Show that XD and AM meet on Gamma

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MathStudent2002

934 posts

MathStudent2002

#1 h Jul 19, 2017, 4:32 PM • 23 Y

Y by Davi-8191, tenplusten, anantmudgal09, doxuanlong15052000, nguyendangkhoa17112003, AlastorMoody, Pluto1708, yushanlzp, MathbugAOPS, Gaussian_cyber, aops5234, Lilathebee, megarnie, Jupiter_is_BIG, tiendung2006, AlienGirl05, Adventure10, Mango247, Rounak_iitr, deplasmanyollari, Sedro, Funcshun840, cursed_tangent1434

Let $ABC$ be a triangle with circumcircle $\Gamma$ and incenter $I$ and let $M$ be the midpoint of $\overline{BC}$ . The points $D$ , $E$ , $F$ are selected on sides $\overline{BC}$ , $\overline{CA}$ , $\overline{AB}$ such that $\overline{ID} \perp \overline{BC}$ , $\overline{IE}\perp \overline{AI}$ , and $\overline{IF}\perp \overline{AI}$ . Suppose that the circumcircle of $\triangle AEF$ intersects $\Gamma$ at a point $X$ other than $A$ . Prove that lines $XD$ and $AM$ meet on $\Gamma$ .

Proposed by Evan Chen, Taiwan

This post has been edited 2 times. Last edited by v_Enhance, May 6, 2019, 1:36 PM

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nikolapavlovic

1246 posts

nikolapavlovic

#2 h Jul 19, 2017, 4:33 PM • 6 Y

Y by MathStudent2002, tenplusten, Lilathebee, megarnie, Adventure10, Mango247

Let $D'$ be the the touch point of $A$ -excircle and let $Q$ be the isogonal conjugate of harmonic conjugate of $D'$ wrt $BC$ .Note that $E,F$ are intouch points of $A$ -mixtlinear incircle(well-known) so by $\sqrt{bc}$ -inversion( $E,F$ got carried to the touch points of the A-excircle) $A,X,Q$ are collinear $\implies$ $X(B,C;Q,D)=(B,C;A,XD\cap \odot ABC,A)$ but $D,D'$ are are isotomic conjugates by Steiner's(and $BD=CD'$ ) $(B,C;Q,D)=\tfrac{AB^2}{AC^2}=(B,C;A,M)$ and hence $XD\cap \odot ABC\equiv M$ .

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v_Enhance

6870 posts

v_Enhance

#3 h Jul 19, 2017, 4:38 PM • 74 Y

Y by MSTang, anantmudgal09, dungnguyentien, tenplusten, Ankoganit, huricane, doxuanlong15052000, samuel, vsathiam, brokendiamond, rkm0959, tritanngo99, e_plus_pi, naw.ngs, Wizard_32, yayups, niyu, nguyendangkhoa17112003, AlastorMoody, RAMUGAUSS, sriraamster, thczarif, HolyMath, Vidush, Aryan-23, myh2910, moab33, GeoMetrix, amar_04, Bassiskicking, rashah76, lahmacun, MathbugAOPS, OlympusHero, Arshia.esl, Siddharth03, srijonrick, hsiangshen, super.shamik, tree_3, v4913, Kanep, TETris5, guptaamitu1, math31415926535, Executioner230607, ike.chen, HamstPan38825, player01, FishHeadTail, Pranav1056, CyclicISLscelesTrapezoid, Lilathebee, megarnie, rayfish, David-Vieta, Taco12, Iora, egxa, IMUKAT, starchan, dgkim, Adventure10, Mango247, nmoon_nya, Rounak_iitr, Kingsbane2139, EpicBird08, Math_legendno12, Stuffybear, Ishaan_Mittal_374877, xyr, Sedro, NicoN9

This problem was proposed by me.

First Solution (Joshua Lee): Letting $D_1$ be the extouch point, we observe that $\measuredangle\left( \overline{XM}, \overline{AD_1} \right) = \measuredangle XMI = \measuredangle XBF = \measuredangle XBA$ since $\overline{IM} \parallel \overline{AD_1}$ , and $\triangle XBF \sim \triangle XMI$ by spiral similarity.

Thus $\overline{XM}$ and $\overline{AD_1}$ meet on $\Gamma$ . Now by the Butterfly Theorem, $\overline{XD}$ and $\overline{AM}$ meet on $\Gamma$ as well.

Second Solution (mine): Let $\omega_A$ denote the $A$ -mixtilinear incircle, tangent to $\overline{AB}$ and $\overline{AC}$ at $F$ and $E$ . Suppose $\omega_A$ is tangent to $\Gamma$ at $T$ . Again let line $AM$ meet $\Gamma$ again at $K$ . Let $D_1$ and $T_1$ be the reflections of $D$ and $T$ with respect to the perpendicular bisector of $\overline{BC}$ .

