Easy Geo Regarding Euler Line

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Easy Geo Regarding Euler Line

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Source: 2021 Taiwan TST Round 2 Independent Study 1-G

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USJL

#1 h Apr 7, 2021, 6:12 AM • 3 Y

Y by Mango247, Mango247, Mango247

Let $ABCD$ be a convex quadrilateral with pairwise distinct side lengths such that $AC\perp BD$ . Let $O_1,O_2$ be the circumcenters of $\Delta ABD, \Delta CBD$ , respectively. Show that $AO_2, CO_1$ , the Euler line of $\Delta ABC$ and the Euler line of $\Delta ADC$ are concurrent.

(Remark: The Euler line of a triangle is the line on which its circumcenter, centroid, and orthocenter lie.)

Proposed by usjl

This post has been edited 2 times. Last edited by USJL, Apr 7, 2021, 6:22 AM

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#2 h Apr 7, 2021, 6:18 AM

Y by

We use Cartesian coordinate.：）
But I use complex number and waste a lot of time.

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20 posts

#3 h Apr 7, 2021, 6:23 AM

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Let H be the orthocenter of ADC, X be the circumcenter of ADC, then its easy to see that there's a homothety that takes triangle $O_1O_2X$ to triangle $ACH$ since the side are parallel and we can use Desargues theorem

The result follows easily

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222 posts

#4 h Apr 13, 2021, 7:06 PM • 2 Y

Y by p_square, guptaamitu1

Solution with p_square, BOBTHEGR8, Arwen713

We use DDIT! Let $O_3$ be the circumcenter of $ABC$ and $G_3$ be the centroid. Also, let $M,N$ be the midpoints of $AB,BC$ . We also know that in a quadrilateral with perpendicular diagonals $AC, BD$ , $AC\cap BD$ has an isogonal conjugate. Let this be $E$ . Also, let $AO_2\cap CO_1=F$ .

Now, applying DDIT on quadrilateral $AMCN$ , we get that there's an involution from $O_3$ with pairs $(O_3G_3, O_3B), (O_3M, O_3N)$ and $(O_3A, O_3C)$ . But, observe that $O_3O_1M$ are collinear and $O_3O_2N$ are collinear and $O_3,B,E$ are collinear. Now, by DDIT on $AO_2CO_1$ , there's an involution swapping $(O_3A,O_3C), (O_3O_1, O_3O_2)$ and $(O_3E, O_3F)$ . But $(O_3O_1,O_3O_2)=(O_3M, O_3N)$ . So, the two mentioned involutions are the same and as $O_3E=O_3B$ , we have $O_3G=O_3F$ but then $F$ lies on $O_3G_3$ i.e. Euler line of $\Delta ABC$ . Similarly, we can get the result for $\Delta ADC$ .

This post has been edited 1 time. Last edited by Rg230403, Apr 13, 2021, 7:07 PM

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#5 h Apr 17, 2021, 4:27 PM • 1 Y

Y by Snark_Graphique

Let $O_A$ be the circumcenter of $\triangle ABD$ and define $O_B,O_C$ and $O_D$ similarly. Let $H_B$ and $H_D$ be the orthocenters of $\triangle ABC$ and $\triangle ADC$ . We will show that quadrilateral $CH_BAH_D$ and $O_AO_BO_CO_D$ are homothetic. Notice that $O_AO_B$ and $CH_B$ are parallel because they are perpendicular to $AB$ . This is true for all six sides including the diagonals, and we're done.

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#6 h Apr 17, 2021, 4:50 PM

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Homothety taking triangle $O_A$ $O_C$ $O_D$ to triangle $CAH_D$ .

This post has been edited 3 times. Last edited by SerdarBozdag, Apr 17, 2021, 4:53 PM
Reason: .

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hakN

#7 h Apr 17, 2021, 5:37 PM • 1 Y

Y by Mango247

Let $O_3$ and $O_4$ be the circumcenters of $\triangle ABC$ and $\triangle ADC$ respectively. Also let $H_1$ and $H_2$ be the orthocenters of $\triangle ABC$ and $\triangle ADC$ respectively.
It is clear that $O_2O_3\perp BC$ and $AH_1\perp BC \implies AH_1\parallel O_2O_3$ and similarly $AH_2\parallel O_2O_4$ .
Let $H_1O_3\cap H_2O_4 = F$ . The homothety centered at $F$ taking $\triangle FO_3O_4$ to $\triangle FH_1H_2$ will take $O_2$ to $AH_1\cap AH_2 = A$ . So $F,O_2,A$ are collinear and similarly $C,O_1,F$ are collinear and thus we are done.

This post has been edited 1 time. Last edited by hakN, Apr 17, 2021, 8:34 PM

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#8 h Jul 31, 2022, 10:42 PM • 1 Y

Y by centslordm

Let $O_B$ and $H_B$ be the circumcenter and orthocenter of $\triangle ABC,$ respectively. Similarly define $O_D$ and $H_D$ for $\triangle ADC.$ Notice $\overline{AC}$ is the radical axis of $(ABC)$ and $(ADC),$ so $\overline{O_BO_D}\perp\overline{AC}$ and $\overline{O_BO_D}\parallel\overline{BD}=\overline{H_BH_D}.$ Also, $\overline{AB}$ is the radical axis of $(ABD)$ and $(ABC),$ so $\overline{O_BO_1}\perp\overline{AB}$ and $\overline{O_BO_1}\parallel\overline{CH_B}.$ Similarly, $\overline{O_CO_1}\parallel\overline{CH_D}$ so $\triangle O_1O_BO_D$ is homothetic to $\triangle CH_BH_D,$ which implies $CO_1$ and the two Euler lines are concurrent. We can analogously show that $AO_2$ passes through this point. $\square$

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#9 h May 30, 2023, 12:03 PM • 1 Y

Y by archp

Let $P,Q$ be the circumcentre and orthocentre of $\triangle ABC$ . Then, $AQ\perp BC$ and $PO_2\perp BC$ because $P,O_2$ lie on perpendicular bisector of $BC$ . Thus, $AQ\parallel PO_2$ and similarly, $CQ\parallel PO_1$ .

