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

(Solution)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
A small [[square (geometry) | square]] is constructed inside a square of [[area]] 1 by dividing each side of the unit square into <math>n</math> equal parts, and then connecting the [[vertex | vertices]] to the division points closest to the opposite vertices. Find the value of <math>n</math> if the the area of the small square is exactly <math>\frac1{1985}</math>.  
+
A small [[square (geometry) | square]] is constructed inside a square of [[area]] 1 by dividing each side of the unit square into <math>n</math> equal parts, and then connecting the [[vertex | vertices]] to the division points closest to the opposite vertices. Find the value of <math>n</math> if the the [[area]] of the small square is exactly <math>\frac1{1985}</math>.  
 
{{image}}
 
{{image}}
 
== Solution ==
 
== 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 = 032</math>.
 +
 
 
== See also ==
 
== See also ==
 
* [[1985 AIME Problems/Problem 3 | Previous problem]]
 
* [[1985 AIME Problems/Problem 3 | Previous problem]]

Revision as of 18:09, 19 January 2007

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 $\frac1{1985}$.


An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.


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 $n = 032$.

See also