[b]p1.[/b] The edges of a tetrahedron are all tangent to a sphere. Prove that the sum of the lengths of any pair of opposite edges equals the sum of the lengths of any other pair of opposite edges. (Two edges of a tetrahedron are said to be opposite if they do not have a vertex in common.)


[b]p2.[/b] Find the simplest formula possible for the product of the following $2n - 2$ factors: $$\left(1+\frac12 \right),\left(1-\frac12 \right), \left(1+\frac13 \right) , \left(1-\frac13 \right),...,\left(1+\frac{1}{n} \right), \left(1-\frac{1}{n} \right)$$. Prove that your formula is correct.



[b]p3.[/b] Solve $$\frac{(x + 1)^2+1}{x + 1} + \frac{(x + 4)^2+4}{x + 4}=\frac{(x + 2)^2+2}{x + 2}+\frac{(x + 3)^2+3}{x + 3}$$


[b]p4.[/b] Triangle $ABC$ is inscribed in a circle, $BD$ is tangent to this circle and $CD$ is perpendicular to $BD$. $BH$ is the altitude from $B$ to $AC$. Prove that the line $DH$ is parallel to $AB$.
[img]https://cdn.artofproblemsolving.com/attachments/e/9/4d0b136dca4a9b68104f00300951837adef84c.png[/img]


[b]p5.[/b] Consider the picture below as a section of a city street map. There are several paths from $A$ to $B$, and if one always walks along the street, the shortest paths are $15$ blocks in length. Find the number of paths of this length between $A$ and $B$.
[img]https://cdn.artofproblemsolving.com/attachments/8/d/60c426ea71db98775399cfa5ea80e94d2ea9d2.png[/img]

[b]p6.[/b] A [u]finite [/u] [u]graph [/u] is a set of points, called [u]vertices[/u], together with a set of arcs, called [u]edges[/u]. Each edge connects two of the vertices (it is not necessary that every pair of vertices be connected by an edge). The [u]order [/u] of a vertex in a finite graph is the number of edges attached to that vertex.
[u]Example[/u]
The figure at the right is a finite graph with $4$ vertices and $7$ edges. [img]https://cdn.artofproblemsolving.com/attachments/5/9/84d479c5dbd0a6f61a66970e46ab15830d8fba.png[/img]
One vertex has order $5$ and the other vertices order $3$.

Define a finite graph to be [u]heterogeneous [/u] if no two vertices have the same order.
Prove that no graph with two or more vertices is heterogeneous.



PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
To solve the equation 

\[
\frac{(x + 1)^2+1}{x + 1} + \frac{(x + 4)^2+4}{x + 4} = \frac{(x + 2)^2+2}{x + 2} + \frac{(x + 3)^2+3}{x + 3},
\]

we start by simplifying each term individually.

1. Simplify the left-hand side:
   \[
   \frac{(x + 1)^2 + 1}{x + 1} = \frac{x^2 + 2x + 1 + 1}{x + 1} = \frac{x^2 + 2x + 2}{x + 1} = x + 1 + \frac{1}{x + 1},
   \]
   \[
   \frac{(x + 4)^2 + 4}{x + 4} = \frac{x^2 + 8x + 16 + 4}{x + 4} = \frac{x^2 + 8x + 20}{x + 4} = x + 4 + \frac{4}{x + 4}.
   \]

2. Simplify the right-hand side:
   \[
   \frac{(x + 2)^2 + 2}{x + 2} = \frac{x^2 + 4x + 4 + 2}{x + 2} = \frac{x^2 + 4x + 6}{x + 2} = x + 2 + \frac{2}{x + 2},
   \]
   \[
   \frac{(x + 3)^2 + 3}{x + 3} = \frac{x^2 + 6x + 9 + 3}{x + 3} = \frac{x^2 + 6x + 12}{x + 3} = x + 3 + \frac{3}{x + 3}.
   \]

3. Substitute the simplified terms back into the original equation:
   \[
   x + 1 + \frac{1}{x + 1} + x + 4 + \frac{4}{x + 4} = x + 2 + \frac{2}{x + 2} + x + 3 + \frac{3}{x + 3}.
   \]

4. Combine like terms:
   \[
   2x + 5 + \frac{1}{x + 1} + \frac{4}{x + 4} = 2x + 5 + \frac{2}{x + 2} + \frac{3}{x + 3}.
   \]

5. Subtract \(2x + 5\) from both sides:
   \[
   \frac{1}{x + 1} + \frac{4}{x + 4} = \frac{2}{x + 2} + \frac{3}{x + 3}.
   \]

6. Rearrange the equation:
   \[
   \frac{4}{x + 4} - \frac{3}{x + 3} = \frac{2}{x + 2} - \frac{1}{x + 1}.
   \]

7. Combine the fractions on each side:
   \[
   \frac{4(x + 3) - 3(x + 4)}{(x + 4)(x + 3)} = \frac{2(x + 1) - 1(x + 2)}{(x + 2)(x + 1)},
   \]
   \[
   \frac{4x + 12 - 3x - 12}{(x + 4)(x + 3)} = \frac{2x + 2 - x - 2}{(x + 2)(x + 1)},
   \]
   \[
   \frac{x}{(x + 4)(x + 3)} = \frac{x}{(x + 2)(x + 1)}.
   \]

8. Since \(x \neq 0\), we can cancel \(x\) from both sides:
   \[
   \frac{1}{(x + 4)(x + 3)} = \frac{1}{(x + 2)(x + 1)}.
   \]

9. Cross-multiply to solve for \(x\):
   \[
   (x + 4)(x + 3) = (x + 2)(x + 1),
   \]
   \[
   x^2 + 7x + 12 = x^2 + 3x + 2,
   \]
   \[
   7x + 12 = 3x + 2,
   \]
   \[
   4x = -10,
   \]
   \[
   x = -\frac{5}{2}.
   \]

10. Verify if \(x = 0\) is also a solution:
    \[
    \frac{(0 + 1)^2 + 1}{0 + 1} + \frac{(0 + 4)^2 + 4}{0 + 4} = \frac{(0 + 2)^2 + 2}{0 + 2} + \frac{(0 + 3)^2 + 3}{0 + 3},
    \]
    \[
    \frac{1 + 1}{1} + \frac{16 + 4}{4} = \frac{4 + 2}{2} + \frac{9 + 3}{3},
    \]
    \[
    2 + 5 = 3 + 4,
    \]
    \[
    7 = 7.
    \]
    Thus, \(x = 0\) is also a solution.

The final answer is \(\boxed{x = 0 \text{ and } x = -\frac{5}{2}}\).