Reflections of lines through reflections of excenters

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Reflections of lines through reflections of excenters

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cjquines0

510 posts

cjquines0

#1 h Jul 19, 2017, 4:37 PM • 5 Y

Y by rkbish, Adventure10, Mango247, Rounak_iitr, Funcshun840

Let $I$ be the incentre of a non-equilateral triangle $ABC$ , $I_A$ be the $A$ -excentre, $I'_A$ be the reflection of $I_A$ in $BC$ , and $l_A$ be the reflection of line $AI'_A$ in $AI$ . Define points $I_B$ , $I'_B$ and line $l_B$ analogously. Let $P$ be the intersection point of $l_A$ and $l_B$ .

Prove that $P$ lies on line $OI$ where $O$ is the circumcentre of triangle $ABC$ .
Let one of the tangents from $P$ to the incircle of triangle $ABC$ meet the circumcircle at points $X$ and $Y$ . Show that $\angle XIY = 120^{\circ}$ .

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SidVicious

584 posts

SidVicious

#2 h Jul 19, 2017, 4:40 PM • 5 Y

Y by Garfield, rkbish, Adventure10, Mango247, Funcshun840

Sketch for a): $AI_{A}'$ passes through $U-$ antigonal conjugate of $I,$ but since isogonal conjugate of $U$ (say $V$ ) is nothing but inverse of $I$ WRT $\odot(ABC) \implies AV \equiv l_a$ passes through inverse of $I$ hence the conclusion.

This post has been edited 1 time. Last edited by SidVicious, Jul 19, 2017, 4:41 PM

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EulerMacaroni

851 posts

EulerMacaroni

#3 h Jul 19, 2017, 5:07 PM • 4 Y

Y by rmtf1111, rkbish, Adventure10, MS_asdfgzxcvb

Lemma: Let $\odot(A), \odot(B), \odot(C)$ be circles in the plane such that $\odot(C)$ is the inverse of $\odot(B)$ with respect to $\odot(A)$ . Then for every point $P$ , $B(A(P))=A(C(P))$ , where $X(P)$ denotes inversion about $X$ .

Notice that $X_A, I_A'$ are $\sqrt{bc}$ inverses since the circumcircle of triangle $AI_A'I_A$ passes through the reflection of $A$ over $BC$ . Then by the above lemma with $\odot(A)$ $\sqrt{bc}$ inversion, $\odot(B)$ inversion in $\odot(ABC)$ , and $P\equiv I_A$ (notice here that inversion in $\odot(C)$ degenerates to reflection in $BC$ ), we know that $X_A$ is the inverse of $I$ with respect to $\odot(ABC)$ , solving part (a).

Hence it suffices to show for part (b) that $\angle XOY=\tfrac{2\pi}{3}$ since quadrilateral $XYOI$ is cyclic. Compute $d(O,XY)=r\cdot \frac{PO}{PI}=r\cdot \frac{R^2}{R^2-OI^2}=\frac{R}{2}$ as desired.

This post has been edited 1 time. Last edited by EulerMacaroni, Jun 12, 2018, 6:24 AM
Reason: clarify notation

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v_Enhance

6870 posts

v_Enhance

#4 h Jul 19, 2017, 5:31 PM • 12 Y

Y by MathStudent2002, WL0410, mathwizard888, sameer_chahar12, rkbish, mijail, v4913, Modesti, HamstPan38825, Adventure10, Mango247, Funcshun840

The following ingenious elementary solution is due to Wanlin Li. We will rephrase part (a) as follows: in triangle $ABC$ with orthic triangle $DEF$ , if we let $A'$ be the reflection of $A$ across $EF$ then we wish to show that the $D$ -isogonal of $\overline{DA'}$ passes through a common point on the Euler line with the analogous isogonals to $\overline{EB'}$ , $\overline{FC'}$ .

Let $O$ , $H$ be as usual. Actually, it is a well-known result that the circumcircles of $AOD$ , $BOE$ , $COF$ meet at a common point $P$ on line $OH$ . So we will show that $\overline{DA}$ bisects $\angle PDA'$ . Indeed, we note that $DHOA'$ is cyclic (by power of a point from $A$ ) and then observe $\measuredangle ADA'= \measuredangle HDA' = \measuredangle HOA = \measuredangle POA = \measuredangle PDA.$

$[asy] size(10cm); pair A = dir(115); pair B = dir(208); pair C = dir(332); pair D = foot(A, B, C); pair E = foot(B, C, A); pair F = foot(C, A, B); pair H = orthocenter(A, B, C); draw(unitcircle, lightblue); pair O = circumcenter(A, B, C); pair Z = foot(A, E, F); pair Ap = 2*Z-A; draw(A--B--C--cycle, lightblue); draw(D--E--F--cycle, lightblue); pair N = midpoint(H--O); pair M = midpoint(A--H); draw(A--D, lightblue); draw(circumcircle(D, E, F), heavycyan); pair P = -O+2*foot(circumcenter(A, O, D), O, H); draw(circumcircle(A, O, D), dashed+red); draw(circumcircle(Ap, O, D), dashed+red); draw(O--P, heavygreen); draw(P--D--Ap, heavygreen); draw(A--Ap, lightblue); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(250)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$H$", H, dir(315)); dot("$O$", O, dir(350)); dot("$A'$", Ap, dir(Ap)); dot("$N$", N, dir(270)); dot("$M$", M, dir(M)); dot("$P$", P, dir(P)); /* TSQ Source: A = dir 115 B = dir 208 C = dir 332 D = foot A B C R250 E = foot B C A F = foot C A B H = orthocenter A B C R315 unitcircle 0.1 lightcyan / lightblue O = circumcenter A B C R350 Z := foot A E F A' = 2*Z-A A--B--C--cycle 0.1 lightcyan / lightblue D--E--F--cycle lightblue R220 N = midpoint H--O R270 M = midpoint A--H A--D lightblue circumcircle D E F 0.1 lightcyan / heavycyan P = -O+2*foot circumcenter A O D O H circumcircle A O D 0.1 lightred / dashed red circumcircle Ap O D 0.1 lightred / dashed red O--P heavygreen P--D--Ap heavygreen A--Ap lightblue */ [/asy]$

As for (b), by the Poncelet Porism from earlier we just need to show that $H$ is the inverse of $P$ with respect to $(DEF)$ (since then $PH \cdot PN$ is equal to the power). But $\overline{MN} \parallel \overline{AO}$ , so $PDNM$ is cyclic; so inversion swaps $\overline{MD}$ and $(PMND)$ as needed.

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v_Enhance

6870 posts

v_Enhance

#5 h Jul 19, 2017, 5:32 PM • 8 Y

Y by anantmudgal09, e_plus_pi, Cindy.tw, rkbish, v4913, HamstPan38825, Adventure10, Mango247

Here is also a solution with barycentric coordinates.

First, we calculate the barycentric coordinates of $P$ . We have $I_A = (-a:b:c)$ and the foot from $I_A$ to $\overline{BC}$ has coordinates $(0:s-b:s-c)$ and so we deduce $\begin{align*} I_A' &= (a^2 : 4(s-a)(s-b) - ab : 4(s-a)(s-c) - ac) \\ &= \left( a^2 : c^2-a^2-b^2+ab : b^2-a^2-c^2+ac \right). \end{align*}$ By now it is evident that $\overline{AI_A'}$ , $\overline{BI_B'}$ , $\overline{CI_C'}$ concur, and taking the isogonal conjugate we obtain $P = \left( - : - : c^2(2S_C-ab) \right).$ Since $O = (- : - : c^2S_C)$ and $I = (a:b:c)$ it follows already that $P \in \overline{OI}$ .

It remains to show $\angle XIY = 120^{\circ}$ . We will actually show the following lemma holds for any chord $XY$ tangent to the incircle:

Lemma: $\angle XIY = 120^{\circ}$ if and only if $\angle XOY = 120^{\circ}$ if and only if $XOIY$ is cyclic.

Proof. By Poncelet Porism we can draw a point $Z$ on the circumcircle such that $\triangle XYZ$ has the same incircle as $\triangle ABC$ . Then $\angle XIY = 90^{\circ} + \frac{1}{2} \angle Z$ and $\angle XOY = 2 \angle Z$ , and all parts are equivalent to $\angle Z = 60^{\circ}$ . $\blacksquare$

So, we would be done upon showing that $PO \cdot PI = PX \cdot PY.$ One can finish by computing lengths, but a trick due to Anant Mudgal is to instead show that $P$ lies on the radical axis of $(AIO)$ and $(ABC)$ .

If $(AIO)$ has equation $-a^2yz-b^2zx-c^2xy + (x+y+z)(vy+wz) = 0$ , we get $\begin{align*} abc &= bv + cw \\ (abc)^2 \cdot (8K^2) &= (16K^2)\left( b^2S_B v + c^2S_C w \right) \end{align*}$ whence Cramer's rule gives $\begin{align*} -v \div w &= \det \begin{bmatrix} c & abc \\ c^2 S_C & \frac{1}{2} (abc)^2 \end{bmatrix} \div \det \begin{bmatrix} b & abc \\ b^2 S_B & \frac{1}{2} (abc)^2 \end{bmatrix} \\ &= \det \begin{bmatrix} c & 2 \\ c^2 S_C & abc \end{bmatrix} \div \det \begin{bmatrix} b & 2 \\ b^2 S_B & abc \end{bmatrix} \\ &= \left( c^2(2S_C-ab) \right) \div \left( b^2(2S_B-ac) \right) \end{align*}$ Hence the radical axis $vy+wz$ passes through $P$ .

This post has been edited 1 time. Last edited by v_Enhance, Sep 28, 2017, 8:10 PM
Reason: Shorten solution using Anant's trick

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anantmudgal09

1979 posts

anantmudgal09

#6 h Jul 19, 2017, 7:31 PM • 6 Y

Y by Yamcha, Ankoganit, MarkBcc168, rkbish, mijail, Adventure10

cjquines0 wrote:

Prove that $P$ lies on line $OI$ where $O$ is the circumcentre of triangle $ABC$ .
Let one of the tangents from $P$ to the incircle of triangle $ABC$ meet the circumcircle at points $X$ and $Y$ . Show that $\angle XIY = 120^{\circ}$ .

