AD is Euler line of triangle IKL

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AD is Euler line of triangle IKL

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Source: IGO 2021 Advanced P5

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#1 h Dec 30, 2021, 5:24 PM • 1 Y

Y by amar_04

Given a triangle $ABC$ with incenter $I$ . The incircle of triangle $ABC$ is tangent to $BC$ at $D$ . Let $P$ and $Q$ be points on the side BC such that $\angle PAB = \angle BCA$ and $\angle QAC = \angle ABC$ , respectively. Let $K$ and $L$ be the incenter of triangles $ABP$ and $ACQ$ , respectively. Prove that $AD$ is the Euler line of triangle $IKL$ .

Proposed by Le Viet An, Vietnam

This post has been edited 4 times. Last edited by VicKmath7, Oct 16, 2023, 5:55 PM

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Project_Donkey_into_M4

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Project_Donkey_into_M4

#2 h Dec 30, 2021, 5:59 PM

Y by

I was really really rude

Sorry @Vickmath

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#4 h Dec 30, 2021, 6:06 PM

Y by

VicKmath7 wrote:

Yes, of course I know, but I believe that everyone who knows what IGO is will understand what A means in this case. Anyways, I will correct it.

Yes but even I got confused at first sight,it's reccomendable to write the full form cuz there are a lot of people who don't know abt IGO

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MP8148

#5 h Dec 31, 2021, 2:09 AM • 2 Y

Y by leon.tyumen, MS_asdfgzxcvb

Not the best solution, but I guess it works. Ignore config issues (I don't think there are any but who knows).

diagram

Let $O,H,G$ denote the circumcenter, orthocenter, and centroid of $\triangle IKL$ , respectively. We will show that

$A$ , $O$ , $H$ are collinear
$A$ , $G$ , $D$ are collinear

which imply the conclusion.

proof that A-O-H
proof that A-G-D

Additional fact: $\overline{KL}$ , $\overline{OX}$ , $\overline{BC}$ seem to concur, although I didn't use/prove it.

