PartialMathematician

The correct answers are x = 3, y = -2, and z = 1.

- PM

I checked both of your answers, and I found out that Guest's answers are correct.

If the first solution confuses you, here is a second solution. 

Create a trapezoid with inscribed circle O exactly like in the solution above, and extend lines AD and BC from the solution above and label the point at where they meet H. Because \(\frac{\overline{GC}}{\overline{BE}} = \frac{1}{9}\), \(\frac{\overline{HG}}{\overline{HE}} = \frac{1}{9}\). Let \(\overline{HG} = x\) and \(\overline{GE} = 8x\).

Because these are radii, \(\overline{GO} = \overline{OE} = \overline{OF} = 4x\). \(\overline{OF} \perp \overline{BH}\) , so \(\overline{OF}^2 + \overline{FH}^2 = \overline{OH}^2\). Plugging in, we get \(4x^2 + \overline{FH}^2 = 5x^2 \), so \(\overline{FH} = 3x\).Triangles OFH and BEH are similar, so \(\frac{\overline{OF}}{\overline{BE}}  = \frac{\overline{FH}}{\overline{EH}}\), which gives us \(\frac{4x}{18} = \frac{3x}{9x}\). Solving for x, we get \(x = 1.5\) and \(4x = \boxed{6}\).

Hope this helps,

- PM

Consider a trapezoidal (label it ABCD as follows) cross-section of the truncate cone along a diameter of the bases:

The radius is \(\boxed{6}\).

- PM

Good afternoon is correct.

CPHill, I have a question about logarithms and functions too. 

Let \(f(n) = \){ \(n^2 + 1\)     if n is odd.

                   {  \(\dfrac{n}{2}\)    if n is even.

For how many integers n from 1 to 100, inclusive, does \(f(f(...f(n)...)) = 1\) for some number of applications of f?

Thanks!

No problem!