Cevian Triangle with Perpendiculars

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

Cevian Triangle with Perpendiculars

G H J

Reveal topic tags

geometry proposed

L

G H BBookmark kLocked kLocked NReply

Source: ELMO 2014 Shortlist G10, by Sammy Luo

The post below has been deleted. Click to close.

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

6877 posts

#1 h Jul 24, 2014, 1:36 PM • 5 Y

Y by narutomath96, swamih, Gaussian_cyber, Adventure10, swynca

We are given triangles $ABC$ and $DEF$ such that $D\in BC, E\in CA, F\in AB$ , $AD\perp EF, BE\perp FD, CF\perp DE$ . Let the circumcenter of $DEF$ be $O$ , and let the circumcircle of $DEF$ intersect $BC,CA,AB$ again at $R,S,T$ respectively. Prove that the perpendiculars to $BC,CA,AB$ through $D,E,F$ respectively intersect at a point $X$ , and the lines $AR,BS,CT$ intersect at a point $Y$ , such that $O,X,Y$ are collinear.

Proposed by Sammy Luo

Z K Y

The post below has been deleted. Click to close.

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

552 posts

XmL

#2 h Jul 24, 2014, 5:42 PM • 4 Y

Y by narutomath96, Adventure10, Mango247, and 1 other user

We know that $AD,BE,CF$ concur at the orthocenter of $DEF$ , denoted by $H$ . Let $X'$ denote the isogonal conjugate of $H$ wrt $ABC$ , If $DEF$ isn't the pedal triangle of $X'$ , then its pedal triangle is homothetic with $DEF$ which is impossible since the lines that connect the corresponding points of the two homothetic triangles don't concur. Hence $X'\equiv X$ . It's a well known property of isogonal conjugates that $O,X,H$ are collinear and $RST$ is the pedal triangle of $H$ .

We will prove that $BS\cap CT=Y'$ lies on line $OXH$ , in which it's clear that the concurrence at $Y$ follows. Apply Pappus' theorem on $CES,BFT$ and we have $H,Y',ET\cap FS$ are collinear. Let $HS,HT\cap (O)=S',T'$ again and since $SS'\cap TT'=O$ , by Pascal theorem we have $O,H,ET\cap FS$ are collinear and hence $O,X,H,ET\cap FS,Y'$ are collinear and we are done.

Z K Y

The post below has been deleted. Click to close.

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

2312 posts

#3 h Oct 23, 2014, 11:52 PM • 5 Y

Y by Gaussian_cyber, Ru83n05, Adventure10, Mango247, and 1 other user

My solution:

Since the lines through $A, B, C$ and perpendicular to $EF, FD, DE$ are concurrent at the orthocenter of $\triangle DEF$ ,
so $\triangle DEF$ and $\triangle ABC$ are orthologic $\Longrightarrow$ the perpendiculars through $D, E, F$ to $BC, CA, AB$ are concurrent at $X$ .

Since $AD, BE, CF$ are concurrent ,
so by Terquem theorem we get $AR, BS, CT$ are concurrent at $Y$ .

Since $\triangle RST$ is the pedal triangle of $X'$ ( $X'$ is the isogonal conjugate of $X$ ) ,
so the line passing $A, B, C$ and perpendicular to $ST, TR, RS$ are concurrent at $X$ ,
hence by Sondat theorem (for $\triangle ABC$ and $\triangle RST$ ) we get $Y, X', X$ are collinear.
ie. $X, O, Y$ are collinear

Q.E.D

Remark:

(1) Easy to see $X' = AD \cap BE \cap BF$ which is the orthocenter of $\triangle DEF$

(2) Another interesting property in this configuration :
Denote $O'$ as the circumcenter of $\triangle ABC$ , then the isogonal conjugate of $Y$ lie on $O'H$ .
(see http://www.artofproblemsolving.com/community/c6h366335 )

This post has been edited 1 time. Last edited by TelvCohl, Apr 13, 2015, 12:53 AM

Z K Y

The post below has been deleted. Click to close.

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

366 posts

#4 h Apr 13, 2015, 12:03 AM • 2 Y

Y by Adventure10, Mango247

Here is my solution.

Let $H$ be the orthocenter of $DEF$ , which clearly lies on $AD,BE,CF$ . By Desargues on $EDF,CH_DB$ , where $H_D$ is the foot of the perpendicular from $D$ to $EF$ (define $H_E,H_F$ similarly), we get that $H_EH_F,EF,CB$ are concurrent and therefore $D(T,E,H,F)$ is a harmonic pencil, and so $TEH_DF$ is a harmonic quadrilateral. Thus if $H_D'$ is the reflection of $H_D$ across the perpendicular bisector of $EF$ (which is also clearly the point diametrically opposite to $D$ ), since $TH_D$ is symmedian of $TEF$ we get that $TH_D'$ is a median of $TEF$ . Thus $T$ lies on $MH_D'$ , where $M$ is the midpoint of $EF$ , but $H$ also lies here, and so these 4 points are collinear. Therefore, $\boxed{\angle HTF=90}$ (since $DH_D'$ is a diameter).