Let $I_A$ be the $A$ -excenter, $P_A$ be the reflection of $I_A$ in $BC$ , $A'$ be the touching point of the $A$ -excircle on $BC$ , $T$ the exsimilicenter of $\odot(I)$ and $\odot(O)$ . Note that $AA', BB', CC'$ concur at the Nagell Point $N$ of triangle $ABC$ and $T, N$ are isogonal conjugates.

Project $-1=(I_A, P_A, A', \infty)$ from $A$ onto $OI$ after reflection in $AI$ . We conclude that if $S=AP_A \cap OI$ then $(O, T, I, S)=-1$ . Evidently, $S$ is the common point of lines $OI, AP_A, BP_B, CP_C$ proving part a.).

Note that $\frac{r}{R-r}=\frac{VI}{IO}=\frac{VS}{SO} \implies OS=\frac{R}{R-2r} \cdot OI=\frac{R^2}{OI},$ since $OI^2=R(R-2Rr)$ where $R, r$ are the circumradius and inradius respectively.

Hence, $\{S, I\}$ are inverses in $\odot(O)$ , consequently, $SX \cdot SY=\text{Pow}(S, \odot(O))=SI \cdot SO,$ hence, $X, I, Y, O$ are concyclic. It suffices to show $\angle XOY=120^{\circ}$ .

To see this, let $d$ be the distance between $O$ and $XY$ ; then, $\frac{d}{r}=\frac{SO}{SI}=\frac{\frac{R^2}{OI}}{\frac{R^2}{OI}-OI}=\frac{R}{2r} \implies d=\frac{R}{2},$ hence, $\angle XOY=120^{\circ}$ as desired. $\blacksquare$

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uraharakisuke_hsgs

365 posts

uraharakisuke_hsgs

#7 h Jul 20, 2017, 7:45 PM • 4 Y

Y by rkbish, Adventure10, Mango247, MS_asdfgzxcvb

My solution
$(a)$ : Let $P_a$ be the inversion image of $I_a'$ from the inversion $I^A_{AB.AC} . R_{AI_a}$ . Let $A'$ reflects with $A$ about $BC$
Because $AA'I_aI_a'$ is an isoceles trapezoid so it's cyclic. Then , through the inversion, we get $P_a$ lies on $OI$ .
Similary we have $P_b,P_c$ also lie on $OI$ . Then, we need to prove $P_a$ is also the inversion image of $I_b'$ through the inversion centre $B$
It's equivalent to prove $\frac{BC}{BI_b'} = \frac{BP}{BA}$ . From above we almost have $\frac{BP}{BA} = \frac{CI_a'}{AI_a'}$ . So, need to prove $\frac{BC}{I_aC} = \frac{BI_b'}{AI_a'}$
We have $\frac{AI_a'}{IO} = \frac{A'I_a}{IO} = \frac{AI_a}{R} \implies AI_a' = \frac{OI.AI_a}{R} \implies \frac{AI_a'}{BI_b'} = \frac{AI_a}{BI_b} = \frac{I_cI_b}{I_cI_a} = \frac{CB}{CI_a}$ as desire
So , $P_a \equiv P_b \equiv P_c \equiv P$ lies on $OI$
$(b)$ From part $(a)$ we have $AI_a.AP = AB.AC$ so $\triangle I_a'AI_a \sim \triangle I_aAP$
Let $AP$ cuts $(O)$ at $G$ , $AI_a'$ cuts $(O)$ at $L$ . Then $\angle OGA = 90 - \angle ALG = \angle AI_a'I_a \implies \triangle OGP \sim \triangle AI_a'I_a \implies \triangle OGP \sim \triangle AIP$ . Then , we have $AI.AO = AG.AP = AX.AY$
So , $XYOI$ is cyclic . Then it's surfice to prove that $\angle XIY = 120$ . Let $M$ be the midpoint of $XY$
We have $\frac{OM}{r} = \frac{OP}{IP} = \frac{OP}{AI_a}.\frac{AI_a}{I_aI_a'}.\frac{I_a'I_a}{IP} = \frac{R}{AI_a'}.\frac{AI_a}{I_aI_a'}.\frac{AI_a'}{AI}= \frac{R}{2r_a}.\frac{r_a}{r} = \frac{R}{2r} \implies OM = \frac{R}{2}$ . Then $\angle XOY = 120 = \angle XIY$

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ABCDE

1963 posts

ABCDE

#8 h Jul 20, 2017, 8:22 PM • 3 Y

Y by tapir1729, rkbish, Adventure10

(a) Let $Q$ be the exsimilicenter between the incircle and circumcircle and $D,E$ be the foot from $A,I_A$ to $BC$ . Note that $A(D,E;I_A,I_A')=-1$ . Reflecting over $AI$ gives $A(O,Q;I,P)=-1$ and similarly we obtain $B(O,Q;I,P)=-1$ , so $P$ lies on line $OIQ$ .

(b) We claim that $I$ and $P$ are inverses about the circumcircle. Let $I'$ be the inverse of $I$ with respect to the circumcircle, so it suffices to show that $(O,Q;I,I')=-1$ . We prove it with the following length chase: Recall from Euler's formula that $OI=\sqrt{R(R-2r)}$ . Also, as $Q$ is the exsimilicenter, we have that $\frac{IQ}{OQ}=\frac rR$ , so $\frac{IO}{OQ}=\frac{R-r}R\implies OQ=\frac{R\sqrt{R(R-2r)}}{R-r}$ so if $Q'$ is the inverse of $Q$ with respect to the circumcircle, $OQ'=\frac{R(R-r)}{\sqrt{R(R-2r)}}$ . We have that $OI'=\frac{R^2}{\sqrt{R(R-2r)}}$ and also that $OI=\frac{R(R-2r)}{\sqrt{R(R-2r)}}$ , so $Q'$ is the midpoint of $II'$ . Inverting about the circumcircle gives $(O,Q;I,I')=-1$ as desired.

Now, by Poncelet's Porism, the tangents to the incircle at $X$ and $Y$ different from $XY$ meet at a point $Z$ on the circumcircle. Note that $O$ and $I$ are the circumcenter and incenter of $XYZ$ . Inverting $XYP$ about the circumcircle gives $XYOI$ cyclic, so $2\angle XZY=\angle XOY=\angle XIY=90^\circ+\frac{\angle XZY}2\implies \angle XZY=60^\circ\implies \angle XIY=120^\circ$ , as deisred.

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navi_09220114

475 posts

navi_09220114

#9 h Jul 26, 2017, 6:01 AM • 3 Y

Y by rkbish, Adventure10, Mango247

a) Let $OI\cap I_aI_a'=X_{40}$ be the Bevan point of $\triangle ABC$ , $M$ be the midpoint of $II_a$ which is also the midpoint of arc $BC$ , $N$ the midpoint of major arc $BAC$ , and $Q$ the inverse point of $I$ with respect to $(ABC)$ . We will prove that $AQ$ and $AI_a'$ are isogonal, which will prove $Q\in \ell_A$ and likewise $\ell_B$ , which will prove $P=Q\in OI$ .

It is well known that $p(I_a,(ABC))=2Rr_a$ . Let us quicky prove this fact again. Let incircle $(I)$ touch $BC$ at $D$ and $DD'$ is diameter in $(I)$ , then if $A$ -excircle touch $BC$ at $E$ then $A, D', E$ are colinear. Let $K$ be a point on $AE$ so that $IK\parallel BC$ , then consider triangle $\triangle D'DE$ , $I$ is midpoint of $DD'$ , hence $K$ must be the midpoint of $D'E$ , so $KE=KD$ . But we also have $BD=CE$ , then $KB=KC$ . In particular, $M, K, N$ colinear. If $X=AM\cap BC, Y=MN\cap BC$ , by Shooting Lemma we have $ANXY$ cyclic, yet $IK\parallel XY$ hence $ANKI$ is cyclic. From here we have $\frac{MI_a}{2R}=\frac{MI}{MN}=\frac{MP}{MA}=\frac{I_aE}{I_aA}=\frac{r_a}{I_aA}\Rightarrow p(I_a,(ABC))=I_aM\cdot I_aA=2Rr_a$ Now $X_{40}I_a=2OM$ and $I_aI_a'=2I_aE=2r_a$ , so $I_aI\cdot I_aA=2I_aM\cdot I_aA=2p(I_a,(ABC))=(2R)(2r_a)=I_aX_{40}\cdot I_aI_a'$ So $AIX_{40}I_a'$ is cyclic. Note that $\angle IQA=\angle OAI=\angle OMA=\angle X_{40}I_aA$ , so $QAX_{40}I_a$ is also cyclic. Then we have $\angle QAI=\angle IX_{40}I_a=\angle IAI_a'$ , so $AQ$ and $AI_a'$ are indeed isogonal.

b) By Poncelet's Theorem, tangents to $(I)$ at $X$ and $Y$ other than line $PXY$ , must intersect at a point on $(ABC)$ , call it $Z$ . Note that $QI\cdot QO=OQ^2-OI\cdot OQ=OQ^2-R^2=p(Q,(ABC))=QX\cdot QY$ Hence $XYOI$ is cyclic. Note that $O, I$ are the circumcenter and incenter for $\triangle XYZ$ as well, so we get $\displaystyle 90+\frac{\angle XZY}{2}=\angle XIY=\angle XOY=2\angle XZY\Rightarrow \angle XZY=60^{\circ}\Rightarrow \angle XIY=120^{\circ}$ . Q.E.D

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math_pi_rate

1218 posts

math_pi_rate

#11 h Jul 15, 2018, 4:01 PM • 3 Y

Y by rkbish, Adventure10, Mango247

Another approach: We start off with the following lemma.

LEMMA Let $I'$ be the inverse of $I$ about $\odot (ABC)$ . Then $I'$ lies on $l'_A$ .

PROOF Let $AI \cap \odot (ABC) = T, I_AI'_A \cap BC = D$ . Also let $H$ be the orthocenter of $\triangle I_ABC$ , and $M$ be the midpoint of $BC$ .

