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"

(→‎Solution 1)
 
(29 intermediate revisions by 9 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== 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 ==
 
== Solution ==
  
−
=== Solution 1 ===
+
<center><asy>
−
We first notice that for the translated square to contain two points with integer coordinates (lattice points) in its interior, these two points must be adjacent. This can be shown by considering the diagonal of <math>S</math>. The diagonal is <math>\sqrt{0.2^2 + 1.4^2} = \sqrt{2}</math>, which is the length of the diagonal of a unit square. Because <math>S</math> square is not parallel to the axis, square <math>T(v)</math> cannot two points that are not adjacent.
+
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>
  
−
Because we have showed that the two lattice points contained in <math>T(v)</math> must be adjacent, let us 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.
+
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>.
 
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>.
+
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>.
  
 
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>.
 
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 if <math>v = (x,y)</math>, where <math>0 \le x \le 2012</math> and <math>0 \le y \le 2012</math>. Therefore, the answer is <math>0.16</math>. <math>\boxed{C}</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>
  
−
Note: the range of <math>x</math> and <math>y</math> in the problem is arbitrary.
+
==Video Solution by Richard Rusczyk==
 +
https://artofproblemsolving.com/videos/amc/2012amc12a/254
  
 
== See Also ==
 
== See Also ==
 
{{AMC12 box|year=2012|ab=A|num-b=22|num-a=24}}
 
{{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"