Reflecting $\ell$ , the line perpendicular to $CB$ that passes through $D$ , through $O$ , it is easily seen that it transforms into line $TH$ . The analogous happens for $E$ and $F$ . Thus the three original lines are concurrent at the reflection of $H$ through $O$ . Therefore, $X$ is the reflection of $H$ through $O$ . We must now prove $Y$ exists and lies on the Euler line of $\triangle DEF$ .

To this end, we invert through $H$ with any radius. Let $Z'$ denote the inverse of $Z$ for any point $Z \neq H$ . Lines $HD,HE,HF$ invert to themselves. Notice that the circle $HEH_DH_FS$ has a diameter $HE$ , and hence its inverse is a line perpendicular to $HE$ (the analogous applies to $HD,HF$ ). From this we gather that $D',E',F'$ are the feet of the altitudes in a triangle, and that $S,R,T$ lie on the sides. Since $DEFSRT$ are concyclic, so are their inverses and so $S', R', T'$ are the midpoints of said triangle, and their circumcenter is the 9-point circle. Finally, since $\angle HSA=\angle HTA=90$ , we gather that $STAH$ are cyclic, so $S',T',A'$ are collinear. From this, $A'=T'S' \cap HD'$ . The analogous occurs for $B',C'$ . From now on, for convenience, let $Z$ denote what we previously denoted by $Z'$ , for any letter $Z$ . Let the big triangle be $XYZ$ .

The question is now to prove that the circumcenters of $HTC,HRA,HSB$ have the Euler line of $XYZ$ as a common radical axis (since $O$ inverted to a point on this line, because it also contains the circumcenter of $DEF$ , and $H$ and $O$ are collinear with this circumcenter by symmetry). Let $G$ be the barycenter of $XYZ$ , let's prove that $G$ has the same power of a point to all 3 circles.

Let $Z_0$ be the intersection of $ZT$ and the 9-point circle. Then $(ZZ_0)(ZT)=(ZS)(ZE)=(ZC)(ZH)$ since $HCSE$ is cyclic because $\angle HCS= \angle HES=90$ . Then $THCZ_0$ are concyclic. But $ZT$ is the median of $XYZ$ , and so $G$ lies on it. Therefore $G$ has the same power of a point to the circumcircle of $THCZ_0$ than to the 9-point circle, and by analogy we are done. $\blacksquare \text{ Q.E.D}$ .

This post has been edited 2 times. Last edited by JuanOrtiz, Apr 13, 2015, 12:09 AM

Z K Y

The post below has been deleted. Click to close.

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

1916 posts

#5 h Jul 7, 2020, 12:20 PM • 6 Y

Y by GeoMetrix, Bumblebee60, Gaussian_cyber, Ru83n05, Mango247, Mango247

ELMOSL 2014 G10 wrote:

We are given triangles $ABC$ and $DEF$ such that $D\in BC, E\in CA, F\in AB$ , $AD\perp EF, BE\perp FD, CF\perp DE$ . Let the circumcenter of $DEF$ be $O$ , and let the circumcircle of $DEF$ intersect $BC,CA,AB$ again at $R,S,T$ respectively. Prove that the perpendiculars to $BC,CA,AB$ through $D,E,F$ respectively intersect at a point $X$ , and the lines $AR,BS,CT$ intersect at a point $Y$ , such that $O,X,Y$ are collinear.

Proposed by Sammy Luo

Clearly $AD,BE,CF$ are concurrent at the Orthocenter $U$ of $\Delta DEF$ . So, $AR,BS,CT$ are concurrent at the Cyclocevian Conjugate $(Y)$ of $U$ WRT $\Delta ABC$ $(\bigstar)$ . Also Clearly $\{\Delta ABC,\Delta DEF\}$ are Orthologic, so the perpendiculars to $BC,CA,AB$ through $D,E,F$ concurs at $X$ . Now notice that the Perpendiculars from $R,S,T$ to $BC,CA,AB$ respectively concurs at a point $X^*$ which is the Isogonal Conjugate of $X$ . Also the perpendiculars from $A,B,C$ to $ST,RT,SR$ are concurrent at $X$ . Hence, $\{\Delta ABC,\Delta RST\}$ are Orthologic with centers of Orthology $\{X,X^*\}$ . Also from $(\bigstar)$ we get that $\{\Delta ABC,\Delta RST\}$ are perspective. So, by Sondat's Theorem $Y\in\overline{XX^*}$ but $O\in\overline{XX^*}$ . Hence. $\overline{O-X-Y}$ . $\blacksquare$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a