Geometry tangent circles

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Stefan4024

129 posts

Stefan4024

#1 h Apr 13, 2016, 11:18 AM • 8 Y

Y by fffggghhh, junioragd, HWenslawski, mathematicsy, Adventure10, Mango247, OronSH, Rounak_iitr

Two circles $\omega_1$ and $\omega_2$ , of equal radius intersect at different points $X_1$ and $X_2$ . Consider a circle $\omega$ externally tangent to $\omega_1$ at $T_1$ and internally tangent to $\omega_2$ at point $T_2$ . Prove that lines $X_1T_1$ and $X_2T_2$ intersect at a point lying on $\omega$ .

This post has been edited 2 times. Last edited by djmathman, Sep 12, 2020, 1:59 AM

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Lawasu

212 posts

Lawasu

#2 h Apr 13, 2016, 12:33 PM • 4 Y

Y by CherryMagic, gandhi, HWenslawski, Adventure10

Let $O,O_1$ and $O_2$ be the centers of $\omega , \omega_1$ and $\omega_2$ . Obviously $O_1X_1O_2Y_2$ is a rhombus. In this figure there are two homotheties: $H_1$ , which sends $\omega_1$ to $\omega$ , and $H_2$ , which sends $\omega_2$ to $\omega$ .
Let $Y_1$ and $Y_2$ denote $H_1(X_1)$ and $H_2(X_2)$ . Obviously these two points are situated on the same side of $T_1T_2$ . Due to the homothety properties we have $OY_1\parallel O_1X_1\parallel O_2X_2\parallel OY_2$ . Now we may conclude that $Y_1\equiv Y_2$ .

This post has been edited 3 times. Last edited by Lawasu, Apr 13, 2016, 3:25 PM

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mihaith

133 posts

mihaith

#3 h Apr 13, 2016, 12:49 PM • 2 Y

Y by Adventure10, Mango247

Apply an inversion $\Psi$ with centre $X_1$ , with an arbitrary radius and denote by $\Psi(X)$ the image of $X$ about the inversion $\Psi$ .The fact that the circles $\omega_1$ and $\omega_2$ have the same radius implies that, after inversion, the images $\Psi(\omega_1),\Psi(\omega_2)$ ( which are lines, because these two circles pass through the pole) form an angle for which the image of the radical axis $\Psi(X_1X_2)=X_1\Psi(X_2)$ is an angle bisector.

Now $\Psi(\omega)$ is a circle which is tangent to $\Psi(\omega_1),\Psi(\omega_2)$ and so from $\Psi(X_2)$ are drawn two equal tangents, which are $\Psi(X_2)\Psi(T_2),\Psi(X_2)\Psi(T_1)$ , so the latter are equal.

Proving the required is equivalent to proving that the intersection of $\Psi(X_1T_1)=X_1\Psi(T_1)$ and $\Psi(\omega)$ lies on the circle $\odot(X_1\Psi(X_2)\Psi(T_2))$ .Denote the intersection mentioned above with $L$ .We have $m(\widehat{\Psi(\omega_1),\Psi(\omega_2)})=$ $=$ $m( \widehat{\Psi(X_2) \Psi(T_2) \Psi(T_1)} )+m(\widehat{ \Psi(X_2) \Psi(T_1) \Psi(T_2)})$ , so $\frac{1}{2} \big( m(\widehat{\Psi(\omega_1),\Psi(\omega_2)}) \big )=$ $=$ $\frac{1}{2} \big ( m(\widehat{\Psi(X_2)\Psi(T_2)\Psi(T_1))}+m(\widehat{\Psi(X_2)\Psi(T_1)\Psi(T_2) } \big )$ .
From here we have $m(\widehat{X_1 \Psi(X_2) \Psi(T_2)} )=m(\widehat{X_1 L \Psi(T_2)})$ The problem is solved.

This post has been edited 5 times. Last edited by mihaith, May 9, 2016, 5:48 PM

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anantmudgal09

1980 posts

anantmudgal09

#4 h Apr 13, 2016, 1:19 PM • 2 Y

Y by Adventure10, Mango247

Hmm, so basically by homothety centred about $T_2$ we only need to prove that lines $X_1T_1$ and $XX_2$ are parallel where $X$ is the second intersection of ray $T_2T_1$ with circle $\omega_1$ .

