perpendicular lines from 3 vertices of convex hexagon are concurrent

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 lines from 3 vertices of convex hexagon are concurrent

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: Polish second round 1995 p2

The post below has been deleted. Click to close.

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

30629 posts

#1 h Jan 19, 2020, 9:51 PM • 2 Y

Y by ImSh95, Adventure10

Let $ABCDEF$ be a convex hexagon with $AB = BC, CD = DE$ and $EF = FA$ .
Prove that the lines through $C,E,A$ perpendicular to $BD,DF,FB$ are concurrent.

Z K Y

The post below has been deleted. Click to close.

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

204 posts

#2 h Dec 15, 2021, 3:24 PM • 1 Y

Y by ImSh95

Let $X$ such that: $AX \perp FB$ .
Let $Y$ such that: $CY \perp BD$
Let $Z$ such that: $EZ \perp DF$
$FA^2-AB^2=FX^2-XB^2 ; DE^2-EF^2=DZ^2-ZF^2 ; BC^2-CD^2=DY^2-YB^2$
$\implies FX^2-XB^2+DZ^2-ZF^2+DY^2-YB^2=0$
By Carnot' Theorem:
The lines through $C,E,A$ perpendicular to $BD,DF,FB$ are concurrent.

Z K Y

The post below has been deleted. Click to close.

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

5417 posts

#4 h May 25, 2022, 6:59 PM • 2 Y

Y by ImSh95, franzliszt

because and only because of aops precal section 13.3 -- repeat, i don't support bashing
actually, i found that the lines through $B$ , $D$ , and $F$ perpendicular to $AC$ , $CE$ , and $EA$ were concurrent at the circumcenter of $ACE$ (since these perpendiculars are in fact perpendicular bisectors due to the length conditions), so in my defense if i had known that orthologic triangles existed before solving this problem, i would've seen the clean solution (see an unofficial thread)
anyway...

