Sondat Theorem about Orthologic and Perspective Triangles

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Sondat Theorem about Orthologic and Perspective Triangles

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Source: P. Sondat 1894

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#1 h Feb 27, 2004, 3:29 PM • 15 Y

Y by jatin, anantmudgal09, Orkhan-Ashraf_2002, 62861, GoJensenOrGoHome, Caspper, JasperL, Wizard_32, Amir Hossein, k12byda5h, Adventure10, lepton727, Mango247, and 2 other users

Given two non-degenerate triangles ABC and A'B'C', the perpendiculars to the lines BC, CA, AB through A', B', C' concur if and only if the perpendiculars to B'C', C'A', A'B' through A, B, C concur. Two triangles ABC and A'B'C' satisfying these relations are called orthologic.

Well, the above theorem is quite easy to prove (I know three different ways: similtude, Steiner aka Carnot theorem on concurrent perpendiculars, Ceva theorem in the sine form).

Okay, this was the warm-up exercise. Now, for two orthologic triangles ABC and A'B'C', let P be the point where the perpendiculars to BC, CA, AB through A', B', C' meet, and let P' be the point where the perpendiculars to B'C', C'A', A'B' through A, B, C meet. These two points P and P' are called the orthologic centers of triangles ABC and A'B'C'.

Now, regard the case when triangles ABC and A'B'C' are not only orthologic, but also perspective: The lines AA', BB', CC' concur at a point Q, and (after the Desargues theorem) the intersections of the lines B'C', C'A', A'B' with the lines BC, CA, AB, respectively, lie on one line l.

The Sondat theorem states:

The points P, P', Q lie on one line perpendicular to l.

Of course, it is enough to show that PQ is perpendicular to l; then, the similar fact for P'Q follows by analogy and the theorem is proven.

Also, with the help of orthology, the theorem to be proven can be reduced to the following fact: If the lines C'A' and CA meet at B", and the lines A'B' and AB meet at C", then the lines AP' and A'P and the perpendiculars from B" to BB' and from C" to CC' meet at one point. Of course, again, it is enough here to prove three of these four lines concurrent, and the fourth one will follow by analogy.

However, with all these proof ideas, I have not found a synthetic proof of the above. (Maybe, even all of my proof ideas are dead end observations.) Does anybody know or can anybody find a good synthetic proof?

An analytical proof for the Sondat theorem can be found in

| Victor Thébault, Perspective and orthologic
| triangles and tetrahedrons, American Mathematical
| Monthly 1952 pp. 24-28.

The proof by Thébault uses the Menelaos and Stewart theorems and is not what I would call elegant. He justifies his proof by its generalization potential (in fact, it goes analogously for tetrahedrons); however, I am mainly interested in a synthetic proof for the planimetric case.

For the Sondat theorem, Thébault refers to

| P. Sondat, L'intermédiaire des mathématiciens, 1894,
| p. 10 [question 38, solved by Sollerstinsky, p. 94].

However, I don't have access to this reference and don't know what the proof given there was like.

Ah, let me finally remark that the Sondat theorem was rediscovered and proved in a similar way to the above Thébault paper by Mitrea and Mitrea in

| D. Mitrea, M. Mitrea, A Generalization of a Theorem
| of Euler, American Mathematical Monthly January
| 1994.

The Sondat theorem generalizes not only the "Theorem of Euler" (meaning the Euler line of a triangle), but also several triangle collinearities.

Finally, let me remark that there was another theorem of Sondat on perspective triangles with mutually perpendicular sidelines, but this is another story...

Darij Grinberg

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pacoga

#2 h Sep 8, 2005, 3:34 PM • 1 Y

Y by Adventure10

I am reading the proof of the Sondat theorem by Mitrea and Mitrea and I don't understand why in page 57 it says

Clearly any two of the vectors $\vec a$ , $\vec b$ , $\vec c$ are linearly independient

Is this true?

If not, how can we terminate the proof in this case?

(The referred article is in The American Mathematical Monthly, vol 101, no. 1, (January 1994), 55-58.