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MatteD

#6 h Dec 31, 2021, 10:52 AM • 4 Y

Y by Acrylic2005, SerdarBozdag, toilaDang, PHSH

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12.cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 4.4, xmax = 9.2, ymin = 0.8, ymax = 4.; /* image dimensions */ pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); pen ubqqys = rgb(0.29411764705882354,0.,0.5098039215686274); pen xfqqff = rgb(0.4980392156862745,0.,1.); /* draw figures */ draw((6.109430702956201,3.762535441352023)--(4.520861536062415,0.9952229780472427), linewidth(1.2) + xdxdff); draw((9.,1.)--(6.109430702956201,3.762535441352023), linewidth(1.2) + xdxdff); draw((6.109430702956201,3.762535441352023)--(6.793968020425234,0.9976472566588956), linewidth(1.2) + xdxdff); draw((6.109430702956201,3.762535441352023)--(5.430792468160217,0.9961934236079506), linewidth(1.2) + xdxdff); draw((6.109430702956201,3.762535441352023)--(5.001628488792055,0.9957357182328481), linewidth(1.2) + ubqqys); draw((6.109430702956201,3.762535441352023)--(7.711717014885521,0.9986260402193841), linewidth(1.2) + ubqqys); draw((5.001628488792055,0.9957357182328481)--(6.6393827585556835,1.9451703298270977), linewidth(1.2) + xfqqff); draw((5.827852937434829,1.7529086067332473)--(7.711717014885521,0.9986260402193841), linewidth(1.2) + xfqqff); draw((6.109430702956201,3.762535441352023)--(6.356672751838788,0.9971808792261161), linewidth(1.2) + ubqqys); draw((6.111695185347248,1.639260673316434)--(6.355540510643266,2.0588182632439107), linewidth(1.2) + green); draw((4.520861536062415,0.9952229780472427)--(5.001628488792055,0.9957357182328481), linewidth(1.2) + xdxdff); draw((5.001628488792055,0.9957357182328481)--(7.711717014885521,0.9986260402193841), linewidth(1.2) + ubqqys); draw((9.,1.)--(7.711717014885521,0.9986260402193841), linewidth(1.2) + xdxdff); draw((4.520861536062415,0.9952229780472427)--(5.827852937434829,1.7529086067332473), linewidth(1.2) + xdxdff); draw((5.827852937434829,1.7529086067332473)--(6.355540510643266,2.0588182632439107), linewidth(1.2) + green); draw((6.355540510643266,2.0588182632439107)--(6.6393827585556835,1.9451703298270977), linewidth(1.2) + green); draw((6.6393827585556835,1.9451703298270977)--(9.,1.), linewidth(1.2) + xdxdff); /* dots and labels */ dot((6.109430702956201,3.762535441352023),linewidth(3.pt) + dotstyle); label("$A$", (6.128176663628198,3.7891605035276705), NE * labelscalefactor); dot((4.520861536062415,0.9952229780472427),linewidth(3.pt) + dotstyle); label("$B$", (4.539303375246712,1.019140670079517), NE * labelscalefactor); dot((9.,1.),linewidth(3.pt) + dotstyle); label("$C$", (9.015890814734231,1.0233440385672987), NE * labelscalefactor); dot((6.355540510643266,2.0588182632439107),linewidth(3.pt) + dotstyle); label("$I$", (6.371972035919536,2.082592897488292), NE * labelscalefactor); dot((6.793968020425234,0.9976472566588956),linewidth(3.pt) + dotstyle); label("$P$", (6.809122358648834,1.0233440385672987), NE * labelscalefactor); dot((5.430792468160217,0.9961934236079506),linewidth(3.pt) + dotstyle); label("$Q$", (5.447230968607561,1.0233440385672987), NE * labelscalefactor); dot((6.356672751838788,0.9971808792261161),linewidth(3.pt) + dotstyle); label("$D$", (6.371972035919536,1.0233440385672987), NE * labelscalefactor); dot((5.827852937434829,1.7529086067332473),linewidth(3.pt) + dotstyle); label("$K$", (5.8465509749468225,1.7799503663680083), NE * labelscalefactor); dot((6.6393827585556835,1.9451703298270977),linewidth(3.pt) + dotstyle); label("$L$", (6.657801093088692,1.9691019483181857), NE * labelscalefactor); dot((7.711717014885521,0.9986260402193841),linewidth(3.pt) + dotstyle); label("$M$", (7.729660057473027,1.0233440385672987), NE * labelscalefactor); dot((5.001628488792055,0.9957357182328481),linewidth(3.pt) + dotstyle); label("$N$", (5.018487382853826,1.019140670079517), NE * labelscalefactor); dot((6.111695185347248,1.639260673316434),linewidth(3.pt) + dotstyle); label("$H$", (6.128176663628198,1.666459417197902), NE * labelscalefactor); dot((6.274258735544592,1.918965733268085),linewidth(3.pt) + dotstyle); label("$G$", (6.300514771627247,1.8850345785625513), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Let $M,N$ denote the reflections of $A$ with respect to $BI,CI$ respectively, and let $G,H$ be the centroid and orthocenter of $\triangle AMN$ .
First, notice that by symmetry $IM=IA=IN$ , so $I$ is the circumcenter of $\triangle AMN$ and $D$ is the midpoint of $MN$ .
Since $\triangle BAC \sim \triangle BPA$ , we have $\frac{BK}{BI}=\frac{BA}{BC}=\frac{BM}{BC}$ , so $MK$ is parallel to $CI$ . Since $CI \perp AN$ , we also have $MK \perp AN$ , and $NL \perp AM$ similarly, so $H=NL \cap MK$ . $KILH$ is a parallelogram, and it is well-known that $\frac{IG}{IH}=\frac{1}{3}$ , so $G$ is the centroid of $\triangle IKL$ . The similarity of $\triangle BAC$ and $\triangle BPA$ also gives $\angle AKI = \angle AKB = \angle BIC$ , and similarly we have $\angle BIC= \angle ILA$ , which is enough to know that $A$ lies on the Euler line of $\triangle IKL$ . Since $G \in AD$ , we have that $AD$ is the Euler line of $\triangle IKL$ . $\blacksquare$

