area wanted, trisecting each side of a square 2015 Fall Animath Cup p2

Art of Problem Solving

AoPS Online

High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

3 M G

BBookmark VNew Topic kLocked

High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

area wanted, trisecting each side of a square 2015 Fall Animath Cup p2

G H J

Reveal topic tags

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.

30629 posts

#1 h Oct 13, 2020, 2:50 AM

Y by

Each side of a unit square is divided into $3$ equal segments. We draw the figure below from this share. What is the area of the gray polygon? Justify your answer.
//cdn.artofproblemsolving.com/images/2/1/2/212837843da588752e1a24c1e8bc375b6fc2bd03.png

//cdn.artofproblemsolving.com/images/2/1/2/212837843da588752e1a24c1e8bc375b6fc2bd03.png

This post has been edited 3 times. Last edited by parmenides51, Oct 13, 2020, 3:10 AM

Z K Y

The post below has been deleted. Click to close.

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

8219 posts

#3 h Oct 13, 2020, 3:38 AM

Y by

Define $\theta$ s.t. $\tan\theta=\frac{1}{2}$
We can see that the angle of the octagon thing is $180-2\theta$

Look at those white right triangle things on the side of the octagon.
The side adjacent to the angle $2\theta$ has length $\frac{1}{6}\sec\theta$ and the side opposite has length $\frac{1}{6}\sec\theta \tan 2\theta$ , so the area of each right triangle is $\frac{1}{72}\sec^2\theta\tan 2\theta$

Each of the small squares has area $\frac{5}{9}$ , so the area of the octagon is $S=\frac{10-\sec^2\theta\tan 2\theta}{18}$

$\sec^2\theta=1+\frac{1}{4}=\frac{5}{4}$
$\tan 2\theta=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}$

$S=\frac{25}{54}$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a