Euler Line Madness

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raxu

398 posts

raxu

#1 h Jun 26, 2015, 2:01 AM • 6 Y

Y by doxuanlong15052000, Awesome_guy, ImSh95, Adventure10, Mango247, ehuseyinyigit

Let ABC be a scalene triangle. Let $K_a$ , $L_a$ and $M_a$ be the respective intersections with BC of the internal angle bisector, external angle bisector, and the median from A. The circumcircle of $AK_aL_a$ intersects $AM_a$ a second time at point $X_a$ different from A. Define $X_b$ and $X_c$ analogously. Prove that the circumcenter of $X_aX_bX_c$ lies on the Euler line of ABC.
(The Euler line of ABC is the line passing through the circumcenter, centroid, and orthocenter of ABC.)

Proposed by Ivan Borsenco

This post has been edited 2 times. Last edited by v_Enhance, Aug 23, 2016, 12:45 AM
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raxu

398 posts

raxu

#2 h Jun 26, 2015, 2:18 AM • 2 Y

Y by ImSh95, Adventure10

Some of the solutions include complex bash and inversions. Below are hints for a synthetic approach (Credits to Zilin Jiang):
Hint 0
Hint 1
Hint 2

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drmzjoseph

445 posts

drmzjoseph

#3 h Jun 26, 2015, 6:57 AM • 4 Y

Y by No16, ImSh95, Adventure10, ehuseyinyigit

The problem is easy without bash, using projections of orthocenter on bisectors, the circle of diameter $HA$ cut $AM_a,AK_a,AL_a$ at $A_1,A_2,A_3$ . also $\angle HAA_2=\angle AL_aB =\angle A_3A_2A \Rightarrow A_3A_2K_aL_a$ is cyclical, then from the link we get $M_aA_2.M_aA_3=M_aA_1,M_aA=M_aL_a.M_aK_a$ i.e. $A_1 \in \odot (AL_aK_a)$ i.e. $A_1 \equiv X_a$ hence $X_a$ belongs to circle of diameter $HG$

Consider $H$ ortocenter and $G$ centroid.

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andria

824 posts

andria

#4 h Jun 26, 2015, 9:47 AM • 4 Y

Y by NHN, ImSh95, Adventure10, Mango247

Here is my solution:
Let $D,E,F$ feet of $A,B,C$ onto $BC,AC,AB$ let $X$ be the projection of $H$ onto the $AM_a$ .
Lemma1: $BC.EF,HX$ are collinear.
Prove: let $EF\cap BC=T$ then $AH$ is polar of $T$ WRT $\odot (BFEC)$ so $AM_a\perp TH$ so the lemma proved.
Lemma 2: the locus of point $X$ such that $\frac{XB}{XC}=\frac{AB}{AC}$ is the $A$ -appolonus circle(a circle with diameter $K_al_a$ )
Prove: it's well known.
Lemma 3: in cyclic quadrilateral $ABCD$ let $AB\cap CD=R$ then $\frac{RD}{RC}=\frac{AD}{AC}.\frac{BD}{BC}$ .
Prove: it's well known.
BACK TO THE MAIN PROBLEM:
$HXM_aD$ is cyclic $\longrightarrow TD.TM_a=TH.TX$ and $(TDBC)=-1\longrightarrow TD.TM_a=TB.TC$ so $TB.TC=TH.TX$ hence $HXCB$ is cyclic.
using Menelaus theorem for $T,E,F$ and $\triangle ABC$ we have $\frac{TB}{TC}=\frac{AE}{CE}.\frac{BF}{AF}$ (1).
Using lemma 3 for cyclic quadrilateral $HXCB$ we have $\frac{TB}{TC}=\frac{HB}{HC}.\frac{XB}{XC}$ (2)
From (1) , (2) we deduce that $\frac{XB}{XC}=\frac{AE}{CE}.\frac{BF}{AF}.\frac{HC}{HB}$
From Menelaus for three concurrent lines $AD,BE,CF$ we have $\frac{AE}{CE}.\frac{BF}{AF}=\frac{BD}{CD}$ in triangle $HBC$ : $\frac{BD}{CD}=\frac{HB}{HC}.\frac{\sin C}{\sin B}$ so $\frac{XB}{XC}=\frac{AE}{CE}.\frac{BF}{AF}.\frac{HC}{HB}=\frac{BD}{CD}\frac{HC}{HB}=\frac{\sin C}{\sin B}=\frac{AB}{AC}$
Hence by lemma (2) because $\frac{XB}{XC}=\frac{AB}{AC}$ we get that $X\in \odot (K_aL_a)$ so $X\equiv X_a$ thus $X_a\in \odot (GH)$ similarly $X_b,X_c\in \odot (GH)$
DONE

