OMG is it mixtilinear again?!

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OMG is it mixtilinear again?!

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Source: 2016 APMC #7

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#1 h Dec 24, 2016, 7:07 PM • 5 Y

Y by aopser123, baopbc, anantmudgal09, ImSh95, Adventure10

Let $\triangle ABC$ be given, it's $A-$ mixtilinear incirlce, $\omega$ , and it's excenter $I_A$ . Let $H$ be the foot of altitude from $A$ to $BC$ , $E$ midpoint of arc $\overarc{BAC}$ and denote by $M$ and $N$ , midpoints of $BC$ and $AH$ , respectively. Suposse that $MN\cap AE=\{ P \}$ and that line $I_AP$ meet $\omega$ at $S$ and $T$ in this order: $I_A-T-S-P$ .
Prove that circumcircle of $\triangle BSC$ and $\omega$ are tangent to each other.
Diagram

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#2 h Dec 25, 2016, 12:53 AM • 9 Y

Y by baopbc, aopser123, gradysocool, thinhrost1, AlastorMoody, ImSh95, appreciatehardwork, Adventure10, Mango247

Notation: Let incircle $(I)$ , touches $BC$ at $D$ , $AI_A$ meet $(ABC)$ again at $K$ , perpendicular to $AI$ at $I$ meet $BC$ at $R$ and let $\Phi$ be inversion WRT $(BIC)$

Lemma 1:
$K$ , $G$ and $R$ are collinear.

Proof:
Since $R$ is on polar of $I$ WRT $\Phi$ , by La Hire, $I$ is on polar of $R$ i.e. $\angle I\Phi(R)K=90^{\circ}\stackrel{ \angle I_A\Psi(G)V=90^{\circ}\Longrightarrow \angle IGK=90^{\circ}}{\Longrightarrow} \text{K, G and R}$ are collinear. $\square$

Lemma 2:
Lines $KR$ and $AD$ concur on $\omega$

Proof:
Consider homothety taking $(ABC)\rightarrow (I)$ to get $\overline{KK}||BC$ and homothety taking $\omega\rightarrow (ABC)$ to get $\overline{TT}|| \overline{KK}|| BC$ , thus the statement follows. $\square$

Lemma 3:
Let $\triangle ABC$ is given, it's orthocenter $H$ and it's orthic triangle $\triangle DEF$ ( $D\in BC$ , $E\in AC$ and $F\in AB$ ).
Let $DK$ be $D-$ altitude in $\triangle ABC$ and let $M$ and $N$ be midpoint of $DK$ and $EF$ , respectively. If $P$ is the midpoint of $BC$ and $MN\cap BC=\{ R \}$ , then $AR$ , $PH$ and $EF$ are concurrent.

Proof:
Bashy proof:
Let's "bu(a)ry" it.

