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 $1$.
Problem
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Tags: geometey, areas, square
hopetosuccess9102
26.09.2020 02:11
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.
natmath
26.09.2020 08:59
@above How did you get $A_1(\frac{5}{12},1)$?
natmath
26.09.2020 09:06
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 $1$.
mathingbingo
26.09.2020 10:26
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 $=1$.