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 == | ||
Revision as of 20:24, 6 January 2018
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 
![[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]](http://latex.artofproblemsolving.com/6/6/4/664c2f1df8ff423eac64185612ee090643239ddd.png)
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,
![\[x^2 + y^2 = \left(\frac{43}{99}\right)^2\]](http://latex.artofproblemsolving.com/5/3/e/53e477338fc6a5946980845f2d103babe0c1b55e.png)
Also,
Substituting,
Thus, the area of the octagon is 
, so 
.
Solution 2
Triangles 
, 
, 
, etc. are congruent, and each area is 
. Since the area of a triangle is 
, 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
