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Find the Length of a Carpet

Date: 4/22/96 at 8:18:2
From: Stefan Isberg
Subject: Find the Length of a Carpet 

You have a rectanguler room with sides of 3 and 5 meters.
You put a carpet on the diagonal (not exactly, of course!)

The carpet's corners must touch each wall in the room, 
so the problem will not be as easy as it seems. 

We have tried to illustrate how the carpet shall lie.  
The carpet lies approximately on the diagonal.

         5 meter
!                         \   !     
!                          \  !    
!                           \ ! 3 meter
!\                            !
! \                           !
!  \                          !

The carpet's width is 1 meter (The carpet is rectanguler) .
What is the exact length of this carpet?

Stefan and Anders, Math students at Umea University.

Date: 5/3/96 at 4:15:3
From: Dr. Alain
Subject: Re: Find the Length of a Carpet 

Draw the room like this:

(-5,0)                                       (0,0)
  |                                        |
  |                                        |
  |                                        |
  |                                        |
  |                                        |
  |                                        |
  |                                        |
  |                                        |
  |                                        |
(-5,-3)                                     (0,-3)  

Say the carpet hits the top line at (-a,0) and the right line at 
(0,-b). Then it must hit the left wall at (-5, -3 + b), and hit the 
bottom wall at (-5 + a, -3).

We know the width of the carpet is 1, so:

a^2 + b^2 = 1.

We also have that the carpet is rectangular, so the line from 
(-5 + a, -3) to (0,-b) makes a right angle with the line from 
(0,-b) to (-a,0). 

This is equivalent to the product of the slope of these two lines 
which is -1.

The slope of the first line is  m1= (3-b)/(5-a).
The slope of the second line is m2= -b/a.

So: m2*m1=-b/a*(3-b)/(5-a) = -1.

Or b^2 - 3b = a^2 - 5a.

So we have 

(1)   a^2 + b^2 =1.
(2)   b^2 - 3b = a^2 - 5a.

These two equations are enough to find a and b. These equations
bring up a fourth degree polynomial equation. It is possible to
find the exact solution to such an equation (when the solution
exists) but the formula for fourth degree polynomials is quite
horible. The best way to treat this polynomial equation is with
numerical methods such as the Newton-Raphson algorithm. 

From the endpoints of the carpet we can find its length, L:

L^2 = (-5 + a)^2 + (3 - b)^2.

Now we have:

L^2 = 25 - 10a + a^2 + 9 - 6b + b^2.

Once you found a and b you can find L, which is the answer to your

-Doctor Alain,  The Math Forum
 Check out our web site!   
Associated Topics:
High School Coordinate Plane Geometry
High School Geometry
High School Triangles and Other Polygons

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