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| ==Problem==
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| 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>
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| ==Solution==
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| Let <math>A</math> have coordinates <math>(a_1, a_2)</math>, <math>B</math> have coordinates <math>(b_1, b_2)</math>,
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| and <math>C</math> have coordinates <math>(c_1, c_2)</math>.
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| Note that all these coordinates are uniformly distributed between
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| <math>-1</math> and <math>1</math>.
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| Thus, we want to find the probability that
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| <cmath>
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| a_1 + b_1 + c_1 \ge 0
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| </cmath>
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| and
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| <cmath>
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| a_2 + b_2 + c_2 \ge 0
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| </cmath>
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| both hold, which are independent events.
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| If <math>a+b+c>0</math>, then <math>(-a)+(-b)+(-c) < 0</math>.
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| Thus, there exists a bijection between when
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| <math>a_1 + b_1 + c_1 \ge 0</math> and when <math>a_1 + b_1 + c_1 \le 0</math>.
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| (The case of <math>a_1 + b_1 + c_1 = 0</math> occurs with probability
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| <math>0</math>).
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| so the probability is <math>\frac{1}{2}</math> for the chance
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| <math>a_1 + b_1 + c_1 \ge 0</math>.
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| Thus, the answer is thus <math>\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}</math>
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