Difference between revisions of "2012 AMC 12A Problems/Problem 23"
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== Problem == | == Problem == | ||
− | Let <math>S</math> be the square one of whose diagonals has endpoints <math>( | + | 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)}\ | + | <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 == | ||
− | === | + | <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> | ||
− | Because | + | 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' | + | 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 | + | 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 == | == 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 be the square one of whose diagonals has endpoints and . A point is chosen uniformly at random over all pairs of real numbers and such that and . Let be a translated copy of centered at . What is the probability that the square region determined by contains exactly two points with integer coefficients in its interior?
Solution
The unit square's diagonal has a length of . Because square is not parallel to the axis, the two points must be adjacent.
Now consider the unit square with vertices and . Let us first consider only two vertices, and . We want to find the area of the region within that the point will create the translation of , such that it covers both and . By symmetry, there will be three equal regions that cover the other pairs of adjacent vertices.
For to contain the point , must be inside square . Similarly, for to contain the point , must be inside a translated square with center at , which we will call . Therefore, the area we seek is Area.
To calculate the area, we notice that Area Area by symmetry. Let . Let be the midpoint of , and along the line . Let be the intersection of and within , and be the intersection of and outside . Therefore, the area we seek is Area. Because all have coordinate , they are collinear. Noting that the side length of and is (as shown above), we also see that , so . If follows that and . Therefore, the area is Area.
Because there are three other regions in the unit square that we need to count, the total area of within such that contains two adjacent lattice points is .
By periodicity, this probability is the same for all and . Therefore, the answer is
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 |
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