By Fact 5 (also known as Incenter Excenter Lemma), $T$ is the center of $\odot (IBI_AC)$ . From now on, WLOG assume $AB \geq AC$ .

By using homotheties and the fact that the $A$ -intouch point and $D$ are isotomic points on $BC$ , it can easily be shown that $AD \parallel IM$ .

Also, $BI \parallel CH$ and $CI \parallel BH$ $\Rightarrow BHCI$ is a parallelogram $\Rightarrow I, M, H$ are collinear

$\Rightarrow IH \parallel AD \Rightarrow \frac{I_AH}{I_AD} = \frac{I_AI}{I_AA} \Rightarrow \frac{I_AH}{\frac{I_AI_A'}{2}} = \frac{2I_AT}{I_AA} \Rightarrow \frac{I_AH}{I_AI_A'} = \frac{I_AT}{I_AA} \Rightarrow TH \parallel AI'_A$ ( $*$ )

And, $I'$ is the inverse of $I$ about $\odot (ABC) \Rightarrow OI \cdot OI' = OA^2 \Rightarrow OA$ is tangent to $\odot (AII')$

$\Rightarrow \angle OI'A = \angle OAI = \angle OTA \Rightarrow OAI'T$ is cyclic $\Rightarrow \angle I'AI = \angle I'OT$ ( $**$ )

Now, Let $TO \cap \odot (ABC) = T_1 \Rightarrow T_1$ is the inverse of $M$ w.r.t. $\odot (IBI_AC) \Rightarrow TT_1 \cdot TM = TI^2$

It is well known that in any triangle the distance of any vertex from the orthocenter is twice the distance of the circumcenter from the opposite side.

$\Rightarrow I_AH = 2TM \Rightarrow 2TO \cdot \frac{I_AH}{2} = TI \cdot TI_A \Rightarrow \frac{TO}{TI} = \frac{I_AT}{I_AH}$

But, $TO \parallel I_AH \Rightarrow \angle OTI = \angle TI_AH \Rightarrow$ Using the above equality, we get $\triangle OTI \sim \triangle TI_AH$

$\Rightarrow \angle IOT = \angle I_ATH \Rightarrow$ Using ( $*$ ) and ( $**$ ), we get $\angle I'AI = \angle I_AAI'_A \Rightarrow I'$ lies on $l'_A$ $\Box$

Return to the problem at hand. From the symmetry of our LEMMA, we get that $I'$ lies on $l'_B$ also $\Rightarrow P = I'$

PART a) This follows from the fact that $P$ is the inverse of $I$ in $\odot (ABC)$

PART b) $PX \cdot PY = Pow_{\odot (ABC)}P = OP^2 - OA^2 = OP^2 - OP \cdot OI = OP(OP-OI) = PO \cdot PI \Rightarrow XYOI$ is cyclic.

Now, Let the other tangents from $X$ and $Y$ to the incircle meet at a point $K$ . Then by Poncelet's Porism, $K$ lies on $\odot (ABC)$ .

$\Rightarrow I$ is the incenter of $\triangle KXY$ , and $O$ is the circumcenter of $\triangle KXY$ .

$\Rightarrow \angle XIY = 90^{\circ}+\frac{\angle XKY}{2}$ and $\angle XOY = 2\angle XKY$

But, As $XYOI$ is cylic, so $\angle XIY = \angle XOY \Rightarrow 90^{\circ}+\frac{\angle XKY}{2} = 2\angle XKY \Rightarrow \angle XKY = 60^{\circ} \Rightarrow \angle XIY = 120^{\circ}$
$\blacksquare$

REMARK 1: I believe that without part b) the problem is much more difficult, as the condition in part b) triggers the fact that $P$ is the inverse of $I$ w.r.t. $\odot (ABC)$ after which everything goes on in a flow (this happened for me atleast, obviously after I noticed the application of Poncelet's Porism).

REMARK 2: The point $P$ is actually the isogonal conjugate of the Bevan Point of $\triangle ABC$ (the circumcenter of the excentral triangle). This easily follows from our Lemma.

Also, My first G7, so yay :trampoline:

This post has been edited 5 times. Last edited by math_pi_rate, Nov 8, 2018, 4:49 AM

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yayups

1614 posts

yayups

#12 h Nov 8, 2018, 12:12 AM • 7 Y

Y by Wizard_32, rkbish, tapir1729, Inconsistent, Adventure10, Mango247, foolish07

(a) Let $I'$ denote the inverse of $I$ in $(ABC)$ . Let $\phi_{bc}$ denote the $\sqrt{bc}$ inversion map, and note that $\phi_{bc}$ is a Mobius transform. Therefore, $\phi_{bc}(I_A')$ is the inverse of $\phi_{bc}(I_A)=I$ in $\phi_{bc}(BC)=(ABC)$ , so $\phi_{bc}(I_A')=I'$ . Therefore, $AI'$ and $AI_A'$ are isogonal in $\angle A$ , and similarly $BI'$ and $BI_A'$ are isogonal in $\angle B$ . Therefore, $P=I'$ .

(b) Let $Z$ denote the intersection of the tangent to the incircle from $X$ and $(ABC)$ . Poncelet Porisim says that the incircle of $XYZ$ is the incircle of $ABC$ . Now, we have $X,Y,P$ collinear, so inverting about $(ABC)$ , we have $OXYI$ concyclic. Thus, $\angle XOY=\angle XIY$ , so
$2Z = \pi-(X/2-Y/2)=\pi/2+Z/2,$ so $Z=\pi/3$ , so $\angle XIY=2\pi/3$ . $\blacksquare$

This post has been edited 1 time. Last edited by yayups, Nov 8, 2018, 12:19 AM

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RC.

439 posts

RC.

#13 h Nov 8, 2018, 2:00 AM • 3 Y

Y by rkbish, Adventure10, Mango247

Taking the advantage of the Bump, here is something (different?)

cjquines0 wrote:

Prove that $P$ lies on line $OI$ where $O$ is the circumcentre of triangle $ABC$ .
Let one of the tangents from $P$ to the incircle of triangle $ABC$ meet the circumcircle at points $X$ and $Y$ . Show that $\angle XIY = 120^{\circ}$ .

https://ds055uzetaobb.cloudfront.net/uploads/15KZA3rNbs-2016-g8.PNG

Since $I_AG= I_A'G \Rightarrow IE= EF=r$ Let $H$ be the reflection of $F$ in $AI$ then $IH = 2\times r.$
So: $\dfrac{IH}{AO} = \dfrac{2\times r}{R} = \dfrac{\text{dist.(I, XY)}}{OM} = \dfrac{PI}{PO}$ From here we draw two conclusions;

a. $\dfrac{PI}{PO}$ is independent of the choice of vertex. This proves the first part.
b. $OM= R/2$ $\Rightarrow \angle XOY = 120^{\circ}$ . Then by Poncelet's Porism let $\Delta XYZ$ be circumscribed upon $(I)$ , we have $\angle XZY = 60^{\circ} \Rightarrow \angle XIY = 90^{\circ}+ 60^{\circ}/2 = 120^{\circ}.$ $\blacksquare$

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MarkBcc168

1594 posts

MarkBcc168

#16 h Jun 16, 2019, 1:53 PM • 5 Y

Y by yayups, gamerrk1004, Gaussian_cyber, Adventure10, Mango247

Homothety solution.

(a) Let $DEF$ be the intouch triangle of $\triangle ABC$ . Note that the homothety at $A$ which maps $A$ -excircle to the incircle also maps the extouch point to the antipode of $D$ . Thus it maps $I_A'$ to point $X$ such that $IX = 2ID$ .

Let $\gamma = \odot(I,2r)$ . Let $X'$ be the reflection of $X$ across $AI$ . Clearly $X,X'\in\gamma$ . Moreover, it's clear by reflection that the tangent to $\gamma$ at $X$ is parallel to the tangent to $\odot(ABC)$ at $A$ . Thus $AX'$ pass through exsimilicenter of $\gamma$ and $\odot(ABC)$ . So $P$ is that exsimilicenter.

(b) By Poncelet's porism, the other tangents from $X,Y$ to $\odot(I)$ meet at $Z\in\odot(ABC)$ . Let the homothety at $P$ which maps $\odot(ABC)\to\gamma$ sends $X\to X'$ and $Y\to Y'$ . Let $U,V$ be the reflections of $X',Y'$ across $XI$ and $YI$ . Then by similar argument above, $IU\perp YZ$ and $IV\perp YZ$ .

Thus if we let $\triangle X_1Y_1Z_1$ be the intouch triangle. Then $3r = UX_1 = AX_1\cos\angle Z$ . But $AX_1 = r\cot\tfrac{\angle Z}{2}$ . Hence $\cos\angle Z\cot\tfrac{\angle Z}{2}=3$ . By monotonicity, the only answer is $\angle Z=60^{\circ}$ so $\angle XIY=120^{\circ}$ , done.

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AlastorMoody

2125 posts

AlastorMoody

#17 h Jan 2, 2020, 6:39 AM • 5 Y

Y by amar_04, gamerrk1004, mueller.25, rkbish, Adventure10

Solution (My First Pen/Paper G7)

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GeoMetrix

924 posts

GeoMetrix

#18 h Mar 18, 2020, 3:40 PM • 4 Y

Y by mueller.25, amar_04, Aryan-23, sameer_chahar12

Here is a solution found with amar_04,mueller.25and Aryan_23
$[asy] size(10cm); pair A = dir(115); pair B = dir(208); pair C = dir(332); pair D=foot(A,B,C); pair F=foot(C,A,B); pair E=foot(B,A,C); pair K=foot(A,E,F); pair H=orthocenter(A,B,C); pair O=circumcenter(A,B,C); pair A1=2*K-A; pair P = -O+2*foot(circumcenter(A, O, D), O, H); draw(circumcircle(A,O,D),dashed); pair N9=(O+H)/2; draw(A--B--C--A); draw(D--E--F--D); draw(A--O--A1); draw(A--H--D); draw(O--H--P); draw(D--P); draw(circumcircle(O,H,D),dashed); dot("$A$", A, dir(80)); dot("$B$", B, dir(150)); dot("$C$", C, dir(330)); dot("$D$", D, dir(300)); dot("$E$", E, dir(390)); dot("$F$", F, dir(110)); dot("$A'$", A1, dir(60)); dot("$H$", H, dir(120)); dot("$O$", O, dir(40)); dot("$P$", P, dir(150)); dot("$N_9$", N9, dir(110)); [/asy]$

Proof: (for part a) Firstly embed the problem w.r.t to the excentral triangle. We obtain the following problem.

