Two lines meeting on circumcircle

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Zhero

2043 posts

Zhero

#1 h Jul 5, 2012, 1:49 AM • 5 Y

Y by nguyendangkhoa17112003, samrocksnature, Adventure10, ehuseyinyigit, Rounak_iitr

Let $ABC$ be a triangle with circumcircle $\omega$ , incenter $I$ , and $A$ -excenter $I_A$ . Let the incircle and the $A$ -excircle hit $BC$ at $D$ and $E$ , respectively, and let $M$ be the midpoint of arc $BC$ without $A$ . Consider the circle tangent to $BC$ at $D$ and arc $BAC$ at $T$ . If $TI$ intersects $\omega$ again at $S$ , prove that $SI_A$ and $ME$ meet on $\omega$ .

Amol Aggarwal.

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Luis González

4148 posts

Luis González

#2 h Jul 5, 2012, 4:38 AM • 8 Y

Y by MinatoF, nguyendangkhoa17112003, thczarif, Siddharth03, samrocksnature, Adventure10, Mango247, and 1 other user

Denote by $\Omega$ the circle internally tangent to $\omega$ at $T$ and tangent to $BC$ at $D.$ $TB,TC$ cut $\Omega$ again at $B',C'.$ Since $T$ is exsimilicenter of $\Omega \sim \omega,$ then $BC \parallel B'C'$ $\Longrightarrow$ arcs $DB'$ and $DC'$ of $\Omega$ are equal, i.e. $TD$ bisects $\angle BTC$ $\Longrightarrow$ $M \in TD.$ $ME$ cuts $\omega$ again at $F$ and $FI_A$ cuts $\omega$ again at $S^*.$ It's well known that $I,I_A$ lie on the circle $(M)$ with center $M$ and radius $MB=MC,$ thus inversion WRT $(M)$ takes $D,E$ into $T,F$ and $I,I_A$ into themselves $\Longrightarrow$ $\odot(IDT)$ and $\odot(I_AEF)$ are orthogonal to $(M)$ $\Longrightarrow$ $\angle MFS^*=\angle EI_AM=\angle DIM=\angle MTS$ $\Longrightarrow$ $S \equiv S^*.$ Lines $ME$ and $SI_A$ intersect at $F \in \omega.$

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TelvCohl

2312 posts

TelvCohl

#3 h Dec 17, 2014, 3:51 PM • 13 Y

Y by Mediocrity, Tommy2000, AlgebraFC, nguyendangkhoa17112003, thczarif, Pluto1708, HolyMath, Nathanisme, samrocksnature, e_plus_pi, AllanTian, Adventure10, Mango247

My solution:

Let $T'=ME \cap (ABC)$ .

From homothety we get $T, D, M$ are collinear .

Since $BD=CE$ ,
so $T'$ is the reflection of $T$ in the bisector of $BC$ .
Since $\triangle MBD \sim \triangle MTB$ ,
so we get $MI^2=MB^2=MD \cdot MT$ .
ie. $\triangle MID \sim \triangle MTI$

Since $\text{Rt} \triangle BDI \sim \text{Rt} \triangle I_AEB$ ,
so we get $DM \cdot ET'=DM \cdot DT=DB \cdot DC=DB \cdot EB=DI \cdot EI_A$ ,
hence combine with $\angle I_AET'=\angle MDI$ we get $\triangle I_AET' \sim \triangle MDI$ ,
so from $\angle ET'I_A=\angle DIM=\angle ITM=\angle ET'S$ we get $T', S, I_A$ are collinear .
ie. $ME \cap I_AS \in (ABC)$

Q.E.D

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Sardor

804 posts

Sardor

#4 h Dec 17, 2014, 5:37 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

Telv Cohl thank you for nice proof..My solution similar to Luiz Gonzalez's solution(but without invesion).

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Seventh

82 posts

Seventh

#5 h Dec 17, 2014, 8:58 PM • 4 Y

Y by samrocksnature, Bubu-Droid, Adventure10, Mango247

Other solution:

Claim 1: Are given two circle internally tangent at a point $T$ , $AB$ a chord of the bigger circle tangent to the smaller circle in a point $C$ . So $TC$ is the internal bisector of $\angle ATB$ .
Prove: A simple homothety with center $T$ carring one circle in the other.

