Guard Dog's Area of Coverage

quantyst · April 2nd, 2009, 9:20 am

One end of a guard dog's leash is tied to the corner of a rectangular building of dimensions AxB (where A<B), and the other end is tied to the dog. Find the total area that the dog can cover if the length of the leash is L. Express the answer as an explicit function of A, B and L.

mahalekrishna · April 2nd, 2009, 10:09 pm

is there any catch in this one ?l>b; aba <=l <b; 0.5(a(l^2-a^2)^0.5 + l^2 taninv(a/(l^2-a^2)^0.5)) 0=< l <a; pi l^2/4 if the dog is tied outsidel>a+b; 0.25pi(3l^2+(l-a)^2 + (l-b)^2) - {donkeywork}; where donkeywork is the overlapping areab<=l<a+b; 0.25pi(3l^2+(l-a)^2 + (l-b)^2)a<=l<b; 0.25pi(3l^2+(l-a)^2)0<=l<a; 0.75pi l^2

Traden4Alpha · April 2nd, 2009, 11:12 pm

Partial solution for the dog's area of coverage, C (assuming the dog cannot enter the building): L <= A: C = c0 = 3/4 * pi * L^2A < L <=B: C = c0+c1, c1 = 1/4 * pi * (L-A)^2B < L <=A+B: C = c0+c1+c2, c2 = 1/4 * pi * (L-B)^2L > A+B: C = c0+c1+c2 - c3, c3 = the area of the ugly overlap between the (L-A) radius quarter circle and the (L-B) radius quarter circle at the corner opposite the leash's tie point.I need to grind out the c3 term which doesn't look hard! EDIT: looks like I cross-posted with mahalekrishna's edits and we have the same solution

daveangel · April 3rd, 2009, 9:34 am

is L < a ?