[b]p1A.[/b] Compute
         $$1 + \frac{1}{2^3} +  \frac{1}{3^3} +  \frac{1}{4^3} +  \frac{1}{5^3} + ...$$
          $$1 -  \frac{1}{2^3} +  \frac{1}{3^3} -  \frac{1}{4^3} +  \frac{1}{5^3} - ...$$

[b]p1B.[/b] Real values $a$ and $b$ satisfy $ab = 1$, and both numbers have decimal expansions which repeat every five digits:
      $$  a = 0.(a_1)(a_2)(a_3)(a_4)(a_5)(a_1)(a_2)(a_3)(a_4)(a_5)...$$
          and
       $$ b = 1.(b_1)(b_2)(b_3)(b_4)(b_5)(b_1)(b_2)(b_3)(b_4)(b_5)...$$
If $a_5 = 1$, find $b_5$.

[b]p2.[/b] $P(x) = x^4 - 3x^3 + 4x^2 - 9x + 5$.  $Q(x)$ is a $3$rd-degree polynomial  whose graph intersects the graph of $P(x)$ at $x = 1$, $2$, $5$, and $10$.  Compute  $Q(0)$.


[b]p3.[/b] Distinct real values $x_1$, $x_2$, $x_3$, $x_4 $all satisfy $ ||x - 3| - 5| = 1.34953$.  Find $x_1 + x_2 + x_3 + x_4$.


[b]p4.[/b] Triangle $ABC$ has sides $AB = 8$,  $BC = 10$, and $CA = 11$.  Let $L$ be the locus of points in the interior of triangle $ABC$ which are within one unit  of either $A$, $B$, or $C$.  Find the area of $L$.


[b]p5.[/b] Triangles $ABC$ and $ADE$ are equilateral, and $AD$ is an altitude of $ABC$.  The area of the intersection of these triangles is $3$.  Find the area of  the larger triangle $ABC$.


PS. You had better use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
1. Recognize that \(a\) and \(b\) are repeating decimals with a period of 5 digits. This implies that both \(a\) and \(b\) can be expressed as fractions with denominators of the form \(99999\), which is \(10^5 - 1\).
2. Let \(a = \frac{x}{99999}\) and \(b = \frac{y}{99999}\), where \(x\) and \(y\) are integers.
3. Given that \(ab = 1\), we have:
   \[
   \left(\frac{x}{99999}\right) \left(\frac{y}{99999}\right) = 1 \implies xy = 99999^2
   \]
4. Since \(a\) is a repeating decimal less than 1, \(x\) must be an integer between 0 and 99999. Similarly, since \(b\) is a repeating decimal greater than 1 but less than 2, \(y\) must be an integer between 99999 and 199998.
5. We need to find \(x\) and \(y\) such that \(xy = 99999^2\). We can start by checking possible values for \(x\) and solving for \(y\):
   \[
   y = \frac{99999^2}{x}
   \]
6. After some trial and error or using a systematic approach, we find that \(x = 73441\) and \(y = 136161\) satisfy the equation \(xy = 99999^2\).
7. Therefore, \(a = \frac{73441}{99999}\) and \(b = \frac{136161}{99999}\).
8. Converting \(b\) to its decimal form:
   \[
   b = \frac{136161}{99999} = 1 + \frac{36162}{99999} = 1.361623616236162\ldots
   \]
9. Given that \(a_5 = 1\), we need to find \(b_5\). Observing the decimal expansion of \(b\), we see that the fifth digit after the decimal point is 2.

The final answer is \( \boxed{ 2 } \).