For this, we see that proving $T_2T_1$ bisects $X_1X_2$ is sufficient. (Since, then, rays $X_1T_2$ and $X_2X$ will be symmetric with respect to the midpoint of segment $X_1X_2$ .)

Now, we let $O$ denote the centre of $\omega_2$ and apply inversion about $T_2$ of radius $\sqrt{T_2X_1.T_2X_2}$ followed by reflection in the bisector of angle $X_1T_2X_2$ . Let $T$ be the intersection of the tangents to $\omega_2$ at points $X_1,X_2$ .

Now, it is clear that circle $\omega_1$ is mapped onto the circumcircle of triangle $X_1OX_2$ and circle $\omega$ is mapped to a line tangent to this new circle parallel to $X_1X_2$ . Thus, point $T_2$ being the tangency point, is mapped onto point $T$ and we see that $T_2T$ is a symmedian. Thus, line $T_2T_1$ is a median in triangle $X_1T_2X_2$ .

Our proof is complete now, by the proposition made in the second paragraph.

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v_Enhance

6882 posts

v_Enhance

#5 h Apr 13, 2016, 2:09 PM • 19 Y

Y by anantmudgal09, CherryMagic, mhq, Stens, serquastic, sualehasif996, reveryu, tapir1729, Vrangr, TheCosineLaw, Aryan-23, HamstPan38825, ike.chen, Adventure10, Mango247, Mogmog8, Ritwin, panche, Math-Problem-Solving

Consider the composition of homotheties $\omega_1 \xrightarrow{T_1} \omega \xrightarrow{T_2} \omega_2.$ This is a negative homothety, and since $\omega_1$ and $\omega_2$ have equal radius, it follows that the composition is just reflection about the midpoint of $\overline{X_1X_2}$ . In particular, it sends $X_1$ to $X_2$ , and thus $\overline{X_1T_1} \cap \overline{X_2T_2}$ is just the image of $X_1$ under the first homothety.

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Math_CYCR

431 posts

Math_CYCR

#6 h Apr 13, 2016, 2:11 PM • 2 Y

Y by fffggghhh, Adventure10

Call $O_1, O_2$ and $O$ the centers of $w_1, w_2$ and $w$ respectively.

And $A \equiv X_1T_2 \cap w$
$B \equiv X_2T_2 \cap w$ .

Notice that $AB \parallel X_1X_2$ . Hence $\triangle T_2AB \sim \triangle T_2X_1X_2$ and $AO \parallel X_1O_2$ which implies $\triangle X_2X_1O_2 \sim \triangle BAO$ . Hence:

$\frac{OB}{O_2X_2} = \frac{AB}{X_1X_2} = \frac{T_2B}{T_2X_2}$ .

Hence $OB \parallel O_2X_2$ .

Since $O_1X_1=O_1X_2=O_2X_1=O_2X_2$ we get $O_1X_1O_2X_2$ is a parallelogram. Hence $X_1O_1 \parallel X_2O_2$

Since $O_1, T_1, O$ are collinear and $O_1X_1 \parallel OB$ by Reim's theorem, we get $X_1, T_1$ and $B$ are collinear.

Done!

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gavrilos

233 posts

gavrilos

#7 h Apr 13, 2016, 2:13 PM • 3 Y

Y by mihaith, aidan0626, Adventure10

Hello.

My solution,without transformations.

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Let $O_1,O_2,O$ be the centers of $(\omega_1),(\omega_2),(\omega )$ respectively,and $M$ the midpoint of $O_1O_2$ .Since $(\omega _1),(\omega _2)$ are equal, $M$ lies on $X_1X_2$ and $X_1X_2$ is the perpendicular bisector of $O_1O_2$ .

Lemma: $T_1,T_2,M$ are collinear.

Proof:Let $r$ be the radius of $(\omega)$ and $R$ that of $(\omega_1),(\omega_2)$ .Obviously, $T_1,O_1,O$ are collinear,and $T_2,O_2,)$ are collinear,since $(\omega)$ is tangent to $(\omega_1),(\omega_2)$ .

