# How far will they have to walk?Sally is standing on the north side of a circular building with a radius of R and Henry is standing on the south side. Sally starts walking north and Henry starts...

How far will they have to walk?

Sally is standing on the north side of a circular building with a radius of R and Henry is standing on the south side. Sally starts walking north and Henry starts walking east, they are walking at the same rate. How far will they have to walk until they can see each other?

*print*Print*list*Cite

### 1 Answer

Let's say the circular building is centered at the origin. Then Sally is standing at (0,R). When she has walked a distance of d she will be standing at (0,R+d). Henry is standing at (0,-R) and when he has walked a distance of d he will be standing at (d,-R). They will be able to see each other as soon as the line between them is tangent line to the circular building. So we are really looking for a distance d such that the line connecting (0,R+d) with (d,-R) is tangent to the circle with radius R, centered at the origin.

Suppose we had such a d. Then we would have the following picture:

(here I made R=2, but you can have R as anything)

Now, all the red lines have length R, and the blue lines have length d

The solid red line is tangent to the solid blue line, so that makes a right angle. Similarly, the diagonal (dashed) red line is tangent to the green line, so we have a right angle there.

What we know:

The solid red line, diagonal-dashed red line, solid blue line, and solid green line form a symmetric kite, so we know the length of the solid green line must be d

The "tight dash" red line, the "tight dash" blue line, the "tight dash" green line, and the diagonal red line form a right triangle, so we can use the Pythagorean theorem to get that the length of the "tight dash" green line is: `sqrt( (R+d)^2 - R^2 )` , or `sqrt(2Rd + d^2) `

Now, all of the lines put together form a (large) right triangle, so we can again use the Pythagorean theorem to get the length of the entire green line: `sqrt( (2R+d)^2 + d^2 )` , or `sqrt(4R^2 + 4Rd + 2d^2)`

Thus the length of the entire green line can be expressed in two ways:

`sqrt(4R^2 + 4Rd + 2d^2)` or `d + sqrt(2Rd + d^2) `

So:

`sqrt(4R^2 + 4Rd + 2d^2) = d + sqrt(2Rd + d^2)`

Square both sides:

`4R^2 + 4Rd + 2d^2 = d^2 + 2Rd + d^2 + 2d sqrt(2Rd + d^2)`

Simplify:

`4R^2 + 4Rd = 2Rd + 2d sqrt(2Rd + d^2)`

`4R^2 + 2Rd = 2d sqrt(2Rd + d^2)`

`2R^2 + Rd = d sqrt(2Rd + d^2)`

`R(2R+d) = d sqrt(d(2R+d))`

Square both sides:

`R^2(2R+d)^2 = d^2 [d(2R+d)]`

Simplify:

`R^2(2R+d) = d^3`

`d^3 - R^2d-2R^3 = 0`

This is called a "depressed cubic". It has a formula you can use to solve it, similar to the quadratic equation.

(see: http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method )

If `d^3 + pd +q = 0` , then a root of the equation is:

`root(3)( -q/2 + sqrt( q^2/4 + p^3/27) ) + root(3)( -q/2 - sqrt( q^2/4 + p^3/27) )`

For us, we get `d=R [root(3)(1+ sqrt(26/27)) + root(3)(1- sqrt(26/27))] ~~ 1.5214 R`