Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Added written solution)
(→‎Solution)
Line 5: Line 5:
 
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("<math>(0,0)</math>", A, SW);
 +
label("<math>(1,0)</math>", B, SE);
 +
label("<math>(1,1)</math>", C, E);
 +
label("<math>(0,1)</math>", D, W);
 +
 +
filldraw(A--H--K--F--cycle, lightgray);
 +
filldraw(K--G--C--J--cycle, lightgray);
 +
 +
dot(K);
 +
dot("<math>P</math>", P, W);
 +
draw(Q -- R, dashed);
 +
 +
label("<math>\frac 38</math>", H--K, E);
 +
label("<math>\frac 58</math>", K--J, W);
 +
label("<math>\frac 7{16}</math>", G--C, E);
 +
label("<math>\frac 38</math>", C--J, N);
 +
label("<math>\frac 1{16}</math>", A--F, dir(160));
 +
 +
[/asy]
  
 
==Video Solution==
 
==Video Solution==

Revision as of 12:43, 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

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

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_2&oldid=124442"