Applying Claim 1 in the problem, we have that $T, D, E$ are collinear. Also is well known that $CE = BD$ , and since $MB = MC$ , follows that $MD = ME$ . Denote $ME \cap \omega = T'$ , and since $MD = ME$ , and the tangent to $\omega$ at $M$ is parallel to $DE$ , follows that $MT' = MT$ .

Now we gonna prove that $T', S$ and $I_A$ are collinear.

Note that $(A, M; I, I_A) = (A, M; T, SI_A \cap \omega)$ , where $(A, M; I, I_A) = \dfrac{AI}{AI_A} = \dfrac{r}{r_A}$ , since $MI = MI_A = MB = MC$ . But also note that $(A, M; T, T') = \dfrac{AT}{AT'}$ , since $MT = MT'$ , where $\dfrac{AT}{AT'} = \dfrac{\text{sen}(\angle TMA)}{\text{sen}(\angle AMT')} = \dfrac{DR}{RE}$ , since $MD = ME$ , where $R = IM \cap DE$ . But $\triangle IDR$ and $\triangle I_AER$ are similar, thus $\dfrac{DR}{RE} = \dfrac{r}{r_A}$ .

Therefore, $(A, M; T, SI_A \cap \omega) = (A, M; T, T')$ , in other words $T' = SI_A \cap \omega$ , as desired.

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anantmudgal09

1980 posts

anantmudgal09

#6 h Sep 27, 2015, 11:12 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

Similar to Telv Cohl's solution.
Assume without loss of generality, $AB <AC$ .(The case $AB=AC$ being trivial.)

Let $\omega$ denote the circumcircle of $\triangle ABC$ and let $ME$ meet $\omega$ again at $X$ .
Now, let $AM$ meet $BC$ at $Y$ .
Clearly, $MI^2= MY.MA=ME.MX=MD.MT$ and so $A,Y,E,X$ are concyclic.
Now applying $\sqrt {bc}$ inversion we get that $X \longrightarrow X'$
where $X'$ is the point of intersection of $BC$ with $T_AM$ where $T_A$ is the touch-point of the $A$ -mixitilinear in-circle with $\omega$ .
Now it is well-known that $X'$ lies on the mixitilinear touch-chord of the $A$ -mixitilinear in-circle and so we conclude that $\angle AXI_A=90$ . Now let $AI_A$ meet $\omega$ again at $S'$ . Then clearly, $S'$ is the antipodal point of $A$ w.r.t $\omega$ .
So it suffices to showing that $T,I,S'$ are col-linear.
This is equivalent to the fact that $\angle MTI=\angle MAS'$ .
But, $\angle MTI=\angle MID=\frac{\angle B-\angle C}{2}=\angle MAS'$ .
Hence the conclusion follows.

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EulerMacaroni

851 posts

EulerMacaroni

#7 h Sep 28, 2015, 12:27 AM • 5 Y

Y by anantmudgal09, Tommy2000, samrocksnature, e_plus_pi, Adventure10

I think this is the simplest solution yet...

Invert about $M$ with radius $MB$ . The problem becomes to show that the circumcircles of $(MDI)$ and $(MEI_a)$ intersect on $BC$ . Let $K$ be the intersection of $(MEI_a)$ with $BC$ . Then since $I_aE\perp EK$ , $\angle I_aEK=\angle I_aMK=\angle AMK=\angle AMS=90^{\circ}$ , where $S$ is the intersection of $MK$ with $\omega$ , so $S$ is the antipode of $A$ in $\omega$ . Now, let $K'$ be the intersection of $(MDI)$ with $BC$ , and $S'$ be the intersection of $MK'$ with $\omega$ . Then $\angle IDK'=\angle IMK'=\angle AMS'=90^{\circ}$ , so $S\equiv S'$ , hence $K\equiv K'$ and we're done.

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infiniteturtle

1131 posts

infiniteturtle

#8 h May 29, 2016, 4:15 AM • 7 Y

Y by v_Enhance, niyu, Kayak, samrocksnature, primesarespecial, Adventure10, Mango247

Every problem needs a nice BARY-BASH! The following computation is very instructive in the ways of bashing.

