[b]p1.[/b] One hundred hobbits sit in a circle. The hobbits realize that whenever a hobbit and his two neighbors add up their total rubles, the sum is always $2007$. Prove that each hobbit has $669$ rubles.


[b]p2.[/b] There was a young lady named Chris, 
Who, when asked her age, answered this:
"Two thirds of its square
Is a cube, I declare."
Now what was the age of the miss?
(a) Find the smallest possible age for Chris. You must justify your answer.
(Note: ages are positive integers; "cube" means the cube of a positive integer.)
(b) Find the second smallest possible age for Chris. You must justify your answer. 
(Ignore the word "young.")


[b]p3.[/b] Show that $$\sum_{n=1}^{2007}\frac{1}{n^3+3n^2+2n}<\frac14$$


[b]p4.[/b] (a) Show that a triangle $ABC$ is isosceles if and only if there are two distinct points $P_1$ and $P_2$ on side $BC$ such that the sum of the distances from $P_1$ to the sides $AB$ and $AC$ equals the sum of the distances from $P_2$ to the sides $AB$ and $AC$.
(b) A convex quadrilateral is such that the sum of the distances of any interior point to its four sides is constant. Prove that the quadrilateral is a parallelogram.
(Note: "distance to a side" means the shortest distance to the line obtained by extending the side.)


[b]p5.[/b] Each point in the plane is colored either red or green. Let $ABC$ be a fixed triangle. Prove that there is a triangle $DEF$ in the plane such that $DEF$ is similar to $ABC$ and the vertices of $DEF$ all have the same color.



PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
To solve the problem, we need to show that the sum of the series 

\[
\sum_{n=1}^{2007}\frac{1}{n^3+3n^2+2n}
\]

is less than \(\frac{1}{4}\).

1. **Factor the Denominator:**
   First, we factor the denominator \(n^3 + 3n^2 + 2n\):
   \[
   n^3 + 3n^2 + 2n = n(n+1)(n+2)
   \]
   Thus, we can rewrite the term as:
   \[
   \frac{1}{n^3 + 3n^2 + 2n} = \frac{1}{n(n+1)(n+2)}
   \]

2. **Partial Fraction Decomposition:**
   We decompose \(\frac{1}{n(n+1)(n+2)}\) into partial fractions:
   \[
   \frac{1}{n(n+1)(n+2)} = \frac{A}{n} + \frac{B}{n+1} + \frac{C}{n+2}
   \]
   Multiplying both sides by \(n(n+1)(n+2)\) and equating coefficients, we get:
   \[
   1 = A(n+1)(n+2) + Bn(n+2) + Cn(n+1)
   \]
   To find \(A\), \(B\), and \(C\), we substitute convenient values for \(n\):
   - For \(n = 0\):
     \[
     1 = A(1)(2) \implies A = \frac{1}{2}
     \]
   - For \(n = -1\):
     \[
     1 = B(-1)(1) \implies B = -1
     \]
   - For \(n = -2\):
     \[
     1 = C(-2)(-1) \implies C = \frac{1}{2}
     \]

   Therefore, the partial fraction decomposition is:
   \[
   \frac{1}{n(n+1)(n+2)} = \frac{1}{2n} - \frac{1}{n+1} + \frac{1}{2(n+2)}
   \]

3. **Sum the Series:**
   We now sum the series:
   \[
   \sum_{n=1}^{2007} \left( \frac{1}{2n} - \frac{1}{n+1} + \frac{1}{2(n+2)} \right)
   \]
   This series has a telescoping effect. Let's write out the first few terms to see the pattern:
   \[
   \left( \frac{1}{2 \cdot 1} - \frac{1}{2} + \frac{1}{2 \cdot 3} \right) + \left( \frac{1}{2 \cdot 2} - \frac{1}{3} + \frac{1}{2 \cdot 4} \right) + \cdots + \left( \frac{1}{2 \cdot 2007} - \frac{1}{2008} + \frac{1}{2 \cdot 2009} \right)
   \]

   Notice that many terms cancel out. Specifically, each \(-\frac{1}{k+1}\) term cancels with a \(\frac{1}{2(k+1)}\) term from the next fraction. After cancellation, we are left with:
   \[
   \frac{1}{2} - \frac{1}{2} + \frac{1}{6} + \frac{1}{4} - \frac{1}{3} + \frac{1}{4} + \frac{1}{6} - \frac{1}{4} + \cdots + \frac{1}{2 \cdot 2009} - \frac{1}{2 \cdot 2008}
   \]

4. **Estimate the Remaining Terms:**
   The remaining terms are:
   \[
   \frac{1}{4} + \frac{1}{2 \cdot 2009} - \frac{1}{2 \cdot 2008}
   \]
   Since \(\frac{1}{2 \cdot 2009} - \frac{1}{2 \cdot 2008}\) is very small, we can approximate the sum:
   \[
   \frac{1}{4} + \text{(small positive value)} - \text{(small positive value)} \approx \frac{1}{4}
   \]

   However, the exact sum is slightly less than \(\frac{1}{4}\).

Conclusion:
\[
\sum_{n=1}^{2007}\frac{1}{n^3+3n^2+2n} < \frac{1}{4}
\]

The final answer is \( \boxed{ \sum_{n=1}^{2007}\frac{1}{n^3+3n^2+2n} < \frac{1}{4} } \)