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#7 h Dec 31, 2021, 2:08 PM • 4 Y

Y by parmenides51, PHSH, hakN, MS_asdfgzxcvb

Here is my simple proof using barycentric coordinates.

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#8 h Jan 7, 2022, 7:10 PM

Y by

Here's a solution me and my friends found at our 3 A.M :alien:

:alien:

Claim 1.In $\triangle ABC$ ,let $P$ be a point such that $\angle PBA=\angle PCA=180-A$ . Then $P$ is on the Euler line of $\triangle ABC$ .
Proof.It's well-known that the locus of points $X$ such that $\angle XBA=\angle XCA$ is a hyperbola passing through $A,B,C,H,A'$ where $H$ is the orthocenter of $ABC$ and $A'$ is the antipod of $A$ in $(ABC)$ . So $P$ is on this hyperbola. Also, by the definition of $P$ , we have $PC \cap AB$ lies on the perpendicular bisector of $AC$ . Now , applying Pascal on the hyperbola $AA'CPHBA$ we'll get that a line through $PC \cap AB$ and $HP \cap AA'$ is perpendicular to $AC$ and since $PC \cap AB$ lies on the perpendicular bisector of $AC$ , $HP \cap AA'$ is in fact the circumcenter of $(ABC)$ . So $H,O,P$ are collinear and we're done.

Claim2 .In $\triangle ABC$ with incenter $I$ , If a circle with center $I$ intersects $AB,BC,CA$ at $X,{Y,Z},T$ , then $XY,ZT,AD$ are concurrent.
Proof. Let the lines through $D$ parallel to $AB,AC$ intersects $XY,ZT$ at $E,F$ respectively. Since $CT=CZ$ and $DF||AC$ , we have $DZ=DF$ , and similarly , $DE=DF$ . On the other hand , since $IY=IZ$ ,it follows that $DY=DZ$ . So $DE=DF$ and $\triangle DEF$ is an isosceles triangle. Also , with a little angle chasing , $AXY$ is an isosceles to , so since $DE||AB$ and $DF||AC$ , $\triangle AXY$ and $\triangle DEF$ are homothetic. the conclusion follows.

Back to the problem.

By Claim 1 , Since $\angle AKI = \angle ALI = 90 - \angle A /2 = 180 - \angle KIL$ , $A$ lies on the Euler line of $\triangle KIL$ .

Now , let the perpendicular bisector of $AI$ intersects $AC,AB$ at $M,N$ respectively. Since $MI=MA$ and $\angle AMI = 2\angle ALI$ , $M$ is the circumcenter of $\triangle ALI$ . Similarly , $N$ is the circumcenter of $\triangle AKI$ . Now , let $M',N'$ be the reflection of $M,N$ to $IL,IK$ respectively . Since $MI=MK$ , $M'I = M'L$ and similar for $N$ . But $M',N'$ lie on $BC$ . So since $I$ is the center of $(MNM'N')$ , and since $MM',NN'$ are the perpendicular bisectors of $IL,IK$ respectively , by claim 2 , the circumcenter of $\triangle IKL$ lies on $AD$ .

So $AD$ is the Euler line of $\triangle IKL$ and we're done.

This post has been edited 1 time. Last edited by alinazarboland, Jan 11, 2022, 8:25 PM

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#9 h Jan 8, 2022, 12:59 PM • 1 Y

Y by amar_04

Proposed by Le Viet An -Vietnam

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#11 h Jan 24, 2022, 4:46 AM • 1 Y

Y by buratinogigle

It seems that this one is not hard to bash.

