C))[1], Q = extension(K,N,A,H); draw(A--B--C--A, heavyblue); draw(circumcircle(B,O,C), heavycyan); draw(circle(N,abs(N-foot(A,B,C))), heavycyan); draw(T1--P--T2, heavycyan+dashed); draw(unitcircle, heavyblue); draw(L--K--P--L, purple); draw(A--H--O, purple); draw(N--(A+L)/2, purple+dashed); dot("$A$", A, dir(120)); dot("$B$", B, dir(190)); dot("$C$", C, dir(350)); dot("$K$", K, dir(270)); dot("$O$", O, dir(135)); dot("$L$", L, dir(90)); dot("$P$", P, dir(135)); dot("$H$", H, dir(330)); dot("$N$", N, dir(265)); dot("$Q$", Q, dir(15)); [/asy]$
Let $\triangle ABC = \mathcal{L}$ with $A$ the third vertex. Denote $O$ , $H$ , $N$ to be the circumcenter, orthocenter, and nine-point center of $\mathcal{L}$ , respectively. Let $K$ be the circumcenter of $\triangle BOC$ , $L$ be the midpoint of arc $BAC$ , $Q = \overline{AH} \cap \overline{KN}$ , and $P = \overline{KN} \cap \overline{AL}$ . It suffices to show that $P$ is the exsimilicenter of $\Gamma$ and $\omega$ .

Note that $OKHQ$ is a parallelogram because $N$ is the midpoint of $\overline{OH}$ and $\overline{AH} \parallel \overline{OK}$ . We calculate $\frac{PN}{PK} = \frac{\tfrac 12 (AQ+LK)}{LK} = \frac{(AH-OK) + (R+OK)}{2(R+OK)} = \frac{R+AH}{2(R+OK)}.$ Recall that the inverse of $K$ with respect to $\Omega$ is $O'$ , the reflection of $O$ over $\overline{BC}$ . This means $AH = OO' = R^2/OK$ . Plugging it back gives $\frac{PN}{PK} = \frac{R+\tfrac{R^2}{OK}}{2(R+OK)} = \frac{\tfrac 12 R}{OK},$ which implies the conclusion. $\blacksquare$

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#8 h Jan 6, 2023, 1:38 PM

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Sol:- Let $I,J$ be their exsimilicenter, insimilicenter. So, $(I,J;N_9,O_a)=-1$ . By $\sqrt \frac{bc}{2}$ inversion $G,G'$ and $H,H'$ are mapped to each other so $GH' \parallel N_9O_a \parallel HG'$ . So we have $\frac{N_9J}{JO_a}=\frac{GN_9}{H'O_a}=\frac{AN_9}{AO_a}$ . So, $AJ,AI$ are internal, external angle bisectors respectively. So, we are done

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#9 h Jan 6, 2023, 2:12 PM • 1 Y

Y by GeoKing

Suppose that $\mathcal{T}$ is $\triangle ABC;$ $\Omega'$ is reflection of $\Omega$ in $BC;$ $O, O', O''$ are center of $\Omega, \Omega', \omega;$ $S$ be midpoint of arc $BAC$ . It's easy to prove that $OS^2 = \overline{OO'} \cdot \overline{OO''},$ then $\dfrac{\overline{SO}}{\overline{OO'}} = \dfrac{\overline{OO''}}{\overline{SO}} = \dfrac{\overline{SO''}}{\overline{SO'}}$ or $S$ is external homothetic center of $\Omega'$ and $\omega$ . But $A$ is external homothetic center of $\Gamma$ and $\Omega'$ so by Monge and D'Alembert theorem, we have the external homothetic center of $\omega$ and $\Gamma$ lies on $AS$

This post has been edited 1 time. Last edited by khanhnx, Jan 6, 2023, 2:27 PM

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