p1. Juan was born before the year $2000$. On August $25$, $2001$ he is as old as the sum of the digits of the year of his birth. Determine your date of birth and justify that it is the only possible solution.


p2. Triangle $ABC$ is right isosceles. The figure below shows two basic ways to inscribe a square in it. The square $ADEF$ is known to have area $2250$. Determine the area of the square $GHIJ$.
[img]https://cdn.artofproblemsolving.com/attachments/9/3/84d0cda6e109aaf01fb2f3de4d1933f0f16f82.jpg[/img]

p3. Given $9$ people, show that there exists a value of $n$ such that with people you can form $n$ groups of  $3$, so that each pair of people is in exactly one of these groups. Also, show one of the possible formations of the groups.


p4. Show that there are no positive integers $a, b, c$ such that  $a^2 + b^2 = 8c + 6$.


p5. In a circular roulette the numbers from $ 1$ to $36$ are randomly placed. Show that necessarily there must be $3$ consecutive numbers whose sum is at least $55$.


p6. Let $ABCD$ be a quadrilateral inscribed in a circle of radius $ r$, such that its diagonals are perpendicular at point $E$. Prove that  $$AE^2 + BE^2 + CE^2 + DE^2 = 4r^2.$$


p7. In a rectangular board of $m$ rows and $n$ columns, place  $ 1$ or $0$ in each of the $mn$ cells, so that the numbers of each row have the same sum $f$ and those of each column have the same sum $c$. 
For a board with $m = 15$, $n = 10$, and for $f = 4$, determine the value of  $c$ that makes this assignment possible and show a way to do this assignment.

PS. Juniors p2, p6, were posted also as [url=https://artofproblemsolving.com/community/c4h2691996p23368695]Seniors[/url] variation of p2,  part of p6.
1. Let's denote: \(AB = AC = b\), \(BC = c\), \(h\) as the altitude of \(\triangle ABC\), and the side lengths of squares \(ADEF\) and \(GHIJ\) as \(a_1\) and \(a\) respectively. The areas of \(\triangle ABC\), \(ADEF\), and \(GHIJ\) are \(A_{ABC}\), \(A_1\), and \(A_2\) respectively.

2. **Claim 1:** \(c = b\sqrt{2}\)

   **Proof:**
   By the Pythagorean Theorem in \(\triangle ABC\):
   \[
   b^2 + b^2 = c^2 \implies 2b^2 = c^2 \implies c = b\sqrt{2}
   \]
   Q.E.D.

3. **Claim 2:** \(h = \frac{b^2}{c}\)

   **Proof:**
   The area of \(\triangle ABC\) can be expressed as:
   \[
   A_{ABC} = \frac{AB \cdot AC}{2} = \frac{BC \cdot h}{2} \implies b^2 = c \cdot h \implies h = \frac{b^2}{c}
   \]
   Q.E.D.

4. **Claim 3:** \(b = 2a_1\)

   **Proof:**
   Since \(\triangle ABC\) is isosceles and right-angled, the altitude from \(A\) to \(BC\) bisects \(BC\). Therefore, the side length of the square \(ADEF\) inscribed in \(\triangle ABC\) is half the length of \(AB\) or \(AC\):
   \[
   b = 2a_1
   \]
   Q.E.D.

5. Given that the area of square \(ADEF\) is \(2250\):
   \[
   A_1 = a_1^2 = 2250 \implies a_1 = \sqrt{2250}
   \]

6. To find the area of square \(GHIJ\), we note that \(\triangle AGJ \sim \triangle ABC\). Therefore, the ratio of the sides of the squares to the sides of the triangle is the same:
   \[
   \frac{h}{c} = \frac{h - a}{a} \implies \frac{h}{a} = \frac{h}{c} + 1
   \]
   Substituting \(h = \frac{b^2}{c}\) and \(c = b\sqrt{2}\):
   \[
   \frac{h}{a} = \frac{\frac{b^2}{c}}{a} + 1 = \frac{b^2}{ac} + 1 = \frac{b^2 + c^2}{c^2}
   \]
   \[
   \frac{b^2}{ac} = \frac{b^2 + c^2}{c^2} = \frac{b^2 + (b\sqrt{2})^2}{(b\sqrt{2})^2} = \frac{b^2 + 2b^2}{2b^2} = \frac{3b^2}{2b^2} = \frac{3}{2}
   \]
   \[
   3ac = 2b^2 \implies a = \frac{2}{3} \cdot \frac{b^2}{c}
   \]
   \[
   A_2 = a^2 = \left(\frac{2}{3} \cdot \frac{b^2}{c}\right)^2 = \left(\frac{2}{3} \cdot \frac{b^2}{b\sqrt{2}}\right)^2 = \left(\frac{2}{3} \cdot \frac{b}{\sqrt{2}}\right)^2 = \frac{4}{9} \cdot \frac{b^2}{2} = \frac{2b^2}{9}
   \]
   \[
   A_2 = \frac{2b^2}{9} = \frac{2 \cdot (2a_1)^2}{9} = \frac{8}{9}a_1^2 = \frac{8}{9} \cdot 2250 = \boxed{2000}
   \]