Embeded Problem wrote:

Let $\triangle{DEF}$ bethe orthic triangle of $\triangle{ABC}$ with $D,E,F$ on $\overline{BC},\overline{CA},\overline{AB}$ respectively. Now let $A'$ be the reflection of $A$ in $\overline{EF}$ and let $\ell$ be the line isogonal to $\overline{DA'}$ w.r.t $\angle FDE$ . Then prove that if $\ell \cap \overline{OH}=P$ then $P \in \odot(AOD)$ .

Observe that to prove $(AODP)$ cyclic we just need to show $(HODA')$ cyclic since $\angle PDA=\angle HDA'$ . But this is trivial after an inversion with radius $\sqrt{AH \cdot AD}$ centred at $A$ since under this inversion $O \mapsto A'$ and $H \mapsto D$ . $\square$ .

Now for the second part. Notice that by poncelets porism we just need to show that $(H,P)$ are inverses w.r.t $\odot(DEF)$ . To prove this notice that it is equivalent to showing $N_9H \cdot N_9P={N_9D}^2$ . Now notice that it's well known that $N_9H^2=\frac{9R^2-(a^2+b^2+c^2)}{4}$ where $a,b,c$ are the sides of the triangle and $R$ is the circumradius. Also we have that $\overline{N_9A}=\frac{R}{2}$ . Now we are all set to length chase. Notice that $\overline{N_9H} \cdot \overline{N_9P}={N_9H}^2+N_9H\cdot HP=\frac{9R^2-(a^2+b^2+c^2)}{4}+\frac{AH \cdot HD}{2}$ . We have to show that this expression equals $N_9D^2=\frac{R^2}{4}$ so we are left to show that $8R^2=a^2+b^2+c^2-2\cdot AH\cdot HD$ .Now we'll evaluate the left hand side.Let $A,B,C$ be the angles at vertex $A,B,C$ . Notice that

$a^2+b^2+c^2-2\cdot AH\cdot HD=4R^2(\sin^2{A}+\sin^2{B}+\sin^2{C})-2\cdot(2R \cos{A})(2R\cos{B}\cos{C})=4R^2(\sin^2{A}+\sin^2{B}+\sin^2{C}-2\cos{A}\cos{B}\cos{C})=8R^2$

where the last part follows from the well knwon identity that $\sin^2A+\sin^2{B}+\sin^2{C}=2+2\cos{A} \cos{B} \cos{C}$ . Hence we are done $\blacksquare$ .

This post has been edited 3 times. Last edited by GeoMetrix, Mar 18, 2020, 4:56 PM

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Flip99

82 posts

Flip99

#19 h Apr 21, 2020, 9:27 AM • 1 Y

Y by rkbish

SidVicious wrote:

How can we prove that $AI_{A}'$ passes through the antigonal conjugate of $I$
?

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Flip99

82 posts

Flip99

#23 h Apr 22, 2020, 6:56 PM • 3 Y

Y by rkbish, Mango247, Mango247

Anyone...

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Flip99

82 posts

Flip99

#24 h Apr 23, 2020, 3:28 PM • 1 Y

Y by rkbish

Anyone...

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algebra_star1234

2467 posts

algebra_star1234

#25 h Jun 13, 2020, 11:21 PM • 3 Y

Y by rkbish, Mango247, Mango247

a) Note that $I_A = (-a:b:c) = (-2a^2:2ab:2ac)$ . Let $D = (0:s-b:s-c)$ be the foot of the altitude from $I_A$ to $BC$ . Note that $D = (0: c^2-(a-b)^2:b^2-(a-c)^2)$ . Then,
$I_A'= 2D - I = (2a^2: 2c^2-2a^2-2b^2+2ab: 2b^2-2a^2-2c^2+2ac)$ $I_A' = (-a^2: 2S_C- ab: 2S_B - ac) .$ Note that clearly $AI_A'$ and $BI_B'$ will meet at
$P'= \left( \frac{1}{2S_A-bc}, \frac{1}{2S_B-ac}, \frac{1}{2S_C-ab} \right)$ The isogonal conjugate of this point has coordinates
$P = (a^2(S_A-bc), b^2(S_B-ac), c^2(S_C - ab) ).$ Since $O = (a^2S_A: b^2S_B: c^2S_C)$ and $I = (a:b:c)$ , we clearly see that $P$ is a linear combination of $O$ and $I$ , so $P$ is on $OI$ .
b) By Poncelet porism, we can find a point $Z$ such that $XZ$ and $YZ$ are tangent to the incircle. Note that $\angle XOY = 2 \angle XYZ$ and $\angle XIY = 90^\circ + \frac12 \angle XYZ$ . When $\angle XOY = \angle XIY$ , we have $\angle XYZ = 60^{\circ}$ and $\angle XIY = 120^{\circ}$ . Therefore, it suffices to show $XOIY$ is cyclic, or that $PO \cdot PI = PX \cdot PY$ . In other words, we just have to show that $P$ is on the radical axis of $(AOI)$ and $(ABC)$ . To find the radical axis, we just have to find the equation of the $(AOI)$ . This is
$-a^2 yz - b^2xz -c^2xy+(x+y+z)(vy+wz)=0$ To find $v$ and $w$ , we plug in $O = (a^2S_A: b^2S_B: c^2S_C)$ and $I = (a:b:c)$ . The equations we obtain are
$-abc(a+b+c) + (a+b+c)(bv+cw) = 0,$ $-a^2b^2c^2 (S_AS_B+S_BS_C + S_CS_A) + (a^2S_A+b^2S_B+c^2S_C) (b^2S_Bv+c^2S_Cw) = 0.$ Note that $S_B + S_C = a^2$ , so we can find that $(a^2S_A+b^2S_B+c^2S_C) = 2(S_AS_B+S_BS_C + S_CS_A)$ . Thus, we can simplify the equations to
$bv+cw = abc, b^2S_Bv + c^2S_Cw = \frac{a^2b^2c^2}{2} .$ The solution to this system is
$w = \frac{ \frac{ab^2}{2} (2S_B - ac) }{bS_B - cS_C}, \qquad v = - \frac{ \frac{ac^2}{2} (2S_C - ab)}{bS_B -cS_C} .$ The equation of the radical axis is $vy+wz = 0$ , and we see that $P$ is clearly on this line, so we are done.

This post has been edited 1 time. Last edited by algebra_star1234, Jun 13, 2020, 11:22 PM

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magicarrow

146 posts

magicarrow

#26 h Oct 12, 2020, 11:23 PM • 2 Y

Y by rkbish, centslordm

Here's a relatively simple elementary solution (a little length computation at the end, but it's really easy):

(1) Set $D$ = foot of angle bisector from $A$ to $BC$ .
(2) Let $X_0$ = second intersection of circle with center $I$ and radius $IL$ and $(ABC)$ ; then $X_0$ is reflection of $L$ over $IO$
(3) Thus $\angle IAO = \angle ILO = \angle IX_0O$ and hence $IAX_0O$ is cyclic.
(4) Let $LO$ intersect circle w/ center $D$ and radius $DI_A$ again at $S$ . Then $\angle ISO=\angle IAO=\angle ILO$ hence $ISL$ is isosceles.
(5) Let reflection of $I_A$ in $BC$ be $T$ , then $ISL$ and $DTI_A$ are homothetic with center $A$ ; hence $A,S,T$ are collinear.
(6) Since $ASIX_0$ is cyclic and $IS=IX_0$ , we see that $\angle SAI=\angle X_0AI$ and $\angle TAI=\angle XAI$ , $X_0,X,A$ are collinear.
(7) Hence the desired concurrency point $P$ is the radical axis of the circumcircles of $AOI$ , $BOI$ , and $COI$ . Furthermore, remark $PI \cdot PO = PX_0 \cdot PA = PO^2-R^2$ and so $PO \cdot OI = R^2$ and $OI = \sqrt{R^2-2Rr}$ , so the remainder of the problem is a (very) easy computation. [EDIT: reading above solutions, for the computation part I forgot to write the important step that $XYOI$ is cyclic but this is clear from PoP with respect to point $P$ ].

This post has been edited 2 times. Last edited by magicarrow, Oct 13, 2020, 12:08 AM

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Idio-logy

206 posts

Idio-logy

#27 h Nov 20, 2020, 9:13 AM • 2 Y

Y by rkbish, Mango247

Here is a hybrid solution.

For part (a), we use barycentrics wrt $\triangle ABC$ . Recall that $I_A = (-a:b:c)$ and the $A$ -extouch point is $(0 : a+c-b : a+b-c)$ , so
$\begin{align*} I_A' &= 2\left(0, \frac{a+c-b}{2a}, \frac{a+b-c}{2a}\right) - \left(-\frac{a}{b+c-a}, \frac{b}{b+c-a}, \frac{c}{b+c-a}\right) \\ &= (a^2 : c^2-a^2-b^2+ab : b^2-a^2-c^2+ac).\end{align*}$ Therefore, by isogonal conjugation, $P$ is of the form $\left(\bullet : \frac{b^2}{c^2-a^2-b^2+ab} : \frac{c^2}{b^2-a^2-c^2+ac}\right)$ where $\bullet$ stands for some real number. Similarly, $P$ is also of the form $\left(\frac{a^2}{c^2-a^2-b^2+ab} : \bullet : \frac{c^2}{a^2-b^2-c^2+bc}\right)$ , so we conclude that
$P = (a^2(a^2-b^2-c^2+bc) : - : -) = (a^2(bc-2S_A) : b^2(ac-2S_B) : c^2(ab-2S_C)).$ (Notice that this also implies that $l_a, l_b, l_c$ concur.) Hence,
$\begin{align*} \left|\begin{matrix} a^2(bc-2S_A) & b^2(ac-2S_B) & c^2(ab-2S_C) \\ a & b & c \\ a^2S_A & b^2S_B & c^2S_C \end{matrix}\right| &= \left|\begin{matrix} a^2bc & b^2ac & c^2ab \\ a & b & c \\ a^2S_A & b^2S_B & c^2S_C \end{matrix}\right| \\ &= abc \left|\begin{matrix} a & b & c \\ a & b & c \\ a^2S_A & b^2S_B & c^2S_C \end{matrix}\right| = 0,\end{align*}$ which implies that $P,I,O$ collinear.