Let $A=(1,0,0),B=(0,1,0),C=(0,0,1)$ . Then $D=(0:s-c:s-b), E=(0:s-b:s-c),I=(a:b:c),I_a=(-a:b:c),M=(-a^2:b^2+bc:c^2+bc).$ All fine and good, but now how do we proceed? We need to think very carefully about how to make this barybash feasible. In particular, the definition of $T$ is nearly impossible to use as is. We need the synthetic observation that $T,D,M$ are collinear by homothety. This makes $T$ a workable point.

So now, we are going to intersect a bunch of lines with the circumcircle. The only way this is remotely possible is by using Vieta. Indeed, this bash will show just how ridiculously overpowered Vieta is...

Let $X=ME\cap \omega$ . Our solution path is as follows: compute $T=MD\cap \omega$ and $X$ by intersecting each line with the circle. Then we'll find $S=XI_a\cap \omega$ by intersecting this line with the circle, and finally check that $S$ lies on $XI$ .

Let's get started! First, the equation of $MD$ is
$\begin{vmatrix} x & y & z \\ -a^2 & b^2+bc & c^2+bc \\ 0 & s-c & s-b \end{vmatrix}=0\iff$ $\iff x(b+c)(b(s-b)-c(s-c))+y(a^2)(s-b)-z(a^2)(s-c)=0$ $\iff x(c^2-b^2)(s-a)+y(a^2)(s-b)-z(a^2)(s-c)=0$ $\iff z=x\cdot \frac{(c^2-b^2)(s-a)}{a^2(s-c)}+y\cdot \frac{(s-b)}{(s-c)}.$ The circumcircle has equation $a^2yz+b^2zx+c^2xy=0$ , so now we'll plug $z$ into the equation: $y^2\cdot \frac{a^2(s-b)}{(s-c)}+x^2\cdot \frac{b^2(c^2-b^2)(s-a)}{a^2(s-c)}+Qxy=0,$ for some function $Q$ of $a,b,c$ . Note that Vieta allows us to not care what $Q$ is! Now if we normalize to $x=1$ and observe that one solution for $y$ is when $M$ is the considered point, i.e. $y=\frac{-b(b+c)}{a^2}$ , we have $\left( \frac{-b(b+c)}{a^2}\right) \left(\frac{a^2(s-b)}{(s-c)}\right)y=\frac{(c^2-b^2)(s-a)}{a^2(s-c)}.$ This reduces to $y=\frac{b(b-c)(s-a)}{a^2(s-b)}.$ Now since $x=1$ we can easily calculate $z$ as $z=\frac{c(c-b)(s-a)}{a^2(s-c)},$ which is acceptable since it has to agree (symmetrically) with $y$ . Now we re-normalize to the cleaner form $T=\left(a^2(s-b)(s-c):b(b-c)(s-a)(s-c):c(c-b)(s-a)(s-b)\right).$ The computation of $X$ is virtually identical; we find $X=\left( a^2(s-b)(s-c):b(c-b)s(s-b):c(b-c)s(s-c) \right).$ (Note to reader: We can also synthetically observe that $T,X$ have the same x-coordinate [when normalized, although in the above forms they still do.] This could be used to find $X$ . The time expenditure of setting up such a bash instead of copying the one we already have for $T$ probably makes it counter-productive.)
Now we need to compute the equations of line $IT,I_aX$ . First, line $IT$ has equation
$\begin{vmatrix} x & y & z \\ a & b & c \\ a^2(s-b)(s-c) & b(b-c)(s-a)(s-c) & c(c-b)(s-a)(s-b) \end{vmatrix}=0$ $\iff x(bc)(s-a)(c-b)a+y(ac)(s-b)(a(s-c)+(b-c)(s-a))-z(ab)(s-c)(a(s-b)+(c-b)(s-a))=0.$ Now, we divide through by $a$ and, more importantly, observe that
$a(s-c)+(b-c)(s-a)=S_c, a(s-b)+(c-b)(s-a)=S_b.$ Thus line $IT$ has equation
$x(bc)(s-a)(c-b)+y(c)(s-b)S_c-z(b)(s-c)S_b=0.$ Again computing $I_aX$ is done in nearly the same way; somewhat surprisingly, we again find a nice simplification, this time
$a(s-b)+(b-c)s=S_c, a(s-c)+(c-b)s=S_b,$ which allows us to reduce the equation of $I_aX$ to
$x(bcs)(b-c)+y(c)(s-c)S_c-z(b)(s-b)S_b=0.$ Now we will compute $S$ as the intersection of $I_aX$ and $\omega$ . Perhaps it seems unintuitive to not intersect the two lines with each other; however, there are good reasons for intersecting with the circle:
(1) There isn't a terribly easy way to intersect two lines,
(2) we have already seen how powerful Vieta is, and
(3) if $S$ turns out to be nasty, it will be much easier to check if it lies on a line rather than if it lies on a circle.
(The choice to use $I_aX$ here instead of $IT$ is essentially arbitrary; their equations are very similar.)

