perpendicular feet

barasawala · Poincare Conjecture Offline Joined: 14 Dec 2006 Posts: 125

Let $AA_{1},BB_{1},CC_{1}$ be the altitudes in acute triangle $ABC$ , and let $X$ be an arbitrary point. Let $M,N,P,Q,R,S$ be the feet of the perpendiculars from $X$ to the lines $AA_{1},BC,BB_{1},CA,CC_{1},AB$ . Prove that $MN,PQ,RS$ are concurrent.

**Posted:** Thu Dec 28, 2006 11:51 pm

darij grinberg

_________________
Now the die is cast, the first step taken, a glimmer of hope lights up our lives
Visions of the past, dreams forsaken forming right under our eyes
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vittasko

**Posted:** Fri Dec 29, 2006 9:50 am

probability1.01

I did not check the above links, but here is a way to prove it:

Dilate with a factor of 2 about X. The image of MN becomes the line $A_{1}X$ reflected about $AA_{1}$ , and analogous results hold for the other two sides. Since the altitudes of ABC are the angle bisectors of its orthic triangle, the images of MN, PQ, and RS concur at the isogonal conjugate of X in the orthic triangle. Thus MN, PQ, and RS themselves must also concur.

**Posted:** Fri Dec 29, 2006 5:16 pm

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Don't hatch your chickens before they count. Otherwise, they will be far behind in middle school math and perform poorly at MathCounts.

the Search for a Blog Name Continues. . .

darij grinberg

**Posted:** Sat Dec 30, 2006 2:21 am

_________________
Now the die is cast, the first step taken, a glimmer of hope lights up our lives
Visions of the past, dreams forsaken forming right under our eyes
We are alive...

vittasko

THE PARACEVIAN PERSPECTOR THEOREM.

– A triangle $\bigtriangleup ABC$ is given and let $H$ be, an arbitrary fixed point, and let $\bigtriangleup HaHbHc$ be, it’s cevian triangle.

- Let $P$ be, an arbitrary also point and $Pa,$ $Pb,$ $Pc,$ the traces on sidelines $BC,$ $CA,$ $AB,$ from the lines through $P$ and parallel to $AHa,$ $BHb,$ $CHc,$ respectively.

- Let $Ap,$ $Bp,$ $Cp$ be, the traces on segment lines $AHa,$ $BHb,$ $CHc,$ from the lines through $P$ and parallel to the sidelines $BC,$ $CA,$ $AB,$ respectively.

- Prove that the lines $ApPa,$ $BpPb,$ $CpPc,$ are concurrent at one point so be it, $Q,$ as the midpoint of the segment between the point $P$ and the $H-Ceva$ conjugate of $P.$

PROOF 1. - $($ In my drawing $AB = 15.3,$ $AC = 14.2,$ $BC = 11.2,$ $AH = 12.8,$ $BH = 4.2,$ $AP = 8.2,$ $CP = 6.8$ $).$

$1)$ – We denote the intersection points of sidelines of $\bigtriangleup ABC,$ from the segment lines $PAp,$ $PBp,$ $PCp,$ as follows:

- As $D,$ $D',$ the intersection points of $AC,$ $AB$ respectively, from the segment line $PAp.$

- As $E,$ $E',$ the intersection points of $AB,$ $BC$ respectively, from the segment line $PBp.$

- As $F,$ $F',$ the intersection points of $BC,$ $AC$ respectively, from the segment line $PCp.$

Also we denote:

- As $A',$ the intersection point of $DE',$ $D'F.$

- As $B',$ the intersection point of $DE',$ $EF'.$

- As $C',$ the intersection point of $EF',$ $D'F.$

Based on the below proposition, we will prove that the lines $AP,$ $BP,$ $CP,$ pass through the points $A',$ $B',$ $C',$ respectively.

$2)$ – PROPOSITION 1. - A triangle $\bigtriangleup ABC$ is given and let $P$ be, an arbitrary point inwardly to it. We denote as $D,$ $D',$ the intersection points of $AC,$ $AB$ respectively, from the line through $P$ and parallel to $BC.$ Also we denote as $E',$ $F,$ the intersection points of $BC,$ from the lines through $P$ and parallel to $AC,$ $AB$ respectively. Prove that the lines $DE',$ $D'F,$ $AP,$ are concurrent.

