Ken's POTW


Two Lines and a Table
  1. Find the perpendicular distance between the two parallel lines: y=mx+a and y=mx+b in terms of a, b, and m.

  2. Put a round table into a (right-angled) corner so that the table top touches both walls and the feet are firmly on the ground. If there is a point on the perimeter of the table which is 10 inches from one wall and 5 inches from the other, what is the diameter of the table?

Source: 1. Original, Ken's POTD 3/24/94 2. Internet newsgroup rec.puzzles, Ken's POTD 5/4/94

Solution:
1. This was solved by several people, including: Wilfred Theunissen, PaulWogJ@aol.com, and Bill Chapp. The answer is: 

   |a-b|
-----------
      2
sqrt(m  +1)
Bill Chapp's Solution:
1. abs((a-b)/sqrt(m^2+1))

   Let line A be y=m*x+a, and
       line B be y=m*x+b.

   Pick point (0,a) on line A.  The line through this point and
   perpendicular to line A is:

       line C, y=(-1/m)*x+a

   Find the intersection of lines B and C at point (xc,yc):

     yc = m*xc + b
     yc = (-1/m)*xc + a

     m*xc + b = (-1/m)*xc + a
     (m + 1/m) * xc = a - b

     xc = (a - b) * (1/(m + 1/m))
        = (m / (m^2 + 1)) * (a - b)

     yc = m * (m / (m^2 + 1)) * (a - b) + b
        = (m^2 / (m^2 + 1)) * a + (1 / (m^2 +1)) * b
        = (1 / (m^2 + 1)) * (m^2*a - b)

   Find distance from (0,a) to (xc,yc):

      D = sqrt((xc-0)^2 + (yc-0)^2)
        = sqrt((m/(m^2+1)*(a-b))^2 + (1/(m^2+1)*(m^2*a+b) - a)^2)
        = sqrt((m/(m^2+1)*(a-b))^2 + (m/(m^2+1)*(m*a + b/m - (m^2+1)/m*a))^2))
        = sqrt((m/(m^2+1))^2 * ((a-b)^2 + ((m^2*a + b - m^2*a -a)/m)^2))
        = abs(m/(m^2+1)) * sqrt((a-b)^2 + ((a-b)/m)^2)
        = abs(m/(m^2+1)) * sqrt((a-b)^2 * ((m^2+1)/m^2))
        = abs((a-b)/(m^2+1)) * sqrt(m^2+1)
        = abs(a-b)/sqrt(m^2+1)
2. This was solved by Wilfred Theunissen, PaulWogJ@aol.com, and Bill Chapp.

The diameter can be either 10 in. or 50 in.

Bill Chapp's Solution: 

   Looking from the top, let the walls be the x and y axes.  Then the
   equation for the circle of the table is:

     (x-R)^2 + (y-R)^2 = R^2   (Eqn 1)

   To have points on the perimeter of the circle 10" from one wall and
   5" from the other, the point (10,5) must be on the circle:

     (10-R)^2 + (5-R)^2 = R^2
     100 - 20R + R^2 + 25 - 10R + R^2 = R^2
     R2 - 30R + 125 = 0
     (R-5)*(R-25) = 0

     Therefore R equals 5 or 25 inches.
     [KD: And D equals 10 or 50 inches.]
PaulWogJ's solution:
I made a radius to the place on the table that was 10 inches from one wall and 5 inches from the other. Then you make a right triangle with the radius just made as the hypotenuse, all of another radius but the ten inches as another, and a line connecting the point in the problem to the other leg to finish the right triangle. By the pythagorean theorem 
R^2 = (R-5)^2 + (R - 10) ^ 2
R^2 = R^2 - 10R + 25 + R^2 - 20R + 100
0 = R^2 - 30R + 125
0 = (R - 5)(R - 25)
R = 5 or 25
D = 2R = 10 or 50 inches.
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