So let's do this! From the equation for $I_aX$ we obtain
$z=x\cdot \frac{cs(b-c)}{(s-b)S_b}+y\cdot \frac{c(s-c)S_c}{b(s-b)S_b}.$ Plugging this into the circumcircle equation we find
$y^2\cdot \left(\frac{a^2c(s-c)S_c}{b(s-b)S_b}\right)+x^2\cdot \left( \frac{b^2cs(b-c)}{(s-b)S_b}\right)+Q\cdot xy=0,$ for some function $Q$ of $a,b,c$ . Normalize to $x=1$ , then one solution of this is $y=\frac{b(c-b)s(s-b)}{a^2(s-b)(s-c)}.$ Therefore the one we care about satisfies
$\frac{b(c-b)s(s-b)}{a^2(s-b)(s-c)}\cdot \frac{a^2c(s-c)S_c}{b(s-b)S_b}\cdot y=\frac{b^2cs(b-c)}{(s-b)S_b,}$ which miraculously simplifies to $y=\frac{-b^2}{S_c}$ . Back substituting, we find $z=\frac{-c^2}{S_b}$ , which is good since $y$ and $z$ are symmetrical. Re-normalizing, we find the cleaner form $S=(-S_bS_c:b^2S_b:c^2S_c).$ We're almost there! We just need to check that this lies on $IT$ , whose equation we already have. Plugging in, we want
$-S_bS_cbc(s-a)(c-b)+b^2c(s-b)S_bS_c-c^2b(s-c)S_bS_c=0.$ We can cancel the $6$ th degree $bcS_bS_c$ from everything, leaving only
$-(s-a)(c-b)+b(s-b)-c(s-c)=0.$ Now factoring $(b-c)$ from this, it becomes
$(b-c)(2s-a-b-c)=0,$ which is clearly true. Hooray!

Most important thing to note from this: Vieta is incredible, making intersecting a line with a circle very very easy (look at how ridiculously easily we compute $S$ .)

Also! I have seen this point $S$ before when bary bashing another problem... bash configuration recognition?

FWIW: This only took about 1.5 pages to solve, and is probably doable in an hour or so (I did it while watching a movie, so it took a while

)

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john111111

105 posts

john111111

#9 h Sep 10, 2016, 8:54 AM • 3 Y

Y by samrocksnature, opIst_w, Adventure10

My solution:
Define $E'$ as the intersection of $ME$ with $\omega$ and it suffices to prove that $E',S,I_A$ are collinear.
By the homothety of center $T$ which sends the circle tangent to $BC$ and $\omega$ to $\omega$ we easily deduce that $T,D,M$ are collinear.
Now we have that $\angle BTM=\angle BAM=\angle MBC$ so $MB$ is tangent to the circumcircle of $TBD$ so $MD \cdot MT=MB^2=MI^2$ so $MI$ is tangent to the circumcircle of $TDI$ so $\angle ITD=\angle DIM$ .
Now invert about the circle $(IBI_AC)$ with center $M$ and radius $MI^2$ .This inversion sends ( $BC$ common chord of $\omega$ and $M$ ) $\omega$ to $BC$ , $BC$ to $\omega$ .Then obviously $T$ goes to $D$ , $E$ goes to $E'$ and $S$ goes to the intersection of $MS$ with $BC$ , let it be $S'$ .
Then since $MS \cdot MS'=MI^2=MT \cdot MD$ , $S'SDT$ is cyclic so $\angle MS'D=\angle DTS= \angle DIM$ and thus $S'MDI$ is also cyclic so $\angle IMS'=IDS'=90$
Now since $\angle S'MI_A=\angle I_AES'=90$ we obtain that $MES'I_A$ is cyclic.
Thus the circle that passes through these 4 points (and $M$ the center of the inversion) after the inversion will become a line passing through the images of $I_A,E,S'$ .But since $I_A$ lies on $M$ it will map to itself and as we mentioned, $S'$ maps to $S$ and $E$ maps to $E'$ .So $I_A,S,E'$ are collinear and we are done.

