[b]p1.[/b] (a) Find three positive integers $a, b, c$ whose sum is $407$, and whose product (when written in base $10$) ends in six $0$'s.
(b) Prove that there do NOT exist positive integers $a, b, c$ whose sum is $407$ and whose product ends in seven $0$'s.


[b]p2.[/b] Three circles, each of radius $r$, are placed on a plane so that the center of each circle lies on a point of intersection of the other two circles. The region $R$ consists of all points inside or on at least one of these three circles. Find the area of $R$.


[b]p3.[/b] Let $f_1(x) = a_1x^2+b_1x+c_1$, $f_2(x) = a_2x^2+b_2x+c_2$ and $f_3(x) = a_3x^2+b_3x+c_3$ be the equations of three parabolas such that $a_1 > a_2 > a-3$. Prove that if each pair of parabolas intersects in exactly one point, then all three parabolas intersect in a common point.


[b]p4.[/b] Gigafirm is a large corporation with many employees.
(a) Show that the number of employees with an odd number of acquaintances is even.
(b) Suppose that each employee with an even number of acquaintances sends a letter to each of these acquaintances. Each employee with an odd number of acquaintances sends a letter to each non-acquaintance. So far, Leslie has received $99$ letters. Prove that Leslie will receive at least one more letter.
(Notes: "acquaintance" and "non-acquaintance" refer to employees of Gigaform. If $A$ is acquainted with $B$, then $B$ is acquainted with $A$. However, no one is acquainted with himself.)


[b]p5.[/b] (a) Prove that for every positive integer $N$, if $A$ is a subset of the numbers $\{1, 2, ...,N\}$ and $A$ has size at least $2N/3 + 1$, then $A$ contains a three-term arithmetic progression (i.e., there are positive integers $a$ and $b$ so that all three of the numbers $a$,$a + b$, and $a + 2b$ are elements of $A$).
(b) Show that if $A$ is a subset of $\{1, 2, ..., 3500\}$ and $A$ has size at least $2003$, then $A$ contains a three-term arithmetic progression.


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
**p2.** Three circles, each of radius \( r \), are placed on a plane so that the center of each circle lies on a point of intersection of the other two circles. The region \( R \) consists of all points inside or on at least one of these three circles. Find the area of \( R \).

1. **Understanding the Configuration:**
   - Each circle has radius \( r \).
   - The centers of the circles form an equilateral triangle with side length \( 2r \).
   - Each circle passes through the centers of the other two circles.

2. **Area Calculation:**
   - The area of each circle is \( \pi r^2 \).
   - Since the circles overlap, we need to subtract the areas of the overlapping regions.

3. **Overlapping Regions:**
   - The overlapping regions are segments of the circles.
   - Each pair of circles intersects at two points, forming a lens-shaped region.
   - The area of the lens-shaped region can be calculated by subtracting the area of the equilateral triangle from the sum of the areas of the circular segments.

4. **Area of the Equilateral Triangle:**
   - The side length of the equilateral triangle is \( 2r \).
   - The height of the equilateral triangle is \( \sqrt{3}r \).
   - The area of the equilateral triangle is \( \frac{\sqrt{3}}{4} \times (2r)^2 = \sqrt{3}r^2 \).

5. **Area of the Circular Segments:**
   - Each circular segment is a sector of the circle minus the area of the equilateral triangle.
   - The angle of each sector is \( 60^\circ \) or \( \frac{\pi}{3} \) radians.
   - The area of each sector is \( \frac{1}{6} \pi r^2 \).
   - The area of the circular segments is \( 3 \times \left( \frac{1}{6} \pi r^2 - \frac{1}{2} \sqrt{3} r^2 \right) = \frac{1}{2} \pi r^2 - \frac{3}{2} \sqrt{3} r^2 \).

6. **Total Area of Region \( R \):**
   - The total area of the three circles is \( 3 \pi r^2 \).
   - Subtract the area of the overlapping regions: \( 3 \pi r^2 - \left( \frac{1}{2} \pi r^2 - \frac{3}{2} \sqrt{3} r^2 \right) = \frac{3}{2} \pi r^2 + \sqrt{3} r^2 \).

The final answer is \( \boxed{r^2\left(\sqrt{3} + \frac{3}{2} \pi\right)} \).