Line passing through a square

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August 25th 2008, 10:31 AM
atreyyu

Line passing through a square

Straight line $l$ passes through the center of a square $ABCD$ , whose area is $1$ . Let $a, b, c, d$ denote the shortest distances between line $l$ and points $A, B, C, D$ respectively. How to prove that $a^2 + b^2 + c^2 + d^2 = 1$ ?
August 25th 2008, 12:51 PM
Plato

Quote:

Originally Posted by atreyyu

Straight line $l$ passes through the center of a square $ABCD$ , whose area is $1$ . Let $a, b, c, d$ denote the shortest distances between line $l$ and points $A, B, C, D$ respectively. How to prove that $a^2 + b^2 + c^2 + d^2 = 1$ ?

We can model the square as: $A:\left( {0,0} \right),\;B:\left( {1,0} \right),\;C:\left( {1,1} \right),\;D:\left( {0,1} \right),\;E:\left( {\frac{1}{2},\frac{1}{2}} \right)$ ; E is the center.
Any line containing the center has this form: $px + qy + r = 0\,\mbox{ so } \,p + q = - 2r$ .
Using the standard rule for distance we get:
$a = \frac{{\left| r \right|}}{{\sqrt {p^2 + q^2 } }}\,,\,b = \frac{{\left| {p + r} \right|}}{{\sqrt {p^2 + q^2 } }}\,,\,c = \frac{{\left| {p + q + r} \right|}}{{\sqrt {p^2 + q^2 } }}\,,\,d = \frac{{\left| {q + r} \right|}}{{\sqrt {p^2 + q^2 } }}\,$ .

Can you put it all together?
August 25th 2008, 03:07 PM
Soroban

Hello, atreyyu!

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

Quote:

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

The vertices are: . $A(0,0),\;B(1,0),\;C(1,1),\;D(0,1)$

The line through center (½, ½) with slope m has the equation:
. . $y - \frac{1}{2} \:=\:m\left(x - \frac{1}{2}\right) \quad\Rightarrow\quad L\!:\:2mx - 2y + 1 - m \:=\:0$

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

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

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

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

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

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

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

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

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

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

. . $= \;\;\frac{4m^2+4}{4(m^2+1)} \;\;=\;\;\frac{4(m^2+1)}{4(m^2+1)} \;\;=\;\;1$
August 25th 2008, 04:20 PM
Plato

Soraban, does your approach work if the line through the center is vertical?
August 26th 2008, 02:02 AM
atreyyu

Quote:

Originally Posted by Soroban

The line through center (½, ½) with slope m has the equation:
. . $y - \frac{1}{2} \:=\:m\left(x - \frac{1}{2}\right) \quad\Rightarrow\quad L\!:\:2mx - 2y + 1 - m \:=\:0$

That's the part I don't get (Worried)... would be great if you could explain it further.
August 26th 2008, 03:29 AM
Soroban

Hello, Plato!

Quote:

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 $\lim_{m\to\infty}\frac{4(m^2+1)}{4(m^2+1)} \;=\;1$

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