We'll use them WRT $\triangle DEF$ .
Let find equation of $AR$ , first.
Note that $B$ and $C$ are excenters of $\triangle DEF\ . \ . \ . \ \bigstar$
We know that Lemoine point $=(d^2, e^2, f^2)$ of $\triangle DEF$ is on $MN$ , thus equation of that line is:
$\begin{align*} &\underline{\begin{rcases} & \ \ \ \ \ \begin{vmatrix} 0 & \ \ \ \frac12 & \ \ \ \frac 12 \\ \\ d^2 & \ \ \ e^2 & \ \ \ f^2 \\ \\ x & y & z \notag \end{vmatrix}=0\Longleftrightarrow d^2(y-z)-e^2x+f^2x=0 \\ & \bigstar \left\{\begin{array}{rl} B=(d,-e, f) \\ C=(d,e,-f) \end{array}\right\} \Longrightarrow BC\equiv \begin{vmatrix} d & \ \ -e & \ \ f \\ \\ d & \ \ e & \ \ -f \\ \\ x & \ \ y & \ \ z \notag \end{vmatrix}=0\Longleftrightarrow z=-y\frac{f}{e} \end{rcases}}\Longrightarrow \left\{\begin{array}{rl} & y=x\frac{e(e-f)}{d^2}\\ \\ & z=-x\frac{f(e-f)}{d^2} \end{array}\right\}\Longrightarrow R=\left(d^2, e(e-f), -f(e-f)\right) \\ & \stackrel{A=(-d,e,f)}{\Longrightarrow} AR\equiv \begin{vmatrix} d^2 & \ \ \ e(e-f) & \ \ \ -f(e-f) \\ \\ -d & \ \ \ e & \ \ \ f \\ \\ x & y & z \notag \end{vmatrix}=0\\ & \Longleftrightarrow AR\equiv x\cdot 2ef(e-f)-y\cdot df(d+f-e)+z\cdot de(d+e-f) =0\ . \ . \ . \ \bigstar \bigstar \end{align*}$
Now let find $P$ i.e. point of intersection of
$\begin{align*} & \underbrace{\text{(perpendicular bisector of }EF)\equiv d^2(z-y)+x(f^2-e^2)=0 \text{ and } BC\equiv e\cdot z+y\cdot f=0 }_{\Big \Downarrow}\\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P=\left(d^2, e(f-e), f(e-f)\right) \end{align*}$
Now note that $H$ is incenter of $\triangle DEF$ , thus $H=(d,e,f)$ , thus
$\begin{align*} \begin{rcases} P=\left(d^2, e(f-e), f(e-f)\right) \\ H=(d,e,f) \end{rcases}& \Longrightarrow PH\equiv \begin{vmatrix} d^2 & \ \ e(f-e) & \ \ f(e-f) \\ \\ d & \ \ e & \ \ f \\ \\ x & \ \ y & \ \ z \notag \end{vmatrix}=0 \\ & \Longleftrightarrow PH\equiv -y\cdot df(d+f-e)+z\cdot de(d+e-f)=0 \ . \ . \ . \ \bigstar \bigstar \bigstar \end{align*}$
Finally:
$\begin{align*} EF\equiv x=0 & \stackrel{\bigstar \bigstar\text{ and }\bigstar \bigstar \bigstar}{\Longrightarrow} \left. \begin{array}{c} \text{AR $\rightarrow$} \\ \\ \text{PH $\rightarrow$} \\ \\ \text{EF $\rightarrow$} \end{array} \right. \begin{vmatrix} 2ef(e-f) & \ \ -df(d+f-e) & \ \ de(d+e-f) \\ \\ 0 & \ \ -df(d+f-e) & \ \ de(d+e-f) \\ \\ 1 & \ \ 0 & \ \ 0 \notag \end{vmatrix}=0 \end{align*}$
Thus $AR$ , $PH$ and $EF$ are concurrent. $\square$

Synthetic proof:
Using notation as in main problem!
Denote point $F$ as on the beginning of Back to main problem part of solution.
First notice that $ID||EM||AN\ . \ . \ . \ \bullet$
Now note that:
$\begin{align*} \text{collinear} \left\{\begin{array}{rl} & I_AD\cap PM\equiv N\\ & I_AI\cap PE\equiv A\\ & DI\cap ME\equiv \infty \end{array}\right\} \stackrel{\text{Desargues: }\triangle I_ADI\stackrel{=}{\wedge}\triangle PME}{\underset{\bullet}{\Longleftrightarrow}} I_AP\cap DM\cap IE\stackrel{\text{Lemma 3}}{\equiv} \{ F \}\ .\ \blacksquare \end{align*}$
DIAGRAM for lemma 3

Lemma 4:
Points $A$ , $D$ and $T$ are collinear.

Proof:
Will post it tommorow, since I'm very tired right now(and I can't remember on instant how I proved it...)...

Lemma 5: If circle that passes through $BC$ is tangent to $\omega$ at $S^*$ , then $I_A$ , $D_1$ , $I$ , $S^*$ and $R$ are concyclic.

Proof:
By Circle tangent to mixtilinear incircle or by we know that $I_A$ , $D$ , $I$ and $S^*$ are concyclic, but as $\angle I_AD_1R=90^{\circ}=\angle I_AIR$ we are done. $\square$

Lemma 6:
Points $E$ , $I$ and $G$ are collinear and $EI\perp GR$ .

Proof:
Since $\Phi(G)=R$ , obviously, $R$ is on the polar of $I$ and on the polar of $E$ WRT $\Phi$ , thus $EI$ is the polar of $R$ and so $E$ , $I$ and $G$ are collinear and $EI\perp GR$ . $\square$

Lemma 7:
Quadrilateral $IGDR$ is cyclic.

Proof:
$\angle IGR\stackrel{\text{Lemma 6}}{=}90^{\circ}=\angle IDR\ . \ \square$
_____________________________________________________________________________________________
Back to the main problem:
Note first that we are using Lemma 3 on $\triangle ABC$ as orthic triangle WRT $\triangle I_AI_BI_C$ , i.e. we have that $\overline{PSTI_A}$ , $\underbrace{\overline{GIE}}_{\text{Lemma 6}}$ and $BC$ concur at e.g. $F$ .
By Lemma 5 we obviously have to prove that $\angle I_ASR=90^{\circ}\text{ and } (R,F;B,C)=-1$ Why??
Now you're asking, why am I now proving, just that $SI_A$ is bisector of $\angle BSC$ , well, because it's harder way, at least longer(I guess

