Difference between revisions of "1999 AIME Problems/Problem 4"
Flyhawkeye (talk | contribs) (→Solution) |
|||
| Line 28: | Line 28: | ||
__TOC__ | __TOC__ | ||
== Solution == | == Solution == | ||
| − | === Solution 1 === | + | === Simple Solution === |
| + | Triangles <math>AOB</math>, <math>BOC</math>, <math>COD</math>, etc. are congruent, 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>. | ||
| + | |||
| + | |||
| + | === Other Solution === | ||
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>. | ||
| Line 43: | Line 47: | ||
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>. | ||
| − | |||
| − | |||
| − | |||
== See also == | == See also == | ||
Revision as of 11:57, 31 July 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
Simple Solution
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 
.
Other Solution
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 
.
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. 
