2023 SSMO Speed Round Problems/Problem 2

Problem

Let $A$ , $B$ , $C$ be independently chosen vertices lying in the square with coordinates $(-1, - 1)$ , $(-1, 1)$ , $(1, -1)$ , and $(1, 1)$ . The probability that the centroid of triangle $ABC$ lies in the first quadrant is $\frac{m}{n}$ for relatively prime positive integers $m$ and $n.$ Find $m+n.$

Solution

Let $A$ have coordinates $(a_1, a_2)$ , $B$ have coordinates $(b_1, b_2)$ , and $C$ have coordinates $(c_1, c_2)$ .

Note that all these coordinates are uniformly distributed between $-1$ and $1$ .

Thus, we want to find the probability that $a_1 + b_1 + c_1 \ge 0$ and $a_2 + b_2 + c_2 \ge 0$ both hold, which are independent events.

If $a+b+c>0$ , then $(-a)+(-b)+(-c) < 0$ . Thus, there exists a bijection between when $a_1 + b_1 + c_1 \ge 0$ and when $a_1 + b_1 + c_1 \le 0$ . (The case of $a_1 + b_1 + c_1 = 0$ occurs with probability $0$ ). so the probability is $\frac{1}{2}$ for the chance $a_1 + b_1 + c_1 \ge 0$ .

Thus, the answer is thus $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$

$1+4=\boxed{5}$

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2023 SSMO Speed Round Problems/Problem 2

Problem

Solution