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

Latest revision as of 20:40, 2 March 2020

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 $n = \boxed{32}$ .

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 $\boxed{32}$ .

Solution 3

Line Segment $DE = \frac{1}{n}$ , so $EC = 1 - \frac{1}{n} = \frac{n-1}{n}$ . Draw line segment $HE$ parallel to the corresponding sides of the small square, $HE$ has length $\frac{1}{\sqrt{1985}}$ , as it is the same length as the sides of the square. Notice that $\triangle CEL$ is similar to $\triangle HDE$ by $AA$ similarity. Thus, $\frac{LC}{HE} = \frac{EC}{DE} = n-1$ , so $LC = \frac{n-1}{\sqrt{1985}}$ . Notice that $\triangle CEL$ is also similar to $\triangle CDF$ by $AA$ similarity. Thus, $\frac{FC}{EC} = \frac{DC}{LC}$ , and the expression simplifies into a quadratic equation $n^2 - n - 992 = 0$ . Solving this quadratic equation yields $n =\boxed{32}$ .

@@ Line 4: / Line 4: @@
 [[Image:AIME_1985_Problem_4.png]]
-== Solution ==
+== Solution 1==
-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>.
+The lines passing through <math>A</math> and <math>C</math> divide the square into three parts, two [[right triangle]]s and a [[parallelogram]].  Using the smaller side of the parallelogram, <math>1/n</math>, as the base, where the height is 1, we find that the area of the parallelogram is <math>A = \frac{1}{n}</math>.  By the [[Pythagorean Theorem]], the longer 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 = \boxed{32}</math>.
+==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.
+<cmath>\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*}</cmath>
+Also,
+<cmath>\begin{eqnarray*}
+\frac{BE}{1}=\frac{EC}{\frac{n-1}{n}}\implies
+EC=\frac{\sqrt{1985}}{1985}(n-1)
+\end{eqnarray*}</cmath>
+Thus,
+<cmath>\begin{eqnarray*}
+(BE)(EC)+\frac{1}{1985}=1\\
+n^{2}-2n+1=1985\\
+n(n-1)=992
+\end{eqnarray*}</cmath>
+Simple factorization and guess and check gives us <math>\boxed{32}</math>.
+== Solution 3 ==
+[[File:AIME_1985_Problem_4_Solution_3_Diagram.png]]
+Line Segment <math>DE = \frac{1}{n}</math>, so <math>EC = 1 - \frac{1}{n} = \frac{n-1}{n}</math>. Draw line segment <math>HE</math> parallel to the corresponding sides of the small square, <math>HE</math> has length <math>\frac{1}{\sqrt{1985}}</math>, as it is the same length as the sides of the square. Notice that <math>\triangle CEL</math> is similar to <math>\triangle HDE</math> by <math>AA</math> similarity. Thus, <math>\frac{LC}{HE} = \frac{EC}{DE} = n-1</math>, so <math>LC = \frac{n-1}{\sqrt{1985}}</math>. Notice that <math>\triangle CEL</math> is also similar to <math>\triangle CDF</math> by <math>AA</math> similarity. Thus, <math>\frac{FC}{EC} = \frac{DC}{LC}</math>, and the expression simplifies into a quadratic equation <math>n^2 - n - 992 = 0</math>. Solving this quadratic equation yields <math>n =\boxed{32}</math>.
 == 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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