$[asy] size(10cm); pair A = dir(115); pair B = dir(210); pair C = dir(330); draw(A--B--C--cycle, red); pair I = incenter(A, B, C); pair O = circumcenter(A, B, C); pair M_A = -A+2*foot(O, A, I); pair M_B = -B+2*foot(O, B, I); pair M_C = -C+2*foot(O, C, I); draw(unitcircle, blue); pair L = dir(90); pair T = -L+2*foot(O, I, L); pair E = extension(T, M_B, A, C); pair F = extension(T, M_C, A, B); pair D = foot(I, B, C); draw(incircle(A, B, C), red); pair M = midpoint(B--C); pair K = -A+2*foot(O, A, M); draw(circumcircle(A, E, F), blue); pair X = -K+2*foot(O, K, D); draw(circumcircle(E, F, T), heavygreen); pair D_1 = 2*M-D; pair T_1 = -A+2*foot(O, A, D_1); draw(A--T, red); draw(A--T_1, red); pair S = -T+2*foot(circumcenter(T, E, F), A, T); pair R = extension(A, X, F, E); draw(R--A, purple); draw(R--S, purple); draw(R--E, purple); draw(R--T, purple); draw(X--T_1, grey+dashed); draw(A--K--X, orange+dotted); draw(B--X--C, lightcyan); draw(I--M, grey+dashed); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I$", I, dir(-100)); dot("$T$", T, dir(T)); dot("$E$", E, dir(E)); dot("$F$", F, dir(230)); dot("$D$", D, dir(D)); dot("$M$", M, dir(M)); dot("$K$", K, dir(K)); dot("$X$", X, dir(X)); dot("$D_1$", D_1, dir(45)); dot("$T_1$", T_1, dir(T_1)); dot("$S$", S, dir(130)); dot("$R$", R, dir(R)); /* TSQ Source: !size(10cm); A = dir 115 B = dir 210 C = dir 330 A--B--C--cycle 0.1 lightred / red I = incenter A B C R-100 O := circumcenter A B C M_A := -A+2*foot O A I M_B := -B+2*foot O B I M_C := -C+2*foot O C I unitcircle 0.1 lightcyan / blue L := dir 90 T = -L+2*foot O I L E = extension T M_B A C F = extension T M_C A B R230 D = foot I B C incircle A B C 0.1 lightred / red M = midpoint B--C K = -A+2*foot O A M circumcircle A E F 0.1 lightcyan / blue X = -K+2*foot O K D circumcircle E F T 0.2 lightgreen / heavygreen D_1 = 2*M-D R45 T_1 = -A+2*foot O A D_1 A--T red A--T_1 red S = -T+2*foot circumcenter T E F A T R130 R = extension A X F E R--A purple R--S purple R--E purple R--T purple X--T_1 grey dashed A--K--X orange dotted B--X--C lightcyan I--M grey dashed */ [/asy]$

It is well known that $\angle BAT = \angle CAD_1$ . (To prove this, notice that an inversion at $A$ with $\sqrt{AB \cdot AC}$ plus a reflection around the angle bisector swaps the $A$ -excircle and $A$ -mixtilinear incircle, so it also swaps $T$ and $D_1$ .) Thus $A$ , $D_1$ , $T_1$ are collinear.

Now we will prove that $X$ , $M$ , and $T_1$ are collinear. Let $R$ be the radical center of $\omega_A$ , $\Gamma$ , and the circumcircle of triangle $AEF$ . Moreover, let $\overline{AT}$ meet the $A$ -mixtilinear circle again at $S$ . We see that $SFTE$ is harmonic quadrilateral, and therefore that $(R, \overline{AT} \cap \overline{EF}; F, E) = -1$ is harmonic. Upon projecting onto $\Gamma$ from $A$ we see that $XBTC$ is a harmonic quadrilateral. Since $\overline{TT_1} \parallel \overline{BC}$ , using perspectivity at $T_1$ we conclude that $X$ , $M$ , $T_1$ are collinear.

Now, applying the Butterfly Theorem, we see that $X$ , $D$ , $K$ are collinear as well, as required.

This post has been edited 4 times. Last edited by dcouchman, Jul 20, 2017, 5:04 PM
Reason: In paragraph 2, replaced "Thus, $T$, $D_1$, $T_1$ are colinear" with "$A$, $D_1$, $T_1$ are colinear."

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ShinyDitto

63 posts

ShinyDitto

#4 h Jul 19, 2017, 4:47 PM • 5 Y

Y by Lilathebee, Adventure10, Mango247, Rounak_iitr, ehuseyinyigit

Thank you @v_Enhance for your document A Guessing Game: Mixtilinear Incircles. When I tried this problem at home, I managed to do it in a very short time (:

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EulerMacaroni

851 posts

EulerMacaroni

#5 h Jul 19, 2017, 5:10 PM • 4 Y

Y by Aryan-23, Steff9, Lilathebee, Adventure10

Let $T$ be the $A$ -mixtilinear touchpoint on $\odot(ABC)$ ; note by $\sqrt{bc}$ inversion that $XBTC$ is harmonic, so projecting through $D$ we want to show that $TD$ passes through the point $A'$ on $\odot(ABC)$ with $AA'\parallel BC$ , which is well-known.

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Anzoteh

124 posts

Anzoteh

#6 h Jul 19, 2017, 5:22 PM • 7 Y

Y by Abderrahmane_Driouch, utlight, Lilathebee, Adventure10, Mango247, sabkx, ehuseyinyigit

As per my personal hobby it's time to trigo-bash the whole thing again

WLOG let $AB<AC$ .

First, well-known spiral similarity property should dictate the similarity of triangles $BXF$ an $CXE$ , so $\frac{CX}{CE}=\frac{BX}{BF}$ .
Also, let's also invoke an identity for triangles (feel free to verify it; I'm not gonna do this):
$\frac{BX}{XC}\cdot \frac{\sin\angle BXD}{\sin\angle CXD}=\frac{BD}{DC}.$ Denoting $N_1$ as the other intersection of $XD$ and $\Gamma$ gives $\frac{\sin\angle BXD}{\sin\angle CXD}=\frac{BN_1}{CN_1}.$
Similarly we have $\frac{AB}{AC}\cdot \frac{\sin\angle ABM}{\sin\angle ACM}=\frac{BM}{CM}=1$ .
ALso let $N_2$ as the other intersection of $AM$ and $\Gamma$ and we have $\frac{\sin\angle ABM}{\sin\angle ACM}=\frac{BN_2}{CN_2}.$
Therefore all we need is $\frac{\sin\angle ABM}{\sin\angle ACM}=\frac{\sin\angle BXD}{\sin\angle CXD}$ ,
and it's not hard to see that $\frac{\sin\angle ABM}{\sin\angle ACM}=\frac{AC}{AB}$ ,
so we are left with proving the fact
$\frac{BF}{EC}\cdot\frac{AC}{AB}=\frac{BD}{DC}$ .