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andria

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andria

#5 h Jun 26, 2015, 1:31 PM • 2 Y

Y by ImSh95, Adventure10

My second solution: (using inversion)
Let $P,Q,X$ projections of $H$ onto $AK_a,AL_a,AM_a$ and let $D,E,F$ feet of $A,B,C$ on $BC,CA,AB$ . Apply an inversion with center $A$ and radius $\sqrt{AH.AD}$ then $X\longleftrightarrow M_a$ and $A$ -appolonus circle is taken to line $PQ$ so in order to prove the problem we have to show that $P,Q,M_a$ are collinear.let $M$ midpoint of $AH$ then because $M_aE=M_aF,ME=MF$ line $MM_a$ is perpendicular bisector of $EF$ (1).let $S,T$ projections of $P$ on $AC,AB$ so $PS=PT$ .Because $AFPE$ is cyclic we have $\angle PES=\angle PFT$ thus $\triangle PES=\triangle PFT\longrightarrow PE=PF$ similarly $QE=QF$ so $PQ$ is perpendicular bisector of $EF$ (2). From (1),(2) we deduce that $P,Q,M_a$ are collinear.
DONE

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v_Enhance

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v_Enhance

#6 h Jun 26, 2015, 4:24 PM • 3 Y

Y by mssmath, ImSh95, Adventure10

The main difficulty of the problem is claiming that $\angle HX_aA = 90^\circ$ (ie that the circumcenter is the midpoint of $HG$ ). To prove this we just compute $X_a = (a^2:2S_A:2S_A)$ from the givens and then observe this lies on the circle with diameter $AH$ .

Unlike the recent TST problem the Euler Line is a red herring.

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v_Enhance

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v_Enhance

#7 h Jun 26, 2015, 4:33 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

Oh, I see the synthetic solution now. Just show that $M_a$ has the same power with respect to the circle with diameter $AH$ and the circle with diameter $K_aL_a$ . Then $AM_a$ is the radical axis, so $X_a$ lies on both circles.

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EulerMacaroni

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EulerMacaroni

#8 h Jun 26, 2015, 5:31 PM • 4 Y

Y by v_Enhance, ImSh95, Adventure10, Mango247

If you've done this problem, you should be able to recognize $X_a$ with some effort, and then the crucial claim about the point lying on both the circle through $AK_aL_a$ and circle with diameter $\overline{AH}$ is pretty easy to make.

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Cezar

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Cezar

#9 h Jun 26, 2015, 5:32 PM • 4 Y

Y by nguyenvanthien63, ImSh95, Adventure10, Mango247

Since $(L_{a}BK_{a}C)=-1$ , we find $M_{a}B^{2}=M_{a}X_{a} \cdot MA$ $\Longrightarrow$ $BC$ is tangent to $\odot ABX_{a}$ and $\odot ACX_{a}$ .
By angle chasing we find $X_{a} \in \odot BHC$ . Let the $R$ be the antipode of $H$ in $\odot BHC$ . By angle chasing we find $RCAB$ is a parallelogram $\Longrightarrow$ $A,M_{a},R$ are collinear.
So $\angle HX_{a}G=90^ {\circ}$ $\Longrightarrow$ $X_{a}\in (HG)$ $\Longrightarrow$ center of $X_{a}X_{b}X_{c}$ is the midpoint of $HG$ .