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lebathanh

464 posts

lebathanh

#10 h Sep 11, 2016, 4:42 PM • 3 Y

Y by samrocksnature, Adventure10, ehuseyinyigit

my solution : lemma for ABC is a triangle have (I) is its incircle and (O) is its circumcircle. ID perpendicular BC . OM is a diameter of (O) MI cut (O) at S.prove SD through the midpoint of BC chord not contain A. (easy). from lemma we have AS is a diameter. AI*Aia =bc=AD*AS(AD is perpendicular BC) => DIM+ASIa=180 we only need prove ID//ME (easy by trigo)

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nikolapavlovic

1246 posts

nikolapavlovic

#11 h Mar 27, 2017, 2:11 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

$ME\cap \odot ABC=\{R\}$ , $ME\cdot MR=MB^2=MD\cdot MT$ , $ME=MD$ $\implies$ $MR=MT$ $\implies$ $BR=CT$ .Let $\odot I$ touch $AB,AC$ in $P,Q$ . $M-D-T$ $\implies$ $\tfrac{TP}{TQ}=\tfrac{TB}{TC}=\tfrac{BD}{CD}=\tfrac{BP}{CQ}$ $\implies$ $T\equiv \odot ABC\cap \odot APQ$ $\implies$ $\angle ATI=90^{\circ}$ and so $S$ is the antipode of $A$ . $SM\perp AM$ , $IM=I_AM$ $\implies$ $\angle ISM=\angle MSI_A$ .Let $SI_A\cap \odot =\{R'\}$ $\implies$ $\angle R'ST=\pi-2\angle TSM$ .
On the other hand $\angle RST=\angle RMT=\pi-2\angle MTR'$ but as $MR=MT$ $\angle TSM=\angle MTR'$ $\implies$ $R\equiv R'$ . $\blacksquare$

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andrei.pantea

172 posts

andrei.pantea

#12 h Mar 27, 2017, 2:48 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

nikolapavlovic wrote:

$M-D-T$ $\implies$ $\tfrac{TP}{TQ}=\tfrac{TB}{TC}=\tfrac{BD}{CD}=\tfrac{BP}{CQ}$

Could you please explain why this is true?

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nikolapavlovic

1246 posts

nikolapavlovic

#13 h Mar 27, 2017, 3:23 PM • 3 Y

Y by andrei.pantea, samrocksnature, Adventure10

$M$ is the midpoint of $\widehat{BC}$ and so by angle bisector theorem $\tfrac{TB}{TC}=\tfrac{BD}{CD}$ but as $BP=BD,CD=CQ$ we have that $T$ is the center of spiral similarity carrying $PQ\mapsto BC$ as $T\in \odot ABC$ .That's why the first eqality holds (it was there so that i can explain why the next paragraph would follow)

This post has been edited 1 time. Last edited by nikolapavlovic, Mar 27, 2017, 3:24 PM

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andrei.pantea

172 posts

andrei.pantea

#14 h Mar 27, 2017, 3:31 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

nikolapavlovic wrote:

$M$ is the midpoint of $\widehat{BC}$ and so by angle bisector theorem $\tfrac{TB}{TC}=\tfrac{BD}{CD}$ but as $BP=BD,CD=CQ$ we have that $T$ is the center of spiral similarity carrying $PQ\mapsto BC$ as $T\in \odot ABC$ .That's why the first eqality holds (it was there so that i can explain why the next paragraph would follow)

Thank you!

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FERZIO

74 posts

FERZIO

#15 h Apr 10, 2017, 3:55 PM • 2 Y

Y by samrocksnature, Adventure10

TelvCohl wrote:

My solution:

Let $T'=ME \cap (ABC)$ .

From homothety we get $T, D, M$ are collinear .

Since $BD=CE$ ,
so $T'$ is the reflection of $T$ in the bisector of $BC$ .
Since $\triangle MBD \sim \triangle MTB$ ,
so we get $MI^2=MB^2=MD \cdot MT$ .
ie. $\triangle MID \sim \triangle MTI$

Since $\text{Rt} \triangle BDI \sim \text{Rt} \triangle I_AEB$ ,
so we get $DM \cdot ET'=DM \cdot DT=DB \cdot DC=DB \cdot EB=DI \cdot EI_A$ ,
hence combine with $\angle I_AET'=\angle MDI$ we get $\triangle I_AET' \sim \triangle MDI$ ,
so from $\angle ET'I_A=\angle DIM=\angle ITM=\angle ET'S$ we get $T', S, I_A$ are collinear .
ie. $ME \cap I_AS \in (ABC)$

Q.E.D

Which app you use to draw this picture
Thx

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thdnder

194 posts

thdnder

#43 h Oct 18, 2023, 2:14 PM

Y by

I didn't want to invert this one, so I found solution without inversion.

It's well-known that $M, D, T$ are collinear.

Claim:
$AS$ is diameter of $\omega$ .

Proof:
Since $MI^2 = MD \cdot MT$ , so $\angle{MDT} = \angle{MIT}$ . Note that $\angle{MDT} = \angle{MDC} + 90^{\circ} = \angle{TAI} + 90^{\circ}$ . Hence $\angle{ATI} = 90^{\circ}$ , so $AS$ is diameter of $\omega$ . $\blacksquare$

Since $\angle{ATI} = 90^{\circ}$ , thus $(ATI)$ and $(BICI_A)$ are tangent to each other. Let $l$ be line passing through $I$ and tangent to $(BICI_A)$ . Then three radical axis of $(ATI), (BICI_A), \omega$ are concurrent, so $l, BC, AT$ are concurrent. Let $R = l \cap BC \cap AT$ . Let $X = ME \cap \omega$ . We'll prove that $X, S, I_A$ are collinear. Since $\angle{AXS} = 90^{\circ}$ , so it's enough to show that $\angle{AXI_A} = 90^{\circ}$ .

Claim:
$A$ is the center of spiral similarity sending $RI$ to $I_AX$ .

Proof:
Since $TM, MX$ are symmetric about perpendicular bisector of $BC$ , so $\angle{RAI} = \angle{XAI}$ , so $\angle{RAB} = \angle{CAX}$ . Since $\angle{ABR} = \angle{AXC}$ , so $A$ is the center of spiral similarity sending $RB$ to $CX$ . Thus $AR \cdot AX = AB \cdot AC$ . It's not hard to check that $AI \cdot AI_A = AB \cdot AC$ , so $\frac{AR}{AI} = \frac{AI_A}{AX}$ and since $\angle{RAI} = \angle{I_AAX}$ , therefore $A$ is the center of spiral similarity sending $RI$ to $I_AX$ . $\blacksquare$

Hence $\angle{AXI_A} = \angle{AIR} = 90^{\circ}$ , as needed. $\blacksquare$

This post has been edited 1 time. Last edited by thdnder, Dec 9, 2023, 6:01 PM

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BVKRB-

322 posts

BVKRB-

#44 h Oct 20, 2023, 1:59 PM

Y by

Well I did read something related to fact 5 inversion in this thread but to my defence I can say that atleast the idea came to me naturally xD
And of course if I don't solve a problem normally I always invert

, also I am so out of touch that I didn't even recognize the sharky devil point

It is obvious that $T=MD \cap \omega$ because of Curvilinear circle lemma from EGMO (or homothety or inversion).

Now invert with respect to $\odot(BICI_A)$ and let the image of $S$ be $S'$ and it becomes obvious that $S$ is actually just the antipode of $A$ (because $MS' \perp AM$ which would be obvious if I had recognized the point :blush:

). Due to the inversion we also obtain that $S'E \perp EI_A$ as $S' \in \overline{BC}$ and hence we obtain that $ES'I_AM$ is cyclic. By inverting back we get that this concyclicity directly implies the question as the intersection point is the inverted image of $E$ $\blacksquare$

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lelouchvigeo

179 posts

lelouchvigeo

#45 h Dec 26, 2023, 10:40 AM • 1 Y

Y by ehuseyinyigit

Very beautiful configuration
Here is a sketch
Observe
- TDM collinear
- S is antipode of A
- Let Excentre of A and S meet the circumcircle at X
Angle EXS=DTH' (Here H' is reflection of orthocenter on the circumcircle)
This is due to BD=CE

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shendrew7

794 posts

shendrew7

#46 h Jan 2, 2024, 1:23 PM

Y by

We note that $T$ is simply $MD \cap (ABC)$ , or the Sharkydevil point. Thus $S$ is the antipode of $A$ , so if we define $K = ME \cap (ABC)$ , it suffices to prove $\angle AKI_A = 90$ .

