computational, areas, square - 2003 Estonia Open Junior 2.4

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

computational, areas, square - 2003 Estonia Open Junior 2.4

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 Sep 25, 2020, 7:43 PM

Y by

Consider the points $A_1$ and $A_2$ on the side $AB$ of the square $ABCD$ taken in such a way that $|AB| = 3 |AA_1|$ and $|AB| = 4 |A_2B|$ , similarly consider points $B_1$ and $B_2, C_1$ and $C_2, D_1$ and $D_2$ respectively on the sides $BC$ , $CD$ and $DA$ . The intersection point of straight lines $D_2A_1$ and $A_2B_1$ is $E$ , the intersection point of straight lines $A_2B_1$ and $B_2C_1$ is $F$ , the intersection point of straight lines $B_2C_1$ and $C_2D_1$ is $G$ and the intersection point of straight lines $C_2D_1$ and $D_2A_1$ is $H$ . Find the area of the square $EFGH$ , knowing that the area of $ABCD$ is $1$ .

Z K Y

The post below has been deleted. Click to close.

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

hopetosuccess9102

192 posts

hopetosuccess9102

#2 h Sep 25, 2020, 11:11 PM

Y by

We have $A(0,1),B(1,1),C(1,0),D(0,0)$ , We can quickly get $A_2(\frac 34,1),B_2(1,\frac 14),C_2(\frac 14,0),D_2(0,\frac 34)$ and $A_1(\frac{5}{12},1),B_1(1,\frac{7}{12}),C_1(\frac{7}{12},0),D_1(0,\frac{5}{12})$ . From here, we have $A_2B_1 \implies y=-\frac 53 x+\frac 94$ $A_1D_2 \implies y=\frac 35 x+\frac 34$ $C_2D_1 \implies y=-\frac 53 x+\frac{5}{12}.$ Solving, we get $E=A_2B_1 \cap A_1D_2 \implies E(\frac{45}{68},\frac{39}{34})$ $H=A_1D_2 \cap C_2D_1 \implies H(-\frac{5}{34},\frac{45}{68}).$ Calculating distance and squaring to find the area, we have $EH^2=\boxed{\frac{121}{136}}.$ Edit: Read the problem wrong. I thought that it said $|AB|=3|A_1A_2|$ , not $|AB|=3|AA_1|$ . This is an incorrect solution.

This post has been edited 6 times. Last edited by hopetosuccess9102, Sep 26, 2020, 6:51 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 Sep 26, 2020, 5:59 AM

Y by

@above How did you get $A_1(\frac{5}{12},1)$ ?

This post has been edited 1 time. Last edited by natmath, Sep 26, 2020, 6:01 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

#4 h Sep 26, 2020, 6:06 AM • 1 Y

Y by Mango247

Note that $A_1A_2=A_2B_1...=\frac{5}{12}$ . This means that those triangle pieces that are added to $ABCD$ to make $EFGH$ are congruent to the corner pieces that are taken away, hence the area is $1$ .

Z K Y

The post below has been deleted. Click to close.

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

40 posts

#6 h Sep 26, 2020, 7:26 AM

Y by

Let the side be $12a$ . Every side is divided into lengths of $4a$ , $5a$ , $3a$ . Since $EFGH$ is a square, upon finding some lenths we find that all the corner triangles are congruent. Therefore same area $\implies$ area $=1$ .

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a