[b]p1.[/b] An organization decides to raise funds by holding a $\$60$ a plate dinner. They get prices from two caterers. The first caterer charges $\$50$ a plate. The second caterer charges according to the following schedule: $\$500$ set-up fee plus $\$40$ a plate for up to and including $61$ plates, and $\$2500$ $\log_{10}\left(\frac{p}{4}\right)$ for $p > 61$ plates.
a) For what number of plates $N$ does it become at least as cheap to use the second caterer as the first?
b) Let $N$ be the number you found in a). For what number of plates $X$ is the second caterer's price exactly double the price for $N$ plates?
c) Let $X$ be the number you found in b). When X people appear for the dinner, how much profit does the organization raise for itself by using the second caterer?


[b]p2.[/b] Let $N$ be a positive integer. Prove the following:
a) If $N$ is divisible by $4$, then $N$ can be expressed as the sum of two or more consecutive odd integers.
b) If $N$ is a prime number, then $N$ cannot be expressed as the sum of two or more consecutive odd integers.
c) If $N$ is twice some odd integer, then $N$ cannot be expressed as the sum of two or more consecutive odd integers.


[b]p3.[/b] Let $S =\frac{1}{1^2} +\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...$

a) Find, in terms of $S$, the value of $S =\frac{1}{2^2} +\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...$

b) Find, in terms of $S$, the value of$S =\frac{1}{1^2} +\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...$

c) Find, in terms of $S$, the value of$S =\frac{1}{1^2} -\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+...$


[b]p4.[/b] Let $\{P_1, P_2, P_3, ...\}$ be an infinite set of points on the $x$-axis having positive integer coordinates, and let $Q$ be an arbitrary point in the plane not on the $x$-axis. Prove that infinitely many of the distances $|P_iQ|$ are not integers.
a) Draw a relevant picture.
b) Provide a proof.


[b]p5.[/b] Point $P$ is an arbitrary point inside triangle $ABC$. Points $X$, $Y$ , and $Z$ are constructed to make segments $PX$, $PY$ , and $PZ$ perpendicular to $AB$, $BC$, and $CA$, respectively. Let $x$, $y$, and $z$ denote the lengths of the segments $PX$, $PY$ , and $PZ$, respectively.
a) If triangle $ABC$ is an equilateral triangle, prove that $x + y + z$ does not change regardless of the location of $P$ inside triangle ABC.
b) If triangle $ABC$ is an isosceles triangle with $|BC| = |CA|$, prove that $x + y + z$ does not change when $P$ moves along a line parallel to $AB$.
c) Now suppose that triangle $ABC$ is scalene (i.e., $|AB|$, $|BC|$, and $|CA|$ are all different). Prove that there exists a line for which $x+y+z$ does not change when $P$ moves along this line.


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
1. To determine the number of plates \( N \) for which it becomes at least as cheap to use the second caterer as the first, we need to compare the costs from both caterers.

   - The first caterer charges $50 per plate.
   - The second caterer charges $500 set-up fee plus $40 per plate for up to and including 61 plates, and $2500 \log_{10}\left(\frac{p}{4}\right)$ for \( p > 61 \) plates.

2. For \( p \leq 61 \):
   - The cost for the first caterer is \( 50p \).
   - The cost for the second caterer is \( 500 + 40p \).

   We set these two costs equal to find the break-even point:
   \[
   50p = 500 + 40p
   \]
   Subtract \( 40p \) from both sides:
   \[
   10p = 500
   \]
   Divide by 10:
   \[
   p = 50
   \]

   Therefore, for \( p = 50 \), the costs are equal. For \( p > 50 \), the second caterer becomes cheaper. Thus, \( N = 51 \).

The final answer is \( \boxed{ N = 51 } \).