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 "2012 AMC 12A Problems/Problem 23"

(Created page with "== Problem 23 == Let <math>S</math> be the square one of whose diagonals has endpoints <math>(0.1,0.7)</math> and <math>(-0.1,-0.7)</math>. A point <math>v=(x,y)</math> is chos...")
 
 
(36 intermediate revisions by 9 users not shown)
Line 1: Line 1:
−
== Problem 23 ==
+
== Problem ==
  
−
Let <math>S</math> be the square one of whose diagonals has endpoints <math>(0.1,0.7)</math> and <math>(-0.1,-0.7)</math>.  A point <math>v=(x,y)</math> is chosen uniformly at random over all pairs of real numbers <math>x</math> and <math>y</math> such that <math>0 \le x \le 2012</math> and <math>0\le y\le 2012</math>.  Let <math>T(v)</math> be a translated copy of <math>S</math> centered at <math>v</math>.  What is the probability that the square region determined by <math>T(v)</math> contains exactly two points with integer coefficients in its interior?
+
Let <math>S</math> be the square one of whose diagonals has endpoints <math>(1/10,7/10)</math> and <math>(-1/10,-7/10)</math>.  A point <math>v=(x,y)</math> is chosen uniformly at random over all pairs of real numbers <math>x</math> and <math>y</math> such that <math>0 \le x \le 2012</math> and <math>0\le y\le 2012</math>.  Let <math>T(v)</math> be a translated copy of <math>S</math> centered at <math>v</math>.  What is the probability that the square region determined by <math>T(v)</math> contains exactly two points with integer coefficients in its interior?
  
−
<math> \textbf{(A)}\ 0.125\qquad\textbf{(B)}\ 0.14\qquad\textbf{(C)}\ 0.16\qquad\textbf{(D)}\ 0.25 \qquad\textbf{(E)}\ 0.32 </math>
+
<math> \textbf{(A)}\ \frac{1}{8}\qquad\textbf{(B) }\frac{7}{50}\qquad\textbf{(C) }\frac{4}{25}\qquad\textbf{(D) }\frac{1}{4}\qquad\textbf{(E) }\frac{8}{25} </math>
  
 +
== Solution ==
 +
 +
<center><asy>
 +
pair A=(0.1,0.7), C=(-0.1,-0.7), B=(-0.7,0.1), D=(0.7,-0.1), X=(1,0), W=(-1,0), Y=(0,1), Z=(0,-1);
 +
draw (A--B--C--D--A);
 +
draw(A--C);
 +
draw(B--D);
 +
draw(W--X);
 +
draw(Y--Z);
 +
label("\((0.1,0.7)\)",A,NE);
 +
label("\((-0.1,-0.7)\)",C,SW);
 +
label("\(x\)",X,NW);
 +
label("\(y\)",Y,NE);
 +
</asy></center>
 +
 +
The unit square's diagonal has a length of <math>\sqrt{0.2^2 + 1.4^2} = \sqrt{2}</math>. Because <math>S</math> square is not parallel to the axis, the two points must be adjacent.
 +
 +
Now consider the unit square <math>U</math> with vertices <math>(0,0), (1,0), (1,1)</math> and <math>(0,1)</math>. Let us first consider only two vertices, <math>(0,0)</math> and <math>(1,0)</math>. We want to find the area of the region within <math>U</math> that the point <math>v=(x,y)</math> will create the translation of <math>S</math>, <math>T(v)</math> such that it covers both <math>(0,0)</math> and <math>(1,0)</math>. By symmetry, there will be three equal regions that cover the other pairs of adjacent vertices.
 +
 +
For <math>T(v)</math> to contain the point <math>(0,0)</math>, <math>v</math> must be inside square <math>S</math>. Similarly, for <math>T(v)</math> to contain the point <math>(1,0)</math>, <math>v</math> must be inside a translated square <math>S</math> with center at <math>(1,0)</math>, which we will call <math>S'</math>. Therefore, the area we seek is Area<math>(U \cap S \cap S')</math>.
 +
 +
To calculate the area, we notice that Area<math>(U \cap S \cap S') = \frac{1}{2} \cdot</math> Area<math>(S \cap S')</math> by symmetry. Let <math>S_1 = (0.1, 0.7), S_2 = (0.7, -0.1), S'_1 = (1.1, 0.7), S'_2 = (0.3, 0.1)</math>. Let <math>M = (0.7, 0.4)</math> be the midpoint of <math>S'_1S'_2</math>, and <math>N = (0.7, 0.7)</math> along the line <math>S_1S'_1</math>. Let <math>I</math> be the intersection of <math>S</math> and <math>S'</math> within <math>U</math>, and <math>J</math> be the intersection of <math>S</math> and <math>S'</math> outside <math>U</math>. Therefore, the area we seek is <math>\frac{1}{2} \cdot</math> Area<math>(S \cap S') = \frac{1}{2} [IS'_2JS_2]</math>. Because <math>S_2, M, N</math> all have <math>x</math> coordinate <math>0.7</math>, they are collinear. Noting that the side length of <math>S</math> and <math>S'</math> is <math>1</math> (as shown above), we also see that <math>S_2M = MS'_1 = 0.5</math>, so <math>\triangle{S'_1NM} \cong \triangle{S_2IM}</math>. If follows that <math>IS_2 = NS'_1 = 1.1 - 0.7 = 0.4</math> and <math>IS'_2 = MS'_2 - MI = MS'_2 - MN = 0.5 - 0.3 = 0.2</math>. Therefore, the area is <math>\frac{1}{2} \cdot</math> Area<math>(S \cap S') = \frac{1}{2} [IS'_2JS_2] = \frac{1}{2} \cdot 0.2 \cdot 0.4 = 0.04</math>.
  