We have $\frac{T_2O}{T_2O_2}\cdot \frac{O_2M}{O_1M}\cdot \frac{O_1T_1}{OT_1}=\frac{r}{R}\cdot \frac{R}{r}=1$ ,and the converse of Menelaus's theorem gives the desired result.

The theorem of medians gives

$OM^2=\frac{2O_2^2+2O_1O^2-O_1O_2^2}{4}=\frac{2(R+r)^2+2(R-r)^2-4O_1M^2}{4}=R^2+r^2-O_1M^2$ .

Hence, PoP gives $MT_1\cdot MT_2=OM^2-r^2=R^2-O_1M^2$ .Also, $MX_1^2=O_1X^2-O_1M^2=R^2-O_1M^2$ .

Thus $MX_2^2=MT_1\cdot MT_2$ ,which gives that $MX_2$ is tangent to the circumcircle of $\triangle{T_1T_2X_2}$ .

It follows that $\angle{T_1T_2X_2}=\angle{T_1X_2X_1} \ (1)$ .

Let $\overline{yT_1y'}$ be the common tangent of $(\omega),(\omega _1)$ and $S\equiv T_2X_2\cap (\omega)$ .

We have $\angle{X_1X_2T_1}=\angle{X_1T_1y}$ and $\angle{T_1T_2S}=\angle{ST_1y'}$ .

Now $(1)$ gives $\angle{X_1T_1y}=\angle{ST_1y'}\Rightarrow X_1,T_1,S$ collinear.

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Tsarik

27 posts

Tsarik

#8 h Apr 13, 2016, 2:48 PM • 2 Y

Y by microsoft_office_word, Adventure10

Let $P$ be the radical center of $\omega,\omega_1,\omega_2$ . Denote by $T$ the reflection of $T_1$ across line $X_1X_2$ (obviously $T \in \omega_2$ ).
Let $X = X_1T_1 \cap \omega$ , by simple angle chasing we get that $\angle T_1T_2X = \angle X_1T_2T$ so we need to show that $T_1T_2$ and $TT_2$ are isogonal lines in $\triangle X_2T_2X_1$ . Using the fact that $PT$ is tangent to $\omega_2$ , we get that $PT = PT_1 = PT_2$ , so $P$ is the circumcenter of $\triangle TT_1T_2$ , in particular $\angle X_1PT = \angle TT_2T_1$ . Using some more angle chasing and since $\angle TT_2X_2 = \angle TX_1X_2$ , we are done.

This post has been edited 1 time. Last edited by Tsarik, Apr 13, 2016, 3:28 PM

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ABCDE

1963 posts

ABCDE

#9 h Apr 13, 2016, 3:40 PM • 2 Y

Y by Adventure10, Mango247

Consider the problem from the perspective of triangle $T_2X_1X_2$ . By an inversion about $X_1$ , it is easy to show that $T_2T_1$ passes through the midpoint of $X_1X_2$ . Let $T_3$ be the second intersection of $T_1T_2$ and $\omega_1$ . Since the circles are congruent, $X_1T_2X_2T_3$ is a parallelogram. Now, the homothety centered at $T_1$ sending $\omega_1$ to $\omega$ sends $T_3X_2$ to a line parallel to it. $T_3$ goes to $T_2$ , and $X_2$ goes to $X_2T_2\cap \omega$ , so $X_2T_2\cap\omega$ lies on $X_1T_1$ as desired.

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EulerMacaroni

851 posts

EulerMacaroni

#12 h Apr 14, 2016, 1:06 AM • 1 Y

Y by Adventure10

Invert about $X_1$ with the radius of the circles. The problem becomes the following:

Let $\omega \equiv \odot(X_1,R)$ be a circle and let $\ell_1, \ell_2$ be lines intersecting inside $\omega$ such that the internal angle bisector of the lines passes through $X_1$ . Define $\gamma$ is a circle tangent to $\ell_1$ and $\ell_2$ such that the tangency point $T_1$ on $\ell_1$ is on the opposite side of $\ell_2$ as $X_1$ and the tangency point $T_2$ on $\ell_2$ is on the same side of $\ell_1$ as $X_1$ . If $X_2\equiv\ell_1\cap\ell_2$ , then $\odot(X_1X_2T_2)$ and line $X_1T_1$ intersect on $\gamma$ .

