Difference between revisions of "1985 AIME Problems/Problem 4"
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== Solution 1== | == 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 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== | ==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. | ||
− | < | + | <cmath>\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*}</cmath> | |
Also, | Also, | ||
− | < | + | <cmath>\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*}</cmath> | |
Thus, | Thus, | ||
− | < | + | <cmath>\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*}</cmath> | |
− | Simple factorization and guess and check gives us <math>\boxed{ | + | 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 == | == See also == |
Latest revision as of 20:40, 2 March 2020
Contents
[hide]Problem
A small square is constructed inside a square of area 1 by dividing each side of the unit square into equal parts, and then connecting the vertices to the division points closest to the opposite vertices. Find the value of if the the area of the small square is exactly .
Solution 1
The lines passing through and divide the square into three parts, two right triangles and a parallelogram. Using the smaller side of the parallelogram, , as the base, where the height is 1, we find that the area of the parallelogram is . By the Pythagorean Theorem, the longer base of the parallelogram has length , so the parallelogram has height . But the height of the parallelogram is the side of the little square, so . Solving this quadratic equation gives .
Solution 2
Surrounding the square with area are right triangles with hypotenuse (sides of the large square). Thus, , where is the area of the of the 4 triangles. We can thus use proportions to solve this problem. Also, Thus, Simple factorization and guess and check gives us .
Solution 3
Line Segment , so . Draw line segment parallel to the corresponding sides of the small square, has length , as it is the same length as the sides of the square. Notice that is similar to by similarity. Thus, , so . Notice that is also similar to by similarity. Thus, , and the expression simplifies into a quadratic equation . Solving this quadratic equation yields .
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 |