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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 = 32</math>.
 
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>.
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 +
==Solution 2==
 +
[[File:Aime.png]]
 +
 +
Surrounding the square with area <math>\frac{1}{1985}</math> are <math>4</math> right triangles with hypotenuse <math>1</math> (sides of the large square). Thus, <math>X + \frac{1}{1985} = 1</math>, where <math>X</math> is the area of the of the 4 triangles.
 +
We can thus use proportions to solve this problem.
 +
<div style="text-align:center;">
 +
<math>\begin{eqnarray*}
 +
\frac{GF}{BE}=\frac{CG}{CB}\implies
 +
\frac{\frac{1}{\sqrt{1985}}}{BE}=\frac{\frac{1}{n}}{1}\implies
 +
BE=\frac{n\sqrt{1985}}{1985}</math>
 +
<div style="text-align:left;">
 +
Also,
 +
<div style="text-align:center;">
 +
<math>\begin{eqnarray*}
 +
\frac{BE}{1}=\frac{EC}{\frac{n-1}{n}}\implies
 +
EC=\frac{\sqrt{1985}}{1985}(n-1)</math>
 +
<div style="text-align:left;">
 +
Thus,
 +
<div style="text-align:center;">
 +
<math>\begin{eqnarray*}
 +
2(BE)(EC)+\frac{1}{1985}=1\\
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2n^{2}-2n+1=1985\\
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n(n-1)=992</math>
 +
Simple factorization and guess and check gives us <math>\boxed{32}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 16:33, 22 April 2011

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 $n = 32$.

Solution 2

Aime.png

Surrounding the square with area $\frac{1}{1985}$ are $4$ right triangles with hypotenuse $1$ (sides of the large square). Thus, $X + \frac{1}{1985} = 1$, where $X$ is the area of the of the 4 triangles. We can thus use proportions to solve this problem.

$\begin{eqnarray*} \frac{GF}{BE}=\frac{CG}{CB}\implies \frac{\frac{1}{\sqrt{1985}}}{BE}=\frac{\frac{1}{n}}{1}\implies BE=\frac{n\sqrt{1985}}{1985}$ (Error compiling LaTeX. Unknown error_msg)

Also,

$\begin{eqnarray*} \frac{BE}{1}=\frac{EC}{\frac{n-1}{n}}\implies EC=\frac{\sqrt{1985}}{1985}(n-1)$ (Error compiling LaTeX. Unknown error_msg)

Thus,

$\begin{eqnarray*} 2(BE)(EC)+\frac{1}{1985}=1\\ 2n^{2}-2n+1=1985\\ n(n-1)=992$ (Error compiling LaTeX. Unknown error_msg) Simple factorization and guess and check gives us $\boxed{32}$.

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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