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`

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