perpendicular feet

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

perpendicular feet

G H J

Reveal topic tags

geometric transformation

geometry solved

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

124 posts

#1 h Dec 29, 2006, 7:51 AM • 2 Y

Y by Adventure10, Mango247

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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

6555 posts

#2 h Dec 29, 2006, 10:35 AM • 2 Y

Y by Adventure10, Mango247

barasawala wrote:

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.

This is Theorem 2 from the note "On the paracevian perspector" on my website (which currently lives at http://www.cip.ifi.lmu.de/~grinberg/ or, archived, at http://sites.google.com/site/darijgrinberg/website ). In that note, I also prove a generalization of your problem (see also http://www.mathlinks.ro/Forum/viewtopic.php?t=14692 for a discussion of that generalization).

I will leave to you the fun of finding the short solution of the above problem which does not apply to the generalization.

Darij

This post has been edited 3 times. Last edited by darij grinberg, Sep 13, 2009, 11:10 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1327 posts

#3 h Dec 29, 2006, 5:50 PM • 2 Y

Y by Adventure10, Mango247

barasawala wrote:

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.

This problem also has been presented last time in Hyacinthos Forum on Aug 8, 2006, by Quang Tuan Bui and there was an extensive discussion, about also the generalization of the problem, as it has been presented first time by Eric Daneels (''paracevian perspector ?''), on Jul 23,2004.

You can see at: http://groups.yahoo.com/group/Hyacinthos/message/10135 and http://groups.yahoo.com/group/Hyacinthos/message/13907

An alternative (than Darij’s in his website) proof, it has already been posted also in Hyacinthos Forum at
http://groups.yahoo.com/group/Hyacinthos/message/14034 and some notes at http://groups.yahoo.com/group/Hyacinthos/message/14088 for a simpler yet proof.

I will post here next time the two alternative synthetic proofs I have in mind of the ''paracevian perspector '' theorem.

Kostas Vittas.

This post has been edited 1 time. Last edited by vittasko, Jun 16, 2011, 8:34 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

probability1.01

2743 posts

probability1.01

#4 h Dec 30, 2006, 1:16 AM • 1 Y

Y by Adventure10

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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

6555 posts

#5 h Dec 30, 2006, 10:21 AM • 2 Y

Y by Adventure10, Mango247

probability1.01 wrote:

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.

This is the simple proof I spoke of

.

The problem also appears as problem 5.81 in Prasolov's Planimetry problems book (see http://www.math.su.se/~mleites/ and http://michaj.home.staszic.waw.pl/prasolov.html for English and http://www.mccme.ru/free-books/ for Russian versions), with the same proof.

Darij

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1327 posts

#6 h Dec 30, 2006, 10:22 PM • 3 Y

Y by jatin, Adventure10, Mango247

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 $PQ,$

Attachments:

This post has been edited 1 time. Last edited by vittasko, Jun 16, 2011, 8:32 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1327 posts

#7 h Dec 31, 2006, 4:57 PM • 3 Y

Y by jatin, Adventure10, Mango247

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.

This post has been edited 1 time. Last edited by vittasko, Jun 16, 2011, 8:31 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

9772 posts

jayme

#8 h Dec 25, 2008, 1:24 PM • 2 Y

Y by Adventure10, Mango247

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

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1327 posts

#9 h May 9, 2019, 3:45 PM • 2 Y

Y by Adventure10, Mango247

I can't understand why some of the last lines of the text of the proof 1 which I have been presented before ( post #6) doesn't viewed correctly.

So, I post here the text of these lost lines of the text, beginning from the point of $(13)$ .

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.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a