Source: Reader Denis Borris, citing Sam Loyd for the first one. 4. Extension from Al Zimmermann and Nick McGrath independently.
For a better understanding of the proposed solution, please find attached the file (030304solPA.pdf) in pdf format. Q1 Let 20+a the distance covered by the motorcyclist from his starting place at the back to the front of the lin and a the distance he covers to the back. Let V and v the respective speeds of the motorcyclist and of the VIP vehicles. We have (20+a)/V = a/v and a/V = (20-a)/v Therefore (20+a)/a = a/(20-a) = V/v ==> a^2 = 200 and a=10*sqrt(2) The policeman has travelled 20 + 2*a = 20*(sqrt(2)+1) = 48,284...miles Q2 Let V and v the respective speeds of the destroyer and of the convoy. The trip travelled by the destroy is identified in red in the attached file. The distances covered by the destroy is equal to: 20+a when the convoy travels a sqrt(400+b^2) when the convoy travels b 20-c when the convoy travels c sqrt(400+d^2) when the convoy travels d with the condition a+b+c+d = 20 Then we have the relations: (20+a)/a = sqrt(400+b^2)/b = (20-c)/c = sqrt(400+d^2)/d = V/v Therefore b=d, c=10*a/(a+10), b^2 = 10*a^2/(a+10) and a is defined by the solutions of the equation a^4 - 40*a^3 - 800*a^2+40 000 = 0 The sole root is equal to a=6,287089... Then b=d=4,92637... c=3,8601651...corresponding to V/v = 4,1811222... The distance travelled by the destroy is equal to 20*V/v = 83,6225...miles Q3 Let again V and v the respective speeds of the destroyer and of the convoy. The trip travelled by the destroy is identified in red in the attached file. The distances covered by the destroy is equal to: sqrt(400+20*sqrt(3)*a+a^2) when the convoy travels a sqrt(400 -20*sqrt(3)*b+b^2) when the convoy travels b sqrt(400+c^2) when the convoy travels c with the condition a+b+c=20 Then we have the relations: sqrt(400+20*sqrt(3)+a^2)/a = sqrt(400 -20*sqrt(3)+b^2)/b = sqrt(400+c^2)/c = V/v Therefore c = 20*a/sqrt(400+20*a*sqrt(3)) =20*b/sqrt(400-20*b*sqrt(3)) and a+b+20*a/sqrt(400+0*a*sqrt(3)) = 20 The unique solution is given by a = 8,579379... b=4,9222023.... c=6,49841837....corresponding to V/v = 3,32605... The distance travelled by the destroy is equal to 20*V/v = 64,7211...miles Q4 Let R=10 the radius of the circle, V and v the respective speeds of the destroyer and of the convoy with r=V/v. The trip travelled by the destroy is identified in red in the attached file. The distances a,b,c,d travelled by the convoy when the destroy is respectively in the first,second,third and firth quadrants are defined by the following identities: a =v*t(a)= sum[ 20*(cos(u)+sqrt(r^2-sin^2(u))/(r^2-1)*du] for u=0 to pi/2 b = v*t(b)= sum[ 20*(-sin(u)+sqrt(r^2-cos^2(u))/(r^2-1)*du] for u=0 to pi/2 c = v*t(c)= sum[ 20*(-cos(u)+sqrt(r^2-sin^2(u))/(r^2-1)*du] for u=0 to pi/2 d = v*t(d)= sum[ 20*(sin(u)+sqrt(r^2-cos^2(u))/(r^2-1)*du] for u=0 to pi/2 with the condition a+b+c+d = 2*R =20 It is easy to check that a=d and b=c An Excel sheet allows to calculate easily the solution a=5,9674... and b=4,0325...corresponding to the ratio r=V/v=3,36836.... So the distance travelled by the destroy is equal to 20*V/v = 67,367...miles