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Difference between revisions of "2022 SSMO Speed Round Problems/Problem 2"

(Created page with "==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)</mat...")
 
 
(2 intermediate revisions by the same user not shown)
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==Problem==
 
==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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A bag is big enough to hold exactly 6 large pencils, 12 medium pencils, or 30 small pencils, with no space left over. Given that there is 1 large pencil and 3 medium pencils currently in the bag, what is the maximum number of small pencils that may be added to the bag? Note that there may still be space left over in the bag.
  
 
==Solution==
 
==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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Each large pencil takes up <math>\tfrac16</math>th of the bag, each medium pencil takes up <math>\tfrac1{12}</math>th of the bag, and each small pencil takes up <math>\tfrac1{30}</math>th of the bag. So, our answer is <cmath>\left\lfloor\frac{1-\frac16-3\cdot\frac1{12}}{\frac1{30}}\right\rfloor=\left\lfloor\frac{\frac7{12}}{\frac1{30}}\right\rfloor=\left\lfloor17.5\right\rfloor=\boxed{17}.</cmath>
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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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~pinkpig

Latest revision as of 16:52, 13 September 2025

Problem

A bag is big enough to hold exactly 6 large pencils, 12 medium pencils, or 30 small pencils, with no space left over. Given that there is 1 large pencil and 3 medium pencils currently in the bag, what is the maximum number of small pencils that may be added to the bag? Note that there may still be space left over in the bag.

Solution

Each large pencil takes up $\tfrac16$th of the bag, each medium pencil takes up $\tfrac1{12}$th of the bag, and each small pencil takes up $\tfrac1{30}$th of the bag. So, our answer is \[\left\lfloor\frac{1-\frac16-3\cdot\frac1{12}}{\frac1{30}}\right\rfloor=\left\lfloor\frac{\frac7{12}}{\frac1{30}}\right\rfloor=\left\lfloor17.5\right\rfloor=\boxed{17}.\]

~pinkpig

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