Fermat points and Euler line

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#1 h Aug 22, 2004, 4:22 PM • 2 Y

Y by Adventure10, Mango247

Prove that $FF' || OH.$ Where $F$ is Fermat point, $F'$ is its isogonal conjugate and $O$ and $H$ are circumcenter and orthocenter of $\triangle ABC.$

This post has been edited 1 time. Last edited by Luis González, Jan 20, 2018, 7:02 PM
Reason: Making proposition clearer and fixing grammar

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#2 h Aug 22, 2004, 5:05 PM • 2 Y

Y by Adventure10, Mango247

Is $F'$ the second fermat point? Because if it is, then it's not true. As far as I know, If $F_1,F_2$ are the first and second Fermat points, and $I_1,I_2$ are the first and second isodynamic points (just think that $I_i$ is the isogonal conjugate of $F_i$ ), then $F_iI_i\|OH$ .

[There was an error, so I've edited it.]

This post has been edited 1 time. Last edited by grobber, Aug 22, 2004, 7:24 PM

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#3 h Aug 22, 2004, 6:00 PM • 2 Y

Y by Adventure10, Mango247

what are isodinamic points? can someone explain a little about it?

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#4 h Aug 22, 2004, 6:51 PM • 2 Y

Y by Adventure10, Mango247

The locus of point $M$ s.t. $\frac{MB}{MC}=\frac{AB}{AC}$ is a circle having the segment bounded by the foot of the $a$ -bisector and the foot of the exterior $a$ -bisector. This circle is called the $a$ -Apollonius circle. Similarly we define the $b$ and $c$ -Apollonius circles. It can be shown that these three circles are coaxal, and the two points through all three of them pass are called the isodynamic points of the triangle $ABC$ .

It's also true that the isodynamic points are the isogonal conjugates of the Fermat points, and this is, I think, more useful in this context. Another interesting property (which is crucial in proving the fact that they're the isogonal conjugates of the Fermat points) is that these two are the only points which have equilateral pedal triangles.

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#5 h Aug 23, 2004, 2:53 PM • 3 Y

Y by AlastorMoody, Adventure10, Mango247

Omid Hatami wrote:

Prove that $FF' || OH$ that $F$ is fermat point and $O$ and $H$ are citcumcenter and orthocenter.

The only thing that makes me wonder is this strange "Very hard" with 9 exclamation signs. Especially this is because my experience shows that a problem you call "hard" (even without any exclamation sign) is already far beyound my brain. And now this one, really strange...

In fact, I assume that you mean by F' the isogonal conjugate of F, i. e. the first isodynamic point of the triangle.

For reasons of convenience, I rename F' as J (just since I am more used to this letter for an isodynamic point) and hence rewrite your problem as follows:

Let F be the first Fermat point and J the first isodynamic point of a triangle ABC. Prove that the line FJ is parallel to the Euler line of triangle ABC.

For the solution, I will use three lemmas:

Lemma 1. If $P_a$ , $P_b$ , $P_c$ are the reflections of a point P in the sides BC, CA, AB of a triangle ABC, then the circumcenter of the triangle $P_a P_b P_c$ is the isogonal conjugate Q of the point P with respect to the triangle ABC, and we have $P_b P_c \perp AQ$ , $P_c P_a \perp BQ$ and $P_a P_b \perp CQ$ .

Lemma 2 (extended Napoleon theorem). If D, E, L are the centers of the equilateral triangles erected outwardly on the sides BC, CA, AB of the triangle ABC, then the triangle DEL is equilateral, and its center is the centroid G of triangle ABC.

Lemma 3. For the first Fermat point F of triangle ABC, we have $EL \perp AF$ , $LD \perp BF$ , $DE \perp CF$ .

Now, consider the reflections X, Y, Z of the point J in the sides BC, CA, AB of triangle ABC. We know that the first Fermat point F and the first isodynamic point J of triangle ABC are mutually isogonal conjugate points, so we can apply Lemma 1 and see that the circumcenter of the triangle XYZ is the isogonal conjugate of the point J, i. e. the point F, and that we have $YZ \perp AF$ , $ZX \perp BF$ and $XY \perp CF$ . Together with $EL \perp AF$ , $LD \perp BF$ , $DE \perp CF$ (from Lemma 3), this yields YZ || EL, ZX || LD and XY || DE. Hence, the triangles XYZ and DEL are homothetic. Since the triangle DEL is equilateral (Lemma 2), it follows that the triangle XYZ is also equilateral. Hence, instead of saying that the circumcenter of the triangle XYZ is the point F, we can simply claim that the center of the triangle XYZ is the point F.