For part (b), we first prove the following claim:

Claim. Let $H$ be the orthocenter of $ABC$ . Then $\angle I_A' A H = \angle AIO$ .
Proof. Let $IO$ cut $I_AI_A'$ at $T$ . It suffices to prove that $A,I,T,I_A'$ are concyclic. Let $M$ be the midpoint of $II_A$ (which is also the midpoint of arc $BC$ ), and let $D$ be the $A$ -extouch point, then it suffices to prove $A,M,D,T$ concyclic. Denote $R$ by the radius of $(ABC)$ , then
$A,M,D,T \text{ concyclic} \iff I_AM \cdot I_AA = I_AD \cdot I_AT \iff I_AO^2 - R^2 = I_AD\cdot 2R$ which is Euler's theorem. $\square$

Back to the original problem; the claim means that $\angle OIA = \angle I_A'AH = \angle PAO$ . Combining this with (a), we see that $OI \cdot OP = R^2 \iff PI \cdot PO = PO^2 - R^2$ . Therefore, points $X,Y,I,O$ are concyclic.

Draw the other tangents to the incircle from $X,Y$ , and they intersect at a point $Z\in(ABC)$ by Poncelet's Porism. But
$\frac{\pi}{2} + \frac{1}{2}\angle XZY = \angle XIY = \angle XOY = 2\angle XZY \iff \angle XZY = 60^{\circ}$ as desired.

This post has been edited 1 time. Last edited by Idio-logy, Nov 20, 2020, 9:13 AM

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Abhaysingh2003

222 posts

Abhaysingh2003

#28 h Nov 20, 2020, 1:56 PM • 1 Y

Y by rkbish

EulerMacaroni wrote:

Lemma: Let $\odot(A), \odot(B), \odot(C)$ be circles in the plane such that $\odot(C)$ is the inverse of $\odot(B)$ with respect to $\odot(A)$ . Then for every point $P$ , $B(A(P))=A(C(P))$ , where $X(P)$ denotes inversion about $X$ .

Awesome Lemmma!!! @above you probably made a typo in the Claim section.

This post has been edited 2 times. Last edited by Abhaysingh2003, Nov 20, 2020, 1:59 PM

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snakeaid

125 posts

snakeaid

#29 h Dec 4, 2020, 4:05 PM • 1 Y

Y by rkbish

(a) Let $I'$ be the inverse of $I$ w.r.t. the circumcircle of $\triangle ABC$ . We claim that $I'$ is the point $P$ .
Let $\triangle I_AI_BI_C$ be the excentral triangle of $\triangle ABC$ , then $\triangle ABC$ is its orthic triangle.
Let $Q$ , $T$ , $M$ be the midpoints of $\overline{I_AI}$ , $\overline{I_BI_C}$ and $\overline{I_AI_B}$ , $N_9$ and $O$ be the circumcenters of $\triangle ABC$ and $\triangle I_AI_BI_C$ , repsectively, $S$ be the foot of the altitude from $I_A$ to $\overline{BC}$ , $r$ be the circumradius of $\triangle ABC$ .
We have
$r^2=IN_9 \cdot I'N_9=I'N_9 \cdot N_9O=QN_9\cdot TN_9 \implies AI'I_AO \; \text{is cyclic} \implies \angle I'AI_A=\angle I'OI_A$ Now we want to show that $\angle I'OI_A=\angle I_A'AI_A$ . To prove it suffices to show that $AIOI_A'$ is cyclic. Notice that
$\angle CMO=\angle CSO=90^{\circ} \implies CMSO \; \text{is cyclic} \implies I_AM \cdot I_AC= I_AS \cdot I_AO$ But at the same time from the NPC theorem $CMQA$ is cyclic $\implies I_AM \cdot I_AC=I_AQ \cdot I_AA$ , thus $I_AS \cdot I_AO=I_AQ \cdot I_AA \implies I_AI_A' \cdot I_AO=2 \cdot I_AS \cdot I_AO=I_AQ \cdot I_AA = 2 \cdot I_AQ \cdot I_AA=I_AI \cdot I_AA$ which implies the desired concyclicity and we're done.

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mathaddiction

308 posts

mathaddiction

#30 h Dec 23, 2020, 7:47 AM • 1 Y

Y by rkbish

$[asy] size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -26.525592989299724, xmax = 10.776742650665065, ymin = -26.25346172940703, ymax = 5.5128382325417835; /* image dimensions */pen zzttqq = rgb(0.6,0.2,0); pen zzttff = rgb(0.6,0.2,1); pen qqwuqq = rgb(0,0.39215686274509803,0); pen fuqqzz = rgb(0.9568627450980393,0,0.6); /* draw figures */draw(circle((-7.931797651140656,-5.983034040808574), 7.785861576925324), linewidth(0.8)); draw(circle((-8.679862694712606,-9.357676963545268), 3.1256543575021842), linewidth(0.8) + blue); draw((-9.173416398051794,1.7031889186404228)--(-12.60246325908955,-12.212361715105814), linewidth(0.8)); draw((-12.60246325908955,-12.212361715105814)--(-4.189336658171343,-12.810451283891439), linewidth(0.8)); draw((-4.189336658171343,-12.810451283891439)--(-9.173416398051794,1.7031889186404228), linewidth(0.8)); draw((-14.931069982417341,-9.393287305254859)--(-1.5288458879161784,-1.5533783868256765), linewidth(0.8) + zzttqq); draw((-1.5288458879161784,-1.5533783868256765)--(-11.727255614378826,-23.104961424325452), linewidth(0.8) + zzttqq); draw((-11.727255614378826,-23.104961424325452)--(-7.931797651140656,-5.983034040808574), linewidth(0.8) + zzttff); draw((-8.287940441067258,-18.14091464261898)--(-12.60246325908955,-12.212361715105814), linewidth(0.8)); draw((-8.287940441067258,-18.14091464261898)--(-4.189336658171343,-12.810451283891439), linewidth(0.8)); draw((-8.287940441067258,-18.14091464261898)--(-9.173416398051794,1.7031889186404228), linewidth(0.8) + qqwuqq); draw((-10.101621013031377,-11.353555998739683)--(-9.173416398051794,1.7031889186404228), linewidth(0.8) + qqwuqq); draw((-8.287940441067258,-18.14091464261898)--(-7.492646900570977,-6.953785506304048), linewidth(0.8) + qqwuqq); draw((-9.173416398051794,1.7031889186404228)--(-7.492646900570977,-6.953785506304048), linewidth(0.8)); draw((-14.931069982417341,-9.393287305254859)--(-7.297079206386332,-13.742980753958786), linewidth(0.8) + zzttqq); draw((-7.931797651140656,-5.983034040808574)--(-7.183732607568706,-2.6083911180718804), linewidth(0.8) + zzttff); draw((-7.492646900570977,-6.953785506304048)--(-7.183732607568706,-2.6083911180718804), linewidth(0.8) + qqwuqq); draw(circle((-16.194025550941486,-4.151526525680533), 9.141479410446038), linewidth(0.8) + linetype("4 4") + fuqqzz); /* dots and labels */dot((-9.173416398051794,1.7031889186404228),dotstyle); label("$A$", (-9.027765935570661,2.019863340460275), NE * labelscalefactor); dot((-12.60246325908955,-12.212361715105814),dotstyle); label("$B$", (-13.904749747156163,-12.611088094296232), NE * labelscalefactor); dot((-4.189336658171343,-12.810451283891439),dotstyle); label("$C$", (-3.854208784091446,-13.566713300620417), NE * labelscalefactor); dot((-7.931797651140656,-5.983034040808574),linewidth(4pt) + dotstyle); label("$O$", (-8.500524442426283,-6.11942720995607), NE * labelscalefactor); dot((-8.679862694712606,-9.357676963545268),linewidth(4pt) + dotstyle); label("$I$", (-9.258434088821327,-9.414686542108436), NE * labelscalefactor); dot((-11.727255614378826,-23.104961424325452),linewidth(4pt) + dotstyle); label("$P$", (-12.487788234330646,-22.859344617290095), NE * labelscalefactor); dot((-8.483901567889932,-13.749295803082124),linewidth(4pt) + dotstyle); label("$M$", (-9.390244462107422,-14.291670353693938), NE * labelscalefactor); dot((-8.287940441067258,-18.14091464261898),linewidth(4pt) + dotstyle); label("$I_{A}$", (-7.347183676172954,-18.674365265456586), NE * labelscalefactor); dot((-7.492646900570977,-6.953785506304048),linewidth(4pt) + dotstyle); label("$I_{A}'$", (-7.347183676172954,-6.679621296421972), NE * labelscalefactor); dot((-7.8902936708191165,-12.547350074461514),linewidth(4pt) + dotstyle); label("$D$", (-7.742614796031238,-12.281562161080995), NE * labelscalefactor); dot((-7.297079206386332,-13.742980753958786),linewidth(4pt) + dotstyle); label("$X$", (-7.1494681162438125,-14.555291100266128), NE * labelscalefactor); dot((-1.5288458879161784,-1.5533783868256765),linewidth(4pt) + dotstyle); label("$Y$", (-0.9214279784758398,-1.2753959916920914), NE * labelscalefactor); dot((-14.931069982417341,-9.393287305254859),linewidth(4pt) + dotstyle); label("$Z$", (-15.914857939769107,-9.81011766196672), NE * labelscalefactor); dot((-10.101621013031377,-11.353555998739683),linewidth(4pt) + dotstyle); label("$H$", (-10.90606375489751,-10.996411021541572), NE * labelscalefactor); dot((-7.183732607568706,-2.6083911180718804),linewidth(4pt) + dotstyle); label("$I'$", (-7.050610336279242,-2.3298789779808486), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$
(a) Redefine $P$ to be the intersection of $l_A$ and $OI$ , we show that $P$ is the image of I under the inversion w.r.t. $(ABC)$ .
Let $I'$ be the reflection of $I$ in $O$ .
CLAIM 1. $A,I',I_A',I$ are concyclic.
Proof.
Notice that $Q,M$ are the midpoints of $II'$ and $II_A$ respectively, therefore,
$I_AI'\times I_AI_A'=2OM\times 2I_AD=2R\times 2I_AB\cos\frac{C}{2}=4R\sin\frac{A}{2}\times I_AA\frac{\cos\frac{C}{2}}{\sin\frac{A}{2}}=I_AM\times I_AA$ as desired. $\blacksquare$
Therefore, if $H$ is the orthocenter of $\triangle ABC$ , using the fact that $AH,AO$ are isogonal, $\angle PAO=\angle I'AH=180^{\circ}-\angle AI_A'I_A=\angle AIO$ as desired.
This proves $(a)$ since this implies $P$ lies on $l_B$ by symmetry.