Now, $\frac{BD}{DC}=\frac{\tan\frac12\angle C}{\tan\frac12\angle B}$ ,
$\frac{AC}{AB}=\frac{\sin\angle B}{\sin\angle C}=\frac{2\sin\frac 12\angle B\cos\frac 12\angle B}{2\sin\frac 12\angle C\cos\frac 12\angle C}$ .
Also $IE=IF$ , and by angle chasing we have $\angle FIB=\angle ICE=\frac12\angle C$ ,
$\angle EIC=\angle IBF=\frac12\angle B$ .
Therefore $BIF$ and $ICE$ similar, yielding $\frac{BF}{EC}=(\frac{BF}{FI})^2=(\frac{\frac12\angle C}{\frac12\angle B})^2$ ,
now it's no longer difficult to prove that $(\frac{\frac12\angle C}{\frac12\angle B})^2\cdot \frac{2\sin\frac 12\angle B\cos\frac 12\angle B}{2\sin\frac 12\angle C\cos\frac 12\angle C}=\frac{\tan\frac12\angle C}{\tan\frac12\angle B}$ .

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anantmudgal09

1979 posts

anantmudgal09

#7 h Jul 19, 2017, 7:34 PM • 9 Y

Y by Yamcha, Ankoganit, e_plus_pi, Wizard_32, GammaBetaAlpha, guptaamitu1, starchan, Adventure10, Mango247

MathStudent2002 wrote:

Let $ABC$ be a triangle with circumcircle $\Gamma$ and incenter $I$ and let $M$ be the midpoint of $\overline{BC}$ . The points $D$ , $E$ , $F$ are selected on sides $\overline{BC}$ , $\overline{CA}$ , $\overline{AB}$ such that $\overline{ID} \perp \overline{BC}$ , $\overline{IE}\perp \overline{AI}$ , and $\overline{IF}\perp \overline{AI}$ . Suppose that the circumcircle of $\triangle AEF$ intersects $\Gamma$ at a point $X$ other than $A$ . Prove that lines $XD$ and $AM$ meet on $\Gamma$ .

Proposed by Evan Chen, Taiwan

Consider the following

Claim: $\frac{BX}{CX}=\left(\frac{BA}{CA}\right) \cdot \left(\frac{BD}{CD}\right)$

(Proof) Note that $X$ is the spiral center for $BE \mapsto CF$ so $\frac{BX}{CX}=\frac{BE}{CF}=\frac{\left(\tfrac{BE}{IE}\right)}{\left(\tfrac{CF}{IF}\right)}=\left(\frac{\sin \tfrac{C}{2}}{\sin \tfrac{B}{2}}\right)^2,$ by applying sine law to $\triangle BEI$ and $\triangle CFI$ . Next, we see that $\left(\frac{BA}{CA}\right) \cdot \left(\frac{BD}{CD}\right)=\left(\frac{\sin C}{\sin B}\right) \cdot \left(\frac{BI\cos \tfrac{B}{2}}{CI\cos \tfrac{C}{2}}\right).$ Apply sine law to $\triangle IBC$ and use the double-angle formula for sines to conclude that both sides are equal. $\blacksquare$

Let $AM$ meet $\Gamma$ again at point $T$ and let $TD$ meet $\Gamma$ again at $Y$ . Project line $BC$ onto circle $\Gamma$ from $T$ to get $(BC, XA)=(BC, DM) \overset{T}{=} (BC, YA),$ so $X=Y$ just as we desired. $\blacksquare$

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ABCDE

1963 posts

ABCDE

#8 h Jul 19, 2017, 11:39 PM • 10 Y

Y by DrMath, anantmudgal09, joshualee2000, kingofgeedorah, mathwizard888, yayups, Kagebaka, Adventure10, Mango247, sabkx

Here is a fun solution I found:

We make use of the following Pingpong Lemma: Let $P_1,P_2,P_3,P_4$ be points on secant $XY$ of circle $\omega$ . Say a point $P$ on $\omega$ is good if $P_1(P_2(P_3(P_4P\cap\omega)\cap\omega)\cap\omega)\cap\omega=P$ . If a point $P$ different from $X$ and $Y$ on $\omega$ is good, then all points on $\omega$ are good.

Let $Y=EF\cap BC$ , $N=AI\cap\Gamma$ , $Z=AM\cap\Gamma$ , $A'\in\Gamma$ satsify $AA'\parallel BC$ , $K$ be the foot of the $A$ -external-angle-bisector, $L=AK\cap\Gamma$ , $T=NY\cap\Gamma$ , and $P=A'M\cap\Gamma$ . First, note that $X$ is the center of spiral similarity taking $LM$ to $AI$ , $X\in(IMN)=(NY)$ , so $Y\in ML$ . We apply the Pingpong Lemma with $M,K,Y,D$ on secant $BC$ of circle $\omega$ . Note that $ABPC$ is harmonic, so $P\in NK$ , and that $T$ is the $A$ -mixtilinear-touchpoint, so $D\in A'T$ . Thus, $A'P,PN,NT,TA'\cap\Gamma=M,K,Y,D$ , so $A'$ is good. Hence, $Z$ is also good. As $ZM\cap\Gamma=A,AK\cap\Gamma=L,LY\cap\Gamma=X$ , we have that $XD\cap\Gamma=Z$ , as desired.

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WizardMath

2487 posts

WizardMath

#9 h Jul 20, 2017, 9:03 AM • 2 Y

Y by e_plus_pi, Adventure10

Cute problem.