Thanks

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#3 h Sep 11, 2005, 4:46 PM • 3 Y

Y by Adventure10 and 2 other users

pacoga wrote:

I am reading the proof of the Sondat theorem by Mitrea and Mitrea and I don't understand why in page 57 it says

Clearly any two of the vectors $\vec a$ , $\vec b$ , $\vec c$ are linearly independient

Is this true?

This is not true, but actually it is not necessary.

Mitrea and Mitrea pretend to require any two of the vectors $\vec a$ , $\vec b$ , $\vec c$ to be linearly independent, but in fact they use only the linear independence of the vectors $\vec a$ and $\vec b$ and the linear independence of the vectors $\vec a$ and $\vec c$ . The linear independence of the vectors $\vec b$ and $\vec c$ is not really used, what becomes clear if you replace all the $\Longleftrightarrow$ signs on page 58 by $\Longleftarrow$ signs (in fact, only the $\Longleftarrow$ direction is necessary for the proof).

Now, you can always reduce the problem to the case when the vectors $\vec a$ and $\vec b$ are linearly independent and the vectors $\vec a$ and $\vec c$ are also linearly independent: If the point O is one of the vertices A, B, C of triangle ABC, then the Sondat theorem is trivial (or doesn't make sense - depends on how exactly you interpret the theorem in this case), so you can assume that the point O does not coincide with any of the vertices A, B, C of triangle ABC; thus, the point O lies on at most one of the three lines BC, CA, AB; so we can WLOG assume that the point O lies neither on the line CA nor on the line AB. Thus, the vectors $\overrightarrow{OA}=\vec a$ and $\overrightarrow{OC}=\vec c$ are linearly independent, and the vectors $\overrightarrow{OA}=\vec a$ and $\overrightarrow{OB}=\vec b$ are also linearly independent. And then the proof works.

Darij

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#4 h Sep 20, 2005, 6:50 PM • 2 Y

Y by Adventure10, Mango247

A beautiful theorem, I didn't know. 2 or 3 months ago, my friend charlydif tell me a nice problem about orthologics triangles: Let a circunference, a triangle and his "polar triangle". Prove that the triangles are perspective. I think this problems is, with a "orthologic vocabulary": if ABC and A'B'C' are orthologics, and P=P' (the orthologics centers are the same), then ABC and A'B'C' are (necessary) perspective. My solution of this uses Menelao's theorem and similar triangles.
My apologies for my pathetic and non-existent english

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#5 h Sep 20, 2005, 8:22 PM • 2 Y

Y by Amir Hossein, Adventure10

Well, I don't know a syntethic proof of Sondat theorem for orthological triangles, but there's a fourth way to prove your warm-up exercise, a vector one :

The perpendiculars from A',B',C' to the sides BC, CA, AB concurr if and only if $\vec{XA'}\cdot \vec{BC}+\vec{XB'}\cdot \vec{AC} + \vec{XC'}\cdot \vec{AB}=0$ for every point $X$ in the plane.
(from now on I'll omit the arrows over the vectors...)
First of all, the expression is indipendent from $X$ : let $Y$ be another point, then
$(XA'-YA')\cdot BC + (XB'-YB')\cdot AC+(XC'-YC')\cdot AB=XY\cdot(AB+BC+CA)=XY\cdot 0=0$
Now, if the perpendiculars concurr, then let $P$ be the point of concurrence; obviously $PA'\cdot BC+PB'\cdot AC + PC'\cdot AB=0$ , so the sum is zero for every point.
Conversely, if the sum is zero for all $X$ , let $T$ be the intersection of the perpendiculars from A', B' to BC, AC; then
$0=TA'\cdot BC+TB'\cdot AC + TC'\cdot AB=TC'\cdot AB$
so TC' is perpendicular to AB and the assertion 1) is proved.

Now, observe that $0=A'A'\cdot BC+A'B'\cdot AC + A'C'\cdot AB=A'B'\cdot AC + A'C'\cdot AB=B'C'\cdot AA+A'B'\cdot AC + A'C'\cdot AB=B'C'\cdot XA+A'B'\cdot XC+A'C'\cdot XB$ so the perpendiculars to the sides of ABC concurr if and only if the perpendiculars to the sides of A'B'C' concurr.

