Two circumcircles are tangent

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Two circumcircles are tangent

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Source: Kazakhstan National 2019, Problem 6

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#1 h Mar 22, 2019, 9:34 AM • 1 Y

Y by Adventure10

The tangent line $l$ to the circumcircle of an acute triangle $ABC$ intersects the lines $AB, BC$ , and $CA$ at points $C', A'$ and $B'$ , respectively. Let $H$ be the orthocenter of a triangle $ABC$ . On the straight lines A'H, B′H and C'H, respectively, points $A_1, B_1$ and $C_1$ (other than $H$ ) are marked such that $AH = AA_1, BH = BB_1$ and $CH = CC_1$ . Prove that the circumcircles of triangles $ABC$ and $A_1B_1C_1$ are tangent.

This post has been edited 1 time. Last edited by rightways, Mar 22, 2019, 9:34 AM

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#2 h Mar 22, 2019, 10:18 AM • 4 Y

Y by Synthetic_Potato, mijail, DanDumitrescu, Adventure10

Let $H_A, H_B, H_C$ be the reflections of $H$ in sides $BC, CA, AB$ , respectively. Observe that $\angle AA_1H=\angle A'HH_A=\angle A'H_AH$ proving that $A_1$ lies on $\odot(AA'H_A)$ . Invert at $H$ mapping $A$ to $H_A$ ; clearly, $\overline{A'B'C'} \mapsto \odot(A_1B_1C_1)$ , so the tangency follows.

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#3 h Sep 27, 2019, 9:31 AM • 1 Y

Y by Adventure10

Nice problem
Let $l$ be tangent to $\odot ABC$ at $P$ . Let $HP$ meet $\odot ABC$ again in $Q$ . Let $D = \odot ABC \cap AH$ . Let $\omega$ be the circle passing through $H$ and $Q$ tangent to $\odot ABC$ . We shall prove that $\odot A_{1}B_{1}C_{1}$ is $\omega$ .
Angle chasing yields that $AA'A_{1}D$ is cyclic.
So $A'H.HA_{1} = AH.HD = PH.HQ$ .
So $PA'QA_{1}$ is cyclic. So $\angle A'PQ = \angle A'A_{1}Q = \angle HA_{1}Q$ . Let $RQ$ be tangent to $\odot ABC$ such that $R$ and $A'$ lie in the same half plane formed by $PQ$ .
Then $\angle HA_{1}Q = \angle A'PQ = \angle RQP = \angle RQH$ .
This means that $QR$ is tangent to $\odot A_{1}HQ$ also and so $\odot A_{1}HQ$ is tangent to $\odot ABC$ $\implies$ $A_{1}$ lies on $\omega$ . Similarly $B_{1} , C_{1}$ also lie on $\omega$ . So $\odot A_{1}B_{1}C_{1}$ is $\omega$ as desired .
Hence proved. Q.E.D.

This post has been edited 3 times. Last edited by Euler365, Sep 27, 2019, 4:14 PM
Reason: Used wrong notation

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#4 h Sep 27, 2019, 3:36 PM • 1 Y

Y by Adventure10

Nice......

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#5 h Apr 6, 2021, 7:38 PM

Y by

Very nice problem!

Let $\ell$ be tangent to $(ABC)$ at $P$ and let $\omega$ be the $9$ -point circle of $ABC$ . Furthermore, if $A_2$ is the midpoint of $HA'$ , $B_2$ is the midpoint of $HB'$ , $C_2$ is the midpoint of $HC'$ and $D= AH \cap BC$ . Since the homothety centered at $H$ with ratio $\frac{1}{2}$ maps $(ABC)$ to $\omega$ , we have that the line $A_2B_2C_2$ is tangent to $\omega$ at the midpoint $Q$ of $HP$ . Now, observe that since $A_2$ is the midpoint of $HA'$ and $\angle HDA'= 90º$ , then $A_2D=A_2H$ . Moreover, $\angle HA_2D= 180º- 2 \angle A_2H= 180º- 2\angle AHA_1= \angle HAA_1$ , because $AH=AA_1$ . This implies that $AA_1DA_2$ is cyclic. $(\star)$
Similarly, $BB_1EB_2$ and $CC_1FC_2$ are cyclic, where $E= BH \cap AC, F= CH \cap AB$ .

