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 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>. |
==Solution== | ==Solution== | ||
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> | 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> | ||
~mn28407 | ~mn28407 | ||
+ | |||
+ | ==Solution 2 (Official MAA)== | ||
+ | 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 | ||
+ | <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>. | ||
+ | <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== | ==Video Solution== | ||
https://youtu.be/x0QznvXcwHY?t=190 | https://youtu.be/x0QznvXcwHY?t=190 | ||
− | ~IceMatrix | + | ~IceMatrix |
+ | |||
+ | |||
+ | ==Video Solution 2== | ||
+ | https://youtu.be/VBwlVbM0GRw | ||
+ | |||
+ | ~avn | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2020|n=II|num-b=1|num-a=3}} | {{AIME box|year=2020|n=II|num-b=1|num-a=3}} | ||
+ | [[Category:Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 17:24, 24 June 2020
Contents
Problem
Let be a point chosen uniformly at random in the interior of the unit square with vertices at
, and
. The probability that the slope of the line determined by
and the point
is greater than or equal to
can be written as
, where
and
are relatively prime positive integers. Find
.
Solution
The areas bounded by the unit square and alternately bounded by the lines through that are vertical or have a slope of
show where
can be placed to satisfy the condition. One of the areas is a trapezoid with bases
and
and height
. The other area is a trapezoid with bases
and
and height
. Then,
~mn28407
Solution 2 (Official MAA)
The line through the fixed point with slope
has equation
. The slope between
and the fixed point exceeds
if
falls within the shaded region in the diagram below consisting of two trapezoids with area
Because the entire square has area
the required probability is
. The requested sum is
.
Video Solution
https://youtu.be/x0QznvXcwHY?t=190
~IceMatrix
Video Solution 2
~avn
See Also
2020 AIME II (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.