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Dukejukem

695 posts

Dukejukem

#10 h Jul 10, 2015, 3:59 AM • 4 Y

Y by karitoshi, ImSh95, Adventure10, Mango247

Let $H, G$ be the orthocenter, centroid, respectively, of $\triangle ABC.$ From the Internal and External Angle Bisector Theorems, we have $K_aB / K_aC = L_aB / L_aC = AB / AC.$ Thus, the division $(B, C; K_a, L_a)$ is harmonic. Hence, it is well-known (Lemma 1.5) that $M_aK_a \cdot M_aL_a = M_aB^2.$

Meanwhile, from Power of a Point, we find $M_aK_a \cdot M_aL_a = M_aX_a \cdot M_aA.$ Therefore, $M_aB^2 = M_aX_A \cdot M_aA$ , which can be rearranged into $M_aX_a / MB = M_aB / M_aA.$ Combining this relation with $\angle X_aM_aB = \angle BM_aA$ , it follows that $\triangle X_aM_aB \sim \triangle BM_aA.$ Similarly, we have $\triangle X_aM_aC \sim \triangle CM_aA.$ By using these similar triangles, we find $\angle BX_aC = \angle BX_aM_a + \angle CX_aM_a = \angle ABM_a + \angle ACM_a = 180^{\circ} - \angle BAC = \angle BHC.$ It follows that $B, C, H, X_a$ are concyclic. Therefore, $\angle HX_aB = \angle HCB = 90^{\circ} - \angle ABM_a.$ Meanwhile, from $\triangle X_aM_aB \sim \triangle BM_aA$ , it follows that $\angle ABM_a = \angle BX_aM_a.$ Therefore, $\angle HX_aB = 90^{\circ} - \angle BX_aM_a \implies \angle HX_aM_a = 90^{\circ}.$ Therefore $HX_a \perp GX_a.$ Similarly, $HX_b \perp GX_b$ and $HX_c \perp GX_c$ , so that $\odot (X_aX_bX_c)$ is just the circle of diameter $\overline{HG}.$ Consequently, its center is the midpoint of $\overline{HG}$ , which clearly lies on the Euler line. $\square$

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Dukejukem

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Dukejukem

#11 h Jul 10, 2015, 3:59 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

For a problem involving the same key point, see 2005 USA TST #6.

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EulerMacaroni

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EulerMacaroni

#12 h Jul 16, 2015, 10:09 PM • 4 Y

Y by jh235, ImSh95, Adventure10, Mango247

Here's a solution which explicitly does the barycentric calculation.

Solution

Remark

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Dukejukem

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Dukejukem

#13 h Nov 1, 2015, 12:26 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

Let $H$ be the orthocenter of $\triangle ABC$ and let $A'$ be the antipode of $A$ WRT $\odot(ABC).$ Let $AM_a$ cut $\odot(ABC)$ for a second time at $X.$

By the Angle-Bisector Theorem, $\tfrac{K_aB}{K_aC} = \tfrac{L_aB}{L_aC} = \tfrac{AB}{AC} \implies (B, C; K_a, L_a) = -1.$ Then by Power of a Point and a well-known result of harmonic divisions (Lemma 1.5), we have $M_aA \cdot M_aX_a = M_aK_a \cdot M_aL_a = -M_aB \cdot M_aC = -M_aA \cdot M_aX.$ Therefore, $M_aX_a = -M_aX \implies M_a$ is the midpoint of $\overline{XX_a}.$ But since $M_a$ is also the midpoint of $\overline{HA'}$ (well-known), it follows that $HXA'X_a$ is a parallelogram. Hence, $\angle HX_aX = \angle A'XX_a = 90^{\circ}.$ Then if $G$ is the centroid of $\triangle ABC$ , we have $\angle HX_aG = 90^{\circ} \implies X_a$ lies on the circle $\omega$ of diameter $\overline{HG}.$ Similarly, $X_b, X_c \in \omega$ , so the circumcenter of $\triangle X_aX_bX_c$ is just the midpoint of $\overline{HG}.$ $\square$

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EulerMacaroni

851 posts

EulerMacaroni

#14 h Nov 1, 2015, 2:20 AM • 4 Y

Y by ImSh95, pikapika007, Adventure10, Mango247

It suffices to show that $X_a$ lies on the circle with diameter $\overline{AH}$ , where $H$ is the orthocenter of $\triangle ABC$ . From this, we derive that $\angle AX_aH=\angle GX_aH$ , where $G$ is the centroid of $\triangle ABC$ , so that the circumcenter of $\triangle X_aX_bX_c$ is simply the midpoint of $\overline{HG}$ .