To prove this, let $K'\equiv X_1T_1\cap \gamma$ . Then
$\angle X_2T_2K'=\angle T_2T_1K'=\angle T_2T_1X_1=\angle X_2X_1T_1=\angle X_2X_1K'$ as $X_1X_2 \parallel T_1T_2$ since they are clearly both perpendicular to the external angle bisector of $\ell_1,\ell_2$ . Hence, $K\equiv K'$ and we're done.

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kapilpavase

595 posts

kapilpavase

#13 h Apr 14, 2016, 6:51 AM • 3 Y

Y by biomathematics, Adventure10, Mango247

By de monge's theorem, $T_1,T_2$ and midpoint of $X_1X_2,$ (the negative centre of homothety mapping $\omega_1$ to $\omega_2$ )say $M$ are collinear. Now consider the composition of homotheties $\omega_1 \xrightarrow{T_1} \omega \xrightarrow{T_2} \omega_2 \xrightarrow{M} \omega_1$ Since this composition has dilation factor $1$ and $\omega_1$ ie fixed, we have that the entire plane is fixed under this transformation. So if $X_1$ was mapped to a point say $X$ on $\omega$ by $\omega_1 \xrightarrow{T_1} \omega$ then $\omega \xrightarrow{T_2} \omega_2$ take it to $X_2$ , so that after $\omega_2 \xrightarrow{M} \omega_1$ takes it to $X_1$ . So $X$ is the required point lying on $\omega$ . Done

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buratinogigle

2405 posts

buratinogigle

#15 h Apr 14, 2016, 5:57 PM • 6 Y

Y by mihaith, uraharakisuke_hsgs, baopbc, doxuanlong15052000, Adventure10, Mango247

Nice problem, I have seen general problem and extension as following

Problem. Let two circles $(K),(L)$ intersect at $A,B$ . $I$ is insimilicenter of $(K),(L)$ . $BI$ cuts $(K),(L)$ again at $M,Q$ . $AI$ cuts $(K),(L)$ again at $N,R$ . A circle $(\omega)$ is tangent to $(K)$ externally at $S$ and is tangent to $(L)$ internally at $T$ .

a) Prove that $NS$ and $AT$ intersect at $E$ on $(\omega)$ . $MS$ and $BT$ intersect at $F$ on $(\omega)$ . $QT$ and $BS$ intersect at $G$ on $(\omega)$ . $RT$ and $AS$ intersect at $H$ on $(\omega)$ .

b) Prove that $FH,EG$ and $AB$ are concurrent.

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r3mark

255 posts

r3mark

#16 h Apr 14, 2016, 10:59 PM • 4 Y

Y by WinterSecret, Nuterrow, Adventure10, Mango247

Wait, is my solution valid?
My Solution

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MathPanda1

1135 posts

MathPanda1

#17 h Apr 16, 2016, 10:33 PM • 2 Y

Y by Adventure10, Mango247

Let tangents to $\omega$ at $T_1$ and $T_2$ meet at $X$ .
Let $\angle X_1T_2X_2 = \theta, \angle X_1T_1X = \angle X_1X_2T_1 = \beta$ , and $\angle X_1T_2X = \angle X_1X_2T_2 = \gamma$ , where we used some tangency formulas. Hence, $\angle T_1X_2T_2 = \gamma - \beta$
Because of same radii, $\angle X_1T_1X_2 = 180 - \theta$ .
Thus, in quadrilateral $XT_1X_2T_2$ , $180 - \theta + \beta = \angle T_1XT_2 + \gamma + \theta + \gamma - \beta$ i.e. $\angle XT_1T_2 = \angle T_1T_2X = \frac{180 - \angle T_1XT_2}{2} = \theta + \gamma - \beta$ , so $\angle T_1T_2X_2 = \gamma + \theta - (\gamma + \theta - \beta) = \beta$ and $\angle X_1T_1T_2 = \beta + \theta + \gamma - \beta = \theta + \gamma$ i.e. if $P$ is the intersection of lines $X_1T_1$ and $X_2T_2$ , then $\angle T_1PT_2 = \theta + \gamma - \beta = \angle T_1T_2X$ , showing that $XT_2$ is tangent to the circumcircle of $T_1T_2P$ i.e. $P$ lies on $\omega$ .

Does this work? Thanks!

This post has been edited 2 times. Last edited by MathPanda1, Apr 16, 2016, 10:34 PM
Reason: Latex

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codyj

723 posts

codyj

#18 h Apr 17, 2016, 6:36 AM • 8 Y

Y by serquastic, I_m_vanu1996, DrMath, Ankoganit, 62861, AlastorMoody, Aryan-23, Adventure10

Problem authors: me + Charles Leytem

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blueprimes

363 posts

blueprimes

#66 h Mar 25, 2024, 5:25 PM

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Consider the negative homothety $\omega_1 \to \omega$ and the positive homothety $\omega \to \omega_2$ . The composition of these two homotheties is just a negative homothety sending $\omega_1$ to $\omega_2$ , or in other words, a reflection through the midpoint of segment $X_1 X_2$ . Now suppose ray $X_1 T_1$ intersects $\omega$ at $P \ne T_1$ , and ray $T_2 P$ intersects $\omega_2$ at $Q$ . It suffices to show that $Q = X_2$ . But $Q$ is just the composition of the two homotheties stated earlier on $X_1$ , so $Q$ is the reflection of $X_1$ over the midpoint of segment $X_1 X_2$ . But this is just $X_2$ . We may conclude.

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peace09

5419 posts

peace09

#67 h Apr 17, 2024, 1:55 PM

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Oops

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dudade

139 posts

dudade

#68 h Jul 28, 2024, 7:31 PM

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Suppose that $P = X_1T_1 \cap \omega$ . Then,
$\angle O_1X_1T_1 = \angle O_1T_1X_1 = \angle PT_1O = \angle T_1PO.$ Therefore, $O_1X_1 \parallel PO$ . Now, suppose $Q = X_2T_2 \cap \omega$ . Then, a homothety centered at $T_2$ sends $\omega \to \omega_2$ , $O \to O_2$ , and $Q \to X_2$ . Thus, $OQ \parallel X_2O_2$ . Since $O_1X_1 \parallel O_2X_2$ , then $OP \parallel OQ$ which holds if $P = Q$ . Thus, $X_1T_1$ and $X_2T_2$ intersect at a point lying on $\omega$ .

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happypi31415

761 posts

happypi31415

#69 h Aug 25, 2024, 3:22 PM

Y by

umm i never used equal radii ? so where did i go wrong, collinearities are preserved under homothety right?

Let the point of intersection from $X_1$ to $\omega$ be called $A$ . We will prove $X_2, A, T_2$ are collinear. Consider the series of homotheties that will bring $\omega_1$ to $\omega$ and $\omega$ to $\omega_2$ . The first homothety, centered at $T_1$ will bring $X_1$ to $A$ , so $X_1, T_1, A$ are collinear. Then the second homothety, centered at $T_2$ , will bring $A$ to $X_2$ so $T_2$ , $A$ , $X_2$ are collinear, so we are done.

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MrPopo9959

2 posts

MrPopo9959

#70 h Sep 20, 2024, 9:04 PM

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I made a huge mistake by thinking the center of $\omega$ lies on the line connecting the centers of $\omega_{1}$ and $\omega_{2}$ . I realized that mistake after looking at someone else's diagram. Plz anyone check if this is correct

