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

(→‎Solution 2)
(→‎Solution 2)
Line 12: Line 12:
 
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.
 
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.
 
We can thus use proportions to solve this problem.
−
<div style="text-align:center;">
+
<cmath>\begin{eqnarray*}
−
<math>\begin{eqnarray*}
 
 
\frac{GF}{BE}=\frac{CG}{CB}\implies
 
\frac{GF}{BE}=\frac{CG}{CB}\implies
 
\frac{\frac{1}{\sqrt{1985}}}{BE}=\frac{\frac{1}{n}}{1}\implies
 
\frac{\frac{1}{\sqrt{1985}}}{BE}=\frac{\frac{1}{n}}{1}\implies
 
BE=\frac{n\sqrt{1985}}{1985}
 
BE=\frac{n\sqrt{1985}}{1985}
−
\end{eqnarray*}</math>
+
\end{eqnarray*}</cmath>
−
<div style="text-align:left;">
 
 
Also,
 
Also,
−
<div style="text-align:center;">
+
<cmath>\begin{eqnarray*}
−
<math>\begin{eqnarray*}
 
 
\frac{BE}{1}=\frac{EC}{\frac{n-1}{n}}\implies
 
\frac{BE}{1}=\frac{EC}{\frac{n-1}{n}}\implies
 
EC=\frac{\sqrt{1985}}{1985}(n-1)
 
EC=\frac{\sqrt{1985}}{1985}(n-1)
−
\end{eqnarray*}</math>
+
\end{eqnarray*}</cmath>
−
<div style="text-align:left;">
 
 
Thus,
 
Thus,
−
<div style="text-align:center;">
+
<cmath>\begin{eqnarray*}
−
<math>\begin{eqnarray*}
 
 
2(BE)(EC)+\frac{1}{1985}=1\\
 
2(BE)(EC)+\frac{1}{1985}=1\\
 
2n^{2}-2n+1=1985\\
 
2n^{2}-2n+1=1985\\
 
n(n-1)=992
 
n(n-1)=992
−
\end{eqnarray*}</math>
+
\end{eqnarray*}</cmath>
−
<div style="text-align:center;">
 
 
Simple factorization and guess and check gives us <math>\boxed{032}</math>.
 
Simple factorization and guess and check gives us <math>\boxed{032}</math>.
  

Revision as of 21:51, 8 March 2015

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 1

The lines passing through $A$ and $C$ divide the square into three parts, two right triangles and a parallelogram. Using the smaller side of the parallelogram, $1/n$, as the base, where the height is 1, we find that the area of the parallelogram is $A = \frac{1}{n}$. By the Pythagorean Theorem, the longer 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 $n = \boxed{032}$.

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} \end{eqnarray*} Also, \begin{eqnarray*} \frac{BE}{1}=\frac{EC}{\frac{n-1}{n}}\implies EC=\frac{\sqrt{1985}}{1985}(n-1) \end{eqnarray*} Thus, \begin{eqnarray*} 2(BE)(EC)+\frac{1}{1985}=1\\ 2n^{2}-2n+1=1985\\ n(n-1)=992 \end{eqnarray*} Simple factorization and guess and check gives us $\boxed{032}$.

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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