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

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Please stop your intentions to discuss the AOIME.
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==Problem==
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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 or equal to <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
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==Solution 2 (Official MAA)==
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The line through the fixed point <math>\left(\frac58,\frac38\right)</math> with slope <math>\frac12</math> has equation <math>y=\frac12 x + \frac1{16}</math>. The slope between <math>P</math> and the fixed point exceeds <math>\frac12</math> if <math>P</math> falls within the shaded region in the diagram below consisting of two trapezoids with area
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<cmath>\frac{\frac1{16}+\frac38}2\cdot\frac58 + \frac{\frac58+\frac7{16}}2\cdot\frac38 = \frac{43}{128}.</cmath>Because the entire square has area <math>1,</math> the required probability is <math>\frac{43}{128}</math>. The requested sum is <math>43+128 = 171</math>.
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<asy>
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defaultpen(fontsize(8pt));
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unitsize(4cm);
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pair A = (0,0);
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pair B = (1,0);
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pair C = (1,1);
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pair D = (0,1);
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pair F = (0, 1/16);
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pair G = (1, 9/16);
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pair H = (5/8, 0);
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pair J = (5/8, 1);
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pair K = IP(H--J, F--G);
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pair P = (13/16, 12/16);
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pair Q = extension(P,K,A,B);
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pair R = extension(K,P,C,D);
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draw(A--B--C--D--cycle);
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label("$(0,0)$", A, SW);
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label("$(1,0)$", B, SE);
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label("$(1,1)$", C, E);
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label("$(0,1)$", D, W);
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filldraw(A--H--K--F--cycle, lightgray);
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filldraw(K--G--C--J--cycle, lightgray);
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dot(K);
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dot("$P$", P, W);
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draw(Q -- R, dashed);
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label("$\frac 38$", H--K, E);
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label("$\frac 58$", K--J, W);
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label("$\frac 7{16}$", G--C, E);
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label("$\frac 38$", C--J, N);
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label("$\frac 1{16}$", A--F, dir(160));
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</asy>
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==Video Solution==
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https://youtu.be/x0QznvXcwHY?t=190
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~IceMatrix
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==Video Solution 2==
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https://youtu.be/VBwlVbM0GRw
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~avn
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==See Also==
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{{AIME box|year=2020|n=II|num-b=1|num-a=3}}
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[[Category:Intermediate Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 16:24, 24 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 or equal to $\frac12$ 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

Solution 2 (Official MAA)

The line through the fixed point $\left(\frac58,\frac38\right)$ with slope $\frac12$ has equation $y=\frac12 x + \frac1{16}$. The slope between $P$ and the fixed point exceeds $\frac12$ if $P$ falls within the shaded region in the diagram below consisting of two trapezoids with area \[\frac{\frac1{16}+\frac38}2\cdot\frac58 + \frac{\frac58+\frac7{16}}2\cdot\frac38 = \frac{43}{128}.\]Because the entire square has area $1,$ the required probability is $\frac{43}{128}$. The requested sum is $43+128 = 171$. [asy] defaultpen(fontsize(8pt)); unitsize(4cm); pair A = (0,0);  pair B = (1,0);  pair C = (1,1);  pair D = (0,1);  pair F = (0, 1/16);  pair G = (1, 9/16);  pair H = (5/8, 0);  pair J = (5/8, 1);  pair K = IP(H--J, F--G);   pair P = (13/16, 12/16);  pair Q = extension(P,K,A,B);  pair R = extension(K,P,C,D);  draw(A--B--C--D--cycle);   label("$(0,0)$", A, SW);  label("$(1,0)$", B, SE);  label("$(1,1)$", C, E);  label("$(0,1)$", D, W);  filldraw(A--H--K--F--cycle, lightgray);  filldraw(K--G--C--J--cycle, lightgray);  dot(K);  dot("$P$", P, W);  draw(Q -- R, dashed);  label("$\frac 38$", H--K, E); label("$\frac 58$", K--J, W);  label("$\frac 7{16}$", G--C, E);  label("$\frac 38$", C--J, N);  label("$\frac 1{16}$", A--F, dir(160));  [/asy]

Video Solution

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

~IceMatrix


Video Solution 2

https://youtu.be/VBwlVbM0GRw

~avn

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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