Equilateral Triangle and Euler Line

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Equilateral Triangle and Euler Line

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#1 h Jul 12, 2024, 4:06 AM • 4 Y

Y by KevinYang2.71, ehuseyinyigit, Rounak_iitr, EpicBird08

Let $D$ , $E$ , and $F$ be points on the perpendicular bisectors of $BC$ , $CA$ , and $AB$ of triangle $ABC$ such that $DEF$ is equilateral. Show that the center of $DEF$ lies on the Euler line of $ABC$ .

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qwerty123456asdfgzxcvb

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qwerty123456asdfgzxcvb

#2 h Jul 12, 2024, 6:27 AM • 4 Y

Y by ehuseyinyigit, RetroTurtle, centslordm, trigadd123

you can reduce this with moving points to "is there a single point on the euler line that's not the circumcenter such that we can construct DEF centered at that point?"

edit: i heard from tapir that napoleons' theorem wins

This post has been edited 1 time. Last edited by qwerty123456asdfgzxcvb, Jul 12, 2024, 11:43 PM

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#4 h Jul 20, 2024, 2:24 PM

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Any other solution?

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#5 h Jul 24, 2024, 12:04 AM • 4 Y

Y by YaoAOPS, RetroTurtle, qwerty123456asdfgzxcvb, EpicBird08

Solution 1: $\triangle {ABC}$ and $\triangle DEF$ are orthologic through the circumcenter $O$ of $\triangle {ABC}$ $\Longrightarrow$ perpediculars from $A,B,C$ to $EF,FD,DE$ concur at the second orthology center $T.$ Thus $\angle (TB,TC)=\angle (DF,DE)=120^{\circ}$ and same for the others $\Longrightarrow$ $T$ is a Fermat point of $\triangle {ABC}.$ Hence if $O_A,O_B,O_C$ are the circumcenters of $\triangle TBC,$ $\triangle TCA,$ $\triangle TAB$ (apices of external/internal Fermat triangles of $\triangle {ABC}$ ), then $\triangle O_AO_BO_C$ is equilateral with center the centroid $G$ of $\triangle {ABC}$ (well-known) and it's homothetic to $\triangle DEF$ through $O$ $\Longrightarrow$ center of $\triangle DEF$ lies on $OG.$

Solution 2 (Overkill): Let $O$ be the circumcenter of $\triangle {ABC}.$ Perpendiculars to $OD,OE,OF$ at $D,E,F$ bound a triangle $\triangle {XYZ}$ $\Longrightarrow$ pedal triangle of $O$ WRT $\triangle XYZ$ is equilateral $\Longrightarrow$ $O$ is an isodynamic point of $\triangle XYZ$ $\Longrightarrow$ center $U$ of $\triangle DEF$ is the midpoint of $OT$ (being $T$ the isogonal conjugate of $O$ WRT $\triangle XYZ$ ). Now it's well-known that $OT$ is parallel to Euler line of $\triangle XYZ$ (see for example Fermat points and Euler line and elsewhere). Since $\triangle XYZ$ is clearly homothetic to $\triangle ABC,$ then it follows that $U$ is on Euler line of $\triangle {ABC}.$

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#6 h Jul 24, 2024, 3:58 PM • 1 Y

Y by qwerty123456asdfgzxcvb

Just notice that DEF and pedal triangle of 1st(2nd) isodynamic point are homothetic

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#7 h Jan 15, 2025, 3:05 AM

Y by

mmp tunnel vision

We will solve this with reference to the medial triangle: that $DEF$ are on the altitudes of a triangle $ABC$ . Note the Euler line is invariant as a homothety at $G$ swaps the two triangles.

Now, for any equilateral triangle that works, we can homothety it at $H$ so one vertex coincides with the reference triangle.

Let us MMP. Suppose our equilateral triangle is $AEF$ , and our reference triangle $ABC$ . Fix $AEF$ . We will move $H$ on a fixed line through $F$ .

https://i.imgur.com/b88FlyP.png

Let the line be $\ell$ . Then, we get $B$ by intersecting the perpendicular to $\ell$ through $A$ with $EH$ . Similarly, $C$ is the intersection of the perpendicular to $EH$ through $A$ with $\ell$ .

Now, the centroid of $ABC$ is the dilation of the midpoint of $BC$ by $\frac{2}{3}$ . Let us check degrees. If $H$ moves linearly, then $B$ and $C$ move linearly too.

I'm too lazy to show that the midpoint is linear, so I assume that it has degree $2$ . Now, $G$ (centroid) is degree $2$ too. Note the centre of $DEF$ is obviously fixed, $H$ is linearly, and $GH$ is euler line so we check $4$ cases.

Consider $H$ as the point $AEHF$ is a kite, this works lol because symmetry.

