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Thread: Line passing through a square

  1. #1
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    Line passing through a square

    Straight line $\displaystyle l$ passes through the center of a square $\displaystyle ABCD$, whose area is $\displaystyle 1$. Let $\displaystyle a, b, c, d$ denote the shortest distances between line $\displaystyle l$ and points $\displaystyle A, B, C, D$ respectively. How to prove that $\displaystyle a^2 + b^2 + c^2 + d^2 = 1$?
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  2. #2
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    Quote Originally Posted by atreyyu View Post
    Straight line $\displaystyle l$ passes through the center of a square $\displaystyle ABCD$, whose area is $\displaystyle 1$. Let $\displaystyle a, b, c, d$ denote the shortest distances between line $\displaystyle l$ and points $\displaystyle A, B, C, D$ respectively. How to prove that $\displaystyle a^2 + b^2 + c^2 + d^2 = 1$?
    We can model the square as: $\displaystyle A:\left( {0,0} \right),\;B:\left( {1,0} \right),\;C:\left( {1,1} \right),\;D:\left( {0,1} \right),\;E:\left( {\frac{1}{2},\frac{1}{2}} \right)$; E is the center.
    Any line containing the center has this form: $\displaystyle px + qy + r = 0\,\mbox{ so } \,p + q = - 2r$.
    Using the standard rule for distance we get:
    $\displaystyle a = \frac{{\left| r \right|}}{{\sqrt {p^2 + q^2 } }}\,,\,b = \frac{{\left| {p + r} \right|}}{{\sqrt {p^2 + q^2 } }}\,,\,c = \frac{{\left| {p + q + r} \right|}}{{\sqrt {p^2 + q^2 } }}\,,\,d = \frac{{\left| {q + r} \right|}}{{\sqrt {p^2 + q^2 } }}\,$.

    Can you put it all together?
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  3. #3
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    Hello, atreyyu!

    My approach is the same as Plato's . . .


    Straight line $\displaystyle L$ passes through the center of a square $\displaystyle ABCD$ with area 1.
    Let $\displaystyle a, b, c, d$ denote the shortest distances between line $\displaystyle L$ and vertices $\displaystyle A, B, C, D$ resp.
    Prove: .$\displaystyle a^2 + b^2 + c^2 + d^2 \:= \:1$
    The vertices are: .$\displaystyle A(0,0),\;B(1,0),\;C(1,1),\;D(0,1)$

    The line through center (, ) with slope m has the equation:
    . . $\displaystyle y - \frac{1}{2} \:=\:m\left(x - \frac{1}{2}\right) \quad\Rightarrow\quad L\!:\:2mx - 2y + 1 - m \:=\:0$


    Formula: The distance from point $\displaystyle P(x_1,y_1)$ to line $\displaystyle L\!:\:ax + by + c \:=\:0$

    . . . . . . . . is given by: .$\displaystyle \frac{|ax_1 + by_1 + c|}{\sqrt{a^2+b^2}} $


    We have: .$\displaystyle a = 2m, b = -2, c = 1-m$

    The distance is: .$\displaystyle \Delta \:=\:\frac{|2mx_1 - 2y_1 + 1 - m|}{\sqrt{(2m)^2 + 2^2}} \;=\;\frac{|(2x_1-1)m + (1-2y_1)|}{2\sqrt{m^2+1}}$


    For $\displaystyle A(0,0)\!:\;\;a \;=\;\frac{|\text{-}m+1|}{2\sqrt{m^2+1}} \;=\;\frac{|m-1|}{2\sqrt{m^2+1}} $

    For $\displaystyle B(1,0)\!:\;\;b \;=\;\frac{|m+1|}{2\sqrt{m^2+1}} $

    For $\displaystyle C(1,1)\!:\;\;c \;=\;\frac{|m-1|}{2\sqrt{m^2+1}} $

    For $\displaystyle D(0,1)\!:\;\;d \;=\;\frac{|\text{-}m-1|}{2\sqrt{m^2+1}}\;=\;\frac{|m+1|}{2\sqrt{m^2+1}} $


    Hence: .$\displaystyle a^2+b^2+c^2+d^2 \;\;=\;\;\frac{(m-1)^2}{4(m^2+1)} + \frac{(m+1)^2}{4(m^2+1)} + \frac{(m-1)^2}{4(m^2+1)} + \frac{(m-1)^2}{4(m^2+1)} $

    . . $\displaystyle = \;\;\frac{(m^2 - 2m + 1) + (m^2 + 2m + 1) + (m^2 - 2m + 1) + (m^2 + 2m + 1)}{4(m^2+1)} $

    . . $\displaystyle = \;\;\frac{4m^2+4}{4(m^2+1)} \;\;=\;\;\frac{4(m^2+1)}{4(m^2+1)} \;\;=\;\;1$

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  4. #4
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    Soraban, does your approach work if the line through the center is vertical?
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  5. #5
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    Quote Originally Posted by Soroban View Post

    The line through center (, ) with slope m has the equation:
    . . $\displaystyle y - \frac{1}{2} \:=\:m\left(x - \frac{1}{2}\right) \quad\Rightarrow\quad L\!:\:2mx - 2y + 1 - m \:=\:0$

    That's the part I don't get ... would be great if you could explain it further.
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  6. #6
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    Hello, Plato!

    Does your approach work if the line through the center is vertical?
    Good question . . . I hadn't considered it.


    I believe it does work . . . since $\displaystyle \lim_{m\to\infty}\frac{4(m^2+1)}{4(m^2+1)} \;=\;1$

    Besides: .$\displaystyle a = b = c = d = \frac{1}{2}$
    . . Therefore: .$\displaystyle a^2+b^2+c^2+d^2 \:=\:1$

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