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Difference between revisions of "1985 AIME Problems/Problem 4"

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== Solution ==
 
== Solution ==
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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 = 032</math>.
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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 11:49, 6 June 2010

Problem

A small square is constructed inside a square of area 1 by dividing each side of the unit square into $n$ equal parts, and then connecting the vertices to the division points closest to the opposite vertices. Find the value of $n$ if the the area of the small square is exactly $\frac1{1985}$.

AIME 1985 Problem 4.png

Solution

The lines passing through $A$ and $C$ divide the square into three parts, two right triangles and a parallelogram. The area of the triangles together is easily seen to be $\frac{n - 1}{n}$, so the area of the parallelogram is $A = \frac{1}{n}$. By the Pythagorean Theorem, the base of the parallelogram has length $l = \sqrt{1^2 + \left(\frac{n - 1}{n}\right)^2} = \frac{1}{n}\sqrt{2n^2 - 2n + 1}$, so the parallelogram has height $h = \frac{A}{l} = \frac{1}{\sqrt{2n^2 - 2n + 1}}$. But the height of the parallelogram is the side of the little square, so $2n^2 - 2n + 1 = 1985$. Solving this quadratic equation gives $n = 32$.

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
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