Denote the feet of the altitudes from $B$ and $C$ as $E$ and $F$ ; note that the circle with diameter $AH$ passes through $E$ and $F$ . Invert about $A$ with radius $\sqrt{AB \cdot AC}$ , composed with a reflection about the $A$ -angle bisector, and denote inverses with a '. We get that $F'$ is the intersection of the line through $B$ perpendicular through $AB$ with $AC$ , and define $E'$ similarly. $K_a'$ is the midpoint of minor arc $BC$ in the circumcircle of $\triangle ABC$ , and $L_a'$ is the midpoint of major arc $BAC$ .

The conclusion we desire is that the $A$ -symmedian, line $E'F'$ , and line $K_a'L_a'$ are concurrent; we claim that the concurrency point is the intersection of the tangents to $(ABC)$ at $B$ and $C$ , and moreover that this point is the circumcenter of $(BCEF)$ . Let $X_a'$ be the intersection of the tangents from $B$ and $C$ ; angle chasing, we get
$\begin{align*} \angle E'BF'&=180^{\circ}-\angle ABF' \\ &=\angle BAF'+\angle BF'A \\ &=\angle BAC+\angle BF'C \\ &= \frac{1}{2}\big(\angle BOC+\angle BX_a'C\big)\\ &=90^{\circ} \end{align*}$ so that $E'$ and $F'$ are indeed are our desired points, hence we are done $.\:\blacksquare\:$

I believe that we can actually use circle $(BEFC)$ to prove that the intersection of the tangents from $B$ and $C$ in $(ABC)$ lies on the $A$ -symmedian.

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danepale

99 posts

danepale

#15 h Feb 18, 2016, 11:29 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

Throughout the solution, let $Q$ denote $Q_a$ for almost every point $Q$ .
Let $H$ be the orthocenter and $\triangle DEF$ the orthic triangle of $\triangle ABC$ . Let $\Omega = (ABC)$ , $\omega = (DEF)$ .

Let $P$ be the intersection of $(AKL)$ and $\Omega$ . $(AKL)$ is the Apollonius circle WRT $BC$ , so $(A, P, B, C)$ is harmonic, which means $AP$ is the $A$ -symmedian.
$AX$ and $AP$ are isogonal conjugates, so we get $KX = KP$ and $LX = LP$ , hence $X$ is the reflection of $P$ in $BC$ .

Now we forget about $K$ and $L$ . Let $AM$ intersect $(AEHF)$ again in $X'$ . It's trivial to prove that the tangents in $E$ and $F$ to $(AEHF)$ intersect in $M$ , hence $(A, X', E, F) = -1$ .
Let $T$ be the reflection of $H$ in $BC$ . By some cross-ratio chasing, $(D, HX' \cap BC, B, C)$ and $(D, TP \cap BC, B, C)$ are both harmonic, so $HX'$ , $TP$ , $BC$ are concurrent in $R$ . Reflecting $TP$ in $BC$ yields $X' \equiv X$ .

Let $\psi$ be the inversion that swaps $\Omega$ and $\omega$ . It preserves cross-ratios, so $X$ maps to $R$ .
$RA \cdot RB = RE \cdot RB$ , so $R_a$ , $R_b$ , $R_c$ lie on the radical axis of $\Omega$ and $\omega$ , which is perpendicular to the Euler line of $\triangle ABC$ .
Under $\psi$ , the Euler line maps to itself, while the line $R_aR_bR_c$ maps to the circumcircle of $\triangle X_aX_bX_c$ .
$\psi$ preserves angles, so the Euler line passes through the center of $(X_aX_bX_c)$ . $\blacksquare$

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Shreyasharma

667 posts

Shreyasharma

#63 h Nov 19, 2023, 10:16 PM • 1 Y

Y by OronSH

One of the problems of all time.

