Difference between revisions of "2023 SSMO Speed Round Problems/Problem 2"

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Thus, the answer is thus <math>\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}</math>
 
Thus, the answer is thus <math>\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}</math>
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<math>1+4=\boxed{5}</math>
 
<math>1+4=\boxed{5}</math>

Latest revision as of 12:24, 5 July 2023

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 $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$

$1+4=\boxed{5}$