Difference between revisions of "1999 AIME Problems/Problem 4"
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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> | 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> | ||
− | + | <asy> | |
+ | //code taken from thread for problem | ||
+ | real alpha = 25; | ||
+ | pair W=dir(225), X=dir(315), Y=dir(45), Z=dir(135), O=origin; | ||
+ | pair w=dir(alpha)*W, x=dir(alpha)*X, y=dir(alpha)*Y, z=dir(alpha)*Z; | ||
+ | draw(W--X--Y--Z--cycle^^w--x--y--z--cycle); | ||
+ | pair A=intersectionpoint(Y--Z, y--z), | ||
+ | C=intersectionpoint(Y--X, y--x), | ||
+ | E=intersectionpoint(W--X, w--x), | ||
+ | G=intersectionpoint(W--Z, w--z), | ||
+ | B=intersectionpoint(Y--Z, y--x), | ||
+ | D=intersectionpoint(Y--X, w--x), | ||
+ | F=intersectionpoint(W--X, w--z), | ||
+ | H=intersectionpoint(W--Z, y--z); | ||
+ | dot(O); | ||
+ | label("$O$", O, SE); | ||
+ | label("$A$", A, dir(O--A)); | ||
+ | label("$B$", B, dir(O--B)); | ||
+ | label("$C$", C, dir(O--C)); | ||
+ | label("$D$", D, dir(O--D)); | ||
+ | label("$E$", E, dir(O--E)); | ||
+ | label("$F$", F, dir(O--F)); | ||
+ | label("$G$", G, dir(O--G)); | ||
+ | label("$H$", H, dir(O--H));</asy> | ||
__TOC__ | __TOC__ | ||
− | == Solution == | + | == Solution 1 == |
− | + | Triangles <math>AOB</math>, <math>BOC</math>, <math>COD</math>, etc. are congruent by symmetry (you can prove it rigorously by using the power of a point to argue that exactly two chords of length <math>1</math> in the circumcircle of the squares pass through <math>B</math>, etc.), and each area is <math>\frac{\frac{43}{99}\cdot\frac{1}{2}}{2}</math>. Since the area of a triangle is <math>bh/2</math>, the area of all <math>8</math> of them is <math>\frac{86}{99}</math> and the answer is <math>\boxed{185}</math>. | |
+ | |||
+ | == Solution 2 == | ||
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>. | ||
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Thus, the area of the octagon is <math>1 - \frac{13}{99} = \frac{86}{99}</math>, so <math>m + n = \boxed{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>. | ||
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== See also == | == See also == | ||
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 12:27, 12 December 2020
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 1
Triangles , , , etc. are congruent by symmetry (you can prove it rigorously by using the power of a point to argue that exactly two chords of length in the circumcircle of the squares pass through , etc.), and each area is . Since the area of a triangle is , the area of all of them is and the answer is .
Solution 2
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 .
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 |
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