Vectors! First, the length conditions translate to
$\begin{align*} &\Vert\vec{AB}\Vert = \Vert\vec{BC}\Vert \iff \Vert\vec{AB}\Vert^2 = \Vert\vec{BC}\Vert^2 \iff \vec{AB}\cdot\vec{AB} = \vec{BC}\cdot\vec{BC} \\ &\Vert\vec{CD}\Vert = \Vert\vec{DE}\Vert \iff \Vert\vec{CD}\Vert^2 = \Vert\vec{DE}\Vert^2 \iff \vec{CD}\cdot\vec{CD} = \vec{DE}\cdot\vec{DE} \\ &\Vert\vec{EF}\Vert = \Vert\vec{FA}\Vert \iff \Vert\vec{EF}\Vert^2 = \Vert\vec{FA}\Vert^2 \iff \vec{EF}\cdot\vec{EF} = \vec{FA}\cdot\vec{FA}, \end{align*}$ which by substituting $\vec{PQ} = \vec{Q}-\vec{P}$ yields
$\begin{align*} &(\vec{B}-\vec{A})^2 = (\vec{C}-\vec{B})^2 \iff \vec{A}\cdot\vec{A}+\vec{B}\cdot\vec{B}-2\cdot\vec{A}\cdot\vec{B} = \vec{B}\cdot\vec{B}+\vec{C}\cdot\vec{C}-2\cdot\vec{B}\cdot\vec{C} \iff \vec{A}\cdot\vec{A}-2\cdot\vec{A}\cdot\vec{B} = \vec{C}\cdot\vec{C}-2\cdot\vec{B}\cdot\vec{C} \\ &(\vec{D}-\vec{C})^2 = (\vec{E}-\vec{D})^2 \iff \vec{C}\cdot\vec{C}+\vec{D}\cdot\vec{D}-2\cdot\vec{C}\cdot\vec{D} = \vec{D}\cdot\vec{D}+\vec{E}\cdot\vec{E}-2\cdot\vec{D}\cdot\vec{E} \iff \vec{C}\cdot\vec{C}-2\cdot\vec{C}\cdot\vec{D} = \vec{E}\cdot\vec{E}-2\cdot\vec{D}\cdot\vec{E} \\ &(\vec{F}-\vec{E})^2 = (\vec{A}-\vec{F})^2 \iff \vec{E}\cdot\vec{E}+\vec{F}\cdot\vec{F}-2\cdot\vec{E}\cdot\vec{F} = \vec{F}\cdot\vec{F}+\vec{A}\cdot\vec{A}-2\cdot\vec{F}\cdot\vec{A} \iff \vec{E}\cdot\vec{E}-2\cdot\vec{E}\cdot\vec{F} = \vec{A}\cdot\vec{A}-2\cdot\vec{F}\cdot\vec{A}. \end{align*}$ Summing,
$\vec{A}\cdot\vec{A}+\vec{C}\cdot\vec{C}+\vec{E}\cdot\vec{E}-2\cdot\vec{A}\cdot\vec{B}-2\cdot\vec{C}\cdot\vec{D}-2\cdot\vec{E}\cdot\vec{F}=\vec{C}\cdot\vec{C}+\vec{E}\cdot\vec{E}+\vec{A}\cdot\vec{A}-2\cdot\vec{B}\cdot\vec{C}-2\cdot\vec{D}\cdot\vec{E}-2\cdot\vec{F}\cdot\vec{A},$ and cancelling out $\vec{E}\cdot\vec{E}+\vec{A}\cdot\vec{A}+\vec{C}\cdot\vec{C}$ and dividing by $2$ gives
$-\vec{A}\cdot\vec{B}-\vec{C}\cdot\vec{D}-\vec{E}\cdot\vec{F} = -\vec{B}\cdot\vec{C}-\vec{D}\cdot\vec{E}-\vec{F}\cdot\vec{A} \iff \vec{B}\cdot\vec{C}+\vec{D}\cdot\vec{E}+\vec{F}\cdot\vec{A}-\vec{A}\cdot\vec{B}-\vec{C}\cdot\vec{D}-\vec{E}\cdot\vec{F} = 0.\qquad(\ast)$ Next, the magical point of concurrency. Let the perpendiculars from $C$ and $E$ to segments $BD$ and $DF$ intersect at $X$ , so that $\overline{BD}\perp\overline{CX}$ and $\overline{DF}\perp\overline{EX}$ , i.e.,
$\begin{align*} & \vec{BD}\cdot\vec{CX} = (\vec{D}-\vec{B})\cdot(\vec{X}-\vec{C}) = \vec{B}\cdot\vec{C}+\vec{D}\cdot\vec{X}-\vec{B}\cdot\vec{X}-\vec{C}\cdot\vec{D} = 0 \\ & \vec{DF}\cdot\vec{EX} = (\vec{F}-\vec{D})\cdot(\vec{X}-\vec{E}) = \vec{D}\cdot\vec{E}+\vec{F}\cdot\vec{X}-\vec{D}\cdot\vec{X}-\vec{E}\cdot\vec{F} = 0. \end{align*}$ Summing,
$(\vec{B}\cdot\vec{C}+\vec{D}\cdot\vec{E})+(\vec{D}\cdot\vec{X}+\vec{F}\cdot\vec{X})-(\vec{B}\cdot\vec{X}+\vec{D}\cdot\vec{X})+(\vec{C}\cdot\vec{D}+\vec{E}\cdot\vec{F}) = 0.\qquad(\dagger)$ Something about the above equation looks eerily familiar...

Going back to $(\ast)$ , if we add $(\vec{D}\cdot\vec{X}+\vec{F}\cdot\vec{X}+\vec{B}\cdot\vec{X})-(\vec{B}\cdot\vec{X}+\vec{D}\cdot\vec{X}+\vec{F}\cdot\vec{X}) = 0$ , we have that
$(\vec{B}\cdot\vec{C}+\vec{D}\cdot\vec{E}+\vec{F}\cdot\vec{A})+(\vec{D}\cdot\vec{X}+\vec{F}\cdot\vec{X}+\vec{B}\cdot\vec{X})-(\vec{B}\cdot\vec{X}+\vec{D}\cdot\vec{X}+\vec{F}\cdot\vec{X})-(\vec{A}\cdot\vec{B}+\vec{C}\cdot\vec{D}+\vec{E}\cdot\vec{F}) = 0.$ Aha! Subtracting $(\dagger)$ from this new $(\ast)$ yields
$\vec{F}\cdot\vec{A}+\vec{B}\cdot\vec{X}-\vec{F}\cdot\vec{X}-\vec{A}\cdot\vec{B} = (\vec{B}-\vec{F})\cdot(\vec{X}-\vec{A}) = \vec{FB}\cdot\vec{AX} = 0,$ which implies that $\overline{FB}\perp\overline{AX}$ , the requested result. $\blacksquare$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a