$2a)$ - PROOF. - We denote as $A',$ the intersection point of $DE',$ $D'F$ and as $P',$ the one of $BC,$ from the segment line $AP.$

From $DD'//BC,$ $\Longrightarrow$ $\frac{PD}{P'C}= \frac{AP}{AP'}= \frac{PD'}{P'B}$

$\Longrightarrow$ $\frac{PD}{(P'C)-(PD)}= \frac{PD'}{(P'B)-(PD')}$ $\Longrightarrow$ $\frac{PD}{P'E'}= \frac{PD'}{P'F}$ $,(1)$ $($ because of $PD = E'C$ and $PD' = FB$ $).$

From $(1)$ $\Longrightarrow$ $\frac{PD}{PD'}= \frac{P'E'}{P'F}$ and so, the points $P,$ $P',$ $A',$ are collinear. Hence the lines $DE',$ $D'F,$ $AP,$ are concurrent and the proof of proposition 1, is completed.

$3)$ – In our configuration now, we conclude that the line $AP,$ passes through the point $A'$ $($ intersection point of $DE',$ $D'F$ $)$ and similarly the lines $BP,$ $CP,$ pass through the points $B',$ $C',$ respectively.

By the same way, from the triangle $\bigtriangleup AHaC$ and the point $P$ inwardly to it, we conclude that the lines $ApPa,$ $DE',$ $AP,$ are concurrent. Hence, the line $ApPa$ passes through the point $A'$ and similarly the lines $BpPb,$ $CpPc,$ pass through the points $B',$ $C',$ respectively.

$4)$ – We denote as $K,$ $L,$ $M,$ the intersection points of the line $CC',$ trom the segment lines $AB,$ $A'Ap,$ $AHa$ respectively and as $N,$ the intersection point of $EE',$ from the line $B'L.$ We will prove that the points $B,$ $M,$ $N,$ are collinear.

It is enough to prove that $(E, N, P, E') = (K, M, P, C).$

We consider the pencils $B'.C'LPT$ and $A'.C'LPT,$ where $T,$ is the intersection point of $CC',$ $A'B'$ and then we have that

$(B'.C'LPT) = (A'.C'LPT)$ $,(2)$ $($ congruent double ratios of these bundles of lines $).$

$(B'.C'LPT) = (E, N, P, E')$ $,(3)$ $($ intersection of $B'.C'LPT,$ from the line $EE'$ $).$

$(A'.C'LPT) = (D', Ap, P, D)$ $,(4)$ $($ intersection of $A'.C'LPT,$ from the line $DD'$ $).$

from $(2),$ $(3),$ $(4),$ $\Longrightarrow$ $(E, N, P, E') = (D', Ap, P, D)$ $,(5).$

The lines $CC',$ $DD',$ intersect the pencil $A.BHaPC$ and so, we have that $(K, M, P, C) = (D', Ap, P, D)$ $,(6).$

From $(5),$ $(6),$ $\Longrightarrow$ $(E, N, P, E') = (K, M, P, C)$ $,(7).$

From $(7),$ we conclude that the lines $EK,$ $MN,$ $CE',$ are concurrent. So, the line $MN$ passes through the vertex $B$ of $\bigtriangleup ABC$ $($ as the intersection point of $EK,$ $CE'$ $).$ Hence, the points $B,$ $M,$ $N,$ are collinear.

$5)$ – We will prove now, that the lines $ApPa,$ $BpPb,$ $CpPc,$ are concurrent. Let $Q$ be, the intersection point of $BpPb,$ $CpPc$ and it is enough to prove that this point lies on $A'L$ $($ because of it has already been proved that the line $ApPa,$ passes through the point $A'$ $).$ We will prove that the pencils $B'.C'LBpA',$ $C'.B'LCpA',$ have congruent double ratios. That is, we will prove that

$(B'.C'LBpA') = (C'.B'LCpA').$

$(B'.C'LBpA') = (E, N, Bp, E')$ $,(8).$

$(C'.B'LCpA') = (F', P, Cp, F)$ $,(9).$

We consider the bundles of lines ( = pencils ) $B.ENBpE'$ and $C.F'PCpF.$ They have the sideline $BC$ of $\bigtriangleup ABC,$ as their common ray and the points $A$ $($ as the intersection point of $BE,$ $CF'$ $),$ $M$ $($ as the intersection point of $BN,$ $CP$ $)$ and $H$ $($ as the intersection point of $BBp,$ $CCp$ $),$ lie on the same line $AHa.$ So, we have that

