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