(b) Suppose the second tangent from $X$ and $Y$ to the incircle of $(ABC)$ intersect at $Z$ , then $Z$ lies on $(ABC)$ by Poncelet Porism. Moreover, since $P$ is the image of $I$ under inverson,
$PI\times PO=PX\times PY$ hence $X,Y,I,O$ are concyclic.
Therefore,
$90^{\circ}+\frac{\angle XZY}{2}=\angle XIY=\angle XOY=2\angle XZY$ which implies $\angle XZY=60^{\circ}$ , and hence $\angle XIY=120^{\circ}$ as desired.

This post has been edited 1 time. Last edited by mathaddiction, Dec 23, 2020, 8:04 AM

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Ali3085

214 posts

Ali3085

#31 h Jan 26, 2021, 1:51 PM • 1 Y

Y by Muaaz.SY

let $P'=IO \cap \ell_a$
let $N,M$ be the midpoints of ars $BC,AC$
(i)
claim: $P'$ is the image of $I_a'$ under $\sqrt{AB.AC}$ inversion
proof:
let $A'$ be the reflection of $A$ wrt $BC$
note that $O^*=A'$ and $I^*=I_a$
since $AI_a'I_aA'$ is cyclic so $I_a'^* \in IO$
$\blacksquare$
note that $I_a'-A'-AI \cap BC$ are collinear so $AONP'$ is cyclic so $BOMP'$ is also cyclic
as above we will have that $P'$ is the image of $I_b'$ under $\sqrt{BC.BA}$ inversion
so $P=P'$
(ii)
note that $IA.IN=IO.IP$ so $P$ is the image of $O$ under the inversion centered at $I$ with radius $\sqrt{IA.IN}$
so $IOXY$ is cyclic
now let $Z$ on $(ABC)$ such that $ZX,ZY$ are both tangent to the incircle (Poncelet Porism)
since $O,I$ are the circumcenter,incenter of $\triangle ZXY$ and $XYOI$ is cylic so $\angle XZY=60$ so $\angle XOY=\angle XIY=120$
and we win

This post has been edited 1 time. Last edited by Ali3085, Jan 26, 2021, 1:51 PM

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hsiangshen

188 posts

hsiangshen

#32 h Apr 20, 2021, 3:07 AM

Y by

Flip99 wrote:

Anyone...

You can use feuerbach hyperbola

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FloorX

19 posts

FloorX

#33 h Apr 25, 2021, 3:56 PM

Y by

This is a direct conclusion from Komal A779

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Flip99

82 posts

Flip99

#34 h Apr 25, 2021, 6:16 PM

Y by

hsiangshen wrote:

Flip99 wrote:

Anyone...

You can use feuerbach hyperbola

How?

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hsiangshen

188 posts

hsiangshen

#35 h Aug 27, 2021, 7:59 AM • 2 Y

Y by YaoAOPS, GeoKing

OK this problem is kinda easy under some dark and poisonous conic geometry...
(a)
Let $Na$ =Nagel point, tangents of incircle= $D,E,F$ as usual, $D'$ be the antipode of $D$ wrt $(I)$ .
Notice that $ANa$ bisects $I_AI_A'\implies AI_A'$ passes through the reflection of $I$ over $D'$ . Then by kariya theorem, if we construct $BI_B',CI_C'$ similarly, they must concur on the Feuerbach hyperbola, call it $X$ .
Now we prove that point is actually the reflection of $I$ over Feuerbach point= $X_{80}$ .
Notice that $A(I_A',I;D',H)=-1=(X,I;Na,H)$ on Feuerbach hyperbola. Therefore projecting from $I$ on $NaH:(IX,II,INa,IH)=-1=(IX\cap NaH,OI\cap NaH;Na,H)$ , so $IX$ bisects $NaH$ . On the other hand, $-1=(XX,IX;XNa,XH)\implies XX||NaH||OI$ . So $(X,I)$ is a pair of antipode wrt Feuerbach hyperbola, $X=X_{80}$ , as required.
Finally, the problem makes the intersection isogonal conjugate so it clearly lies on the ispgonal conjugate of Feuerbach hyperbola= $OI$ .
(b) We know that the isogonal conjugate of $P$ is the antigonal conjugate of $I$ . One of the equivalent definition of antigonal is the isogonal, inverse, isogonal of a point.(well-known) So $I,P$ are inverses wrt the circumcurcle. Then the rest can be done by poncelet or something like that but I haven't finished it.

This post has been edited 2 times. Last edited by hsiangshen, Aug 27, 2021, 8:00 AM

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Kei0923

94 posts

Kei0923

#36 h Sep 24, 2021, 11:52 AM • 1 Y

Y by Ywgh1

Let $V$ be the reflection of $I$ wrt $O$ and $M$ be the midpoint of arc $BC$ . Suppose that the A-excircle of $\triangle ABC$ touch the side $BC$ at $K$ .

(a)
lemma. $AIVI'_A$ is cyclic.
proof. From Euler's formula, we have $I_AO^2=OM^2+2OM\cdot I_AK.$ Note that $I_AV=2OM, I_AI'_A=2I_AK$ and $I_AI=2I_AM$ then
$I_AM\cdot I_AA=(OM^2+2OM\cdot I_AK)-OM^2=2OM\cdot I_AK\Longrightarrow I_AI\cdot I_AA=4OM\cdot I_AK=I_AV\cdot I_AI'_A,$ implying that $AIVI'_A$ is cyclic.

We get $\measuredangle I'_AAM=\measuredangle I_AVI=\measuredangle MOI$ from lemma.
Let $l_A\cap (ABC)=S$ and $l_B\cap (ABC)=T,$ then $\measuredangle MAS=\measuredangle MOI$ . By simple angle chasing, it follows that $\measuredangle IOS=\measuredangle MOI$ so $ASIO$ is cyclic. Similarly, $BTIO$ is cyclic thus $P$ be the radical center of $(ABC), (ASIO), (BTIO)$ and we have the desired result.

(b) Poncelet porism means that there exists point $Z$ on $(ABC)$ such that the incenter of $\triangle XYZ$ is $I$ .
Because of $P$ lies on radical axis of $(ABC)$ and $(ASOI),$ then $XYOI$ is also cyclic. Hence
$\angle XIY=90^\circ+\displaystyle\frac{1}{2}\angle XZY=\angle XOI=2\angle XZY\Longrightarrow \angle XZY=60^\circ.$ Therefore we conclude $\angle XIY=120^\circ.$

I used the same lemma at Taiwan TST 2019 6.

This post has been edited 6 times. Last edited by Kei0923, Jan 9, 2022, 4:32 AM

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Inconsistent

1455 posts

Inconsistent

#37 h May 29, 2022, 4:03 AM

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Projective, projective, projective. Give me a break.

a) Let the medial triangle of the excentral triangle be $Q_A, Q_B, Q_C$ . Let the projections of $I_A, I_B, I_C$ onto $\triangle ABC$ be $R_A, R_B, R_C$ and let $S_A, S_B, S_C$ be the midpoints of $I_A, I_B, I_C$ with $I$ . Let $O$ be the circumcenter of $(ABC)$ and $O'$ be the circumcenter of $(I_AI_BI_C)$ . Then since $(ABCQ_AQ_BQ_C)$ is the nine-point circle of the excentral triangle, $AI_A \cdot S_AI_A = CI_A \cdot Q_CI_A = OI_A \cdot R_AI_A$ so $\triangle I_AR_AA \sim \triangle I_A S_AO'$ . Hence by similarity, $\angle I_A'AI_A = \angle O'IA = \angle OIA - \angle OAI$ so if $A'$ is $\ell_A \cap IO$ then $\angle OAA' = OIA$ so $A'$ is the inverse of $I$ , $I'$ with respect to $(ABC)$ so we are done by symmetry.

b) By Brokard's theorem, $X' = YI \cap (ABC)$ is the reflection of $X$ over $IO$ , so by Poncelet's porism we have that $\frac{\angle XIY}{2} = \angle XX'I = \angle IX'Y = 2(\angle XIY - \frac{\pi}{2})$ so $\angle XIY = 120^{\circ}$ .