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PHSH

#12 h Jul 14, 2022, 12:42 PM

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motivation

We have that $APB$ is opositive similar to $CAB,$ hence ${KB}{/IB} = {AB}{/BC}.$ Hence, if we define $X$ as the reflection of $A$ over line $BI,$ we will have $KX \parallel IC.$ Similarly, $Y$ is the reflection of $A$ with respect to $CI,$ and $L$ is the intersection of $CI$ and the parallel of $BI$ through $Y.$ We are going to take this redefinition later on.

$[asy] import olympiad; import graph; import geometry; size(250); pair A,B,C,O,I,D,X,Y,K,L,H,I,Bt,Mt,Ct,Nt; A = dir(-240.54); B = dir(-170.37); C = dir(-10.43); O = (0,0); I = incenter(A,B,C); D = foot(I,B,C); X= 2*foot(A,B,I) - A; Y = 2*foot(A,C,I) - A; Bt= B*0.5 + I*0.5; Mt = extension(I,Y,C,Bt); L =extension(B,Mt,C,I); Ct= C*0.5 + I*0.5; Nt = extension(I,X,B,Ct); K = extension(C,Nt,B,I); dot("$K$",K,dir(120)); dot("$L$",L, dir(60)); dot("$Y$", Y, dir(-90)); dot("$X$", X, dir(X)); dot("$B$",B, dir(-90)); dot("$C$",C, dir(-90)); dot("$I$", I, dir(I)); dot("$D$", D, dir(D)); draw(circle(I,distance(I,D)),dotted + green); draw(A--B--C--cycle,gray); draw(B--I,blue); draw(C--I,blue); draw(K--X, dotted+magenta); draw(Y--L,dotted+magenta); dot("$A$",A, dir(A)); [/asy]$

Note now, that on triangle $\triangle AXY,$ we have that $BI$ and $CI$ are the perpendicular bissectors of $AX$ and $AY,$ respectively. Furthermore, $YL$ and $XK$ are altitudes of $\triangle AXY.$ So the new problem reads as follows

Let $\triangle AXY$ be an triangle with circumcenter $I$ . Point $K$ is the intersection of the $X$ -altitude with the perpendicular bissector of $AX;$ similarly point $L$ is the intersection of the $Y$ -altitude with the perpendicular bissector of $AY.$ If $D$ is the midpoint of side $XY,$ show that $AD$ is the euler line of thriangle $IKL.$

So let $H$ be the orthocenter of $\triangle AXY.$ Also let $M$ and $N$ be the midpoints of sides $AX$ and $AY,$ respectively, and let $E$ and $F$ be the foots of $X$ and $Y$ respectively. Also let $T$ be the orthocenter of $IKL$

$[asy] import olympiad; import graph; import geometry; size(250); pair A,B,C,O,I,D,X,Y,K,L,H,I,Bt,Mt,Ct,Nt; A = dir(-240.54); B = dir(-170.37); C = dir(-10.43); O = (0,0); I = incenter(A,B,C); D = foot(I,B,C); X= 2*foot(A,B,I) - A; Y = 2*foot(A,C,I) - A; Bt= B*0.5 + I*0.5; Mt = extension(I,Y,C,Bt); L =extension(B,Mt,C,I); Ct= C*0.5 + I*0.5; Nt = extension(I,X,B,Ct); K = extension(C,Nt,B,I); pair M,N,E,F,H,T; M = A*0.5 + X*0.5; N = A*0.5 + Y*0.5; E = foot(X,A,Y); F = foot(Y,A,X); H = extension(E,X,F,Y); T = extension(A,D,K,A+K-Y); draw(A--D, dotted); draw(K--T--L,magenta); draw(circumcircle(A,F,N), dotted + green); draw(circumcircle(A,E,M),dotted + blue); draw(A--X--Y--cycle,gray); draw(M--I--K--cycle,orange); draw(N--I--L--cycle,orange); draw(X--E,red); draw(F--Y,red); dot("$A$", A, dir(A)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$D$",D, dir(-90)); dot("$M$",M, dir(30)); dot("$N$", N, dir(N)); dot("$E$",E, dir(E)); dot("$F$",F, dir(-10)); dot("$I$",I, dir(90)); dot("$K$",K,dir(260)); dot("$L$",L,-dir(L)); dot("$H$",H,dir(-90)); dot("$T$",T, dir(45)); [/asy]$
First, note that the centroid of $\triangle IKL$ lies on $AD.$ Indeed, by definition $IKHL$ is an parallelogram, so the centroid of $\triangle IKL$ divides the segment $IH$ in the ratio $1{/3};$ hence the centroid of $IKL$ coincides with the centroid of $AXY$ which indeed lies on the $A$ -median, $AD$ .