Shooting Lemma along with Incenter-Excenter tells us
$ME \cdot MK = MC^2 = MI_A^2 \implies \triangle MEI_A \sim \triangle MI_AK,$ which indeed gives
$\measuredangle AKI_A = \measuredangle AKM + \measuredangle MKI_A = \measuredangle (AM, BC) + \measuredangle EI_AM = 90. \quad \blacksquare$

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john0512

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john0512

#47 h Apr 26, 2024, 3:07 AM

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Note that by homothety $T$ , $D$ , $M$ are collinear, which means that $T$ is the $A$ -Sharkydevil point. This also means that $T$ lies on $(AI)$ , so $S$ is the antipode of $A$ on $(ABC)$ . Furthermore, $MD$ and $ME$ are reflections across the perpendicular bisector of $BC$ , so if $ME$ intersects the circumcircle at $K$ , then $K$ is the reflection of the $A$ -sharkydevil point across the perpendicular bisector of $BC$ . It suffices to show that $K$ , $S$ , and $I_a$ are collinear.

To do this, we perform a $\sqrt{bc}$ inversion followed by a reflection across the bisector. The image of $I_a$ is $I$ . The image of $S$ is the foot from $A$ to $BC$ , which we call $X$ . Since $TBCK'$ is an isosceles trapezoid, $AT$ is isogonal to $AK$ , which means that the image of $K$ lies on $AT$ , which means it is the intersection of $AT$ with $BC$ . However, by Radical Axis Theorem on $(ABC)$ , $(AI)$ , and $(BIC)$ , we have that $AT$ , $II$ , and $BC$ concur, so $K'$ lies on $II$ . Since $(AI)$ and $(BIC)$ are tangent, we have $\angle K'IA=90$ . Since $\angle K'XA=90$ as well, $K'XIA$ is cyclic, so we are done.

remark: After the antipode remark, an alternative way to finish would be to note that $ME\cdot MK=MD\cdot MT=MB^2=MI_a^2$ by the tangent circles configuration, after which angle chasing suffices as we can show $\angle AKI_a=90$ . The idea here is that this angle that we want to equal 90 is composed of an angle that we know very well (since it is inscribed in $AM$ ) and a more strange angle, suggesting that the strange angle has a nice value as well. We can then actually work out what that angle must be, and from our angle chasing realize that it's equal to $\angle MI_aE$ and we immediately know what to do from here.

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Aiden-1089

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Aiden-1089

#48 h Apr 26, 2024, 8:55 AM

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Note that $T,D,M$ are collinear, so $T$ is the $A$ -Sharky-Devil point. It follows that $S$ is the antipode of $A$ in $\omega$ .
Let $ME$ intersect $\omega$ again at $X$ , then $X$ is the reflection of $T$ across the angle bisector of $BC$ .
Let $S'=MS \cap BC$ . Inverting about $(BIC)$ , $X$ goes to $E$ and $S$ goes to $S'$ . It suffices to show that $M,E,S',I_A$ are concyclic. But $\measuredangle I_AMS' = \measuredangle AMS = 90^{\circ} = \measuredangle I_AES'$ , so we are done. $\square$

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Shreyasharma

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Shreyasharma

#49 h Apr 27, 2024, 12:53 AM

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Let $X = \overline{AI} \cap \overline{BC}$ . Perform a $\sqrt{bc}$ inversion followed by reflection across $\overline{AI}$ and note,

$E$ is sent to the mixintillinear touch point, which we call $K$ .
$I$ and $I_a$ swap positions.
$X$ and $M$ swap positions.
$S$ maps to the foot from $A$ to $\overline{BC}$ , say $F$ .