We claim that $(X_AX_BX_C)$ is centered at the midpoint of $HG$ , where $H$ and $G$ are the orthocenter and centroid.

To do this we will show that $HX_A \perp X_AG$ or equivalently that $X_A$ is the $A$ humpty point. To see this let $H_A$ be the $A$ humpty point. It is well known that $H_A$ lies on $AM$ . Then by power of a point it suffices to show that,
$\begin{align*} M_AK_A \cdot M_AL_A = M_AH_A \cdot M_AA. \end{align*}$ However, letting $E = BH \cap AC$ we know that
$\begin{align*} M_AH_A \cdot M_AA &= M_AE^2\\ &= M_AB \cdot M_AC\\ &= MK_A \cdot M_AL_A \end{align*}$ where the first line follows from the Three Tangents lemma. Thus in a similar fashion we can show $X_B$ and $X_C$ lie on the circle with diameter $GH$ . Therefore their circumcenter lies on the Euler Line as desired.

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shendrew7

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shendrew7

#64 h Nov 25, 2023, 3:10 AM

Y by

The $A$ -Humpty point can be characterized as the point of intersection of the $A$ -Apollonius circle and the $A$ -median distinct from $A$ , which is exactly $X_a$ . We further note the $A$ -Humpty point is also the foot from $H$ to the $A$ -median, implying
$\angle HX_aA = \angle HX_aG = 90,$
so $X_a$ lies on the circle with diameter $HG$ , or the Euler line, and symmetrically for $X_b$ and $H_c$ . $\blacksquare$

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Scilyse

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Scilyse

#65 h Jan 22, 2024, 9:54 AM

Y by

Note that $(AK_a L_a)$ is the $A$ -Apollonian circle. Let $H$ be the orthocenter of triangle $ABC$ , and further let $X_a'$ , $X_b'$ and $X_c'$ be the feet from $H$ to $AM_a$ , $BM_b$ , $CM_c$ respectively. Now it is well-known that $X_a'$ , $X_b'$ and $X_c'$ lie on the $A$ -, $B$ - and $C$ -Apollonian circles respectively, and since there are at most two intersections of a circle and a line, we know that $X_a = X_a'$ , $X_b = X_b'$ and $X_c = X_c'$ . Thus $HG$ is the diameter of $(X_a X_b X_c)$ where $G$ is the centroid of triangle $ABC$ , so the circumcenter of triangle $X_a X_b X_c$ is the midpoint of segment $\overline{HG}$ , as desired.

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dolphinday

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dolphinday

#66 h Feb 8, 2024, 4:36 AM • 1 Y

Y by ehuseyinyigit

$(AK_aL_a)$ is the $A$ -Apollonius circle of $\triangle ABC$ , which makes $X_a$ the $A$ -Humpty point. It's well known that $AX_a \perp X_aH$ , and by definition $AX_a$ , $BX_b$ , $CX_c$ concur at $G$ . This implies that $GH$ is the diameter of $(X_aX_bX_c)$ , so the center is the midpoint of $GH$ , as desired.

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Aiden-1089

277 posts

Aiden-1089

#67 h Apr 23, 2024, 6:43 AM

Y by

Let $\omega$ be the orthocentroidal circle (circle with diameter $GH$ , where $G$ is the centroid and $H$ is the orthocentre). Its centre clearly lies on the Euler line, so it suffices to show that $X_a,X_b,X_c$ lies on $\omega$ .

Note that $(AK_aL_a)$ is the $A$ -Apollonius circle, so $X_a=(AK_aL_a) \cap AM_a$ is the $A$ -Humpty point. It is well-known that $HX_a \perp AM$ , so $\angle HX_aG = 90^{\circ}$ , from which it follows that $X_a$ lies on $\omega$ . Similarly, $X_b,X_c$ lie on $\omega$ , so we are done.