Now it is sufficient to show that $T$ lies on $AD.$ On one hand, $AD$ is the $A$ -median, so $T$ lies on $AD$ if and only if $[ATY]=[ATX].$ On the other hand, by construction $TK \parallel AY$ and $TL \parallel AX$ and hence $[ATY] = [AKY]$ and $[ATX] =[ALX].$ Then it reduces to $[AKY] = [ALX] \iff KE \cdot AY = LF \cdot AX.$
So let $b =AX, c =AY, p = XE, q = YF.$ Since $AEKM$ is cyclic by construction, it follows by power of a point that
$XH \cdot XE = XM \cdot XA = \frac{b^2}{2} \Rightarrow HE = \frac{2p^2 -b^2}{2p} \Rightarrow KE\cdot AY = \frac{(2p^2-b^2)c}{2p}$ Similarly, $LF \cdot AX = \frac{(2q^2-c^2)b}{2q}$ and it is sufficient to show
$(2p^2 -b^2)cq =(2q^2-c^2)bp$ which is obviously since $p = b\cdot \sin \angle XAY$ and $q = c \cdot \sin \angle XAY.$

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#13 h Jul 31, 2022, 6:18 AM • 1 Y

Y by Mango247

official solution !

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#14 h Sep 26, 2022, 2:43 PM

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Consider point $R, S$ on side $AB, AC$ s.t. $RIA=SIA=90$ , then i claim that $(RKIA)$ cyclic:
$\angle AKI=\beta/ 2+\gamma /2=90-\alpha /2 = \angle ARI$ similarly the other one.now said $O_b, O_c$ the midpoint of $AR, AS$ we have that the axis of $KI, LI$ pass through this points.
Now consider point $X, Y$ as the second intersection of $(RKIA),(SLIA)$ with $AC, AB$ . I claim that the height from $K$ and $L$ in triangle $\triangle KIL$ intersect side $AC, AB$ on $X,Y$ :
$\angle IKX=\angle AIX=\alpha/2=\angle KIL-90$ Now consider the point $E, F$ as the projection on the side $AB, AC$ of $I$ , then we have that the bisector $BI\perp ED$ and so $ED$ is parallel to the axis of $KI$ , and the height of $L$ wrt. $IK$ . Now the problem can be finished by noting that $AX/AY=AO_b/AO_c=AE/AF=1$ and so by homothety in $A$ the point $O,H,D$ are aligned.

This post has been edited 2 times. Last edited by FairyBlade, Sep 26, 2022, 2:45 PM

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#15 h Oct 5, 2022, 10:18 PM

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Let $G, O, H$ be the centroid, circumcenter, orthocenter of $\triangle IKL$ .
Claim 1: $A \in OH$ .
proof
Claim 2: $G \in AD$ .
proof
Combining these two claims clearly solves the problem.