$(B.ENBpE') = (C.F'PCpF)$ $,(10).$

From $(10)$ $\Longrightarrow$ $(E, N, Bp, E') = (F', P, Cp, F)$ $,(11).$

From $(8),$ $(9),$ $(11),$ $\Longrightarrow$ $(B'.C'LBpA') = (C'.B'LCpA')$ $,(12).$

From $(12)$ and because of $B'C'$ is the common ray of pencils $B'.C'LBpA'$ and $C'.B'LCpA',$ we conclude that the points $L$ $($ as the intersection point of $B'L,$ $C'L$ $),$ $Q$ $($ as the intersection point of $B'Bp,$ $C'Cp$ $)$ and $A'$ $($ as the intersection point of $B'A',$ $C'A'$ $),$ are collinear. Hence, the lines $ApPa,$ $BpPb,$ $CpPc,$ are concurrent at one point, so be it $Q.$

$6)$ – We will prove now, that the concurrency point $Q,$ of the lines $ApPa,$ $BpPb,$ $CpPc,$ is the midpoint of the segment between the point $P$ and the $H-Ceva$ conjugate of $P.$

This result has been mentioned by Quang Tuan Bui, at last discussion in Hyacinthos Forum http://groups.yahoo.com/group/Hyacinthos/message/13938 and first time by Bernard Gibert http://groups.yahoo.com/group/Hyacinthos/message/10136

Let $S$ be, the intersection point of $AA',$ $B'C'.$ So, from the parallelogram $AEPF',$ we have that $PS = SA.$ Hence, the line through vertex $A$ of $\bigtriangleup ABC$ and parallel to $B'C',$ intersects the line $PB'$ at a point so be it $B'',$ such that $PB' = B'B''.$

Similarly the line through vertex $C$ of $\bigtriangleup ABC$ and parallel to $A'B',$ intersects the line $PB',$ at the same point $B'',$ because of $PT = TC,$ from the parallelogram $CDPE'.$

Hence, the lines through vertices $A,$ $B,$ $C,$ of $\bigtriangleup ABC$ and parallel to $B'C',$ $A'C,$ $A'B'$ respectively, intersect each other at points $A'',$ $B'',$ $C'',$ lie on the lines $AA',$ $BB',$ $CC'$ respectively. The triangle $\bigtriangleup A''B''C'',$ is the anticevian triangle of $\bigtriangleup ABC,$ with respect to the point $P.$

$6a)$ – We will prove now, that $B''Hb\parallel BpPb.$ In triangle $\bigtriangleup B'BBp$ and from $PPb\parallel BBp$ we have that

$\frac{PPb}{BBp}= \frac{B'P}{B'B}$ $\Longrightarrow$ $\frac{B'B''}{B'B}= \frac{BpHp}{BBp}$ $,(13)$ $($ because of $B'P = B'B''$ and $PPb = BpHb$ $).$

From $(13),$ $\Longrightarrow$ $B''Hb\parallel B'Bp$ and so, we conclude that the line $B''Hb$ connecting the vertices $B'',$ $Hb,$ of the triangles $\bigtriangleup A''B''C'',$ $\bigtriangleup HaHbHc$ respectively, intersects the segment line $PQ,$ at a point so be it $R,$ such that $PQ = QR$ $($ because of $PB' = B'B''$ $).$

By the same way we can prove that the lines $A''Ha,$ $C''Hc,$ are parallel to $A'Ap,$ $C'Cp$ respectively and intersect the segment line $PQ,$ at the sane point $R.$ So, the triangles $\bigtriangleup A''B''C'',$ $\bigtriangleup HaHbHc,$ are perspective with perspector the point $R,$ lies on the segment line $PQ$ and such that $PQ = QR.$

Because of the triangle $\bigtriangleup A''B''C'',$ is the anticevian triangle of $\bigtriangleup ABC,$ with respect to the point $P$ $($ since $\bigtriangleup ABC$ is the cevian triangle of $\bigtriangleup A''B''C''$ wrt $P$ $),$ we conclude that the point $Q$ $($ the concurrency point of $ApPa,$ $BpPb,$ $CpPc$ $),$ is the midpoint of the segment between the point $P$ and the point $R,$ as the $H-Ceva$ conjugate of $P$ and the proof is completed.

