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

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== See also ==
 
== See also ==
* [[1985 AIME Problems/Problem 3 | Previous problem]]
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{{AIME box|year=1985|num-b=3|num-a=5}}
* [[1985 AIME Problems/Problem 5 | Next problem]]
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* [[AIME Problems and Solutions]]
* [[1985 AIME Problems]]
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* [[American Invitational Mathematics Examination]]
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* [[Mathematics competition resources]]
  
 
[[Category: Intermediate Geometry Problems]]
 
[[Category: Intermediate Geometry Problems]]

Revision as of 14:32, 6 May 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 $\frac1{1985}$. 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 $n = 032$.

See also

1985 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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