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#16 h Nov 17, 2022, 8:44 PM • 1 Y

Y by GeoKing

I will share my proof of showing that centroid of $\Delta IKL$ lies on $AD$
This proof is very interesting and beautiful and as i have seen no one else posted it before
Also i found this without geogebra :gleam:

:gleam:

Points $P, Q$ are basically irrelevant, $K$ and $L$ are just points on $BI$ and $CI$ such that $\angle KAB=\frac{\angle C}{2}$ and $\angle LAC=\frac{\angle B}{2}$
As $\angle KAB=\angle ACL$ and $\angle KBA=\angle LAC$ we see that $\Delta KBA\sim \Delta LAC$
Because these triangles are oriented the same way we can now apply the gliding principle
Let $M, N, T$ be midpoints of $AC, AB, KL$
By the gliding principle $\Delta TNM\sim \Delta KBA\sim \Delta LAC$
Let $X=MT\cap AK$ and $Y=NT\cap AL$
$\angle YNM=\angle TNM=\angle LAC=\angle YAM$ so $ANYM$ cyclic
similarly $ANXM$ cyclic
Hence $ANXYM$ cyclic

Consider a homothety at $A$ with scaling ratio $2$
$N\mapsto B, M\mapsto C$ and hence $(ANXYM)\mapsto (ABC)$
Now let $AX$ and $AY$ intersect $(ABC)$ at $R$ and $S$
$X\mapsto R, Y\mapsto S$
Let $J=BS\cap CR$
$T\mapsto J$ so $A-T-J$ and $T$ is midpoint of $AJ$
$\angle JBC=\angle SBC=\angle SAC=\angle ABI=\angle IBC$
similarly $\angle JCB=\angle ICB$
This means that $J$ is the reflection of $I$ in $BC$ , hence $I-D-J$ and $D$ is midpoint of $IJ$
Let $G=IT\cap AD$
Consider triangle $\Delta IJA$
$T$ is midpoint of $AJ$ , $D$ is midpoint of $IJ$ hence $G$ is centroid of $\Delta IJA$
Now $\frac{GT}{GI}=\frac{1}{2}$
As $T$ is midpoint of $KL$ as well we see that $G$ is centroid $\Delta IKL$ as needed

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#17 h May 29, 2023, 3:50 PM

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Let $K' = \overline{CI} \cap \overline{AK}$ , and $L' = \overline{BI} \cap \overline{AL}$ . Note that $\triangle ABC \sim \triangle PBA \sim \triangle QAC$ , so $\triangle BIC \sim \triangle BKA \sim \triangle ALC$ . In particular, $\triangle IKK'$ and $\triangle ILL'$ are isosceles, and so $K$ , $L$ , $K'$ and $L'$ are concyclic.

By EGMO Theorem 10.5, the Euler line of $\triangle IKL$ must then be the radical axis $\ell$ of circles with diameters $KK'$ and $LL'$ , call them $\omega_K$ and $\omega_L$ . Obviously $A$ must lie on $\ell$ since $KK'L'L$ is cyclic, so we just need to show that $D$ has equal powers to both circles.

Let $\omega_K$ intersect $\overline{BI}$ and $\overline{CI}$ at $W$ and $X$ . Define $Y$ and $Z$ similarly for $\omega_L$ . We claim that $D$ lies on $\overline{WX}$ . Let $M_B$ be the midpoint of minor arc $AC$ . By the incenter lemma, $M_BI = M_BC$ , so $\triangle IWX \sim \triangle IK'K \sim \triangle IM_BC$ , so $\overline{WX} \parallel \overline{M_BC}$ . Let $I'$ be the reflection of $I$ over $\overline{BC}$ . Then $\triangle BI'C$ and $\triangle BKA$ are spirally similar, so $\triangle BKI'$ and $\triangle BAC$ are spirally similar as well. In particular, $\angle BKI' = \angle BAC = \angle BM_BC$ , so $\overline{DW} \parallel \overline{I'K} \parallel \overline{M_BC}$ . This shows the collinearity.