My solution.
The parallel through A cuts the circumcircle again at A'. AM meets the circumcircle again at Z. Then By reflection in the perpendicular bisector of BC, noting that the A median and the A symmedian are isogonal, A'B'C' is harmonic. By a $\sqrt{bc}$ inversion at $A$ , noting that AT is isogonal to the A Nagel cevian and that X goes to the harmonic conjugate of the excircle touchpoint with BC, we have that XTBC is harmonic. Inversion at the incircle touchpoint swapping B and C preserves cross ratios of point s on the circumcircle, and thus we are done, by noting that the line TA' meets BC at the incircle touchpoint, from a well known property that if D is the incircle touchpoint, then $\angle DTB = \angle ABC$ and angle chasing.

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wu2481632

4233 posts

wu2481632

#10 h Jul 20, 2017, 3:28 PM • 2 Y

Y by e_plus_pi, Adventure10

long and convoluted

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MarkBcc168

1594 posts

MarkBcc168

#11 h Jul 21, 2017, 10:36 AM • 8 Y

Y by v_Enhance, govind7701, e_plus_pi, Quidditch, Iora, Adventure10, Mango247, H_Taken

Can't believe that no one has posted this solution yet.

Bary Bash

This post has been edited 1 time. Last edited by MarkBcc168, Jan 2, 2018, 4:09 AM

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dungnguyentien

105 posts

dungnguyentien

#12 h Jul 21, 2017, 4:26 PM • 3 Y

Y by buratinogigle, Elyson, Adventure10

Another solution for this problem.

Attachments:

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TYERI

7 posts

TYERI

#13 h Jul 27, 2017, 8:53 PM • 3 Y

Y by mhq, Adventure10, Mango247

Let $N$ be the midpoint of arc $BAC$ , $N'$ be the midpoint of another arc $BC$ , $D_1$ , $A_1$ be the reflections of points $D$ , $A$ across the line $MN$ . By easy angle-chasing problems condition is reduced to $\angle DXM=\angle IMA$ . It's well-known that $MI \parallel D_1A$ , so $\angle IMA=\angle MAD_1 = \angle DA_1M$ , so it's enough to proof that $XDMA_1$ is cyclic.
Let $L= BC \cap EF$ . Consider spiral similarity which takes $BC$ to $EF$ . It's well-known that $X$ is it's center. It also maps $N$ into $A$ and $M$ into $I$ so $X,I,M,L$ are concylic. Let it map $A_1$ into $A'$ . By spiral similarity $\angle AXN=\angle NXA_1=\angle AXA'$ , so $X$ , $A'$ , $N$ are collinear.
Let $R=BE \cap CF$ , $R'=AR \cap BC$ . By Ceva's theorem $\frac{AF\cdot BR'\cdot CE}{FB\cdot R'C\cdot EA}=1$ , so $\frac{XB}{XC}=\frac{BF}{CE}=\frac{BR'}{R'C}$ , so $XR'$ is a bisector of $\angle BXC$ ,so it's passes through $N'$ . Projecting harmonic quadrilateral $BNCN'$ from $X$ we get that $L\in XN$ .
From the above $\angle A_1DM=\angle AD_1M=\angle IML= \angle IXA'=\angle MXA_1$ , so $A_1,D,M,X$ are concylic.

This post has been edited 2 times. Last edited by TYERI, Jul 22, 2024, 10:41 AM

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suli

1498 posts

suli

#14 h Aug 9, 2017, 5:45 AM • 4 Y

Y by absurdist, gemcl, Adventure10, kinnikuma

This is a great example of circular reasoning: XD
Solution

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trumpeter

3332 posts

trumpeter

#15 h Sep 7, 2017, 9:41 PM • 2 Y

Y by Adventure10, Mango247

Solution

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akasht

83 posts

akasht

#81 h Feb 23, 2024, 10:53 AM

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Let $K$ be the excircle touchpoint on $BC$ . $E,F$ are the touchpoints of the $A$ -mixtillinear incircle. Hence under $\sqrt{bc}$ -inversion, we see that $X'$ is the harmonic conjugate of $K$ . Using the inversion distance formula, we have:
$\frac{XB}{XC} = \frac{\frac{r^2}{AX' \cdot AC}}{\frac{r^2}{AX' \cdot AB}} \cdot \frac{X'C}{X'B} = \frac{AB}{AC} \cdot \frac{CK}{BK} = \frac{AB}{AC} \cdot \frac{BD}{CD}$ But by ratio lemma, $\frac{XB}{XC} \cdot \frac{\sin \angle BXD}{\sin \angle CXD} = \frac{BD}{CD} \text{ and } \frac{AB}{AC} \cdot \frac{\sin \angle BAM}{\sin \angle CAM} = 1$ therefore $\frac{\sin\angle BXD}{\sin \angle CXD} = \frac{\sin \angle BAM}{\sin \angle CAM}$ , combined with $\angle BXD + \angle CXD = \angle BAM + \angle CAM$ this implies that $\angle BXD = \angle BAM$ and $\angle CXD = \angle CAM$ , which means $XD$ and $AM$ concur on $(ABC)$ .

This post has been edited 4 times. Last edited by akasht, Feb 23, 2024, 10:59 AM

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YaoAOPS

1500 posts

YaoAOPS

#82 h Mar 31, 2024, 6:20 AM

Y by

Worse ping pong app.

Let $S$ be the Sharkey-Devil point, and let $T = AS \cap BC \cap EF$ . Let $N$ be the midpoint of arc $\widehat{BC}$ , let $L$ be the midpoint of $\widehat{ABC}$ .

Claim: $T$ lies on $AX'$ .
Proof. Well known, but note that $X$ is the center of the spiral sim from $FI$ to $BM$ . As such, $TXIM$ is cyclic which gives that $\angle AXM_A$ is a right angle. $\blacksquare$
We now ping pong on $D, M, T, SX \cap BC$ with $S \to M_A \to N \to X \to S$ , so it follows that $X \to XD \cap \Gamma \to A \to S \to X$ is a valid cycle.