)...
Note that $E$ is the midpoint of arc $BC$ in $\triangle BGC$ , not containing $G$ , thus $GF$ is bisector of $\angle BGC$ , but as $G$ is exsimilicenter of $\omega$ and $(ABC)$ we conclude that $(GB\cap \omega\equiv B_1\ C_1\equiv GC\cap \omega)|| BC$ , thus as $E$ is midpoint of $\overarc{BC}$ WRT $\triangle BGC$ (not containing $G$ ), we get that $\{V\}\equiv \overline{GIE}\cap \omega$ is midpoint of arc $B_{1}C_{1}$ WRT $\triangle B_1GC_1$ (not containing $G$ ). . . $\spadesuit$ . But also note that since by Lemma 4 and Lemma 6 we know $TG\perp IG$ , we get that line $GT$ is external angle bisector of $\angle BGC$ and so $T$ is the midpoint of arc $B_1GC_1$ WRT $\omega$ ( $\spadesuit \spadesuit$ ), thus from $\spadesuit\Longrightarrow VT\perp BC\perp ID\Longrightarrow VT||ID\ . \ . \ . \ \clubsuit$ So we have to prove:
$\begin{align*} \angle I_ASR=90^{\circ}\stackrel{\text{Lemma 6}}{=}\angle RGI & \stackrel{\text{Lemma 3}}{\Longleftrightarrow} \angle FGR=90^{\circ}=\angle FSR \\ &\Longleftrightarrow \angle GSF=\angle GRF \\ &= \angle GRD\\ & \stackrel{\text{Lemma 7}}{=} \angle GID \\ &\stackrel{\clubsuit}{=} \angle GVT\\ &= \angle GST\\ &\stackrel{\text{Lemma 3}}{=} \angle GSF \end{align*}$ Also we have to prove:
$\begin{align*} -1=S(R,F;B,C)=G(R,F;B,C)\stackrel{\spadesuit\Longrightarrow \angle BGF=\angle CGF}{\Longleftrightarrow} \angle RGF=90^{\circ} \end{align*}$ which is true by Lemma 6. $\blacksquare$

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This post has been edited 4 times. Last edited by mihajlon, Dec 25, 2016, 7:31 PM

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#3 h Dec 25, 2016, 2:38 AM • 11 Y

Y by baopbc, buratinogigle, mihajlon, Hermitianism, thinhrost1, worstsol, enhanced, lminsl, ImSh95, Adventure10, Mango247

Redefine $S$ as the point such that $\odot (BSC)$ is tangent to $\omega$ at $S$ and $P$ as the intersection of $AE,$ $MN.$ Thus we'll prove that $I_A,$ $P,$ $S$ are collinear. Let $I$ be the incenter of $\triangle ABC$ and let $J$ $\in$ $IN$ be the tangency point of $\odot (I_A)$ with $BC.$ Let $V$ $\in$ $EI$ be the tangency point of $\omega$ with the circumcircle of $\triangle ABC$ and let $D$ be the intersection of $BC,$ $EI.$ Since $\frac{I_AA}{I_AI} = \frac{JN}{JI} = \frac{1}{2} \cdot \frac{\text{dist}(A,BC)}{\text{dist}(E,BC)} \cdot \frac{\text{dist}(E,BC)}{\text{dist}(I,BC)} = \frac{PA}{PE} \cdot \frac{DE}{DI} \ ,$ so by Menelaus' theorem for $\triangle AIE$ and $D,$ $P,$ $I_A$ we get $I_A,$ $D,$ $P$ are collinear $\qquad$ $(\ddagger).$

On the other hand, the pole $X$ of $SV$ WRT $\omega$ lies on $BC,$ so notice the circle with center $X$ and radius $XD$ $=$ $XS$ $=$ $XV$ is the V-apollonius circle of $\triangle BVC$ we get $SD$ is the bisector of $\angle BSC$ $\Longrightarrow$ $I_A,$ $D,$ $S$ are collinear (well-known), hence combining $(\ddagger)$ we conclude that $I_A,$ $D,$ $P,$ $S$ are collinear.

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#4 h Dec 25, 2016, 3:53 AM • 4 Y

Y by baopbc, Lin_yangyuan, ImSh95, Adventure10

It has a generalization for cyclic quadrilateral.

Let $ABCD$ be cyclic quadrilateral inscribed in circle $(O)$ . Circle $(K)$ is tangent to segment $PC,PD$ and $(O)$ internally. A circle passes through $CD$ and is tangent to $(K)$ at $X$ . $J$ is $P$ -excenter of triangle $PCD$ . $PH$ is altitude of $PCD$ . $M,N$ are midpoint of $CD,AH$ . $MN$ cuts $JX$ at $Q$ . Prove that $PQ$ bisects major arc $CD$ of $(O)$ .

