[b]p1.[/b] Sarah needed a ride home to the farm from town. She telephoned for her father to come and get her with the pickup truck. Being eager to get home, she began walking toward the farm as soon as she hung up the phone. However, her father had to finish milking the cows, so could not leave to get her until fifteen minutes after she called. He drove rapidly to make up for lost time.
They met on the road, turned right around and drove back to the farm at two-thirds of the speed her father drove coming. They got to the farm two hours after she had called. She walked and he drove both ways at constant rates of speed.
How many minutes did she spend walking?


[b]p2.[/b] Let $A = (a,b)$ be any point in a coordinate plane distinct from the origin $O$. Let $M$ be the midpoint of $OA$, and let $P$ be a point such that $MP$ is perpendicular to $OA$ and the lengths $\overline{MP}$ and $\overline{OM}$ are equal. Determine the coordinates $(x,y)$ of $P$ in terms of $a$ and $b$. Give all possible solutions.


[b]p3.[/b] Determine the exact sum of the series
$$\frac{1}{1 \cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \frac{1}{3\cdot 4\cdot 5} +  ... + \frac{1}{98\cdot 99\cdot 100}$$


[b]p4.[/b]  A six pound weight is attached to a four foot nylon cord that is looped over two pegs in the manner shown in the drawing. At $B$ the cord passes through a small loop in its end. The two pegs $A$ and $C$ are one foot apart and are on the same level. When the weight is released the system obtains an equilibrium position. Determine angle $ABC$ for this equilibrium position, and verify your answer. (Your verification should assume that friction and the weight of the cord are both negligible, and that the tension throughout the cord is a constant six pounds.)
[img]https://cdn.artofproblemsolving.com/attachments/a/1/620c59e678185f01ca8743c39423234d5ba04d.png[/img]


[b]p5.[/b] The four corners of a rectangle have the property that when they are taken three at a time, they determine triangles all of which have the same perimeter. We will consider whether a set of five points can have this property.
Let $S = \{p_1, p_2, p_3, p_4, p_5\}$ be a set of five points. For each $i$ and $j$, let $d_{ij}$ denote the distance from $p_i$ to $p_j$. Suppose that $S$ has the property that all triangles with vertices in $S$ have the same perimeter.
(a) Prove that $d$ must be the same for every pair $(i,j)$ with $i \ne j$.
(b) Can such a five-element set be found in three dimensional space? Justify your answer.


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
To find the exact sum of the series 
\[ \frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{3 \cdot 4 \cdot 5} + \cdots + \frac{1}{98 \cdot 99 \cdot 100}, \]
we will use partial fraction decomposition and the telescoping series technique.

1. **Partial Fraction Decomposition:**
   We start by decomposing the general term \(\frac{1}{x(x+1)(x+2)}\) into partial fractions. We assume:
   \[
   \frac{1}{x(x+1)(x+2)} = \frac{A}{x} + \frac{B}{x+1} + \frac{C}{x+2}.
   \]
   Multiplying both sides by \(x(x+1)(x+2)\) gives:
   \[
   1 = A(x+1)(x+2) + Bx(x+2) + Cx(x+1).
   \]
   Expanding and collecting like terms, we get:
   \[
   1 = A(x^2 + 3x + 2) + B(x^2 + 2x) + C(x^2 + x).
   \]
   \[
   1 = (A + B + C)x^2 + (3A + 2B + C)x + 2A.
   \]
   Equating coefficients, we obtain the system of equations:
   \[
   A + B + C = 0,
   \]
   \[
   3A + 2B + C = 0,
   \]
   \[
   2A = 1.
   \]
   Solving these, we find:
   \[
   A = \frac{1}{2},
   \]
   \[
   3 \left(\frac{1}{2}\right) + 2B + C = 0 \implies \frac{3}{2} + 2B + C = 0,
   \]
   \[
   \frac{1}{2} + B + C = 0.
   \]
   Substituting \(A = \frac{1}{2}\) into the equations, we get:
   \[
   B + C = -\frac{1}{2},
   \]
   \[
   \frac{3}{2} + 2B + C = 0 \implies 2B + C = -\frac{3}{2}.
   \]
   Solving these, we find:
   \[
   B = -1, \quad C = \frac{1}{2}.
   \]
   Therefore, the partial fraction decomposition is:
   \[
   \frac{1}{x(x+1)(x+2)} = \frac{1}{2x} - \frac{1}{x+1} + \frac{1}{2(x+2)}.
   \]

2. **Telescoping Series:**
   We rewrite the series using the partial fractions:
   \[
   \sum_{x=1}^{98} \left( \frac{1}{2x} - \frac{1}{x+1} + \frac{1}{2(x+2)} \right).
   \]
   Grouping terms, we get:
   \[
   \sum_{x=1}^{98} \left( \frac{1}{2x} - \frac{1}{2(x+1)} \right) - \sum_{x=1}^{98} \left( \frac{1}{x+1} - \frac{1}{2(x+2)} \right).
   \]
   This can be written as:
   \[
   \frac{1}{2} \left( \sum_{x=1}^{98} \left( \frac{1}{x} - \frac{1}{x+1} \right) \right) - \left( \sum_{x=1}^{98} \left( \frac{1}{x+1} - \frac{1}{x+2} \right) \right).
   \]
   The series telescopes, leaving:
   \[
   \frac{1}{2} \left( 1 - \frac{1}{99} \right) - \left( \frac{1}{2} - \frac{1}{100} \right).
   \]
   Simplifying, we get:
   \[
   \frac{1}{2} \left( 1 - \frac{1}{99} \right) - \left( \frac{1}{2} - \frac{1}{100} \right) = \frac{1}{2} \left( 1 - \frac{1}{99} \right) - \frac{1}{2} + \frac{1}{100}.
   \]
   \[
   = \frac{1}{2} - \frac{1}{198} + \frac{1}{100}.
   \]
   \[
   = \frac{1}{2} - \frac{1}{198} + \frac{1}{100} = \frac{99}{198} - \frac{1}{198} + \frac{1}{100} = \frac{98}{198} + \frac{1}{100}.
   \]
   \[
   = \frac{49}{99} + \frac{1}{100} = \frac{4900 + 99}{9900} = \frac{4999}{9900}.
   \]

The final answer is \(\boxed{\frac{4999}{9900}}\).