Now, consider the inversion $\Phi$ centered at $H$ with radius $\sqrt{-HA.HD}$ , hence $\Phi(\omega)= (ABC)$ , and from $(\star)$ , we have that $\Phi(A_1)=A_2, \Phi(B_1)=B_2, \Phi(C_1)=C_2$ , and since line $A_2B_2C_2$ is tangent to $\omega$ , we have that its image is tangent to the image of $\omega$ , which is $(ABC)$ . Hence, $(A_1B_1C_1)$ is tangent to $(ABC)$ , as desired.

$\blacksquare$

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#6 h Nov 27, 2022, 10:05 AM • 1 Y

Y by farhad.fritl

Different and long solution.
Let's first define new points. $D,E,F,K$ be feet of perpendiculars from $A,B,C,H$ to $BC,CA,AB,l$ , respectively. $O$ be center of $(ABC)$ and $G$ be tangency point of $l$ to $(ABC)$ . $T=GH\cap (ABC)$ and $O'=HK\cap (ADK)$ . $H_J$ be $JH\cap (ABC)$ for $J\in\{A,B,C\}$ . And let $\omega=(ABC)$ .
It is well known that $DH=DH_A$ and similarly others. Since $O'H\cdot O'K=AH\cdot HD=BH\cdot HE=CH\cdot HF$ , $O'$ lies on $(BEK)$ and $(CFK)$ also.
Claim 1: $H,A_1,B_1,C_1$ all lie on a circle with center $O'$ .
Proof: Let $AO'\cap HA'=X_A$ . Since $\angle HDA'=\angle HKA'=90$ , $(HDKA')$ is cyclic $\implies \angle X_AA'D=\angle HA'D=\angle HKD=\angle O'KD=\angle O'AD=\angle X_AAD \implies AX_ADA'$ is cyclic $\implies \angle AX_AA'=\angle ADA'=90 \implies O'A$ is perpendicular bisector of $A_1H \implies O'H=O'A_1$ . Similarly, we get $O'B_1=O'H=O'C_1$ and the result follows $\square$ .
Claim 2: $T-O'-O$ are collinear.
Proof: Since $O'H\perp l\perp OG$ , we get $O'H\parallel OG$ . So, it is enough to prove that $\frac{TG}{TH}=\frac{OG}{O'H}$ .
Let $\angle OGH=\angle GHK=\alpha, HG=a,OG=b$ and $\cos(\alpha)=c$ . Then we get $Pow(H,\omega)=OG^2-OH^2=OG^2-(HG^2+OG^2-2HG\cdot OG\cdot \cos(\alpha))=2HG\cdot OG\cdot \cos(\alpha)-HG^2$ . Then we have that $\frac{TG}{TH}=1+\frac{HG}{TH}=1+\frac{HG^2}{Pow(H,\omega)}=\frac{2HG\cdot OG\cdot \cos(\alpha)}{Pow(H,\omega)}=\frac{2OG\cdot HG\cdot \cos(\alpha)}{HA\cdot HH_A}={2OG\cdot HK}{2HA\cdot HD}=\frac{OG\cdot HK}{HA\cdot HD}=\frac{OG\cdot HK}{HO'\cdot HK}=\frac{OG}{HO'}$ $\square$ .
Finishing: $\angle O'TH=\angle OTH=\angle OGH=\angle O'HT \implies O'T \implies T,H,A_1,B_1,C_1$ lie on a circle with center $O'$ . Since $T-O'-O$ are collinear, we get that this circle tangents $\omega$ $\blacksquare$ .

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This post has been edited 2 times. Last edited by Jalil_Huseynov, Nov 27, 2022, 7:54 PM

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#7 h Jan 28, 2023, 9:34 PM • 1 Y

Y by Mango247

an older source here by LeVietAn

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