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#5 h Dec 29, 2016, 10:42 PM • 3 Y

Y by buratinogigle, ImSh95, Adventure10

Nice proof by $\textbf{TelvCohl}$ . So in fact it also proves generalization by $\textbf{buratinogigle}$ .

$\textbf{Proof of generalization :}$

$\textbf{Lemma 1:}$

For any triangle $ABC$ with incenter $I$ and $A$ - excenter $I_a$ . Let $AH$ be perpendicular from $A$ to $BC$ . Denote by $M$ and $N$ , midpoints of $BC$ and $AH$ , respectively. Consider point $W$ such that $WB = WC$ , let $U= WI\cap BC$ , $V = I_aU\cap WA$ . Then point $V$ lies on $MN$ .

$\textbf{Proof:}$

Point $W$ lie on perpendicular bisector of segment $BC$ . So we can vary point $W$ and get points $W_t$ . Easy to check that if we wary point $W_t$ then locus of corresponding points $V_t$ is linear. Now consider $W_0 := M$ , $W_{\infty}$ - infinite point on perpendicular bisector of segment $BC$ . So $V_0 = M$ and $V_{\infty} = N$ and $V_t\in V_0V_{\infty} = MN, \forall t$ . $\Box$

$\textbf{Lemma 2:}$

Consider triangle $ABC$ and circle $\omega$ which is tangent to $AB, AC$ . Let circles $\psi_1, \psi_2$ goes through points $B, C$ and tangent to $\omega$ at $T_1, T_2$ respectively. Let $I, I_a$ be in-center and $A$ - ex-center of $ABC$ . Consider midpoints $W_1, W_2$ of arc $BC$ for circles $\psi_1, \psi_2$ respectively. Then points $W_1, T_1, I$ lies on same line $l_1$ and points $W_2, T_2, I_a$ lies on same line $l_2$ and $l_1\cap l_2\in BC$ . (see attached picture)

$\textbf{Proof:}$

Fact that points $W_1, T_1, I$ lies on same line $l_1$ and points $W_2, T_2, I_a$ lies on same line $l_2$ is well-known. Now lemma 2 follows from next well-known fact : Consider circles $\psi_1, \psi_2$ and let circle $\omega$ tangent to $\psi_1, \psi_2$ at $T_1, T_2$ respectively. Let $W_1, W_2$ midpoints of arcs $BC$ of $\psi_1, \psi_2$ . Then lines $T_1W_1$ , $T_2W_2$ intersects on radical line of $\psi_1, \psi_2$ . (And points $T_1, T_2, W_1, W_2$ lie on same circle) $\Box$

Main problem is combination of lemmas 1, 2. $\Box$

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#7 h Dec 30, 2016, 9:24 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

P.S In fact one can get proof of lemma 2 applying APMC 2016 #4

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DSD

#8 h May 28, 2017, 4:17 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

mihajlon wrote:

Lemma 4:
Points $A$ , $D$ and $T$ are collinear.

Proof:
Will post it tommorow, since I'm very tired right now(and I can't remember on instant how I proved it...)...

??????

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#9 h Jun 3, 2018, 9:41 AM • 4 Y

Y by ImSh95, appreciatehardwork, Adventure10, Mango247

Here is a more human solution. First, rewrite the problem as the following

Quote:

Let $\triangle ABC$ be a triangle with excenter $I_A$ , $A$ -mixtilinear incircle $\omega$ , circumcircle $\Omega$ . Let $H$ be the foot of altitude from $A$ to $BC$ , $L$ be the midpoint of arc $BAC$ . Let $M, N$ be the midpoints of $BC, AH$ , $P=AL\cap MN$ . Finally, let $S$ be the point not on $\Omega$ such that $\odot(BSC)$ is tangent to $\omega$ . Prove that $I_A, S, P$ are colinear.

Let $K$ be the midpoint of arc $BC$ not containing $A$ , $T$ be the tangency point of $\Omega$ and $\omega$ . Let $I$ be the incenter of $\triangle ABC$ .

Step 1: We prove that $A, S, T, I_A$ are concyclic.

Perform an inversion at $A$ , radius $\sqrt{AB\cdot AC}$ followed by reflection across $AI$ and denote image by asterisk. Clearly we have

$\omega^*$ is the $A$ -excircle of $\triangle ABC$ .
$T^*$ is the tangency point of $\omega^*$ and $BC$ .
$\odot(BS^*C)$ is tangent to $\omega^*$ .
We have to prove that $I, S^*, T^*$ are colinear.