Consider $H$ as $F$ .

https://i.imgur.com/vIMq6C7.png

We'll still show that the result is true, just using $O$ , circumcentre of $ABC$ to prove euler line instead out of convenience.

Let $O'$ be circumcentre of $AEF$ , then, $\angle AFO'=30^\circ$ . Now, from $\angle BFA=120^\circ$ by our assumption, we have $\angle C=60^\circ$ hence $\angle BOA=120^\circ$ . Thus $BFOA$ is cyclic and $\angle AFO=\angle ABO=30^\circ$ , hence $F=H, O,O'$ are collinear.

Now, take $H=AE\cap \ell$ . This works because now, as $FO'\perp HA$ , and we easily check that $\angle HAC$ is right, dilate $FO'$ to $AC$ . Note that centroid and circumcentre coincide on equilateral triangle, so the image of $O'$ divides $AC$ in two to one split, but this is precisely where the centroid of $ABC$ is, noting $B$ is sent to $A$ . (use the ratio definition again)

Finally, send $H$ to $\infty_\ell$ . $C$ is now the foot of $A$ onto $\ell$ , and $B$ is the foot of $A$ onto $E\infty_\ell$ . It suffices to show that the centroid of $ABC$ is the foot of $O'$ onto the perpendicular to $\ell$ through $A$ . Let us project these ratios through $\infty_\ell$ , so equivalently, if $C'=AE\cap \ell$ , and $O_1$ is $\infty_\ell O'\cap AE$ , then we want $O_1$ to be the centroid of $AEC'$

https://i.imgur.com/RvEvgbO.png

Using vector definition of centroid, we just need $AO_1=\frac{AE+AC'}{3}$ . But construct $A'$ as reflection of $A$ over $EF$ , and $O_2$ as the intersection of $\infty_\ell$ with $AE$ . One easily checks $AA'=3AO'$ , and hence $AO_2=3AO_1$ . but by parallelograms, as $A'F\parallel AE$ , and $FC'\parallel A'O_2$ , $O_2C'=A'F$ . Hence, $3AO_1=AO_2=C'O_2+AC'=A'F+AC'=AE+AC'$ .

Thus, we have checked our $(1+2+0)+1$ cases, and we conclude our mmp and the problem is true for every $H$ !

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#8 h Jan 15, 2025, 3:58 AM

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https://www.desmos.com/calculator/jqqv1abttq

We coord bash with the above setup, using the orthocentric phrasing as I have done in my previous phrasing. When we wish to show the concurrence, then dilate by $3$ . Now, we want the following points collinear:
$\left(\frac{k-\sqrt{3}}{k+\frac{1}{c}}-\frac{c+\sqrt{3}}{c+\frac{1}{k}},\frac{2\sqrt{3}+c-k}{kc+1}\right)\quad \left(\frac{3k+3c}{k-c},\frac{6ck}{k-c}+3\sqrt{3}\right)\quad \left(0,2\sqrt{3}\right)$ This is equivalent to
$\frac{\frac{6ck}{k-c}+\sqrt{3}}{\frac{3k+3c}{k-c}}=\frac{\frac{2\sqrt{3}+c-k}{kc+1}-2\sqrt{3}}{\frac{k-\sqrt{3}}{k+\frac{1}{c}}-\frac{c+\sqrt{3}}{c+\frac{1}{k}}}$ Simplifying,
$\frac{6ck+\sqrt{3}\left(k-c\right)}{3k+3c}=\frac{\left(\frac{2\sqrt{3}+c-k}{kc+1}-2\sqrt{3}\right)}{\frac{ck-\sqrt{3}c-ck-\sqrt{3}k}{ck+1}}$ Or
$\frac{6ck+\sqrt{3}\left(k-c\right)}{3k+3c}=-\frac{2\sqrt{3}+c-k-2\sqrt{3}ck-2\sqrt{3}}{\sqrt{3}\left(k+c\right)}$ Rearranging
$6ck+\sqrt{3}\left(k-c\right)=-\sqrt{3}\left(c-k-2\sqrt{3}ck\right)$ Which is true upon expansion.

Hence we are done.

This post has been edited 1 time. Last edited by TestX01, Jan 15, 2025, 3:59 AM
Reason: link to desmos

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#9 h Jan 15, 2025, 4:20 AM

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Let $ABC$ be a triangle with incentre $I$ , and $D,E,F$ be an equilateral triangle on the angle bisectors of $ABC$ . Prove the centre of $DEF$ is on the $OI$ line ( $O$ is circumcentre)

As I have shown above, the same is true for altitudes, where the centre lies on $OH$ line. What other triangle centers satisfy a similar property?

Also, for my first solution, we reduce the degree of $G$ by just adding the linear terms individually to be linear

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