Difference between revisions of "2005 AIME I Problems/Problem 14"
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The coordinates of <math>E</math> are thus <math>(2, - 2)</math>. | The coordinates of <math>E</math> are thus <math>(2, - 2)</math>. | ||
− | Now, since <math>E</math> and <math>C</math> are on the same side, we find the slope of the sides going through <math> | + | Now, since <math>E</math> and <math>C</math> are on the same side, we find the slope of the sides going through <math>E</math> and <math>C</math> to be <math>\frac { - 2 - 0}{2 - 8} = \frac {1}{3}</math>. Because the other two sides are perpendicular, the slope of the sides going through <math>B</math> and <math>D</math> are now <math>- 3</math>. |
Let <math>A_1,B_1,C_1,D_1</math> be the vertices of the square so that <math>A_1B_1</math> contains point <math>A</math>, <math>B_1C_1</math> contains point <math>B</math>, and etc. Since we know the slopes and a point on the line for each side of the square, we use the point slope formula to find the linear equations. Next, we use the equations to find <math>2</math> vertices of the square, then apply the distance formula. | Let <math>A_1,B_1,C_1,D_1</math> be the vertices of the square so that <math>A_1B_1</math> contains point <math>A</math>, <math>B_1C_1</math> contains point <math>B</math>, and etc. Since we know the slopes and a point on the line for each side of the square, we use the point slope formula to find the linear equations. Next, we use the equations to find <math>2</math> vertices of the square, then apply the distance formula. |
Latest revision as of 13:11, 18 February 2018
Problem
Consider the points and There is a unique square such that each of the four points is on a different side of Let be the area of Find the remainder when is divided by .
Solution
Solution 1
Consider a point such that is perpendicular to , intersects , and . E will be on the same side of the square as point .
Let the coordinates of be . Since is perpendicular to , and , we have and The coordinates of are thus .
Now, since and are on the same side, we find the slope of the sides going through and to be . Because the other two sides are perpendicular, the slope of the sides going through and are now .
Let be the vertices of the square so that contains point , contains point , and etc. Since we know the slopes and a point on the line for each side of the square, we use the point slope formula to find the linear equations. Next, we use the equations to find vertices of the square, then apply the distance formula.
We find the coordinates of to be and the coordinates of to be . Applying the distance formula, the side length of our square is .
Hence, the area of the square is . The remainder when is divided by is .
Solution 2
Let denote a normal vector of the side containing . Note that intersect and hence must be opposite vertices of the square. The lines containing the sides of the square have the form , , , and . The lines form a square, so the distance between and the line through equals the distance between and the line through , hence , or . We can take and . So the side of the square is , the area is , and the answer to the problem is .
See also
2005 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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