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Aiden-1089

277 posts

Aiden-1089

#83 h Apr 1, 2024, 2:05 AM

Y by

Let $T$ be the $A$ -mixtilinear intouch point, $U$ be the $A$ -extouch point. First we claim that $(B,C;X,T)=-1$ .
Proof: Taking $\sqrt{bc}$ inversion, the $A$ -mixtilinear incircle is taken to the $A$ -excircle. Since $E$ and $F$ lie on the $A$ -mixtilinear incircle, $E', F',T'$ are the tangency points between the $A$ -excircle and lines $AB$ , $AC$ , $BC$ respectively, so $T'=U$ . Thus, $X'$ is the intersection between $BC$ and $E'F'$ . It suffices to show that $(B,C;X',U)=-1$ , or that lines $AU, BF', CE'$ are concurrent, which is well-known.

Let $\perp_{BC}$ denote the perpendicular bisector of $BC$ , $A'$ be the reflection of $A$ across $\perp_{BC}$ . We claim that $T,D,A'$ are collinear.
Proof: Let $V$ be the second intersection between line $AU$ and $\Gamma$ . Since $\angle BAT = \angle CAU$ , $V$ is the reflection of $T$ across $\perp_{BC}$ . It is well-known that $DM=UM$ , so reflecting the line $AUV$ gives the line $TDA'$ .

Now we can prove the desired result. Let $P$ be the second intersection between line $XD$ and $\Gamma$ , $\infty_{BC}$ denote the point at infinity along $BC$ .
Note that $-1=(C,B;X,T) \stackrel{D}{=} (B,C;P,A') \stackrel{A}{=} (B,C;AP \cap BC,\infty_{BC})$ from which is follows that $A,M,P$ are collinear, so we are done.

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EpicBird08

1740 posts

EpicBird08

#84 h May 9, 2024, 12:51 AM

Y by

i am a geo(gebra) main :skull:

Reflect $D$ across $M$ to get $D',$ and let $X'$ be such that $BXX'C$ is an isosceles trapezoid. Let $AM$ intersect $\Gamma$ at point $G,$ and let $G'$ be such that $BG'GC$ is an isosceles trapezoid. We desire to show that $X', D', G'$ are collinear. It is well-known that $(B,C;A,G') = -1,$ so if we let $Z = G'D' \cap \Gamma \ne G',$ then $(B,C;A,G') \stackrel{D'}{=} (C,B;AD' \cap \Gamma, Z) = -1.$ Hence we would like to show that $(C,B; AD' \cap \Gamma, X') = -1.$ Now, let $T$ be the $A$ -mixtilinear touchpoint (that is, $(EFT)$ is tangent to $\Gamma$ ). It is well-known (say by force-overlaid inversion) that $AT$ and $AD$ are isogonal in $\angle BAC,$ so reflecting our cross ratio across the perpendicular bisector of $BC,$ our problem has reduced to showing that $(B,C;X,T) = -1.$

By the Radical Center Theorem on $\Gamma, (AEF),$ and $(TEF) = \omega,$ we see that $AX$ and $EF$ meet on the tangent to $\Gamma$ at $T,$ which is thereby the tangent to $\omega$ at $T.$ Thus projecting from $A$ onto $EF$ gives
$(B,C;X,T) \stackrel{A}{=} (F,E;AX \cap EF, AT \cap EF) = (F,E; TT \cap EF, AT \cap EF) \stackrel{T}{=} (F,E;T,AT \cap \omega).$ However, since $AF$ and $AE$ are both tangent to $\omega,$ this cross ratio equals $-1,$ and we conclude.

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P2nisic

406 posts

P2nisic

#85 h Jun 27, 2024, 4:50 PM

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MathStudent2002 wrote:

Let $ABC$ be a triangle with circumcircle $\Gamma$ and incenter $I$ and let $M$ be the midpoint of $\overline{BC}$ . The points $D$ , $E$ , $F$ are selected on sides $\overline{BC}$ , $\overline{CA}$ , $\overline{AB}$ such that $\overline{ID} \perp \overline{BC}$ , $\overline{IE}\perp \overline{AI}$ , and $\overline{IF}\perp \overline{AI}$ . Suppose that the circumcircle of $\triangle AEF$ intersects $\Gamma$ at a point $X$ other than $A$ . Prove that lines $XD$ and $AM$ meet on $\Gamma$ .

Proposed by Evan Chen, Taiwan

Let $S$ be the Sharky-Devil point then we have that $AS,GN,BC,TL$ are concur.
Also $A',D,T$ are collinear
Now it is enought to prove that $(G,T/B,C)=-1$
$(G,T/B,C)=(J,Q/B,C)=(S,R/B,C)=-1$
The last one is true from here: USA TSTST 2020 P2

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bjump

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bjump

#86 h Jul 28, 2024, 9:20 PM

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OTIS Walkthrough Writeup:
Note that by EGMO Lemma 4.40 in $A$ -mixtilinear incircle touches $\overline{AC}$ , and $\overline{AB}$ , at $E$ , and $F$ respectively. Let $T$ denote the mixtilinear intouch point, and let $D_1$ denote the reflection of $D$ about $M$ then $\overline{AT}$ , and $\overline{AD_1}$ are isogonal. by EGMO 2013/5. Let ray $AD_{1}$ meet $\Gamma$ again at $T_1$ . Note that $\overline{TT}$ , $\overline{AX}$ , $\overline{EF}$ concur at the radical center of $(ABC)$ , $(AEF)$ , and $(TEF)$ , call it $R$ . Let $\overline{AT}$ meet the $A$ -Mixtilinear incircle again at $S$ . Note that $-1=(ST; FE) \stackrel{T}= (R, \overline{AT} \cap \overline{EF}; EF) \stackrel{A}= (XT ; BC)$ . So quadrilateral $XTBC$ is harmonic. Notice that since $-1 \stackrel{T_1}= (BC; \overline{TT_1} \cap \overline{BC}, \overline{T_1X} \cap \overline{BC})= (BC; \infty \overline{T_1X} \cap \overline{BC})$ Then $\overline{T_1X} \cap \overline{BC} = M$ . Then by butterfly theorem we can conclude.

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navi_09220114

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navi_09220114

#87 h Sep 9, 2024, 3:46 PM

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Suppose I don't want to draw the mixtilinear incircle, how can I solve this?

Construct $D$ such that $ADBC$ is an isosceles trapezoid. We know that $E$ and $F$ are the tangency of the mixtilinear incircle, so if $X'$ is the image of $X$ under $\sqrt{bc}$ -inversion, then $X'$ is $BC$ intersect the polar of $A$ -excircle wrt to $A$ .

However, if we consider the polar of incircle wrt to $A$ and intersect with $BC$ , call this intersection $T$ , then $T$ and $X'$ are symmetric wrt to $M$ , because their harmonic conjugates wrt to $B$ and $C$ are the intouch point and extouch point to $BC$ respectively, which is known to be symmetric wrt to $M$ .

So if we consider $X''$ where $XX''BC$ is an isosceles trapezoid, then $AX$ and $AX''$ are isogonal, so $A$ , $X''$ , $X'$ are colinear. Reflecting across the perpendicular bisector of $BC$ , then $D$ , $X$ , $T$ are colinear. But $(B,C;D,T)=-1$ and if $AM$ intersect $(ABC)$ at $Y$ , then $DBYC$ is harmonic. So $X$ , $D$ , $Y$ are colinear.

QED

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navi_09220114

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navi_09220114

#88 h Sep 9, 2024, 4:09 PM

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Another idea: Denote $f(P)=BP/CP$ for any point $P$ . You want $f(X)f(Y)=f(D)$ , where $Y$ is $AM$ intersect $(ABC)$ .

But $f(Y)f(A)=f(M)=1 \Rightarrow f(Y)=\frac{1}{f(A)}$ . So you want $f(X)=f(A)f(D)$

But $f(X)=\frac{BF}{CE}=\frac{b-\frac{bc}{s}}{c-\frac{bc}{s}}=\frac{bs-bc}{cs-bc}=\frac{b}{c}\cdot \frac{s-c}{s-b}=f(A)f(D)$

QED

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L13832

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L13832

#89 h Oct 20, 2024, 10:26 AM

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MathStudent2002 wrote:

Let $ABC$ be a triangle with circumcircle $\Gamma$ and incenter $I$ and let $M$ be the midpoint of $\overline{BC}$ . The points $D$ , $E$ , $F$ are selected on sides $\overline{BC}$ , $\overline{CA}$ , $\overline{AB}$ such that $\overline{ID} \perp \overline{BC}$ , $\overline{IE}\perp \overline{AI}$ , and $\overline{IF}\perp \overline{AI}$ . Suppose that the circumcircle of $\triangle AEF$ intersects $\Gamma$ at a point $X$ other than $A$ . Prove that lines $XD$ and $AM$ meet on $\Gamma$ .

Proposed by Evan Chen, Taiwan

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draw((-3,4)--(0.4162895927601813,-5.5656108597285066), linewidth(0.6) + ccqqqq); draw(circle((-1.8096925236655736,-0.593546563480146), 2.4064534365198544), linewidth(0.6)); draw(circle((-1.3975860533458082,-2.1839173694797096), 3.239612138309303), linewidth(0.6) + linetype("4 4")); draw((-3,4)--(-2.921444247121779,-5.042753323490155), linewidth(0.6)); draw((-3,4)--(1.9214442471217796,-5.042753323490155), linewidth(0.6) + blue); draw((-1.8096925236655734,-0.593546563480146)--(-0.5,-3), linewidth(0.6)); draw((-5.385721977079134,1.1216413791852862)--(1.9214442471217796,-5.042753323490155), linewidth(0.6) + blue); draw((-8.174322675961044,-2.242787895789885)--(-3,4), linewidth(0.6)); draw((-8.174322675961044,-2.242787895789885)--(0.8931656580669505,0.10683434193304954), linewidth(0.6)); draw((-8.174322675961044,-2.242787895789885)--(-2.921444247121779,-5.042753323490155), linewidth(0.6)); draw((-2.921444247121779,-5.042753323490155)--(2,4), linewidth(0.6) + linetype("4 4")); draw((-3,4)--(-0.5,-5.6478150704935), linewidth(0.6)); draw((-1.8096925236655739,1.812906873039708)--(-1.8096925236655732,-3), linewidth(0.6) + linetype("4 4")); draw((-5,-3)--(4,-3), linewidth(0.6)); draw((-5.385721977079134,1.1216413791852862)--(-5,-3), linewidth(0.6) + zzttqq); draw((-5,-3)--(-2.921444247121779,-5.042753323490155), linewidth(0.6) + zzttqq); draw((-2.921444247121779,-5.042753323490155)--(4,-3), linewidth(0.6) + zzttqq); draw((4,-3)--(-5.385721977079134,1.1216413791852862), linewidth(0.6) + zzttqq); /* dots and labels */ dot((-3,4),linewidth(3pt) + dotstyle); label("$A$", (-3.203011823665007,4.185250207562497), NE * labelscalefactor); dot((-5,-3),linewidth(3pt) + dotstyle); label("$B$", (-5.407158125968947,-3.139297504709032), NE * labelscalefactor); dot((4,-3),linewidth(3pt) + dotstyle); label("$C$", (4.155445831718917,-3.2071173909337682), NE * labelscalefactor); dot((-1.8096925236655734,-0.593546563480146),linewidth(3pt) + dotstyle); label("$I$", (-1.7448842698331692,-0.494321941944313), NE * labelscalefactor); dot((-0.5,-3),linewidth(3pt) + dotstyle); label("$M$", (-0.33762163066988427,-2.88497293136627), NE * labelscalefactor); dot((-1.8096925236655732,-3),linewidth(3pt) + dotstyle); label("$D$", (-1.8974790138388267,-3.5123068789450818), NE * labelscalefactor); dot((0.8931656580669505,0.10683434193304954),linewidth(3pt) + dotstyle); label("$E$", (1.0357310653810325,0.14996697719068255), NE * labelscalefactor); dot((-4.512550705398097,-1.2939274688933415),linewidth(3pt) + dotstyle); label("$F$", (-4.915463950839607,-1.4438003490906224), NE * labelscalefactor); dot((-5.385721977079134,1.1216413791852862),linewidth(3pt) + dotstyle); label("$X$", (-5.712347613980262,1.0485804696684395), NE * labelscalefactor); dot((0.4162895927601813,-5.5656108597285066),linewidth(3pt) + dotstyle); label("$G$", (0.5101269471393236,-5.902957868367039), NE * labelscalefactor); dot((-0.5,-5.6478150704935),linewidth(3pt) + dotstyle); label("$M_A$", (-0.7614959195744881,-6.07250758392888), NE * labelscalefactor); dot((-2.921444247121779,-5.042753323490155),linewidth(3pt) + dotstyle); label("$T$", (-3.33865159611448,-5.343443807012964), NE * labelscalefactor); dot((0.8096925236655732,-3),linewidth(3pt) + dotstyle); label("$X_A$", (0.8492263782630067,-2.7832431020291657), NE * labelscalefactor); dot((1.9214442471217796,-5.042753323490155),linewidth(3pt) + dotstyle); label("$T'$", (2.0360743871958977,-5.343443807012964), NE * labelscalefactor); dot((-8.174322675961044,-2.242787895789885),linewidth(3pt) + dotstyle); label("$J$", (-8.357323176744991,-2.0711342966694337), NE * labelscalefactor); dot((-2.970880788076214,0.6480131530350209),linewidth(3pt) + dotstyle); label("$K$", (-2.897822335653692,0.9129406972189669), NE * labelscalefactor); dot((2,4),linewidth(3pt) + dotstyle); label("$A'$", (2.069984330308266,4.1004753497815765), NE * labelscalefactor); dot((-1.8096925236655739,1.812906873039708),linewidth(3pt) + dotstyle); label("$D'$", (-1.7448842698331692,1.9132840190338285), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Let $T$ be the $A-$ Mixtillinear touch-point so $\odot(TEF)$ is the $A-$ Mixtillinear Incircle. Also let $D'$ be $D-$ Antipode and finally consider the reflections of $D,T$ over the perpendicular bisector of $BC$ to be $X_A, T'$ .
$\textbf{Claim I:}$ It is enough to show that $AX_A$ and $XM$ concur. (hint given to me by $\textbf{kotmhn}$ .)
$\textbf{Proof:}$ Butterfly Theorem.