This post has been edited 1 time. Last edited by Inconsistent, May 29, 2022, 4:19 AM
Reason: edit

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crazyeyemoody907

450 posts

crazyeyemoody907

#38 h Sep 14, 2022, 8:16 PM • 1 Y

Y by teasaffrontaffy

$[asy] //nya pens size(7cm); pen darkgrn, thickgrn; darkgrn=RGB(13,83,76); thickgrn=darkgrn+linewidth(1); pen lightblue, lightgrn; lightblue=RGB(210, 210, 252); lightgrn=RGB(184, 214, 212); // draw pair A,B,C; A=(17,9); B=(0,0); C=(14,0); pair O,I,Ia,Ia1; O=circumcenter(A,B,C); I=incenter(A,B,C); Ia=2*circumcenter(B,I,C)-I; Ia1=2*foot(Ia,B,C)-Ia; pair P=O+(distance(O,A)*distance(O,A)/distance(O,I))*unit(I-O); // inverse of I filldraw(A--I--P--cycle,lightgrn,darkgrn); filldraw(A--Ia--Ia1--cycle,lightblue,blue); draw(A--B--C--A); draw(circumcircle(A,B,C),dotted); draw(O--I,darkgrn); draw(A--O,dashdotted); //label label("$A$",A,dir(70)); label("$B$",B,-dir(30)); label("$C$",C,dir(-90)); label("$I$",I,-1.5*dir(0)); label("$O$",O,-dir(30)); label("$I_a$",Ia,dir(-90)); label("$I_a'$",Ia1,dir(130)); label("$P$",P,dir(-90)); [/asy]$
Redefine $P$ as the inverse of $I$ ; it's clear via Poncelet spam that this point satisfies the second part.
For the first part we assert more strongly that:

Claim: $\triangle AI_aI_a'\overset+\sim \triangle API$ .

Proof: One of the few uses of SAS similarity? By angle chasing, $\angle I_a=\angle P$ follows easily.
To finish, we show $I_aI_a'/I_aA=IP/AP$ ; indeed, the first ratio equals $2\cos\angle BI_aC=2\sin\frac A2$ because of similar triangles; thus, we're left to length chase $IP/AP$ ; this becomes
$\frac{OP}{AP}-\frac{OI}{OA}\frac{OA}{AP}=\frac{OI}{AI}-\frac{OI^2}{OA\cdot AI}=\frac{R}{AI}-\frac{R^2-2rR}{R\cdot AI}$ $=\frac{R-(R-2r)}{AI}=2\sin\frac A2,$ so the ratios are equal, as needed. $\qquad\qquad\square$ .
The claim clearly implies the isogonality.
Remark

This post has been edited 1 time. Last edited by crazyeyemoody907, Sep 14, 2022, 8:16 PM

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VicKmath7

1385 posts

VicKmath7

#39 h Jun 15, 2023, 2:13 PM

Y by

Solution for a)

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HamstPan38825

8857 posts

HamstPan38825

#40 h Nov 9, 2023, 12:10 AM • 1 Y

Y by GeoKing

Writeup from an OTIS walkthrough.

Let $V$ be the circumcenter of triangle $I_AI_BI_C$ .

Part (a) Consider the circumcircles of $AVI_A$ and cyclic permutations. As $I$ is their radical center and they are coaxial, their second intersection point $P$ must lie on $\overline{IV}$ .

On the other hand, $I_AI_A' \cdot AV = 2\cos I_A R \cdot I_AA = I_AI \cdot I_AA$ with reference triangle $I_AI_BI_C$ , thus $AIVI_A'$ is cyclic. To finish, set $P' = \overline{IV} \cap (AVI_A)$ . Then $\angle P'AI_A = \angle IVI_A = \angle I_AAI_A'$ thus $P$ lies on $\ell_A$ . This finishes the first part.

Part (b) Let $Z$ be the point on $(ABC)$ such that $XYZ$ and $ABC$ have the same incircle which exists by Poncelet porism.

Claim. $XIOY$ is cyclic.

Proof. It suffices to show that $P$ and $I$ are inverses with respect to $(ABC)$ . To see this, set $M$ to be the midpoint of $\overline{II_A}$ ; then $M = (POA) \cap (ABC)$ by Reim's theorem.

Then as $OM=OA$ , $\angle MPO = \angle APO$ and $OI \cdot OP = OM^2$ . Hence $PI \cdot PO = PO^2 - R_{ABC}^2 = OX \cdot OY.$
Finally, $\angle XIY = 90^\circ +\frac Z2 = 2Z = \angle XOY$ , thus $\angle XIY = 120^\circ$ .

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HoRI_DA_GRe8

588 posts

HoRI_DA_GRe8

#45 h Nov 11, 2023, 11:51 AM • 2 Y

Y by amar_04, GeoKing

500th post :trampoline:

Also this took less than a hour with GGB (would have done it freehand but I've already wasted so much time in life

)

2016 ISL G7 wrote:

Prove that $P$ lies on line $OI$ where $O$ is the circumcentre of triangle $ABC$ .
Let one of the tangents from $P$ to the incircle of triangle $ABC$ meet the circumcircle at points $X$ and $Y$ . Show that $\angle XIY = 120^{\circ}$ .

Part (a) Let $Be$ be the Bevan point (circumcentre of the excentral triangle) of $\triangle ABC$ .Clearly as $I_AI'_A \perp BC,I_BI'_B \perp CA$ , the lines $I_AI'_A,I_BI'_B$ are isogonal to $I_AI,I_BI$ respectively.Since $I$ is the orthocentre of the excentral triangle it's isogonal
conjugate is $Be$ .So $Be \equiv I_AI'_A \cap I_BI'_B$ $\square$

Now follow up with a $\sqrt{I_AI.I_AA}$ inversion centered at $I_A$ .If $I_AI'_A \cap BC=F$ and $J$ be the reflection of $I_A$ over the Bevan point, then $F$ and $J$ gets swapped under this inversion.
Now,
$I_AI.I_AA=I_AF.I_AJ=2I_AF.\frac{I_AJ}{2}=I_AI'_A.I_ABe$ Which gives that $BeIAI'_A$ and similarly $BeIBI'_B$ are cyclic .Now if $BeI \cap \ell_A=P'$ , then
$\angle P'AI_A=\angle I_AAI'_A=\angle IAI'_A=\angle I_ABeI=\angle I_ABeP'$ Thus $I_ABeAP'$ is cyclic,this gives the following
$P'I.IBe=AI.I_A=BI.II_B$ which implies that $I_BBeBP'$ is cyclic as well.Now it is not hard to prove that $P' \in \ell_B$ by angle chase,thus $P \equiv P'$ .Since $Be,I,O$ are collinear, we have that $P \in OI$ finishing the first part $\blacksquare$

Part (b)We use the well known property that $O$ is the midpoint of $IBe$ and the midpoint of $II_A$ (say $S$ ) lie on the circumcircle of $\triangle ABC$ .
From the cyclic quadrilaterals we obtained in Part (a) we have ,
$PI.2IO=PI.IBe=IA.II_A=2IA.IS\implies PI.IO=IA.IS=\text{Pow}_{\odot(\triangle ABC)}(I)$ Now we present a claim ;
Claim : $P$ is the inverse of $I$ w.r.t $\odot(\triangle ABC)$ .
Proof : We invert around $I$ with radius $\sqrt{\text{Pow}_{\odot(\triangle ABC)}(I)}$ .Clearly $P,O$ get swapped and $I$ and $P_{\infty,PI}$ get swapped.Now we let $PI \cap \odot(\triangle ABC)\equiv M,N$ .Clearly $M,N$ get swapped under the inversion .Since inversion preserves cross ratios, we have
$(M,N;P,I)=(M,N;O,P_{\infty,PI})=-1$ This proves that $P$ and $I$ are inverses w.r.t $\odot(\triangle ABC)$ $\blacksquare$

Now note that
$OI.OP=R^2 \implies OP=\frac{R^2}{OI} \implies \frac{PI}{PO}=1-\frac{OI}{OP}=1-\frac{R^2}{OI^2}=1-\frac{R^2-2Rr}{R^2}=\frac{2r}{R}$ Where $r,R$ are circumradius and inradius of $\triangle ABC$ respectively.Now let $XY$ touch the incircle at $H$ and the midpoint of $XY$ be $L$ .We have $\triangle PIH \sim \triangle POL$ .Thus ,
$OL=IH.\frac{PO}{PI}=\frac{R}{2r}.r=\frac{R}{2} \implies \angle XOY=120^{\circ}$ One final observation, since $I,P$ are inverses we have $PX.PY=PO.PI$ which gives $XYOI$ is cyclic and thus we get that $\angle XIY=\angle XOY=120^{\circ}$ This took more time to write than to solve.So much skill issue :wallbash_red:

$\blacksquare$

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kamatadu

465 posts

kamatadu

#46 h Dec 26, 2023, 7:01 AM • 2 Y

Y by HoRI_DA_GRe8, GeoKing

I spent almost over 4 hours trying to complecks bash part (a) but in vain. Then I noticed the use of inversion distance formula and bruh moment.

Why is this $C$ centered considering the hardness of the problem. It makes it even more harder (at least for me ;-; ). So of course, we $A$ -center the problem. The statement thus reads.

A-centered problem statement wrote:

Let $I$ be the incenter of a non-equilateral triangle $ABC$ , $I_B$ be the $B$ -excenter, $I'_B$ be the reflection of $I_B$ in $CA$ , and $l_B$ be the reflection of line $BI'_B$ in $BI$ . Define points $I_C$ , $I'_C$ and line $l_C$ analogously. Let $P$ be the intersection point of $l_B$ and $l_C$ .

Prove that $P$ lies on line $OI$ where $O$ is the circumcenter of triangle $ABC$ .
Let one of the tangents from $P$ to the incircle of triangle $ABC$ meet the circumcircle at points $X$ and $Y$ . Show that $\angle XIY = 120^{\circ}$ .

$[asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions are done using bubu-asy.py. This adds the dps, xmin, linewidth, fontsize and directions. */ pair A = (-28.83788,16.19840); pair B = (-55.62356,-38.63888); pair C = (40,-40); pair I = (-21.42183,-17.61800); pair O = (-7.65223,-28.11036); pair I_B = (68.23612,37.48706); pair I_C = (-82.14257,4.50850); pair I_A = (4.71935,-136.81900); pair P = (-118.47836,56.33847); pair I_Ap = (7.48934,57.78395); pair Ap = (-30.40953,-94.21634); import graph; size(12cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); real xmin = -5, xmax = 5, ymin = -5, ymax = 5; draw(A--B, linewidth(0.5)); draw(B--C, linewidth(0.5)); draw(C--A, linewidth(0.5)); draw(circle(O, 49.11312), linewidth(0.5)); draw(B--I_B, linewidth(0.5)); draw(C--I_C, linewidth(0.5)); draw(P--O, linewidth(0.5) + linetype("4 4") + red); draw(I_A--A, linewidth(0.5)); draw(circle(I, 21.50552), linewidth(0.5)); draw(Ap--I_A, linewidth(0.5) + blue); draw(A--I_Ap, linewidth(0.5) + blue); draw(P--A, linewidth(0.5)); dot("$A$", A, dir(90)); dot("$B$", B, SW); dot("$C$", C, SE); dot("$I$", I, NE); dot("$O$", O, SE); dot("$I_B$", I_B, NE); dot("$I_C$", I_C, NW); dot("$I_A$", I_A, SE); dot("$P$", P, NW); dot("$I_A'$", I_Ap, NE); dot("$A'$", Ap, NW); [/asy]$

We first solve part (a). I claim that all $\left\{l_A,l_B,l_C\right\}$ pass through the inverse of $I$ w.r.t. $\odot(ABC)$ . Note that this actually finishes.

