View source for 2023 SSMO Speed Round Problems/Problem 2

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

==Solution==
Let <math>A</math> have coordinates <math>(a_1, a_2)</math>, <math>B</math> have coordinates <math>(b_1, b_2)</math>,
and <math>C</math> have coordinates <math>(c_1, c_2)</math>.

Note that all these coordinates are uniformly distributed between
<math>-1</math> and <math>1</math>. 

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

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

Thus, the answer is thus <math>\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}</math>