But this is extraversion of ISL 2002 G7 so we are done.

Step 2: Let $X=TI\cap BC$ . We prove that $S, I_A, X$ are colinear.

By radical axis, tangents to $\omega$ at $S, T$ intersect on a point $V\in BC$ . Since $TI$ bisects $\angle BTC$ , we get $V$ is the center of $T$ -apollonius circle of $\triangle BTC$ . So $VS=VT=VX$ , which implies
$\angle TSX = 90^{\circ} - \angle VXT = \angle TLK = \angle TAI_A = \angle TSI_A$ so we are done.

Step 3: We conclude the problem.

First, let $D= AI\cap BC, E=AL\cap BC$ and $Y=I_AL\cap BC$ , project
$-1=(A, H; N, {\infty}_{AH})=(A, E; P; L) = (D, E; I_AP\cap BC, Y)$ so it suffices to prove that $(D, E; X, Y)=-1$ or equivalently $\angle BAX=\angle CAY$ .

The quickest way to do this is to let $I_B, I_C$ be the excenters of $\triangle ABC$ . By dual of Desargues Involution Theorem on $BICI_A$ and $L$ , we get an involution swapping $(LA, LA), (LB, LC), (LX, LY)$ , projecting this onto $BC$ gives isogonality w.r.t. $\angle BAC$ hence we are done.

This post has been edited 2 times. Last edited by MarkBcc168, Mar 9, 2020, 1:22 PM

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#10 h Sep 22, 2018, 3:54 AM • 2 Y

Y by ImSh95, Adventure10

We redefine $S$ as satisfying that $(BSC)$ is tangent to $\omega$ and redefine $P$ as $I_AS \cap AE$ , we then apply a $\sqrt{bc}$ inversion to everything. In the inverted diagram we get a triangle $ABC$ with incenter $I$ , circumcircle $\Gamma$ , $A$ -excircle $\Omega$ , $A$ -excenter $I_A$ . Let $D$ and $D_1$ be the projections of $I$ and $I_A$ onto $BC$ respectively, and let $ID_1$ cut $\Omega$ again at $S$ . We claim that $S$ is indeed the inverse of $S$ from the original diagram. It suffices to prove that $(BSC)$ is tangent to $\Omega$ . Let $K$ be the midpoints of $ID_1$ and $D_1L$ respectively. Notice that $I, B, I_A, L, C$ are all on the circle with diameter $II_A$ and thus

$D_1B \cdot D_1C = DI_1 \cdot DL = \frac{D_1I \cdot DS}{2} = D_1K \cdot DS$
So $B, K, C, S$ are concyclic. However, because lines $DI$ and $D_1I_A$ are symmetric about the perpendicular bisector of $BC$ we have $KB = KC$ , so the tangent to $(BKCS)$ at $K$ is parallel to $BC$ . It follows that $S$ is the exsimilicenter of $\Omega$ and $(BKCS)$ , so these circles are tangent at $S$ , and so $S$ indeed corresponds to $S$ from the original diagram.

Now let $M$ and $M_A$ be the midpoints of minor and major arcs $\overarc{BC}$ respectively, and let $M_AD_1$ cut $\Gamma$ again at $X$ . Let $Y$ be the reflection of $D_1$ about $M_A$ and let $AM_A \cap BC = Q$ . Notice that $Y, I, D$ are collinear. Now notice

$DS \cdot DI = 2DL \cdot DI = 2DB \cdot DC = 2DX \cdot DM_A = DY \cdot DX$
So $A, I, X, S$ are concyclic. Notice also that $IY \perp BC$ and $M_AM \perp BC$ , so $IY \parallel MM_A$ and

$\angle IAX = \angle MAX = \angle MM_AX = \angle IYX$
So $I, A, X, S, Y$ are all concyclic. Let $P'$ be the reflection of $Q$ about $M_A$ and notice that $QDP'Y$ is a parallelogram as $M_AQ = M_AP'$ and $M_AD = M_AY$ . Notice then that $P'Y$ is parallel to $BC$ , which is parallel to the tangent to $\Gamma$ at $M_A$ . Therefore by Reim's theorem on circles $(AIXS)$ and $\Gamma$ with lines $AM_A$ and $XM_A$ we have that $A, I, X, S, P', Y$ are all concyclic. Notice now that in the original diagram $I_AS$ maps into the circumcircle of $AIS$ , so we in fact have that $P'$ is the inverse of $P$ . It follows that

$AP' \cdot AP = AB \cdot AC \implies AP = \frac{AB \cdot AC}{AP'} = \frac{AB \cdot AC}{AM_A + M_AP'} = \frac{AB \cdot AC}{AM_A + M_AQ}$
So it suffices to show that $P_0 = MN \cap AE$ from the original diagram satisfies this metric relation, which follows from straightforward calculations.