So we prove that $l_A$ passes through $I^*$ . Redefine $P$ as the image of $I_A'$ under $\sqrt{bc}$ -inversion. Note that we need to prove that $I^*$ lies on the isogonal of the line $AI_A$ . Thus if we show that $P$ is the image of $I$ under $\mathbf{I}(\odot(O,OA))$ , then we are done.

Let $A'$ be the reflection of $A$ over $BC$ . It is well known that under $\sqrt{bc}$ -inversion, $O\leftrightarrow A$ , $I\leftrightarrow I_A$ . Note that $AA'I_AI_A'$ is an isosceles trapezium and so, it is cyclic. Thus after inverting, we get that $\overline{P-I-O}$ are collinear.

Now using inversion distance formula, we get that,
$OP = \dfrac{AB\cdot AC}{AA'\cdot AI_A'}\cdot A'I_A' = \dfrac{AB\cdot AC}{AA'\cdot AI_A'}\cdot AI_A.$ We also get,
$OI = \dfrac{AB\cdot AC}{AA'\cdot AI_A}\cdot A'I_A.$
Multiplying the above identities, we get,
$OP\cdot OI = \left(\dfrac{AB\cdot AC}{AA'}\right)^2 = \left(\dfrac{AB\cdot AC}{\dfrac{AB\cdot AC}{AO}}\right)^2 = AO^2.$
This concludes that $P$ is indeed the image of $I$ under $\mathbf{I}(\odot(O,OA))$ and we are done. :yoda:

$[asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions are done using bubu-asy.py. This adds the dps, xmin, linewidth, fontsize and directions. */ pair A = (-39.92648,31.75707); pair B = (-55.62356,-38.63888); pair C = (40,-40); pair I = (-25.10426,-14.53006); pair O = (-7.42914,-12.43734); pair P = (-175.32738,-32.31636); pair X = (-60.67652,0.75126); pair Y = (30.61622,27.08194); pair L = (-27.31701,-63.56166); pair M = (-22.63116,40.27054); import graph; size(15cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); real xmin = -5, xmax = 5, ymin = -5, ymax = 5; draw(A--B, linewidth(0.5)); draw(B--C, linewidth(0.5)); draw(C--A, linewidth(0.5)); draw(circle(O, 54.85638), linewidth(0.5)); draw(P--O, linewidth(0.5) + linetype("4 4") + red); draw(circle(I, 24.54075), linewidth(0.5)); draw(P--A, linewidth(0.5)); draw(P--Y, linewidth(0.5)); draw(X--L, linewidth(0.5)); draw(Y--L, linewidth(0.5)); draw(circle(M, 54.85638), linewidth(0.5) + blue); dot("$A$", A, NW); dot("$B$", B, SW); dot("$C$", C, SE); dot("$I$", I, dir(90)); dot("$O$", O, dir(90)); dot("$P$", P, NW); dot("$X$", X, dir(180)); dot("$Y$", Y, NE); dot("$L$", L, dir(270)); dot("$M$", M, dir(90)); [/asy]$

Since $\overline{P-X-Y}$ are collinear, thus after inverting $\mathbf{I}(\odot(O,OA))$ , we get that $OIXY$ is cyclic.

Now let the tangent to the incircle from $X$ intersect $\odot(ABC)$ at $L\;(\neq Y)$ . Thus by Poncelet Porism, we know that $\triangle LYX$ has the same incircle. Now let $M$ be the midpoint of the arc $\widehat{XY}$ not containing $L$ . Thus $I$ is the incenter of $\triangle LYX$ .

Thus by Fact 5, we get that $MX = MY = MI$ which means that $M$ is the center of $\odot(OIXY)$ . This means $MX=MO$ . But on the other hand, $OX = OM$ . This forces that $\triangle OMX$ is an equilateral triangle. This implies that $\angle MOX = 60^\circ$ . Similarly we also get that $\angle YOM = 60^\circ$ . Adding these two values, we get that $\angle YOX = 120^\circ$ . Now since $OIXY$ is cyclic, we finally get $\angle XIY = \angle XOY = 120^\circ$ and we are done. :yoda:

This post has been edited 2 times. Last edited by kamatadu, Dec 26, 2023, 7:01 AM

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GeoKing

515 posts

GeoKing

#47 h Jul 2, 2024, 5:47 AM

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Sol:- Part (a):- It suffices to show that $AI_A',BI_B',CI_C'$ concur at a point on feuerbach hyperbola $h$ . Let $H,I,N_a$ be orthocenter,incenter, nagel point of $\Delta ABC$ . We know that $H,I,N_a \in h$ . Let $Q$ be the unique point on $h$ such that $-1=(N_a,H;I,Q)_h$ . $D$ be the tangency point of $A$ excircle with $BC$ . $(N_a,H;I,Q)_h=-1=(D,\infty_{I_AI_A'};I_A,I_A') \stackrel{A}{=}(N_a,H;I,AI_A' \cap h)_h \implies A-Q-I_A'$ are collinear. In similar way $Q \in BI_B'$ and $Q \in CI_C'$ . Hence $P$ ,the isogonal conjugate of $Q$ lies on $OI$ .

https://cdn.discordapp.com/attachments/1247512024687181896/1257557976634359931/image.png?ex=6684d7a1&is=66838621&hm=6fc5746f839a843225d6e5159f328373b099aabf229a9c3bd71825eb1036d195&

Part(b):- Let $I_C$ be $C$ excenter, $M$ be the midpoint of arc $BC$ not containing $A$ . $I'$ be the bevan point (circumcenter of $\Delta I_AI_BI_C$ ) ,we know $O$ is midpoint of $II'$ and $I' \in I_AD$ . Note that $\Delta I_ABCMD \stackrel{-}{\sim} \Delta I_AI_BI_CI'A \implies MDI'A$ cyclic ,since $MD \parallel II'$ by reims we obtain $II_A'I'A$ cyclic.Now we have $\measuredangle PAI=\measuredangle I_AAI_A'$ and $\measuredangle AIP=\measuredangle AI_A'I'=\measuredangle AI_A'I_A \implies \Delta PAI \stackrel{+}{\sim} \Delta I_AAI_A' \implies API_AI'$ is cyclic. Since $OM \parallel I_AI'$ by reims theorem we obtain $APMO$ cyclic.By shooting lemma $OI \cdot OP=OA^2$ , thus by converse of shooting lemma we obtain $OIXY$ concyclic.Let the second tangents from $X,Y$ to incircle meet $(ABC)$ at $Z$ (due to poncelet porism). Let $\angle XZY= x$ . Then $90^\circ +\frac{x}{2}=\angle XIY=\angle XOY=2x \implies x=60^\circ \implies \angle XIY=120^\circ.$

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This post has been edited 1 time. Last edited by GeoKing, Jul 2, 2024, 5:50 AM
Reason: diag

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jp62

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jp62

#48 h Jul 7, 2024, 12:44 PM • 1 Y

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Complex bash for (a)
Part (b)

(Note: in solving this, part (b) was used to identify the point $P$ )

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Ilikeminecraft

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Ilikeminecraft

#49 h Feb 8, 2025, 7:54 AM

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used otis walkthrough. still impossible :wallbash:

Let $I_A,I_B,I_C$ be the excenters of the respective points. Let $V$ be the circumcenter of $I_AI_BI_C.$
Claim: $(I_AAV),(I_BBV), (I_CCV)$ form a pencil.
Proof: Note that $I_AI\cdot IA = I_BI\cdot IB = I_CI\cdot IC$ by cyclic quads in the ortho config. Since $I\neq V,$ our claim is done.

Let the second intersection point of these 3 circles be $P \neq V.$
Claim: $V, I, P$ are collinear.
Proof: Note that $I_A, V, I_A'$ are collinear by considering the fact that the circumcenter and orthocenter are isogonal conjugates. Let $S_{IA}, E_{IA}$ be the antipode of $I_A$ in $(I_AI_BI_C)$ and foot from $V$ to $BC.$ By 90 degrees, we deduce that $BI_CS_{IA}E_{IA}$ is cyclic. Hence, we deduce that $I_AB\cdot I_AI_C = I_AS_{IA} \cdot I_AE_{IA} = 2I_AV\cdot \frac12 I_AI_A' = I_AV\cdot I_AI_A'.$ However, $I_AB\cdot I_AI_C = I_AI\cdot I_AA,$ so $VIAI_A'$ is cyclic. Finally, $\angle IAI_A' = \angle I_A VP = \angle PAI_A,$ so $P$ is the point in the desired problem.

Finally, since $I, O, V$ are collinear by Euler line(orthocenter, 9 point center, circumcenter, respectively), we are done with part a.

Let $M$ be the midpoint of $\overline{I_AI}.$
We have that $PAOM$ is concyclic since $PI\cdot OI = PI\cdot VI\frac{1}{2} = IA\cdot I_AI\frac{1}{2} = IA\cdot IM.$ Also note that $M$ lies on the 9 point circle. Thus, $\angle OPA = \angle OMA = \angle OAM,$ so $(IAP)$ is tangent to $OA$ , so $OI \cdot OP = OA^2$ , so $P, I$ are images under inversion with respect to circumcircle of $ABC$ , so $PI\cdot PO = PX\cdot PY,$ so $XYIO$ is cyclic. Now, by poncolet porism, we deduce that arc $XY$ is 120, but this finishes.

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