Where is Tony's cheese? The bottom right corner of a 3x3 grid :0 We’re doing weekly posts of problems until the start of the in-person tournament on April 13th (Sign up [b][url=https://www.stanfordmathtournament.com/competitions/smt-2024]here[/url][/b]!) Registrations for our synchronous Online tournament are also open [b][url=https://www.stanfordmathtournament.com/competitions/smt-2024-online]here[/url][/b]. Also if you’re coming in-person and want to get some pre-competition action, submit your solutions to PoTW [url=https://forms.gle/2FpDoHFeVHg2aqWS6]here[/url], and the top 3 people with the most correct solutions will receive surprise prize at the tournament :)

[b]Easy:[/b] Consider the curves $x^2 + y^2 = 1$ and $2x^2 + 2xy + y^2 - 2x -2y = 0$. These curves intersect at two points, one of which is $(1, 0)$. Find the other one. 

[b]Hard:[/b] Tony the mouse starts in the top left corner of a $3x3$ grid. After each second, he randomly moves to an adjacent square with equal probability. What is the probability he reaches the cheese in the bottom right corner before he reaches the mousetrap in the center.

Again feel free to reach out to us at [url=mailto:stanford.math.tournament@gmail.com]stanford.math.tournament@gmail.com[/url] with any questions, or ask them in this thread!

Best,

[url=https://www.stanfordmathtournament.com/our-team]The SMT Team[/url]

To solve the hard problem, we need to calculate the probability that Tony the mouse reaches the cheese in the bottom right corner of a \(3 \times 3\) grid before he reaches the mousetrap in the center. We will label the points on the grid as follows:

\[
\begin{array}{ccc}
1 & 2 & 3 \\
4 & T & 5 \\
6 & 7 & C \\
\end{array}
\]

Here, \(T\) represents the mousetrap and \(C\) represents the cheese. We will denote \(p(n)\) as the probability that Tony reaches the cheese before the trap starting from point \(n\).

Given:
- \(p(T) = 0\) (probability of reaching the cheese from the trap is zero)
- \(p(C) = 1\) (probability of reaching the cheese from the cheese is one)

We need to set up linear equations for the probabilities \(p(1), p(2), p(3), p(4), p(5), p(6), p(7)\).

1. **Setting up the equations:**

   - For point 1:
     \[
     p(1) = \frac{1}{2} p(2) + \frac{1}{2} p(4)
     \]

   - For point 2:
     \[
     p(2) = \frac{1}{3} p(1) + \frac{1}{3} p(3) + \frac{1}{3} p(T)
     \]
     Since \(p(T) = 0\), this simplifies to:
     \[
     p(2) = \frac{1}{3} p(1) + \frac{1}{3} p(3)
     \]

   - For point 3:
     \[
     p(3) = \frac{1}{2} p(2) + \frac{1}{2} p(5)
     \]

   - For point 4:
     \[
     p(4) = \frac{1}{3} p(1) + \frac{1}{3} p(5) + \frac{1}{3} p(6)
     \]

   - For point 5:
     \[
     p(5) = \frac{1}{4} p(3) + \frac{1}{4} p(4) + \frac{1}{4} p(7) + \frac{1}{4} p(T)
     \]
     Since \(p(T) = 0\), this simplifies to:
     \[
     p(5) = \frac{1}{4} p(3) + \frac{1}{4} p(4) + \frac{1}{4} p(7)
     \]

   - For point 6:
     \[
     p(6) = \frac{1}{2} p(4) + \frac{1}{2} p(7)
     \]

   - For point 7:
     \[
     p(7) = \frac{1}{3} p(5) + \frac{1}{3} p(6) + \frac{1}{3} p(C)
     \]
     Since \(p(C) = 1\), this simplifies to:
     \[
     p(7) = \frac{1}{3} p(5) + \frac{1}{3} p(6) + \frac{1}{3}
     \]

2. **Solving the system of equations:**

   We now have a system of linear equations:
   \[
   \begin{cases}
   p(1) = \frac{1}{2} p(2) + \frac{1}{2} p(4) \\
   p(2) = \frac{1}{3} p(1) + \frac{1}{3} p(3) \\
   p(3) = \frac{1}{2} p(2) + \frac{1}{2} p(5) \\
   p(4) = \frac{1}{3} p(1) + \frac{1}{3} p(5) + \frac{1}{3} p(6) \\
   p(5) = \frac{1}{4} p(3) + \frac{1}{4} p(4) + \frac{1}{4} p(7) \\
   p(6) = \frac{1}{2} p(4) + \frac{1}{2} p(7) \\
   p(7) = \frac{1}{3} p(5) + \frac{1}{3} p(6) + \frac{1}{3}
   \end{cases}
   \]

   By solving this system, we find:
   \[
   p(1) = \frac{1}{7}
   \]

Conclusion:
\[
\boxed{\frac{1}{7}}
\]