Since $WXYZ$ is cyclic, it then follows that $D$ has equal power to both $\omega_K$ and $\omega_L$ . Therefore, $D$ lies on $\ell$ as desired.

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#18 h Sep 15, 2024, 8:49 AM

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Let $P,Q$ be the $A-$ mixtilinear touch points to $AB,AC$ respectively. Let $E,F$ be the tangency points of the incircle with $AC,AB$ . Let $O,H$ be the circumcenter and orthocenter of $\triangle IKL$ .
Claim: $K=BI\cap (API)$ and $L=CI\cap (AQI)$ .
Proof:
$\measuredangle KAP=\measuredangle KAB=\frac{\measuredangle C}{2}=\measuredangle BIP=\measuredangle KIP$ $\measuredangle QAL=\measuredangle CAL=\frac{\measuredangle B}{2}=\measuredangle QIC=\measuredangle QIL$ These yield the desired result. $\square$

Now we present a lemma.
Lemma: $ABC$ is a triangle whose circumcenter is $O$ . Let $O'$ be the reflection of $O$ with respect to $BC$ . The tangent to $(ABC)$ at $A$ intersect $BC$ at $D$ . Points $E,F$ are taken on $O'C,O'B$ such that $EA=EC$ and $FA=FB$ . Then, $D,E,F$ are collinear.
Proof: Let $X,Y$ be the intersection of the tangent to $(ABC)$ at $A$ with the tangent at $C,B$ respectively. By appyling Desargues theorem on $YOX$ and $BO'C,$ since $YB,OO',XC$ are concurrent, we get that these triangles are perpective. Hence $YO\cap BO'=F,OX\cap O'C=E,XY\cap BC=D$ are collinear. $\square$
Claim: $A,O,D$ are collinear.
Proof: Invert the diagram from $I$ with radius $ID$ . Note that $K^*=IB\cap A^*P^*$ and $L^*=IC\cap A^*Q^*$ . $O^*$ is the reflection of $I$ according to $K^*L^*$ . Let $EF$ intersect $BC$ at $T$ . $M,N,W$ are the midpoints of $IE,IF,IT$ respectively. $M,N,W$ are collinear. Let $K_1,L_1$ be the reflections of $I$ to $K^*,L^*$ . By applying the lemma on $\triangle DEF,$ we get that $T,K_1,L_1$ are collinear. Take the homothety centered at $I$ with ratio $\frac{1}{2}$ . Then, $W,K^*,L^*$ are collinear. Since $K^*L^*$ is the perpendicular bisector of $IO^*,$ we have $WT=WI=WO^*$ . Hence $O^*$ is on $(TIDA^*)$ which proves the claim. $\square$
Claim: $A,H,D$ are collinear.
Proof: Let $(AIP)\cap AC=R$ and $(AIQ)\cap AB=S$ .
$\measuredangle HKI=90-(180-\measuredangle CIB)=\frac{\measuredangle A}{2}=\measuredangle RKI$ Thus, $H$ lies on $KR$ . Similarily, $H$ lies on $LS$ . So $H=KR\cap LS$ .
Applying Desargues theorem on $RHS$ and $EDF,$ since $RH\cap ED,RS\cap EF,HS\cap DF$ is the line at infinity $A,H,D$ are collinear as desired. $\blacksquare$

This post has been edited 1 time. Last edited by bin_sherlo, Sep 15, 2024, 9:21 AM

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#19 h May 1, 2025, 9:00 AM

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Wow, what an incredibly cool geo problem! I've just skimmed through the other posts and seems like my solution is unique.

