octagon [with equal angles and rational sides]

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

octagon [with equal angles and rational sides]

G H J

Reveal topic tags

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.

373 posts

Armo

#1 h Apr 17, 2005, 7:20 PM • 2 Y

Y by Adventure10, Mango247

Consider an octagon with equal angles and with rational sides. Prove that it has a center of symmetry.

Z K Y

The post below has been deleted. Click to close.

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

785 posts

#2 h Apr 17, 2005, 8:02 PM • 2 Y

Y by Adventure10, Mango247

Take octogon $A_1A_2A_3A_4A_5A_6A_7A_8$ such that $\overline{A_1A_2}$ is parallel to the $x$ -axis. Let $d_n$ denote the length of the edge connecting vertices $A_n$ and $A_{n+1}$ where $A_{9} = A_1$ . Clearly, $d_1 + d_2 \cdot \frac{1}{\sqrt{2}} + d_3 \cdot 0 - d_4 \cdot \frac{1}{\sqrt{2}} - d_5 - d_6 \cdot \frac{1}{\sqrt{2}} + d_7 \cdot 0 + d_8 \cdot \frac{1}{\sqrt{2}}$
$= (d_1 - d_5) + \frac{1}{\sqrt{2}} \cdot \left(d_2 - d_4 - d_6 + d_8\right) = 0$ . Since the $d_i$ are rational, it is clear that $d_1 = d_5$ . Then we have that $A_1A_2A_5A_6$ is a parallelogram, thus, its diagonals bisect each other. Simiarly, all the diagonals are current at a point $P$ where they bisect each other. $P$ is the desired point of symmetry.

Z K Y

The post below has been deleted. Click to close.

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

741 posts

enescu

#3 h Apr 17, 2005, 9:31 PM • 2 Y

Y by Adventure10, Mango247

Again, an algebraic approach: let $a_{1},\ldots,a_{8}$ be the side lengths of the equilateral octogon. Then, if
$\alpha=\cos\frac{2\pi}{8}+i\sin\frac{2\pi}{8}=\cos\frac{\pi}{4}+i\sin\frac {\pi}{4}$
then
$a_{1}+a_{2}\alpha+\ldots+a_{8}\alpha^{7}=0.$
Since $\alpha^{4}=-1,$ we obtain
$\left( a_{1}-a_{5}\right) +\left( a_{2}-a_{6}\right) \alpha+\left(a_{3}-a_{7}\right) \alpha^{2}+\left( a_{4}-a_{8}\right) \alpha^{3}=0,$
therefore $\alpha$ is a common root of the polynomials
$f\left( x\right) =x^{4}+1$
and
$g\left( x\right) =\left( a_{1}-a_{5}\right) +\left( a_{2}-a_{6}\right) x+\left( a_{3}-a_{7}\right) x^{2}+\left( a_{4}-a_{8}\right) x^{3}.$
If $\deg g>0,$ then the $\gcd\left( f,g\right)$ is a nonconstant polynomial with rational coefficients, implying that $f$ is reducible in $\mathbb{Q}% \left[ X\right] ,$ a contradiction. It results that the opposite sides of the octogon are congruent and the conclusion follows easily.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a