# Difference between revisions of "2012 AMC 12A Problems/Problem 23"

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+ | 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. | ||

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+ | 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. | ||

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+ | 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>. | ||

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+ | 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>y</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>. | ||

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

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+ | 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>. | ||

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+ | Note: the values of <math>x</math> and <math>y</math> in the problem is arbitrary. |

## Revision as of 19:56, 18 February 2012

## 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

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 . The diagonal is , which is the length of the diagonal of a unit square. Because square is not parallel to the axis, square cannot two points that are not adjacent.

Because we have showed that the two lattice points contained in must be adjacent, let us 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 that contains two adjacent lattice points within is .

By periodicity, this probability is the same if , where and . Therefore, the answer is . .

Note: the values of and in the problem is arbitrary.