We start with the following lemma.
Lemma: Let $\triangle ABC$ be a triangle with $\angle BAC>90^\circ$ and $D$ be a point such that $\angle DBA=\angle DCA=180^\circ-\angle BAC$ . Then $D$ is on the Euler line of triangle $ABC$ .
Proof: Let $DB$ and $DC$ intersect $(ABC)$ again in $E, F$ and let $H, O$ be the orthocenter and circumcenter of $\triangle ABC$ respectively. Then its obvious by the angle condition that $AB\parallel CE$ and $AC\parallel BF$ . Let $HB, HC$ intersect $(ABC)$ again in $X, Y$ . Then $XB\perp AC$ so $XB\perp BF$ . Pascal on $EBXFCY$ now gives $D, H, O$ collinear so $D$ is on the Euler line.

CLaim: $A$ is on the Euler line of $\triangle IKL$ .
Proof: Angle chase $\angle AKI=180^\circ-\angle AKB=180^\circ-(90^\circ+\frac{\angle APB}{2})=90^\circ-\frac{\angle BAC}{2}=180^\circ-\angle KIL$ Analogously $\angle ALI=180^\circ-\angle KIL$ . Applying the above lemma with $IKL$ as reference triangle we see that $A$ is on the Euler line of $\triangle IKL$ which finishes the proof of this claim.

Now denote the circumcircles of triangles $DKP$ and $DLQ$ as $\omega_1$ and $\omega_2$ respectively. Let $\omega_1$ and $\omega_2$ intersect again in $S$ and let $BK$ intersect $\omega_1$ again in $X$ and $CL$ intersect $\omega_2$ again in $Y$ . We calculate $\angle XDC=\angle XDP=180^\circ-\angle BKP=90^\circ-\frac{\angle ACB}{2}$ This implies $DX\perp CI$ . So $X$ is on the $C$ -intouch chord of $\triangle ABC$ , since it is also on line $BI$ we see that $X$ is the Iran lemma point and $\angle AXB=90^\circ$ . We calculate $\angle IAX=\angle AIB-90^\circ=\frac{\angle ACB}{2}$ so $\angle CAX=\frac{\angle BAC-\angle BCA}{2}=\frac{\angle CAP}{2}$ so $AX$ is the angle bisector of $\angle PAC$ . If we let $AX, CI$ intersect each other at $E$ then $E$ is the incenter of $\triangle ACP$ . Now we see $\angle EPK=90^\circ=\angle EXP$ so $E$ is on $\omega_1$ and it is the antipode of $K$ . If we define $F$ similarly as the incenter of $\triangle ABQ$ then we now have $\angle EXF=\angle EXK=90^\circ=\angle LYF=\angle EYF$ , so $EFYX$ is cyclic. This implies that $AE\cdot AX=AF\cdot AY$ and therefore $A$ is on the radax of $\omega_1$ and $\omega_2$ .

Let $O$ be the circumcenter of $\triangle IKL$ and $KO, LO$ intersect $\omega_1, \omega_2$ in $M, N$ respectively. Angle chase gives $\angle NDY=\angle NLY=\angle OLI=90^\circ-\angle LKI$ and similarly $\angle MDX=90^\circ-\angle KLI$ . $AXDY$ is a parallelogram so $\angle XDY=\angle XAY=180^\circ-\angle XIY=180^\circ-\angle KIL$ Adding these angles gives $\angle MDN=\angle MDX+\angle XDY+\angle YDN=90^\circ-\angle LKI+90^\circ-\angle LKI+180^\circ-\angle KIL=180^\circ$ so $M, D, N$ are collinear. Now $\angle NMO=\angle DMK=\angle DPK=\frac{\angle BAC}{2}$ and by symmetry $\angle MNO=\frac{\angle BAC}{2}$ , so $|OM|=|ON|$ . Also clearly $|OK|=|OL|$ so $O$ is on the radax of $\omega_1$ and $\omega_2$ . But we already know $A$ and $D$ are on that radax so $A, O, D$ are collinear. Also we already had $A$ on the Euler line so this implies $D$ is on the Euler line also and that is exactly what we wanted to prove.

This post has been edited 1 time. Last edited by ErTeeEs06, May 1, 2025, 9:03 AM
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