Difference between revisions of "2020 AIME II Problems/Problem 2"

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==Problem==
 
==Problem==
 
Let <math>P</math> be a point chosen uniformly at random in the interior of the unit square with vertices at <math>(0,0), (1,0), (1,1)</math>, and <math>(0,1)</math>. The probability that the slope of the line determined by <math>P</math> and the point <math>\left(\frac58, \frac38 \right)</math> is greater than <math>\frac12</math> can be written as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
 
Let <math>P</math> be a point chosen uniformly at random in the interior of the unit square with vertices at <math>(0,0), (1,0), (1,1)</math>, and <math>(0,1)</math>. The probability that the slope of the line determined by <math>P</math> and the point <math>\left(\frac58, \frac38 \right)</math> is greater than <math>\frac12</math> can be written as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
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==Solution==
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The areas bounded by the unit square and alternately bounded by the lines through <math>\left(\frac{5}{8},\frac{3}{8}\right)</math> that are vertical or have a slope of <math>1/2</math> show where <math>P</math> can be placed to satisfy the condition. One of the areas is a trapezoid with bases <math>1/16</math> and <math>3/8</math> and height <math>5/8</math>. The other area is a trapezoid with bases <math>7/16</math> and <math>5/8</math> and height <math>3/8</math>. Then, <cmath>\frac{\frac{1}{16}+\frac{3}{8}}{2}\cdot\frac{5}{8}+\frac{\frac{7}{16}+\frac{5}{8}}{2}\cdot\frac{3}{8}=\frac{86}{256}=\frac{43}{128}\implies43+128=\boxed{171}</cmath>
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~mn28407
  
 
==Video Solution==
 
==Video Solution==

Revision as of 03:05, 8 June 2020

Problem

Let $P$ be a point chosen uniformly at random in the interior of the unit square with vertices at $(0,0), (1,0), (1,1)$, and $(0,1)$. The probability that the slope of the line determined by $P$ and the point $\left(\frac58, \frac38 \right)$ is greater than $\frac12$ can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

The areas bounded by the unit square and alternately bounded by the lines through $\left(\frac{5}{8},\frac{3}{8}\right)$ that are vertical or have a slope of $1/2$ show where $P$ can be placed to satisfy the condition. One of the areas is a trapezoid with bases $1/16$ and $3/8$ and height $5/8$. The other area is a trapezoid with bases $7/16$ and $5/8$ and height $3/8$. Then, \[\frac{\frac{1}{16}+\frac{3}{8}}{2}\cdot\frac{5}{8}+\frac{\frac{7}{16}+\frac{5}{8}}{2}\cdot\frac{3}{8}=\frac{86}{256}=\frac{43}{128}\implies43+128=\boxed{171}\] ~mn28407

Video Solution

https://youtu.be/x0QznvXcwHY?t=190

~IceMatrix

See Also

2020 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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