This caracterization of orthology (does this word exists?) gives also a proof of Sondet, but, obviously, not a synthetical one.

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#6 h Sep 21, 2005, 6:06 PM • 1 Y

Y by Adventure10

For warm-up exercise.....
ABC and A'B'C', P is the intersection of perpendiculars from A', B', C' to BC, CA, AB and P' is the intesection of the perpendiculars from A and C to B'C' and A'B'. We want prove: PB and A'C' are perpendiculars.

Well, let A''B''C'' the translation of ABC, in the direction of P'P ... it's clear that the perpendiculars from A', B' and C' to BC, CA and AB are 'concurrentes' in P. Well, the idea is to prove that A''B''C'' and ABC are orthologics (with the same center), 'es decir': prove that PB' is perpendicular to C'A'...
Let X, Y, Z the proyections from P to B''C'',C''A'',A''B''.
ZYB'A' and YB'C'X are cyclics (it's simple). Then, P is in the radical axis of this circles. Then, PA'.PZ=PC'.PX --> A'ZXC' is cyclic, too, and the warm-up exercise is prove.

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hucht

#7 h Nov 12, 2005, 2:49 AM • 2 Y

Y by Adventure10, Mango247

Could anybody give me the complete Theory about orthological triangles, I need this too much. Thanks

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RSM

#8 h Apr 14, 2012, 6:59 PM • 2 Y

Y by qzc, Adventure10

Problem:-
Given two triangles $ABC$ and $A'B'C'$ so that they are orthologic and perspective. Then prove that the two centers of orthology and the perspector are collinear.

Proof:-

Consider a point $P$ , such that pedal triangle(call it $P_aP_bP_c$ ) of $P$ wrt $ABC$ is homothetic to $A'B'C'$ . The perspector of $ABC$ and $A'B'C'$ be $Q$ . Through $P_a,P_b,P_c$ draw parallels to $AQ,BQ,CQ$ . Suppose, they concurr at $Q'$ . If $P*$ is the isogonal conjugate of $P$ wrt $ABC$ , then we will prove that $Q'P\parallel P*Q$ .Suppose, $H$ and $H'$ are the orthocenters of $ABC$ and $A'B'C'$ . Using the notation of Property 1 here, for the triangles $ABC$ and $P_aP_bP_c$ we have $\Gamma(ABC)$ is the conic passes through $H,P*$ . And $\Gamma(P_aP_bP_c)$ passes through $H',P$ . Since both of them are rectangular hyperbola and they have parallel asymptotes, so they are homothetic. Now draw parallels to $P_aP_b,P_aP_c$ through $B,C$ . Suppose, they meet at $A_2$ . Then note that $P*$ is the orthocenter of $A_2BC$ . So $A_2P*\perp BC$ . Suppose, the homothety that maps $\Gamma(ABC)$ to $\Gamma(P_aP_bP_c)$ maps $A,P*,Q,A_2$ to $A_1,P_1,Q_1,A_2'$ . Note that, tangent at $P_a$ on $\Gamma(P_aP_bP_c)$ is parallel to $A_1A_2'$ . So applying Pascal's theorem on $P_aP*A_2'A_1PP_a$ we get that $P_aP*\parallel PA_1$ . Now applying Pascal's theorem on $PA_1QP*P_aQ'$ we get that $PQ'\parallel P*Q$ . So the lemma is proved.

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#9 h Jun 27, 2018, 11:38 AM • 17 Y

Y by Amir Hossein, yayups, lminsl, huajd, amar_04, AmirKhusrau, Eliot, Aryan-23, Gaussian_cyber, A-Thought-Of-God, Kanep, i3435, lepton727, Adventure10, Mango247, GeoKing, OronSH

Here is my proof of Sondat's Theorem.

Theorem Let triangles $ABC$ and $A_1B_1C_1$ . Suppose that

Perpendiculars from $A, B, C$ to $B_1C_1, C_1A_1, A_1B_1$ are concurrent at $P$
Perpendiculars from $A_1, B_1, C_1$ to $BC, CA, AB$ are concurrent at $P_1$
Lines $AA_1, BB_1, CC_1$ are concurrent at $Q$ .

