Difference between revisions of "1985 AIME Problems/Problem 4"
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== Solution == | == Solution == | ||
− | The lines passing through <math>A</math> and <math>C</math> divide the square into three parts, two [[right triangle]]s and a [[parallelogram]]. The area of the [[triangle]]s together is easily seen to be <math>\frac{n - 1}{n}</math>, so the area of the parallelogram is <math>A = \frac{1}{n}</math>. By the [[Pythagorean Theorem]], the base of the parallelogram has [[length]] <math>l = \sqrt{1^2 + \left(\frac{n - 1}{n}\right)^2} = \frac{1}{n}\sqrt{2n^2 - 2n + 1}</math>, so the parallelogram has height <math>h = \frac{A}{l} = \frac{1}{\sqrt{2n^2 - 2n + 1}}</math>. But the height of the parallelogram is the side of the little square, so <math>2n^2 - 2n + 1 = 1985</math>. Solving this [[quadratic equation]] gives <math>n = | + | The lines passing through <math>A</math> and <math>C</math> divide the square into three parts, two [[right triangle]]s and a [[parallelogram]]. The area of the [[triangle]]s together is easily seen to be <math>\frac{n - 1}{n}</math>, so the area of the parallelogram is <math>A = \frac{1}{n}</math>. By the [[Pythagorean Theorem]], the base of the parallelogram has [[length]] <math>l = \sqrt{1^2 + \left(\frac{n - 1}{n}\right)^2} = \frac{1}{n}\sqrt{2n^2 - 2n + 1}</math>, so the parallelogram has height <math>h = \frac{A}{l} = \frac{1}{\sqrt{2n^2 - 2n + 1}}</math>. But the height of the parallelogram is the side of the little square, so <math>2n^2 - 2n + 1 = 1985</math>. Solving this [[quadratic equation]] gives <math>n = 32</math>. |
== See also == | == See also == |
Revision as of 12:49, 6 June 2010
Problem
A small square is constructed inside a square of area 1 by dividing each side of the unit square into equal parts, and then connecting the vertices to the division points closest to the opposite vertices. Find the value of if the the area of the small square is exactly .
Solution
The lines passing through and divide the square into three parts, two right triangles and a parallelogram. The area of the triangles together is easily seen to be , so the area of the parallelogram is . By the Pythagorean Theorem, the base of the parallelogram has length , so the parallelogram has height . But the height of the parallelogram is the side of the little square, so . Solving this quadratic equation gives .
See also
1985 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |