Tangent circles and collinearity

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Source: Problem 6, Brazilian MO 2014

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236 posts

#1 h Nov 3, 2014, 8:52 PM • 4 Y

Y by Davi-8191, PRMOisTheHardestExam, Adventure10, Mango247

Let $ABC$ be a triangle with incenter $I$ and incircle $\omega$ . Circle $\omega_A$ is externally tangent to $\omega$ and tangent to sides $AB$ and $AC$ at $A_1$ and $A_2$ , respectively. Let $r_A$ be the line $A_1A_2$ . Define $r_B$ and $r_C$ in a similar fashion. Lines $r_A$ , $r_B$ and $r_C$ determine a triangle $XYZ$ . Prove that the incenter of $XYZ$ , the circumcenter of $XYZ$ and $I$ are collinear.

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#2 h Nov 3, 2014, 10:26 PM • 6 Y

Y by shinichiman, andria, mathisreal, primesarespecial, Adventure10, Mango247

Let $\omega$ touch $BC,CA,AB$ at $D,E,F.$ Internal common tangents $\ell_A,\ell_B,\ell_C$ of $\omega$ with $\omega_A,\omega_B,\omega_C$ obviously bound a triangle $\triangle D'E'F'$ with incircle $\omega$ homothetic to $\triangle DEF.$ $J \equiv DD' \cap EE' \cap FF'$ is the homothetic center and this homothety carries $\triangle DEF$ with circumcenter $I$ to $\triangle D'E'F'$ with circumcenter $K$ $\Longrightarrow$ $J,I,K$ are collinear. Hence, the circumcenter and incenter of any triangle homothetic to $\triangle D'E'F'$ under a homothety with center $J$ will lie on the fixed line $JIK.$ Indeed, $\triangle XYZ$ is homothethic to $\triangle D'E'F'$ through $J$ (as $r_A,r_B,r_C$ are the reflections of $EF,FD,DE$ across $\ell_A,\ell_B,\ell_C,$ then $X,Y,Z$ are the reflections of $D,E,F$ on $D',E',F',$ respectively).

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#4 h Nov 9, 2014, 6:04 PM • 3 Y

Y by karitoshi, Adventure10, Mango247

Remark: Actually, it is true that $DD'$ is perpendicular on $EF$ , and from there it is easy because it is a well-known result that in any $\triangle ABC$ with $\triangle DEF$ , its tangency triangle, the circumcenter of $\triangle ABC$ , the incenter of $\triangle ABC$ and the orthocenter of $\triangle DEF$ are collinear.

If i will have enough time, i will post a full solution...

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#5 h Jan 22, 2015, 6:52 PM • 2 Y

Y by Adventure10, Mango247

Homothety! That is enough!

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#6 h Jan 24, 2015, 2:23 AM • 4 Y

Y by buratinogigle, baby_shark, Adventure10, Mango247

Let $O,I'$ be circumcenter and incenter of $\triangle XYZ$ , respectively. $\omega$ touches $BC,CA,AB, \omega_A, \omega_B, \omega_C$ at $D,E,F,M,N,P$ , respectively.
Since $\triangle DEF$ and $\triangle XYZ$ are two non-congruent triangles with parallel corresponding sides, therefore there exists unique homothety with center $H$ that maps $\triangle DEF$ to $\triangle XYZ$ . Hence, this homothety sends circumcenter of $\triangle DEF$ to circumcenter of $\triangle XYZ$ (or $H,I,O$ are collinear) and sends incenter $I_1$ of $\triangle DEF$ to incenter $I'$ of $\triangle XYZ$ (or $H,I_1,I'$ are collinear).

Now we will prove $I_1,I,I'$ are collinear.

First, we will prove $XM$ is bisector of $\angle YXZ$ . Indeed, denote by $t'$ the line tangent to $\omega$ parallel to $B_2C_1$ such that $\omega$ lies between $t'$ and $B_2C_1$ . Denote by $L$ the point where $t'$ is tangent to $\omega$ . Homothety with center $P$ that maps $\omega_C$ to $\omega$ sends $B_2C_1$ to $t'$ and hence $B_2$ to $L$ implying that $B_2,P,L$ are collinear. Similarly, we get $L,N,C_1$ are collinear. In other words, $B_2P,C_1N$ intersect at $L \in \omega$ . On the other hand, we also have $B_2P,C_1N$ are the bisector or $\angle XB_2C_1, \angle B_2C_1X$ , respectively. This follows $L$ is the incenter of $\triangle XB_2C_1 \qquad (1)$ .

We have $\begin{aligned} \angle B_2LD & =90^{\circ}- \angle LB_2D=90^{\circ}- \tfrac 12 \angle B_1B_2C \\ & = 90^{\circ}- \tfrac 12 \left( 90^{\circ}- \tfrac 12 \angle ACB \right) \\ & = \tfrac 12 \left( 90^{\circ}+ \tfrac 12 \angle ACB \right) \\ & = \tfrac 12 \angle AIB = \tfrac 12 \angle MIN= \angle MLN. \end{aligned}$
This follows $LM,LD$ are isogonal conjugates wrt $\angle B_2LC_1 \qquad (2)$ .

From $(1)$ and $(2)$ we get $X,M,L$ are collinear or $LM$ is the bisector of $\angle B_2XC_1$ . Therefore, the problem about collinearity of $I_1,I,I'$ can be shown as similar problem: Let $O,I$ be circumcenter and incenter of $\triangle ABC$ , respectively. $AI,BI,CI$ cut $(O)$ at $D,E,F$ , respectively. $DD',EE',FF',AA',BB',CC'$ are diameter of $(O)$ . Prove that $A'D',B'E',C'F'$ are concurrent at $K$ and $I,O,K$ are collinear.
To prove this, note that $AD \parallel D'A'$ , take $K'$ such that $O$ is midpoint of $IK'$ then $K' \in D'A'$ . Similarly, we have $K' \in B'E',C'F'$ .

Back to the main problem, by applying above problem to $\triangle DEF$ , we can easily have $I,I_1,I'$ are collinear.

Thus, we have $\overline{H,I,O}, \overline{H,I',I_1}$ and $\overline{I,I',I_1}$ so $I,O,I'$ are collinear.

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1595 posts

#7 h Mar 15, 2019, 10:57 AM • 1 Y

Y by Adventure10

Here is my solution sketch. No diagram as it's too difficult to draw.

First, I will re-label some points. Let $\omega_A$ touches $AB$ , $AC$ at $A_b$ , $A_c$ . Define $B_a$ , $B_c$ , $C_a$ , $C_b$ similarly. Let $DEF$ be the intouch triangle. Let $T_a$ , $T_b$ , $T_c$ be the contact point between $\omega_A$ , $\omega_B$ , $\omega_C$ and $\omega$ .

Since $\triangle DEF$ and $\triangle XYZ$ are homothetic and $I$ is circumcenter of $\triangle DEF$ , it suffices to show that center of homothety lies on $OI$ -line of $\triangle XYZ$ . In fact, we claim that this point is isogonal conjugate of Nagel point of $\triangle XYZ$ (i.e. point $X_{56}$ ). It's well known that this point lies on $OI$ -line.

By Reim's theorem, quadriltateral with sides parallel to $DE, DF, EF, BC$ are concyclic. In particular, $B_c, C_b, Y, Z$ are concyclic. Thus it suffices to show that $D$ is ex-touch point of $\triangle XB_cC_b$ .

Note by homothety that line $B_cT_b$ and $C_bT_c$ meet at antipode $D'$ of $D$ w.r.t. $\omega$ . We claim that $DD'$ is excenter of $\triangle XB_cC_b$ . Indeed, just note that $T_b$ , $T_c$ are midpoint of arc $DF$ and $DE$ of $\omega$ thus the result easily follows from a simple angle chasing. Finally, note that $DD'\perp B_cC_b$ so we are done.

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#8 h Aug 12, 2020, 2:15 AM

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Denote the center of homothety between the intouch triangle $DEF$ and $XYZ$ as $K$ . Note that it suffices to show that $K$ lies on the $OI$ line of $\triangle DEF$ .

If we let the inradius of $ABC$ be $r$ and the radius of $\omega_A$ be $r_A$ , we have $\frac{r_A-r}{r_A+r}=\sin\frac{A}{2}\implies R=r\frac{\sin\frac{A}{2}+1}{1-\sin\frac{A}{2}}$ . Now, the distance between lines $EF$ and $YZ$ can be computed by $(R+r)\cos^2\frac{A}{2}=\frac{2r}{1-\sin\frac{A}{2}}\cdot\left(1-\sin^2\frac{A}{2}\right)=2r\left(1+\sin\frac{A}{2}\right)$ If we denote the intersection of the $D$ external bisector with $\omega$ as $A'$ , note $\delta(A',BC)=A'I+\delta(I,BC)=r\left(1+\sin\frac{A}{2}\right)$ . Hence, $YZ$ is the reflection of $BC$ over $A'$ .

We can now finish with barycentric coordinates wrt triangle $DEF$ . Relabeling it as $ABC$ , we see that $A'$ lies on the circumcircle $a^2yz+b^2xz+c^2xy=0$ as well as on line $\frac{y}{z}=-\frac{b}{c}$ , so $A'=(a^2,bc-b^2,bc-c^2)$ . Reflecting the midpoint $M=(0,bc-S_a,bc-S_a)$ over $A'$ to $M'$ , we see that $M'$ has homogenized $x$ coordinate of $\frac{2a^2}{2bc-b^2-c^2+a^2}$ . So, $YZ$ has equation $YZ:x=\frac{2a^2}{2bc-b^2-c^2+a^2}=\frac{a^2}{2(s-b)(s-c)}$ We have similar results for $XZ,XY$ , so $K$ has coordinates $(a^2(s-a),b^2(s-b),c^2(s-c))$ , and it suffices to show that $\begin{vmatrix}a&b&c\\a^2(s-a)&b^2(s-b)&c^2(s-c)\\a^2S_a&b^2S_b&c^2S_c\end{vmatrix}=abc\begin{vmatrix}1&1&1\\a(s-a)&b(s-b)&c(s-c)\\aS_a&bS_b&cS_c\end{vmatrix}=0$ Simplifying the right matrix and grouping by $S_a$ , we get the expression $a(b-c)(s-a)S_a+b(c-a)(s-b)S_b+c(a-b)(s-c)S_c=0$ Now, note that the LHS is at most a 5th degree polynomial. At the same time, it is $0$ when any two of the variables are equal to each other or when any of them are $0$ . So, it is divisible by $abc(a-b)(b-c)(c-a)$ , which is a 6th degree polynomial, and hence must be the 0 polynomial, as desired.

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