## Escorting Shapes

1. A police motorcyclist is escorting a 20 mile line of VIP vehicles. He drives at constant speed from his starting place at the back to the front of the line, and immediately returns with the same speed to the back of the line. He arrives at the back just as the line has moved 20 miles. How many miles did the policeman travel?
2. A destroyer is escorting a convoy of ships 20 miles long and 20 miles wide. The convoy is in the shape of a square. Starting from the rear left corner of the square, it steams to the front; then steams across the front of the square to the front right; then back down the right side and finally across the rear back to the left rear corner, arriving there just as the convoy has moved 20 miles. How many miles did the destroyer travel?
3. Same problem as previous, except the convoy is in the shape of an equilateral triangle, side length = 20. (There are three ways to consider this problem: point forward, point rear, and point side.)
4. Same problem as previous, except the convoy is in the shape of a circle.

Source: Reader Denis Borris, citing Sam Loyd for the first one. 4. Extension from Al Zimmermann and Nick McGrath independently.

Solutions were received from Al Zimmermann, Jimmy Chng Gim Hong, Philippe Fondanaiche, Claudio Baiocchi, Jeremy Galvagni, and Denis Borris. Jeremy Galvagni's solutions can be found here. [Update 18June2007: Sorry the embedded pics are unreadable.  I've lost the original.]  Philippe Fondanaiche's description of the solution is below.
```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
```

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