−
== Solution ==
+
Because there are three other regions in the unit square <math>U</math> that we need to count, the total area of <math>v</math> within <math>U</math> such that <math>T(v)</math> contains two adjacent lattice points is <math>0.04 \cdot 4 = 0.16</math>.
 +
 
 +
By periodicity, this probability is the same for all <math>0 \le x \le 2012</math> and <math>0 \le y \le 2012</math>. Therefore, the answer is <math> 0.16 = \boxed{\frac{4}{25} \textbf{(C)} }</math>
 +
 
 +
==Video Solution by Richard Rusczyk==
 +
https://artofproblemsolving.com/videos/amc/2012amc12a/254
 +
 
 +
== See Also ==
 +
{{AMC12 box|year=2012|ab=A|num-b=22|num-a=24}}
 +
{{MAA Notice}}

Latest revision as of 19:37, 19 January 2021

Problem

Let $S$ be the square one of whose diagonals has endpoints $(1/10,7/10)$ and $(-1/10,-7/10)$. A point $v=(x,y)$ is chosen uniformly at random over all pairs of real numbers $x$ and $y$ such that $0 \le x \le 2012$ and $0\le y\le 2012$. Let $T(v)$ be a translated copy of $S$ centered at $v$. What is the probability that the square region determined by $T(v)$ contains exactly two points with integer coefficients in its interior?

$\textbf{(A)}\ \frac{1}{8}\qquad\textbf{(B) }\frac{7}{50}\qquad\textbf{(C) }\frac{4}{25}\qquad\textbf{(D) }\frac{1}{4}\qquad\textbf{(E) }\frac{8}{25}$

Solution

[asy] pair A=(0.1,0.7), C=(-0.1,-0.7), B=(-0.7,0.1), D=(0.7,-0.1), X=(1,0), W=(-1,0), Y=(0,1), Z=(0,-1); draw (A--B--C--D--A); draw(A--C); draw(B--D); draw(W--X); draw(Y--Z); label("\((0.1,0.7)\)",A,NE); label("\((-0.1,-0.7)\)",C,SW); label("\(x\)",X,NW); label("\(y\)",Y,NE); [/asy]

The unit square's diagonal has a length of $\sqrt{0.2^2 + 1.4^2} = \sqrt{2}$. Because $S$ square is not parallel to the axis, the two points must be adjacent.

Now consider the unit square $U$ with vertices $(0,0), (1,0), (1,1)$ and $(0,1)$. Let us first consider only two vertices, $(0,0)$ and $(1,0)$. We want to find the area of the region within $U$ that the point $v=(x,y)$ will create the translation of $S$, $T(v)$ such that it covers both $(0,0)$ and $(1,0)$. By symmetry, there will be three equal regions that cover the other pairs of adjacent vertices.

For $T(v)$ to contain the point $(0,0)$, $v$ must be inside square $S$. Similarly, for $T(v)$ to contain the point $(1,0)$, $v$ must be inside a translated square $S$ with center at $(1,0)$, which we will call $S'$. Therefore, the area we seek is Area$(U \cap S \cap S')$.

To calculate the area, we notice that Area$(U \cap S \cap S') = \frac{1}{2} \cdot$ Area$(S \cap S')$ by symmetry. Let $S_1 = (0.1, 0.7), S_2 = (0.7, -0.1), S'_1 = (1.1, 0.7), S'_2 = (0.3, 0.1)$. Let $M = (0.7, 0.4)$ be the midpoint of $S'_1S'_2$, and $N = (0.7, 0.7)$ along the line $S_1S'_1$. Let $I$ be the intersection of $S$ and $S'$ within $U$, and $J$ be the intersection of $S$ and $S'$ outside $U$. Therefore, the area we seek is $\frac{1}{2} \cdot$ Area$(S \cap S') = \frac{1}{2} [IS'_2JS_2]$. Because $S_2, M, N$ all have $x$ coordinate $0.7$, they are collinear. Noting that the side length of $S$ and $S'$ is $1$ (as shown above), we also see that $S_2M = MS'_1 = 0.5$, so $\triangle{S'_1NM} \cong \triangle{S_2IM}$. If follows that $IS_2 = NS'_1 = 1.1 - 0.7 = 0.4$ and $IS'_2 = MS'_2 - MI = MS'_2 - MN = 0.5 - 0.3 = 0.2$. Therefore, the area is $\frac{1}{2} \cdot$ Area$(S \cap S') = \frac{1}{2} [IS'_2JS_2] = \frac{1}{2} \cdot 0.2 \cdot 0.4 = 0.04$.

Because there are three other regions in the unit square $U$ that we need to count, the total area of $v$ within $U$ such that $T(v)$ contains two adjacent lattice points is $0.04 \cdot 4 = 0.16$.

By periodicity, this probability is the same for all $0 \le x \le 2012$ and $0 \le y \le 2012$. Therefore, the answer is $0.16 = \boxed{\frac{4}{25} \textbf{(C)} }$

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2012amc12a/254

See Also

2012 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 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=2012_AMC_12A_Problems/Problem_23&oldid=142842"