Then it suffices to show that $(AFI)$ and $(AKX)$ concur on $\overline{BC}$ . Now let $\overline{AT}$ meet $\overline{BC}$ at $Y$ . However as $YT \cdot YA = YB \cdot YC$ as $ABTC$ cyclic, we find $\overline{IY} \perp \overline{AI}$ . Then inverting about $(BIC)$ note that $Y$ and $K$ swap positions and hence $M$ , $K$ and $Y$ collinear. Now note that $A$ and $D$ swap under this transformation and hence $AM \cdot MD = MY \cdot MK$ which gives our first desired concylicity. The other follows as $\angle AFY = \angle AIY = 90$ . Thus we are done.

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Eka01

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Eka01

#50 h Aug 3, 2024, 6:22 AM • 1 Y

Y by Sammy27

Finally got around to writing solutions.
Click to reveal hidden text

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kotmhn

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kotmhn

#51 h Nov 5, 2024, 2:03 PM

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I don't know why this is in spi sim when its trivial by inversion.

Invert at $M$ with radius $MB$ .
Under this
$\begin{align*} & I \leftrightarrow I\\ & I_a \leftrightarrow I_a\\ & T \leftrightarrow D\\ & S \leftrightarrow MS\cap BC\\ & ME \leftrightarrow ME\\ & I_aS \leftrightarrow (I_aMS^{*})\\ \end{align*}$ The problem reduces to showing that $ME \cap (I_aMS^{*}) \in BC$
Then observe that $M,D,I,S^{*}$ is cyclic(due to inversion).
Hence
$\angle S^{*}MI_a = \angle S^{*}MD = 90 = \angle S^{*}EI_a$ So $E \in (I_aMS^{*})$ hence we are done as $E\in BC$ .

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cj13609517288

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cj13609517288

#52 h Nov 14, 2024, 10:04 PM

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Let $P$ and $Q$ be the feet of the perpendiculars from $I$ to $AB$ and $AC$ , and let $N$ be the antipode of $M$ .

Let $T'$ be the reflection of $T$ over the perpendicular bisector of $BC$ . We claim that this is the desired point.

First, note that by the shooting lemma we get $TDM$ collinear. Reflecting this over the perpendicular bisector of $BC$ shows $T'EM$ collinear. So it suffices to show that $I_AST'$ collinear. Then we want
$-1=(TT';NM)\stackrel{S}{=}(II_A;(NS\cap AM)M).$ Thus we want $NS\parallel AM$ . This is equivalent to $S$ being the $A$ -antipode, which is equivalent to $\angle ATI=90^{\circ}$ . By the shooting lemma,
$MI^2=MB^2=(MD)(MT)\Longrightarrow \angle ITM=\angle DIM=\angle AMN=\angle ATN.$ So $\angle NTM=\angle ATI$ too. Yay. $\blacksquare$

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bin_sherlo

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bin_sherlo

#53 h Dec 5, 2024, 11:54 AM

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Denote by $AT\cap BC=R,ME\cap (ABC)=K$ .
By shooting lemma $T$ is $A-$ sharky devil point hence $S$ is the antipode of $A$ on $(ABC)$ . $A$ Under $\sqrt{bc}$ inversion and reflection over the angle bisector of $\measuredangle CAB$ , $I_A,K$ swap with $I,R$ and $S$ swaps with the foot of the altitude from $A$ to $BC$ which all lie on the circle with diameter $AR$ as desired. $\blacksquare$

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Mathandski

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Mathandski

#54 h Dec 17, 2024, 6:27 PM

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Invert >>>> spiral

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Mathandski

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Mathandski

#56 h Dec 17, 2024, 10:03 PM

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Alternate spiral solution

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sky.mty

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sky.mty

#57 h Jan 2, 2025, 2:32 PM

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easy one

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Aiden-1089

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Aiden-1089

#58 h Jan 2, 2025, 6:21 PM

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Solution using projective.

By shooting lemma $T-D-M$ collinear, so $T$ is the $A$ -Sharky-Devil point, so $S$ is the $A$ -antipode in $(ABC)$ .
Now let $N$ be the midpoint of arc $BAC$ , then clearly $AM//SN$ .
We have $-1=(I,I_A;M,\infty_{AM}) \stackrel{S}{=} (T,SI_A \cap (ABC);M,N)$ , so $SI_A$ intersects $(ABC)$ at the reflection of $T$ across $\perp_{BC}$ , which lies on $ME$ since $BD=CE$ . $\square$

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