Now, since the triangles XYZ and DEL are homothetic, there exists a homothety h mapping the triangle XYZ to the triangle DEL. This homothety h maps the center F of the triangle XYZ to the center of the triangle DEL, hence to the centroid G of triangle ABC (because of Lemma 2). Now let $J_1$ be the image of the point J in the homothety h. Then, since a homothety maps lines to parallel lines, we have $DJ_1 \parallel XJ$ . But $XJ \perp BC$ (since the point X is the reflection of the point J in the line BC). Thus, $DJ_1 \perp BC$ . But the point D is the center of the equilateral triangle constructed outwardly on the side BC of triangle ABC, and hence lies on the perpendicular bisector of this side BC. Thus, the line $DJ_1$ , passing through D and being perpendicular to BC, must be the perpendicular bisector of this side BC. In other words, the point $J_1$ lies on the perpendicular bisector of the side BC. Similarly, the same point $J_1$ lies on the perpendicular bisectors of the other two sides of triangle ABC. And this shows that our point $J_1$ coincides with the circumcenter O of triangle ABC. Hence, the image of the point J in the homothety h is the point O.

So we have seen that the homothety h takes the points F and J to the points G and O, respectively. Therefore, GO || FJ. But the line GO is just the Euler line of triangle ABC, and thus we see that the line FJ is parallel to the Euler line of triangle ABC.

Qed..

See also Hyacinthos message #7957 for some related results.

PS. I had also posted some explanations about the isodynamic points on http://www.mathlinks.ro/Forum/viewtopic.php?t=6489 .

Darij

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Peter

#6 h Aug 23, 2004, 8:07 PM • 2 Y

Y by Adventure10, Mango247

darij grinberg wrote:

The only thing that makes me wonder is this strange "Very hard" with 9 exclamation signs. Especially this is because my experience shows that a problem you call "hard" (even without any exclamation sign) is already far beyound my brain. And now this one, really strange...

Perhaps that just means you're better in geometry, darij?

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#7 h Aug 25, 2004, 1:48 PM • 2 Y

Y by Adventure10, Mango247

Believe me the problem is very difficult.Of course your geometry is excellent.
$F'$ also is a point in triangle that:
$<F'BC=FBA , <F'CB=FCA$

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#8 h Aug 25, 2004, 2:48 PM • 3 Y

Y by AlastorMoody, Adventure10, Mango247

Omid Hatami wrote:

Believe me the problem is very difficult.Of course your geometry is excellent.

Thanks, I'm always glad to hear this...

but I remember there are some much harder problems about the Fermat and isodynamic points.

Actually, a triangle ABC has two Fermat points, two Napoleon points and two isodynamic points. These six points altogether are called FNI points (for: Fermat, Napoleon, Isodynamic). It turns out that any two of these six FNI points are collinear with (at least) one other interesting triangle center; hence, altogether, we get $\displaystyle \binom{6}{2}=15$ interesting collinearities, the so-called FNI collinearities. For a list of these collinearities, you can look at my Hyacinthos message #6602, or you can also consult four Forum Geometricorum papers by the late Lawrence S. Evans (paper 1, paper 2, paper 3, paper 4).

The problem with all the FNI collinearities is that they are quite easy to show using barycentric coordinates, but most of them haven't been proven synthetically yet.

The problem you posted, namely to show that the line FF' is parallel to the Euler line of triangle ABC, is number X in my list (actually, in the list, I don't say "the line is parallel to the Euler line", but I say "the line passes through the Euler infinity point"; actually, this Euler infinity point is just the infinite point (on the projective plane) which is common to all lines parallel to the Euler line). Of course, the collinearity number XI is analogous. The collinearity IX has a not-so-difficult synthetic proof, too. But does anybody have synthetic proofs to the other 15 - 3 = 12 FNI collinearities?

And actually, once the day will come and all 15 collinearities will be shown, the time will be ready for a synthetic proof of the Lester circle theorem. So you see, there are lots of VERY hard problems in geometry left to be solved...

Omid Hatami wrote:

$F'$ also is a point in triangle that:
$<F'BC=FBA , <F'CB=FCA$

Yes, that's exactly the definition of F' as the isogonal conjugate of F.

Darij

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jayme

#9 h Jan 22, 2008, 9:16 AM • 1 Y

Y by Adventure10

Dear all,
another proof has been put on my website in an article intilted "La fascinante figure de Cundy" (volume 2 (2008)).
http://perso.orange.fr/jl.ayme/
Sincerely
Jean-Louis

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