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kotmhn

57 posts

kotmhn

#68 h Aug 10, 2024, 4:05 PM

Y by

very beautiful problem.
here is my solution:
Claim: Let $H_a$ be the $A$ humpty point of a circle and $O,N$ denote the circumcenter and 9 point center respectively then $\frac{H_{a}O}{H_{a}N} = 2$ .
Proof: observe that $\frac{HO}{HN} = \frac{GO}{GN} = 2$ where $H,G$ are the orthocenter and centroid respectively.We also know that $\angle HH_{a}N = 90^{\circ}$ . So $H_a$ lies on $(HG)$ and $(HG)$ is the Apollonian circle of $O,N$ with ratio 2. Hence the assertion is true.
PS: we actually only need the concylic part but the result itself is cool so i added it

Claim : the circumcenter of $(H_{a}H_{b}H_{c})$ lies on the euler line.
proof: $(H_{a}H_{b}H_{c}) \equiv (HG)$ from the above claim so the circumcenter is on $HG$ which is the euler line.

Claim: $X_i \equiv H_i$ where $i= a,b,c$ .
proof: as $(BC;L_{a}K_{a})=-1$ so from P1 we have $M_{a}A^2 = ML_{a} \cdot MK_{a} = MX_{a} \cdot A$ . Done.
This completes the proof.

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thdnder

194 posts

thdnder

#69 h Aug 12, 2024, 9:57 AM

Y by

It's clear that $(B, C; K_A, L_A) = -1. \implies M_AB^2 = M_AC^2 = M_AK_A \cdot M_AL_A = M_AX_A \cdot M_AA$ . Thus, $(ABX_A), (ACX_A)$ are tangent to $BC \implies X_A$ is the $A$ -Humpty point. Analogously, $X_B, X_C$ are $B$ -Humpty and $C$ -Humpty points, respectively. Since $X_A$ is $A$ -Humpty point, we deduce that $HX_A$ is perpendicular to $AM_A$ , where $H$ is the orthocenter of $ABC$ . Therefore, $\angle HX_AG = 90^{\circ} \implies X_A$ lies on the circle with diameter $HG$ , where $G$ is the centroid of $ABC$ . Similarly, $\angle HX_BG = \angle HX_CG = 90^{\circ}$ . Hence, $HGX_AX_BX_C$ is cyclic and its diameter is $HG \implies$ center of $(X_AX_BX_C)$ lies on $HG$ . $\blacksquare$

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bebebe

984 posts

bebebe

#70 h Aug 14, 2024, 12:53 PM

Y by

From the internal external angle bisectors, $(L_a, K_a; B, C)=-1,$ so we know $M_a B^2 = M_aK_a \cdot M_aL_a.$ By PoaP, $M_a X_a \cdot M_a A = M_aK_a \cdot M_aL_a= M_a B^2,$ which implies $A$ and $X_a$ are inverses wrt the circle with diameter $BC,$ say $w_{bc}.$

Letting $H_c = w_{bc} \cap AB$ and $H_b = w_{bc} \cap AC,$ we notice $\angle BH_cC = \angle BH_bC = 90,$ so $CH_c \cap BH_b = H,$ the orthocenter. By Brocard, $H \in polar(A).$ However we know $polar(A)$ also passes through the inverse of $A,$ which is $X_a$ . Thus, $HX_a=polar(A)$ and $\angle HX_aA=90.$

Notice that $G \in AM_a,$ so $\angle HX_aG=90.$ Hence $X_a, X_b, X_c$ all lie on circle with diameter $HG,$ and we are done.

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Eka01

204 posts

Eka01

#71 h Aug 20, 2024, 1:50 PM • 1 Y

Y by AaruPhyMath

It is well known that $(AK_aL_a)$ is the $A$ apollonius circle so it cuts the $A$ median at the $A$ humpty point which is $X_a$ and similarly $X_b$ and $X_c$ are humpty points. It is trivial to see that $X_i \in (HG)$ where $i \in$ $\{a,b,c\}$ and $G$ is centroid of $\Delta ABC$ because the humpty point is defined as the perpendicular from $H$ to the corresponding median so we get the desired result.