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#11 h Oct 21, 2018, 6:50 AM • 5 Y

Y by TheDarkPrince, zacchro, ImSh95, Adventure10, Mango247

Problem: Let $ABC$ be a triangle with $A$ -mixtilinear incircle $\Omega$ , $A$ -excenter $J$ , $A$ -altitude $AH$ and let $M$ be the midpoint of $BC$ , $N$ the midpoint of $AH$ and $MN$ meet the external bisector of angle $BAC$ at $P$ . Point $S$ lies on $\Omega$ such that $\odot(BSC)$ is tangent to $\Omega$ . Prove that line $PS$ passes through $J$ .

Solution. Let $M_A$ be the arc midpoint of $BAC$ and $I$ be the incenter of $ABC$ . Let $X=IM_A \cap BC$ and $Y=IM_A \cap \odot(BIC)$ . Since $M_AB, M_AC$ are tangents to $\odot(BIC)$ , we conclude $(YI;XM_A)=-1$ . Project through $J$ onto $AM_A$ ; let $XJ \cap AM_A=P'$ and $AM_A \cap BC=A_1$ , so since $YJ, AM_A, BC$ concur, we have $(AA_1;P'M_A)=-1$ . Project through $M$ onto $AH$ , so $A$ is fixed, $A_1 \mapsto H$ and $M_A \mapsto \infty$ , so $P'$ lies on $MN$ . Thus, $P' \equiv P$ .

We prove $XJ$ passes through $S$ . Let $T_1$ be the touch-point of $\Omega$ with $\odot(ABC)$ ; $D$ be the projection of $I$ on $BC$ , $K$ be the arc midpoint of $BC$ , $T_2$ lie on $\odot(ABC)$ such that $(BC;T_1T_2)=-1$ . Let $T=T_1T_1 \cap T_2T_2$ , clearly $T$ lies on $BC$ . Now by radical axis theorem on $\odot(BSC), \odot(ABC)$ and $\Omega$ , we conclude that $TS$ is tangent to $\Omega$ . In particular, $\odot(T_1ST_2)$ is an Appolonian circle of $BC$ . Consequently, $XS$ bisects angle $BSC$ . Now $Z=AD \cap KT_1$ is the "bottom-point" of $\Omega$ . Thus, $S$ lies on $XZ$ . It suffices to prove $J$ lies on $XZ$ .

Finally, we apply Menelaus's Theorem in $\triangle T_1IK$ for collinearity of $X,Z,J$ to get $\frac{T_1X}{XI} \cdot \frac{IJ}{JK} \cdot \frac{KZ}{ZT_1}=-1 \iff \frac{T_1X}{XI} \cdot 2 \cdot \tfrac{1}{2}\frac{IM_A}{IT_1}=-1 \iff \frac{T_1X}{XI}=\frac{M_AI}{IT_1}$ which is obvious since $(YI;XM_A)=-1 \implies T_1I^2=T_1X \cdot T_1M_A$ and we're done.

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jbaca

#12 h Nov 18, 2018, 5:55 PM • 2 Y

Y by ImSh95, Adventure10

Let $T_A$ be the common point of $\omega$ and $(BAC)$ , and $L$ the point where their external tangent meets $BC$ . Suppose that $S'$ is the second intersection point of the circle centered at $L$ with radius $LT_A$ and $\omega$ . Since $S'L^2 = T_AL^2 = LB\cdot LC$ , $S'L$ is tangent to $(BS'C)$ , so it suffices to argue that $S = S'$ .

Let $K = \overline{ET_A}\cap \overline{ BC}$ and $T$ the symmetric of $K$ wrt. $L$ . It's easy to prove that $T$ lies on $(T_AKS')$ , so this circle has diameter $KT$ . Let $O_A$ be the center of $\omega$ . It's not hard to show that $A$ lies on the $T$ -polar respect to $\omega$ . Furthermore, because $(T,K;B,C) = -1$ , $AK$ coincides with the $T$ -polar; thus, if $R = \overline{AK}\cap \overline{O_AT}$ , we get $\angle TRK = 90^\circ$ and hereby $R$ is on $(T_AKS')$ . Due to the orthogonality of $\omega$ and the latter, we infer that $S'T_ATR$ is harmonic. Therefore,
$(I_A, I;\overline{BC}\cap \overline{AI},A) = -1 = (S', T_A; T,R) \overset{K}{=} (\overline{S'K} \cap \overline{AI}, I;\overline{BC} \cap \overline{AI},A)$ which forces to obtain that $S',\ K$ and $I_A$ are collinear. On the other hand, let $Q =\overline{AE} \cap \overline{BC}$ and see that,
$(\overline{BC}\cap \overline{AI},A; I_A, I) = -1 = (H,A;N, P_{\infty}) \overset{M}{=} (Q,A; P,E) \overset{K}{=} (\overline{BC} \cap \overline{ AI},A;\overline{KP}\cap \overline{AI}, I)$ implying that $I_A = \overline{KP} \cap \overline{AI}$ . In conclusion, $P,\ S',\ K$ and $I_A$ are collinear, hence $S = S'$ .