$\textbf{Claim II:}$ $AX, EF$ and tangent at $T$ to $\odot(TEF)$ concur at $J$ .
$\textbf{Proof:}$ Radax on $\odot(ABC), \odot(TEF)$ and $\odot(AXE)$ .

$\textbf{Claim III:}$ $\overline{A-D'-X_A-T'}\iff \overline{A'-D-T}$ where $A'$ is the $A-$ Antipode.
$\textbf{Proof:}$ $\sqrt{bc}$ inversion takes $T\to X_A$ and also because $\angle BAT=\angle CAX_A$ .

$\textbf{Claim IV:}$ $X$ is the miquel point of $BEFC$ .

$\textbf{Claim V:}$ $XTBC$ is harmonic
$\textbf{Proof:}$ $-1=(TK; EF) \stackrel{T}= (\overline{AT} \cap \overline{EF},J; EF) \stackrel{A}= (XT ; BC)$ .

$\textbf{Claim VI:}$ $\overline{X-M-T'}$
$\textbf{Proof:}$ $\angle AT'X=\angle ABX=\angle IMX=\angle AT'M$ .
Remark

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shendrew7

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shendrew7

#90 h Nov 28, 2024, 3:52 AM

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The key is to draw in the $A$ -mixtilinear touch point $T$ , and let $\gamma = (TEF)$ be the $A$ -mixtilinear incirle. Also define $K = XD \cap \Gamma$ , $L = TD \cap \Gamma$ as the reflection of $A$ over the perpendicular bisector of $BC$ from mixtilinear properties, and $A' = KM \cap \Gamma$ which we would like to show is simply $A$ .

First notice radical axis theorem on $(AEF)$ , $(ABC)$ , and $\gamma$ tells us $AX$ , $EF$ , and the tangent at $T$ concur, say at point $R$ . Since $TA$ is a symmedian in $\triangle TEF$ , and $TR$ is the tangent to $(TEF)$ , we get
$\begin{align*} -1 &= (R, AT \cap ER; FE) \\ &\overset{A}{=} (XT;BC) \\ &\overset{D}{=} (KL;CB) \\ &\overset{M}{=} (A',LM \cap \Gamma;BC) \\ &\overset{L}{=} (LA' \cap BC, M; BC). \end{align*}$
Thus $LA'$ is parallel to $BC$ , giving the desired $A = A'$ . $\blacksquare$

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kotmhn

57 posts

kotmhn

#91 h Dec 15, 2024, 11:17 AM

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Let $AM \cap \Gamma = M'$ and $T_A$ be the A-mixti touch point.
Claim: $T_A,D,M,M'$ are concyclic.
Proof:
$\measuredangle TDM = -\measuredangle TDB = \measuredangle DBT + \measuredangle BTD = \measuredangle DBT + \measuredangle ABC = \measuredangle AM'T$ $\blacksquare$
Now spi sim concludes.

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cursed_tangent1434

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cursed_tangent1434

#92 h Jan 6, 2025, 3:18 AM

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The following solution is atrociously long and unnecessarily complicated. This problem reminded me heavily of 21CHNTST42 which I suppose is because it's the exact same configuration, just with a bunch of unnecessary points deleted. Let $T$ denote the $A-$ mixtillinear intouch point , $X_a$ the $A-$ extouch point and $L$ and $N$ the major and minor $BC$ arc midpoints in $\Gamma$ respectively.

