[b]p1.[/b] Let $A$ be the point $(-1, 0)$, $B$ be the point $(0, 1)$ and $C$ be the point $(1, 0)$ on the $xy$-plane. Assume that $P(x, y)$ is a point on the $xy$-plane that satisfies the following condition $$d_1 \cdot d_2 = (d_3)^2,$$
where $d_1$ is the distance from $P$ to the line $AB$, $d_2$ is the distance from $P$ to the line $BC$, and $d_3$ is the distance from $P$ to the line $AC$. Find the equation(s) that must be satisfied by the point $P(x, y)$.


[b]p2.[/b] On Day $1$, Peter sends an email to a female friend and a male friend with the following instructions:
$\bullet$ If you’re a male, send this email to $2$ female friends tomorrow, including the instructions.
$\bullet$ If you’re a female, send this email to $1$ male friend tomorrow, including the instructions.
Assuming that everyone checks their email daily and follows the instructions, how many emails will be sent from Day $1$ to Day $365$ (inclusive)?


[b]p3.[/b] For every rational number $\frac{a}{b}$ in the interval $(0, 1]$, consider the interval of length $\frac{1}{2b^2}$ with $\frac{a}{b}$ as the center, that is, the interval $\left( \frac{a}{b}- \frac{1}{2b^2}, \frac{a}{b}+\frac{1}{2b^2}\right)$ . Show that $\frac{\sqrt2}{2}$ is not contained in any of these intervals.



[b]p4.[/b] Let $a$ and $b$ be real numbers such that $0 < b < a < 1$ with the property that
$$\log_a x + \log_b x = 4 \log_{ab} x - \left(\log_b (ab^{-1} - 1)\right)\left(\log_a (ab^{-1} - 1) + 2 log_a ab^{-1} \right)$$
for some positive real number $x \ne 1$. Find the value of $\frac{a}{b}$.


[b]p5.[/b] Find the largest positive constant $\lambda$ such that $$\lambda a^2 b^2 (a - b)^2 \le (a^2 - ab + b^2)^3$$ is true for all real numbers $a$ and $b$.


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
1. First, we need to find the equations of the lines \(AB\), \(BC\), and \(AC\).

   - The line \(AB\) passes through points \((-1, 0)\) and \((0, 1)\). The slope of \(AB\) is:
     \[
     m_{AB} = \frac{1 - 0}{0 - (-1)} = 1
     \]
     Thus, the equation of the line \(AB\) is:
     \[
     y = x + 1 \implies x - y + 1 = 0
     \]

   - The line \(BC\) passes through points \((0, 1)\) and \((1, 0)\). The slope of \(BC\) is:
     \[
     m_{BC} = \frac{0 - 1}{1 - 0} = -1
     \]
     Thus, the equation of the line \(BC\) is:
     \[
     y = -x + 1 \implies x + y - 1 = 0
     \]

   - The line \(AC\) passes through points \((-1, 0)\) and \((1, 0)\). The slope of \(AC\) is:
     \[
     m_{AC} = \frac{0 - 0}{1 - (-1)} = 0
     \]
     Thus, the equation of the line \(AC\) is:
     \[
     y = 0
     \]

2. Next, we need to find the distances \(d_1\), \(d_2\), and \(d_3\) from point \(P(x, y)\) to the lines \(AB\), \(BC\), and \(AC\) respectively.

   - The distance from \(P(x, y)\) to the line \(AB\) is:
     \[
     d_1 = \frac{|x - y + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|x - y + 1|}{\sqrt{2}}
     \]

   - The distance from \(P(x, y)\) to the line \(BC\) is:
     \[
     d_2 = \frac{|x + y - 1|}{\sqrt{1^2 + 1^2}} = \frac{|x + y - 1|}{\sqrt{2}}
     \]

   - The distance from \(P(x, y)\) to the line \(AC\) is:
     \[
     d_3 = |y|
     \]

3. According to the given condition, we have:
   \[
   d_1 \cdot d_2 = (d_3)^2
   \]
   Substituting the distances, we get:
   \[
   \left(\frac{|x - y + 1|}{\sqrt{2}}\right) \left(\frac{|x + y - 1|}{\sqrt{2}}\right) = y^2
   \]
   Simplifying this equation, we obtain:
   \[
   \frac{|x - y + 1| \cdot |x + y - 1|}{2} = y^2
   \]
   Multiplying both sides by 2, we get:
   \[
   |x - y + 1| \cdot |x + y - 1| = 2y^2
   \]

The final answer is \(\boxed{|x - y + 1| \cdot |x + y - 1| = 2y^2}\)