Finally, let $X=BC\cap B_1C_1, Y=AC\cap A_1C_1$ and $Z=AB\cap A_1B_1$ . Then $P, P_1, Q$ are colinear and perpendicular to $\overline{XYZ}$ .

Proof : We begin with two lemmas.

Lemma 1 : Given two (not homothetic) triangles $\triangle ABC$ and $\triangle DEF$ . Let $P$ be a point such that the line passing through $D,$ $E,$ $F$ and parallel to $AP,$ $BP,$ $CP$ , respectively are concurrent. Then $P$ lies on a circumconic of $\triangle ABC$ or the line at infinity.

Proof of Lemma 1 : By suitable translation, we can let $D=A$ . Let $K = BP\cap AE$ and $L =CP\cap AF$ and the concurrency point be $Q$ . Then $\Delta EFQ$ and $\Delta KLP$ are homothetic at center $A$ , which implies $KL\parallel EF$ .

Let $K, L$ move along $AE, AF$ such that $KL\parallel BC$ and fix three such points $(K_1, K_2, K_3), (L_1, L_2, L_3)$ . Let $P_i$ be the resulting $P$ from $K_i, L_i$ . Notice that
$B(P_1, P_2; P_3, P) = (K_1, K_2 ; K_3, K) = (L_1, L_2 ; L_3, L) = C(P_1, P_2; P_3, P)$ so $B, C, P_1, P_2, P_3, P$ lies on a conic which implies $P$ lies on a fixed conic. This conic clearly pass through $A$ so we are done.

Lemma 2 : Let $ABC$ be a triangle. Let $P, Q$ be two points on a rectangular circum-hyperbola $\mathcal{H}$ . Let the perpendicular from $P$ to $AQ$ meet $BC$ at $A_1$ and define $B_1, C_1$ similarly. Then $A_1B_1C_1$ is lies on a line perpendicular to $PQ$ .

Proof of Lemma 2 : As $\mathcal{H}$ is rectangular, the orthocenters $H_B$ and $H_C$ of $\triangle BPQ$ and $\triangle CPQ$ are the second intersections of $\mathcal{H}$ with $PB_1$ and $PC_1,$ respectively.

By Pascal theorem for $PH_CCABH_B,$ the intersections $C_1 \equiv PH_C \cap AB,$ $B_1 \equiv PH_B \cap AC$ and $CH_C \cap BH_B$ (point at infinity of $\perp PQ$ ) are collinear, which implies $B_1C_1\perp PQ$ . Hence, we conclude that $A_1,B_1,C_1$ lie on a perpendicular to $PQ.$

Back to the main theorem

By Lemma 1, we see that points $P, Q$ and orthocenter $H$ of $\Delta ABC$ lies on a circumconic of $\Delta ABC$ . So this conic must be rectangular hyperbola.

Let the line perpendicular from $Q$ to $BP, CP$ intersects $AC, AB$ at $Y_1, Z_1$ respectively. Notice that $\Delta DYZ$ and $\Delta QY_1Z_1$ are homothetic at center $A$ . But by Lemma 2, $Y_1Z_1\perp PQ$ so $YZ\perp PQ$ , which is enough to conclude the main theorem.

This post has been edited 4 times. Last edited by MarkBcc168, Jun 16, 2022, 2:16 AM

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jayme

#10 h Jun 27, 2018, 11:53 AM • 4 Y

Y by Amir Hossein, WizardMath, Adventure10, Mango247

Dear Mathlinkers,
you can see also my proof dating from 2007 on

http://jl.ayme.pagesperso-orange.fr/Docs/Le%20theoreme%20de%20Sondat.pdf

Sincerely
Jean-Louis

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#11 h Jun 28, 2018, 8:12 AM • 2 Y

Y by Adventure10, Mango247

Also see the proof of Theorem 1.2 here.

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This post has been edited 1 time. Last edited by shalomrav, May 2, 2021, 3:18 PM

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