Difference between revisions of "1999 AIME Problems/Problem 4"
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== Problem == | == Problem == | ||
− | The two [[square]]s shown share the same [[center]] <math> | + | The two [[square]]s shown share the same [[center]] <math>O_{}</math> and have sides of length 1. The length of <math>\overline{AB}</math> is <math>43/99</math> and the [[area]] of octagon <math>ABCDEFGH</math> is <math>m/n,</math> where <math>m_{}</math> and <math>n_{}</math> are [[relatively prime]] [[positive]] [[integer]]s. Find <math>m+n.</math> |
[[Image:AIME_1999_Problem_4.png]] | [[Image:AIME_1999_Problem_4.png]] | ||
+ | __TOC__ | ||
== Solution == | == Solution == | ||
+ | === Solution 1 === | ||
Define the two possible [[distance]]s from one of the labeled points and the [[vertex|corners]] of the square upon which the point lies as <math>x</math> and <math>y</math>. The area of the [[octagon]] (by [[subtraction]] of areas) is <math>1 - 4\left(\frac{1}{2}xy\right) = 1 - 2xy</math>. | Define the two possible [[distance]]s from one of the labeled points and the [[vertex|corners]] of the square upon which the point lies as <math>x</math> and <math>y</math>. The area of the [[octagon]] (by [[subtraction]] of areas) is <math>1 - 4\left(\frac{1}{2}xy\right) = 1 - 2xy</math>. | ||
By the [[Pythagorean theorem]], | By the [[Pythagorean theorem]], | ||
− | + | <cmath>x^2 + y^2 = \left(\frac{43}{99}\right)^2</cmath> | |
Also, | Also, | ||
− | + | <cmath>\begin{align*}x + y + \frac{43}{99} &= 1\ | |
− | + | x^2 + 2xy + y^2 &= \left(\frac{56}{99}\right)^2\end{align*}</cmath> | |
Substituting, | Substituting, | ||
− | + | <cmath>\begin{align*}\left(\frac{43}{99}\right)^2 + 2xy &= \left(\frac{56}{99}\right)^2 \ | |
− | + | 2xy = \frac{(56 + 43)(56 - 43)}{99^2} &= \frac{13}{99} \end{align*}</cmath> | |
− | Thus, the area of the octagon is <math>1 - \frac{13}{99} = \frac{86}{99}</math>, so <math>m + n = 185</math>. | + | Thus, the area of the octagon is <math>1 - \frac{13}{99} = \frac{86}{99}</math>, so <math>m + n = \boxed{185}</math>. |
− | == Solution 2 == | + | === Solution 2 === |
− | + | Each of the triangle <math>AOB</math>, <math>BOC</math>, <math>COD</math>, etc. are congruent, and their areas are <math>((43/99)\cdot(1/2))/2</math>, since the area of a triangle is <math>bh/2</math>, so the area of all <math>8</math> of them is <math>\frac{86}{99}</math> and the answer is <math>\boxed{185}</math>. | |
− | Each of the triangle <math>AOB</math>, <math>BOC</math>, <math>COD</math>, etc. are congruent, and their areas are <math>((43/99)\cdot(1/2))/2</math>, since the area of a triangle is bh/2, so the area of all 8 of them is 86/ | ||
== See also == | == See also == |
Revision as of 12:16, 26 April 2008
Problem
The two squares shown share the same center and have sides of length 1. The length of is and the area of octagon is where and are relatively prime positive integers. Find
Contents
[hide]Solution
Solution 1
Define the two possible distances from one of the labeled points and the corners of the square upon which the point lies as and . The area of the octagon (by subtraction of areas) is .
By the Pythagorean theorem,
Also,
Substituting,
Thus, the area of the octagon is , so .
Solution 2
Each of the triangle , , , etc. are congruent, and their areas are , since the area of a triangle is , so the area of all of them is and the answer is .
See also
1999 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 |