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MP8148

#13 h Oct 7, 2020, 3:14 AM • 1 Y

Y by ImSh95

diagram

Let $T$ be the $A$ -mixtilinear contact point, $D$ be the antipode of $E$ , $K_1 = \overline{TD} \cap \omega$ , $K_2 = \overline{TE} \cap \omega$ , and redefine $S \ne T$ to be the point on $\omega$ such that $(BSC)$ is tangent to $\omega$ .

By homotheties at $S$ and $T$ , $L = \overline{SK_1} \cap \overline{ED}$ lies on $(BSC)$ with $\overline{K_1K_2} \parallel \overline{EL}$ . It follows that $ESTL$ is cyclic and then by radical axis $\overline{SL}$ , $\overline{ET}$ , and $\overline{BC}$ concur at a point $R$ .

Step 1: Proving $I_A \in \overline{SL}$ .
hidden for length

Step 2: Proving $P \in \overline{SL}$ .
hidden for length

Combining these results finishes the problem.

This post has been edited 2 times. Last edited by MP8148, Oct 12, 2020, 3:33 AM
Reason: typo

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#14 h Jun 30, 2023, 2:51 AM • 1 Y

Y by ImSh95

TelvCohl wrote:

Redefine $S$ as the point such that $\odot (BSC)$ is tangent to $\omega$ at $S$ and $P$ as the intersection of $AE,$ $MN.$ Thus we'll prove that $I_A,$ $P,$ $S$ are collinear. Let $I$ be the incenter of $\triangle ABC$ and let $J$ $\in$ $IN$ be the tangency point of $\odot (I_A)$ with $BC.$ Let $V$ $\in$ $EI$ be the tangency point of $\omega$ with the circumcircle of $\triangle ABC$ and let $D$ be the intersection of $BC,$ $EI.$ Since $\frac{I_AA}{I_AI} = \frac{JN}{JI} = \frac{1}{2} \cdot \frac{\text{dist}(A,BC)}{\text{dist}(E,BC)} \cdot \frac{\text{dist}(E,BC)}{\text{dist}(I,BC)} = \frac{PA}{PE} \cdot \frac{DE}{DI} \ ,$ so by Menelaus' theorem for $\triangle AIE$ and $D,$ $P,$ $I_A$ we get $I_A,$ $D,$ $P$ are collinear $\qquad$ $(\ddagger).$

On the other hand, the pole $X$ of $SV$ WRT $\omega$ lies on $BC,$ so notice the circle with center $X$ and radius $XD$ $=$ $XS$ $=$ $XV$ is the V-apollonius circle of $\triangle BVC$ we get $SD$ is the bisector of $\angle BSC$ $\Longrightarrow$ $I_A,$ $D,$ $S$ are collinear (well-known), hence combining $(\ddagger)$ we conclude that $I_A,$ $D,$ $P,$ $S$ are collinear.

Is this Protassov Problem?

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CyclicISLscelesTrapezoid

372 posts

CyclicISLscelesTrapezoid

#15 h Sep 10, 2023, 5:28 PM • 1 Y

Y by centslordm

~~Solved with~~ Carried by v4913. Let $\infty_\ell$ denote the point at infinity on a line $\ell$ .

Let $D$ be the antipode of $E$ in the circumcircle of $ABC$ , and let $Q$ be the $A$ -mixtilinear intouch point, which lies on $\overline{EI}$ . Redefine $T$ as the intersection of $\overline{I_AP}$ and $\overline{QD}$ . Let $\omega$ intersect $\overline{BC}$ at $X_1$ and $X_2$ .

Claim 1: $T$ is the midpoint of arc $X_1QX_2$ .