First, note that $X$ is the Miquel point of quadrilateral $BCEF$ by definition. Note that since $AL \perp AI \perp EF$ , lines $\overline{AL}$ and $\overline{EF}$ are parallel. Also it is well known that points $T$ , $I$ and $L$ are collinear. We start off with the following lemma.

Claim : Lines $\overline{LX}$ , $\overline{EF}$ , $\overline{BC}$ and $\overline{TN}$ concur, at say $Q$ .

Proof : Let $Q = EF \cap BC$ , and let $L' = QX \cap \Gamma$ . Then note,
$\measuredangle L'QF = \measuredangle XQF = \measuredangle XBF = \measuredangle XBA = \measuredangle XL'A$ which implies that $AL' \parallel EF$ so $L' \equiv L$ which implies that line $\overline{LX}$ also passes through $Q$ . Now note that $XQIMN$ is cyclic with diameter $QN$ due to all the right angles. Let $T'$ be the reflection of $T$ across the perpendicular bisector of segment $BC$ . Then since arcs $TN$ and $T'N$ are equal , lines $\overline{AT}$ and $\overline{AT'}$ are isogonal. Thus points $A$ , $X_a$ and $T'$ are collinear. Now letting $Q' = TN \cap BC$ note,
$\measuredangle MQ'N = \measuredangle NLT = \measuredangle NAT = \measuredangle T'AN = \measuredangle MIN$ since it is well known that $AX_a \parallel IM$ by the Midpoint of the Altitude Lemma. But this implies that $Q'IMN$ is cyclic, which indicates that $Q' \equiv Q$ so $\overline{TN}$ also passes through $Q$ proving the claim.

Now note that by Pascal's Theorem on hexagon $NNXLLT$ , points $Y=NX \cap LT$ , $Q = XL \cap TN$ and $\infty = LL \cap NN$ are collinear. Thus, $Y$ must also lie on line $\overline{BC}$ . Further, due to the right angles $MYTN$ is also cyclic. Now let $R = XD \cap \Gamma$ . Then, we can observe the following.

Claim : Points $T$ , $D$ , $M$ and $R$ are concyclic.

Proof : This is a direct angle chase. Note,
$\measuredangle TMD = \measuredangle TMY = \measuredangle TNY = \measuredangle TNX = \measuredangle TRX = \measuredangle TRD$ which implies the claim.

Further note,
$\measuredangle XT'A = \measuredangle XNA = \measuredangle XNI = \measuredangle XMI$ which implies that points $X$ , $M$ and $T'$ are collinear. Thus, lines $\overline{AX_a}$ and $\overline{XM}$ concur on $\Gamma$ . Now, since $D$ and $X_a$ are well known to be reflections of each other across $M$ , an application of the Butterfly Theorem suffices. But the following finish is also possible (and is the one I found).

Let $A'$ and $X'$ denote the reflections of $A$ and $X$ across the perpendicular bisector of segment $BC$ . Reflecting the previous observation across the perpendicular bisector of segment $BC$ it follows that $\overline{A'D}$ and $\overline{X'M}$ concur on $\Gamma$ (at $T$ ). But then, since arcs $AX$ and $AX'$ are equal,
$\measuredangle DRM = \measuredangle DTM = \measuredangle A' TX' = \measuredangle XT'A = \measuredangle XRA = \measuredangle DRA$ which implies that points $A$ , $R$ and $M$ are collinear. Thus, lines $\overline{AM}$ and $\overline{XD}$ intersect at $R$ , which lies on $\Gamma$ as desired.

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ihategeo_1969

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ihategeo_1969

#93 h Jan 8, 2025, 6:34 AM • 1 Y

Y by cursed_tangent1434

We define some new points

$\bullet$ $T$ is the $A$ -Mixtilinear intouch point of $\triangle ABC$ .
$\bullet$ $\triangle D'E'F'$ is the $A$ -extouch triangle of $\triangle ABC$ .
$\bullet$ $Y=\overline{AM} \cap (ABC)$ .
$\bullet$ $\ell$ is the perpendicular bisector of $\overline{BC}$ .
$\bullet$ $A'$ is reflection of $A$ across $\ell$ .

Claim: $(X,T;B,C)=-1$ .
Proof: By $\sqrt{bc}$ inverting at $A$ in $\triangle ABC$ , then $X^*=\overline{BC} \cap \overline{E'F'}$ .

So we need to prove that $(B,C;D,X^*)=-1$ which is just because $\overline{AD'}$ , $\overline{BE'}$ , $\overline{CF'}$ concur by Ceva. $\square$

Well known that $T$ , $D$ , $A'$ are collinear (just reflect across $\ell$ ). And so we have $-1=(X,T;B,C) \overset{D}= (\overline{DX} \cap (ABC),A';C,B)$ See that $Y$ is the reflection of the $A$ -harmonic conjugate of $\overline{BC}$ across $\ell$ and so we win.

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onyqz

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onyqz

#94 h Feb 6, 2025, 1:05 PM

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storage
solution

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ErTeeEs06

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ErTeeEs06

#95 h Mar 27, 2025, 8:47 PM • 1 Y

Y by Funcshun840

Let $T$ be the tangency point of the mixtilinear incircle and $(ABC)$ . Then invert around $T$ with radius $TI$ and reflect in line $TI$ . This is the $\sqrt{bc}$ inversion in $\triangle TBC$ . I've seen before that $X$ and $M$ are also inverses of each other. Now it follows that $\triangle TAX\sim \triangle TMD$ and we are done by spiral similarity.

This post has been edited 2 times. Last edited by ErTeeEs06, Mar 27, 2025, 8:48 PM
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