Let $O_{1}$ and $O_{2}$ be the centers of the circles $\omega_{1}$ and $\omega_{2}$ and let $O$ be the center of the rhombus $O_{1}X_{1}O_{2}X_{2}$ . By Monge d'Alembert $O$ , $T_{1}$ and $T_{2}$ are collinear. Let $N=\overline{X_{1}T_{1}} \cap \overline{X_{2}T_{2}}$ , $M=\overline{X_{2}T_{1}} \cap \overline{X_{1}T_{2}}$ and let $\overline{T_{2}T_{1}O}$ intersect $\omega_{2}$ again at $Y$ . Applying Ceva on $\triangle T_{2}X_{1}X_{2}$ gives us $q=\frac{T_{2}M}{MX_{1}}=\frac{T_{2}N}{NX_{2}}$ Now applying Van Aubel's Theorem gives us $\frac{T_{2}T_{1}}{T_{1}Y}=\frac{1}{2}\frac{T_{2}T_{1}}{T_{1}O}=\frac{1}{2}2q=q$ Hence the homothety that maps $T_{2}MN$ to $T_{2}X_{1}X_{2}$ is the same as the homothety that maps $\omega$ to $\omega_{2}$ , therefore $\omega$ is the circumcircle of $T_{2}MN$ as desired.

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fearsum_fyz

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fearsum_fyz

#71 h Oct 17, 2024, 10:37 AM

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Consider the homothety $\mathcal{H}_1$ centered at $T_1$ sending $\omega_1$ to $\omega$ and the homothety $\mathcal{H}_2$ centered at $T_2$ sending $\omega$ to $\omega_2$ . Their composition $\mathcal{H}$ (taking $\omega_1$ to $\omega_2$ ) is a homothety centered at $M$ with ratio $- \frac{r_{\omega}}{\cancel{r_{\omega_1}}} \cdot \frac{\cancel{r_{\omega_2}}}{r_{\omega}} = -1$ .
Let $X$ be the image of $X_1$ under $\mathcal{H}_1$ . Since $\mathcal{H}_1$ is the homothety taking $\omega_1$ to $\omega$ , $X$ must be on $\omega$ . Further, since $T_1$ is the center of $\mathcal{H}_1$ , we have that $X_1$ , $T_1$ , and $X$ are collinear.
Then the image of $X$ in $\mathcal{H}_2$ must be the image $X_2$ of $X_1$ in $\mathcal{H}$ , implying that $T_2, X, X_2$ are collinear.
Hence $X$ is the intersection of $X_1T_1$ and $X_2T_2$ .

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redred123

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redred123

#72 h Dec 30, 2024, 9:11 PM

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Maximilian113

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Maximilian113

#73 h Jan 15, 2025, 12:34 AM

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Let $O_1, O_2, O_3$ be the centers of $w_1, w_2, w$ respectively. Define $A_1=X_1T_1 \cap w, A_2= X_2T_2 \cap w.$ Notice that from the homothety $w_1 \rightarrow w,$ $O_3A_1 \parallel X_1O_1.$ From the homothety $w_2 \rightarrow w,$ $O_3A_2 \parallel O_2X_2.$ But by symmetry $O_1X_1 \parallel O_2X_2$ so it follows that $O_3A_1 \parallel O_3A_2 \implies A_1=A_2.$ QED

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Saucepan_man02

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Saucepan_man02

#74 h Feb 17, 2025, 1:32 AM

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Redefine the problem as:

In $\triangle ABC$ , with orthocenter $H$ , consider a point $T_1 \in (BHC)$ such that there exists a circle $\omega$ tangent to $(ABC)$ at $A$ and $(BHC)$ at $T_1$ . Let $X_1 = BT_1 \cap AC, X_2 = CT_1 \cap AB$ . Prove that: $X_1, X_2 \in \omega$ .

Re-define $T_1$ as $A$ HM point of $\triangle ABC$ . Notice that: for $X_1 = BT_1 \cap AC, X_2 = CT_1 \cap AB$ , $X_1 X_2 \parallel BC$ .
Let $\ell_1, \ell_2$ denote the tangents to $(AX_1 X_2), (BT_1C)$ at point $T_1$ . Since $\angle (\ell_1, X_2 T_1) = \angle X_1 X_1 T_1 = \angle T_1 BC = \angle (\ell_2 , CT_1)$ . Thus, two lines pass though same point ( $T_1$ ) and make same angle with line $CX_2$ , thus $\ell_1 = \ell_2$ or $(AX_1 X_2), (BT_1 C)$ are tangent to each other at $T_1$ , and we are done.

