# Interesting Problem

• Jul 11th 2006, 02:15 AM
nath_quam
Interesting Problem
The unit square ABCD where A(1,0), B(1,1), C(2,1), D(2,0)

i) A line, l , passing through the origin with gradient m, cuts the sides AB and CD at P and Q respectively
is the answer to this part
$\displaystyle m < \frac{1}{2}\$

i am unsure on how to approach these two parts any help would be most appreciated.
ii) For what value(s) of m does the line, l, divide the area of the square in the ratio 2:1??

iii) Another like, k, passes through the origin with gradient, n, and cuts the square through the sides AB and Bc at S and T respectively.

Show that it is not possible for k to divide the area of the square in the ratio 2:1

Thanks Nath
• Jul 11th 2006, 03:17 AM
malaygoel
Quote:

Originally Posted by nath_quam
The unit square ABCD where A(1,0), B(1,1), C(2,1), D(2,0)

i) A line, l , passing through the origin with gradient m, cuts the sides AB and CD at P and Q respectively
is the answer to this part
$\displaystyle m < \frac{1}{2}\$

Here is first.
Let the coordinates of P be (1,a) and Q be (2,b).Let O be the origin.
m is the slope.
$\displaystyle m=\frac{a}{1}=\frac{b-a}{1}$
it gives $\displaystyle m = a = \frac{b}{2}$
Since B lies on CD $\displaystyle b < 1$
Hence$\displaystyle m<\frac{1}{2}$
• Jul 11th 2006, 03:25 AM
malaygoel
Quote:

Originally Posted by nath_quam
The unit square ABCD where A(1,0), B(1,1), C(2,1), D(2,0)

i) A line, l , passing through the origin with gradient m, cuts the sides AB and CD at P and Q respectively
is the answer to this part
$\displaystyle m < \frac{1}{2}\$

i am unsure on how to approach these two parts any help would be most appreciated.
ii) For what value(s) of m does the line, l, divide the area of the square in the ratio 2:1??

I will continue with the notation of the previous post.
Let the area of trapezium ADQP be x
Then there are two cases:
$\displaystyle \frac{x}{1-x}=2,\frac{1}{2}$
$\displaystyle x=\frac{2}{3},\frac{1}{3}$
From the diagram, $\displaystyle x=\frac{1}{2}(a+b)=\frac{3a}{2}$
Hence values of slope are $\displaystyle m=\frac{2}{9},\frac{4}{9}$

Keep Smiling
Malay
• Jul 11th 2006, 03:37 AM
nath_quam
Thanks Malay i am understanding any idea on the last?
• Jul 11th 2006, 03:39 AM
malaygoel
Quote:

Originally Posted by nath_quam
The unit square ABCD where A(1,0), B(1,1), C(2,1), D(2,0)

iii) Another like, k, passes through the origin with gradient, n, and cuts the square through the sides AB and Bc at S and T respectively.

Show that it is not possible for k to divide the area of the square in the ratio 2:1

Join O and C.
The square is divided into two parts.
Area of the smaller part(containing the vertex B) is 1/4.
Now if the line passes through AB and BC, the triangle formed by it has area less than 1/4. But for the ratio to be 2:1, it has to be either 1/3 or 2/3(see previous post), which is not possible since 1/3>1/4 and 2/3 >1/4.

Keep Smiling
Malay
• Jul 11th 2006, 03:46 AM
nath_quam
Quote:

Originally Posted by malaygoel
$\displaystyle \frac{x}{1-x}=2,\frac{1}{2}$
$\displaystyle x=\frac{2}{3},\frac{1}{3}$
From the diagram, $\displaystyle x=\frac{1}{2}(a+b)=\frac{3a}{2}$
Hence values of slope are $\displaystyle m=\frac{2}{9},\frac{4}{9}$

Keep Smiling
Malay

i get sorta lost here u jump steps too quickly would u be able to explain it
• Jul 11th 2006, 04:07 AM
malaygoel
Quote:

Originally Posted by nath_quam
i get sorta lost here u jump steps too quickly would u be able to explain it

I used two equations of the previous post.
b=2a
and m=a

Keep Smiling
Malay
• Jul 16th 2006, 04:03 AM
nath_quam
Quote:

Originally Posted by malaygoel
Here is first.
Let the coordinates of P be (1,a) and Q be (2,b).Let O be the origin.
m is the slope.
$\displaystyle m=\frac{a}{1}=\frac{b-a}{1}$
it gives $\displaystyle m = a = \frac{b}{2}$
Since B lies on CD $\displaystyle b < 1$
Hence$\displaystyle m<\frac{1}{2}$

and also greater than zero??
• Jul 16th 2006, 04:16 AM
malaygoel
Quote:

Originally Posted by nath_quam
and also greater than zero??

Yes.
You could see it this way also: Line rises towards right, hence it has a positive slope. It cannot be zero because it will imply that line doesn't rise towards right.

Keep Smiling
Malay
• Jul 16th 2006, 06:29 AM
Rich B.
Show that it is not possible for k to divide the area of the square in the ratio 2:1

Hi:

If the line passes through side(BC), then its slope must be greater than 1/2. But, as Malay has demonstrated, the slope must have value 2/9 or 4/9, neither of which is greater 1/2.

Regards,

Rich B.