We also have $O_1O_2\perp BD$ because $BD$ is the radical axis of $(ABD), (CBD)$ and $AC\perp BD$ . Therefore $AC\parallel O_1O_2$ so $\triangle QAC$ and $\triangle PO_2O_1$ are homothetic, implying that $PQ$ (the Euler line of $\triangle ABC$ ), $AO_2$ and $CO_1$ concur. Similarly, the Euler line of $\triangle ADC$ concurs with $AO_2$ and $CO_1$ too.

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#10 h May 24, 2024, 3:34 AM

Y by

wow

First, we will prove the following lemma, which is the main idea of this solution.

Let two lines $\ell_1$ and $\ell_2$ meet at $O$ , and consider fixed points $A$ and $C$ . Let $\ell$ be a variable line parallel to $AC$ , and have $\ell$ intersect $\ell_1$ and $\ell_2$ at $O_1$ and $O_2$ respectively. Then, the intersection of $AO_2$ and $CO_1$ , which we call $P$ , lies on a fixed line through $O$ as $\ell$ varies.

By similar triangles, we have $P=\frac{A(O_1O_2)+O_2(AC)}{O_1O_2+AC}$ $P-O=\frac{O_1O_2(A-O)+AC(O_2-O)}{O_1O_2+AC}.$
Note that both $O_1O_2$ and $O_2-O$ are proportional to the distance from $O$ to $\ell$ , but $A-O$ and $AC$ stay constant. Thus, the direction of the vector in the numerator does not change as $\ell$ changes, showing the claim.

Let's now assess the situation in the problem. In triangle $\triangle ABC$ , there is some line $\ell$ parallel to $AC$ that represents the perpendicular bisector of $BD$ . Then, let $\ell_1$ and $\ell_2$ be the perpendicular bisectors of $AB$ and $CB$ respectively. By our lemma, we have that the intersection of $AO_2$ and $CO_1$ , which we call $P$ , lies on a fixed line. By taking $\ell$ to pass through $O$ , we see that this line passes through $O$ . Furthermore, by taking $\ell$ to be the $B$ -midline of $\triangle ABC$ , we see that this line passes through $G$ . Thus, $P$ lies on the Euler line of $\triangle ABC$ , and similarly it also lies on the Euler lie of $\triangle ADC$ , done.

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cursed_tangent1434

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cursed_tangent1434

#11 h Nov 27, 2024, 5:11 AM • 1 Y

Y by Shreyasharma

Neat problem! Solved with deduck. Pretty much the same idea as the above solutions, but posting anyways. We denote by $O_b$ , $O_d$ , $H_b$ and $H_d$ respectively, the circumcenters and orthocenters of $\triangle ABC$ and $\triangle ACD$ . The following claim is the pith of the problem.

Claim : Triangles $\triangle O_1O_2O_b \sim \triangle ACH_b$ and $\triangle O_1O_2O_d \sim \triangle ACH_d$ are pairwise homothetic.
Proof : Consider circles $(ABD)$ and $(ABC)$ . Since the line joining the centers of two circles is perpendicular to their radical axis, $O_1O_b \perp AB$ . Similarly, $O_1O_d \perp AD$ , $O_2O_b \perp BC$ , $O_2O_d \perp CD$ and $O_1O_2 \perp BD$ . Now, by the definition of the orthocenter we also know that $H_bC \perp AB$ , $H_bA \perp BC$ , $H_dA \perp CD$ and $H_dC \perp AD$ . Thus, $O_1O_b \parallel CH_b$ , $O_2O_b \parallel AH_b$ , $O_1O_b \parallel AH_d$ , $O_2O_d \parallel CH_d$ and $O_1O_2 \parallel AC$ . Thus, triangles $\triangle O_1O_2O_b$ and $\triangle ACH_b$ and triangles $\triangle O_1O_2O_d$ and $\triangle ACH_d$ have pairwise parallel sides, making these two pairs of triangles homothetic.

Now, since it is well known that lines joining corresponding vertices of homothetic triangles are concurrent, $AO_2$ , $CO_1$ , $O_bH_b$ and $O_dH_d$ all concur, as desired.

Remarks : This is a really interesting configuration. Let $P$ and $Q$ denote the intersections of the diagonals of quadrilaterals $ABCD$ and $O_1O_bO_2O_d$ . Via the same homothety argument it follows that $PQ$ also passes through the aforementioned concurrence point. Further, if $M_Q$ is the Miquel Point of $\triangle ABCD$ , $M_Q$ also lies on line $\overline{PQ}$ . Further, with some work one can also show that quadrilaterals $(M_QAQO_1)$ and similar are all cyclic.

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Ritwin

#12 h Mar 1, 2025, 8:03 AM

Y by

It suffices to show $K = AO_2 \cap CO_1$ lies on the Euler line of $ABC$ . Let $ABC$ have circumcenter $O$ , centroid $G$ , and medial triangle $XYZ$ . Note that $O_1O_2$ is the perpendicular bisector of $BD$ , so we have $O_1O_2 \parallel AC \parallel XZ$ .

This means triangles $AXO_1$ and $CZO_2$ are centrally perspective. Desargues's theorem says they're also axially perspective, meaning $K$ , $O$ , and $G$ are collinear. This is what we wanted. $\blacksquare$

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