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Avron

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Avron

#75 h Mar 9, 2025, 6:53 PM

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Let the centers of $\omega, \omega_1, \omega_2$ be $O,O_1,O_2$ and let $X_iT_i$ meet $\omega$ at $K_i$ . Then the homothety centered at $T_i$ taking $\omega$ to $\omega_i$ takes $O$ to $O_i$ and $K_i$ to $X_i$ so we get $OK_1||O_1X_1$ and $OK_2||O_2X_2$ but clearly $O_1X_1||O_2X_2$ so $K_1=K_2$ and we're done.

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cj13609517288

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cj13609517288

#76 h Mar 16, 2025, 4:46 PM

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Note that the transformation that is the composition of three homotheties, from a point on $\omega_1$ to a point on $\omega$ (by $T_1$ ) to a point on $\omega_2$ (by $T_2$ ) then back onto a point on $\omega_1$ (by reflecting over the midpoint of $X_1X_2$ ), has scale factor $1$ . Since it preserves $\omega_1$ , it must be the identity. In particular, that means that $X_1T_1\cap\omega$ is the same as $X_2T_2\cap\omega$ . $\blacksquare$

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eg4334

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eg4334

#77 h Apr 10, 2025, 12:47 AM

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Let $X_iT_i$ intersect $\omega$ at $F_i$
The homothety at $T_2$ taking $\omega$ to $\omega_2$ implies $F_2O || O_2X_2$ . Similarly, the negative homothety at $T_1$ implies $F_1O || O_1X_2$ . But $O_1X_1 || X_2O_2$ by symmetry to $F_1 = F_2$ and we are done.

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Ilikeminecraft

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Ilikeminecraft

#78 h Apr 25, 2025, 7:38 AM

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Let $A = \overline{X_1T_1} \cap\overline{X_2T_2}.$ By applying a positive homothety about $T_2$ from $O$ to $O_2,$ we get that $\omega \to \omega_1$ , and $A$ maps to $X_2.$ By applying a $-1$ homothety about $P,$ we see that $\omega_1\to\omega_2$ , and $X_2$ maps to $X_1.$ Finally, by taking a negative $T_1$ homothety from $O_2$ to $O,$ we see that $\omega_2 \to\omega,$ and $X_2$ maps to $X_1.$ Thus, we have that $X_1, T_1, A$ are collinear, which finishes the problem.

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zuat.e

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zuat.e

#79 h Apr 29, 2025, 9:53 PM

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Let $Y=X_1T_1\cap \omega$ and $X_3$ be the intersection of $X_1X_2$ and the line joining the centers of $\omega_1$ and $\omega_2$ . It suffices to show $Y\in X_2T_2$ .

Let $A,B,C$ be the homotheties centered at $X_3, T_2,T_1$ , which send $\omega_1\mapsto \omega_2, \omega\mapsto \omega_2, \omega_1\mapsto \omega$ , respectively, then $B(C(\omega_1))=A(\omega_1)$ :
$C(X_1)=Y$ and $B(Y)=Z$ , which is the intersection of $T_2Y$ with $\omega_1$ , while $A(X_1)=X_2$ , hence $Z=X_2$ , implying $X_2-Y-T_2$ collinear, as desired.

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zuat.e

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zuat.e

#80 h Apr 29, 2025, 9:54 PM

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Let $Y=X_1T_1\cap \omega$ and $X_3$ be the intersection of $X_1X_2$ and the line joining the centers of $\omega_1$ and $\omega_2$ . It suffices to show $Y\in X_2T_2$ .

Let $A,B,C$ be the homotheties centered at $X_3, T_2,T_1$ , which send $\omega_1\mapsto \omega_2, \omega\mapsto \omega_2, \omega_1\mapsto \omega$ , respectively, then $B(C(\omega_1))=A(\omega_1)$ :
$C(X_1)=Y$ and $B(Y)=Z$ , which is the intersection of $T_2Y$ with $\omega_1$ , while $A(X_1)=X_2$ , hence $Z=X_2$ , implying $X_2-Y-T_2$ collinear, as desired.

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