1985 AIME Problems/Problem 4

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 $\frac1{1985}$ .

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

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 $\boxed{032}$ .

1985 AIME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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1985 AIME Problems/Problem 4

Contents

Problem

Solution 1

Solution 2

See also