Proof: Let $\overline{AE}$ intersect $\overline{BC}$ at $X$ , and let $U$ be the reflection of $I$ over $Q$ . Since $\angle IQD=90^\circ$ , we have $DI=DU$ , so $BICI_AU$ is cyclic with circumcenter $D$ by the incenter excenter lemma. We have $\angle EAI_A=\angle EUI_A=90^\circ$ , so $EAUI_A$ is cyclic. By the radical axis theorem on the circumcircles of $ABC$ , $BICI_AU$ , and $EAUI_A$ , we obtain that $X$ lies on $\overline{I_AU}$ . Since $\overline{DQ}$ is the $I$ -midline of $II_AU$ , we have $\overline{I_AX} \parallel \overline{DQ}$ . Let $\overline{EI_A}$ and $\overline{DQ}$ intersect at $K$ . Then, we have $-1=(A,H;N,\infty_{\overline{AD}})\overset{M}{=}(A,X;P,E)\overset{I_A}{=}(D,\infty_{\overline{DQ}};T,K),$ so $D$ is the midpoint of $\overline{TK}$ . Notice that $DIT \cong DI_AK$ , so $\overline{IT} \parallel \overline{EI_A}$ . We finish proving the claim in two ways.

Proof 1 (v4913): Let $L$ be the center of $\omega$ , and let $\omega$ touch $\overline{AB}$ at $T_1$ . We have $\frac{IL}{LT}=\frac{IL}{LT_1}=\sin \angle IT_1L=\sin \angle BAD=\frac{DB}{DE}=\frac{DI_A}{DE}.$ Also notice that $\angle LIT=\angle DI_AE$ . Since $\angle EDI_A$ is obtuse, we can use SSA similarity to conclude that $LIT \sim DI_AE$ , so $\overline{LT} \parallel \overline{DE}$ . A homothety centered at $Q$ sends the circumcenter of $ABC$ to $L$ and $D$ to $T$ , so $T$ is the midpoint of arc $X_1QX_2$ . $\square$

Proof 2 (me): Let $A_1$ be the $A$ -intouch point of $ABC$ , and let the homothety taking the incircle of $ABC$ to the $A$ -excircle map $A_1$ to $A_1'$ . Let $V$ be the midpoint of $\overline{A_1A_1'}$ . Notice that $V$ lies on $\overline{DE}$ and $\overline{I_AV} \parallel \overline{BC}$ . Thus, $EAVI_A$ is cyclic and its circumcenter is the midpoint of $\overline{EI_A}$ , which we call $O_1$ . Since $\overline{DO_1}$ is the $I_A$ -midline of $I_AIE$ , we have $\overline{DO_1} \parallel \overline{EI}$ , so $\overline{DO_1} \perp \overline{DQ}$ . Since $DT=DK$ , we know that $A$ , $T$ , and $V$ are collinear by the butterfly theorem. Homothety at $A$ shows that $\overline{AA_1}$ passes through the midpoint of arc $X_1QX_2$ , and homothety at $Q$ shows that $\overline{QD}$ passes through the midpoint of arc $X_1QX_2$ , so $T$ must be this arc midpoint. $\square$

Claim 2: $\overline{PI_A}$ , $\overline{EI}$ , and $\overline{BC}$ concur.

Proof: Since the corresponding sides of $EUI_A$ and $IQT$ are parallel, the triangles are homothetic. Notice that $\overline{EI}$ and $\overline{I_AT}$ intersect at their center of homothety $Y$ . We have $\tfrac{YE}{YU}=\tfrac{YI}{YQ} \implies YE \cdot YQ=YI \cdot YU$ , so $Y$ lies on the radical axis of the circumcircles of $ABC$ and $BICU$ , which is $\overline{BC}$ . $\square$

Notice that $\overline{QE}$ passes through the antipode of $T$ in $\omega$ , which we call $T'$ . If $S$ is the second intersection of $\overline{PT}$ with $\omega$ , then $-1=(T,T';X_1,X_2)\overset{Y}{=}(S,Q;X_2,X_1),$ so the tangents to $\omega$ at $S$ and $Q$ intersect at a point $R$ on $\overline{BC}$ . We have $RB \cdot RC=RQ^2=RS^2$ by power of a point, so the circumcircle of $BSC$ is tangent to $\omega$ , as desired. $\square$

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#16 h Mar 27, 2024, 7:47 PM

Y by

Cute!
Sketch:
Redefine $S$ as the point where $(BSC)$ is tangent to mixti incircle, We show that $I_A,D,P,S$ are collinear, To prove $I_A,D,S$ collinear, just consider the pole of $TS$ where $T$ is mixti touchpoint and appolinous finishes, to prove $I_a,D,P$ collinear, just length chase with parallel lines $AH,I_AJ$ where $J$ is excircle touchpoint, and notice it is equivalent to menelaus in $AEI$

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