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joshualiu315

2513 posts

joshualiu315

#72 h Aug 21, 2024, 5:30 AM • 1 Y

Y by dolphinday

Only consider $X_a$ initially. Notice that $(AK_aL_a)$ forms the exact configuration of an Apollonian circle; that is, we have

$\frac{XB}{XC} = \frac{AB}{AC},$
for all points $X$ on $\overline{BC}$ . Note that the $A$ -Humpty point $X_a'$ in $\triangle ABC$ can be defined as the point on the $A$ -median satsifying $\tfrac{X_a'B}{X_a'C} = \tfrac{AB}{AC}$ . Hence, $X_a' = X_a$ , meaning that $HX_a \perp GX_a$ (where $H$ is the orthocenter and $G$ is the centroid). An analogous argument for $X_b$ and $X_c$ implies that $(X_aX_bX_c)$ has diameter $\overline{HG}$ . $\square$

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Kappa_Beta_725

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Kappa_Beta_725

#73 h Aug 22, 2024, 7:47 AM

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On my way to another one-line solution. Point on the median? It has to be humpty!

The intersection of $(AK_a L_a)$ (the Apollonian circle) with the median gives the Humpty point $X_a$ , which by definition lies on the perpendicular from $AH$ . Therefore $X_aX_bX_c$ lie on the circle with diameter $HG$ . $\blacksquare$

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cj13609517288

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cj13609517288

#74 h Oct 17, 2024, 2:51 PM

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What?
Used the 10\% hint on ARCH.

It turns out that $X$ is exactly the $A$ -Humpty point, so $\angle HXG=90^{\circ}$ therefore the circumcenter of $X_aX_bX_c$ is exactly the midpoint of $GH$ .

But note that $X$ lies on $AM$ and the $A$ -apollonius circle of $ABC$ , so we are done. $\blacksquare$

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cj13609517288

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cj13609517288

#75 h Jan 5, 2025, 8:48 PM

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Re-solved because I somehow forgot that I solved this.

Note that $(AKL)$ is the circle through $A$ that has a constant $PB/PC$ . Thus $X_A$ is exactly the $A$ -Humpty point.

Thus the reflection of $H$ over $X$ , say $Y$ , has $GH=GY$ . Similarly for the other two, so $(Y_aY_bY_c)$ has circumcenter $G$ , so $(X_aX_bX_c)$ has circumcenter $(G+H)/2$ . $\blacksquare$

Remark. A few hours later, I discovered that the $A$ -, $B$ -, and $C$ -Ex Points are collinear by some harmonic stuff. Now the negative inversion centered at $H$ swapping $A$ and the foot of the $A$ -altitude and similar maps the Ex points to the Humpty points, so indeed we get that the Humpty points are also concyclic with $H$ . Thus the three perpendicular bisectors involving $H$ must concur, and at that point it's easy to see what's happening.

This post has been edited 3 times. Last edited by cj13609517288, Jan 6, 2025, 1:12 AM

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lelouchvigeo

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lelouchvigeo

#76 h Jan 7, 2025, 2:16 PM

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Storage

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Ilikeminecraft

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Ilikeminecraft

#77 h Mar 28, 2025, 5:22 AM

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Claim: $X_a$ is the $A$ -humpty point.
Proof: Let $X_a’$ lie on $(ABC)$ such that $ABX_A’C$ forms a harmonic quadrilateral. Using well-known properties, $X_a’K_a$ bisects angle $BX_a’C.$ Clearly, $(K_aL_a;BC) = -1$ from right angles and bisector on point $A.$ Thus, $\angle K_aX_a’L_a = 90.$ It is well known $X_a’$ is the reflection of the Humpty point across $BC.$ Thus, the Humpty point lies on $(AK_aL_a).$ This finishes the desired claim.

It trivially follows that the circumcenter is the midpoitn of $GH.$

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