$\bullet$ This proof is dedicated to Babis Stergiu.

Kostas Vittas.

PS. I will post here next time, the simpler proof of the ''paracevian perspector'' theorem, which has been arisen from the discussion in Hyacinthos Forum, last summer.

**Posted:** Sat Dec 30, 2006 2:22 pm

vittasko

PROOF 2. - $($ see the figure t=126138(c) $).$

It has already been proved that the segment lines $PAp,$ $PBp,$ $PCp,$ as parallel to the sidelines $BC,$ $AC,$ $AB$ of $\bigtriangleup ABC$ respectively, intersect them at pairs of points $D,$ $D'$ and $E,$ $E'$ and $F,$ $F',$ such that the intersection points $A'\equiv DE'\cap D'F,$ $B'\equiv DE'\cap EF',$ $C'\equiv EF'\cap D'F,$ lie on the segment lines $AP,$ $BP,$ $CP,$ respectively.

The lines through vertices $A,$ $B,$ $C,$ of $\bigtriangleup ABC$ and parallel to $B'C',$ $A'C',$ $A'B'$ respectively, intersect each other at the points $A'',$ $B'',$ $C'',$ also lie on the segment lines $AP,$ $BP,$ $CP,$ respectively.

So, we have the configuration of the triangle $\bigtriangleup ABC,$ as the Cevian triangle of $\bigtriangleup A''B''C'',$ with respect to the point $P$ $($ Hence the triangle $\bigtriangleup A''B''C'',$ is the Anticevian triangle of $\bigtriangleup ABC,$ wrt $P$ $),$

Because of the segment lines $AHa,$ $BHb,$ $CHc$ are concurrent at one point $($ here the point $H$ $)$ and also the segment lines $A''A,$ $B''B,$ $C''C$ are concurrent at one point $($ here the point $P$ $)$ , based on the well known proposition 2, we conclude that the segment lines $A''Ha,$ $B''Hb,$ $C''Hc,$ are concurrent at one point, so be it $R,$ which in a such our configuration, is called the $H-Ceva$ conjugate of $P,$ with respect to the triangle $\bigtriangleup ABC.$

It has already been proved $($ see the proof 1 $),$ that the segment lines $ApPa,$ $BpPb,$ $CpPc,$ pass through the points $A',$ $B',$ $C',$ as the midpoints of the segments $PA'',$ $PB'',$ $PC''$ respectively.

Because of $ApPa\equiv A'Ap\parallel A''Ha$ and $BpPb\equiv B'Bp\parallel B''Hb,$ and $CpPc\equiv C'Cp\parallel C''Hc,$ we conclude that the segment lines $ApPa,$ $BpPb,$ $CpPc,$ intersect the segment $PR,$ at the same point, so be it $Q,$ as it’s midpoint. That is the segment lines $ApPa,$ $BpPb,$ $CpPc,$ are concurrent at the point $Q,$ as the midpoint of the segment between $P$ and $R$ as the $H-Ceva$ conjugate of $P$ with respect to the triangle $\bigtriangleup ABC$ and the proof is completed.

$\bullet$ This proof is dedicated to Vladimir Zajic.

Kostas Vittas.

PS. PROPOSITION 2. - ( well known ) - A triangle $\bigtriangleup ABC$ is given and let $\bigtriangleup DEF$ be, the Cevian triangle of an arbitrary point $P,$ in the plane. If $D',$ $E',$ $F',$ are three points on the sidelines $EF,$ $DF,$ $DE$ of $\bigtriangleup DEF$ respectively, such that the segment lines $DD',$ $EE',$ $FF',$ to be concurrent at one point, prove that the segment lines $AD',$ $BE',$ $CF',$ are also concurrent at one point.

**Posted:** Sun Dec 31, 2006 8:57 am

jayme · Yang-Mills Theory Offline Joined: 20 Nov 2007 Posts: 640

Dear Mathlinkers,
a new proof of "The paracevian perspector" has been put on my website
http://perso.orange.fr/jl.ayme vol. 4
Sincerely
Jean-Louis

**Posted:** Thu Dec 25, 2008 5:24 am