Difference between revisions of "2006 AIME II Problems/Problem 1"
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<math>x=46</math></div> | <math>x=46</math></div> | ||
| − | Therefore, <math>AB</math> is <math>\boxed{ | + | Therefore, <math>AB</math> is <math>\boxed{46}</math>. |
== Solution 2 == | == Solution 2 == | ||
Revision as of 22:06, 1 April 2019
Contents
Problem
In convex hexagon 
, all six sides are congruent, 
and 
are right angles, and 
and 
are congruent. The area of the hexagonal region is 
Find 
.
Solution
Let the side length be called 
, so 
.
The diagonal 
. Then the areas of the triangles AFB and CDE in total are 
,
and the area of the rectangle BCEF equals 
Then we have to solve the equation
.
Therefore, is 
.
Solution 2
Because 
, 
, 
, and are congruent, the degree-measure of each of them is 
. Lines 
and 
divide the hexagonal region into two right triangles and a rectangle. Let 
. Then 
. Thus
![\begin{align*} 2116(\sqrt2+1)&=[ABCDEF]\\ &=2\cdot {1\over2}x^2+x\cdot x\sqrt2=x^2(1+\sqrt2), \end{align*}](http://latex.artofproblemsolving.com/7/b/4/7b4eac99a9d419ae9b76ff89eae925b572923827.png)
so 
, and 
.
![[asy] pair A,B,C,D,E,F; A=(0,0); B=(7,0); C=(13,6); E=(6,13); D=(13,13); F=(0,7); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); draw(A--B--C--D--E--F--cycle,linewidth(0.7)); label("{\tiny $A$}",A,S); label("{\tiny $B$}",B,S); label("{\tiny $C$}",C,E); label("{\tiny $D$}",D,N); label("{\tiny $E$}",E,N); label("{\tiny $F$}",F,W); [/asy]](http://latex.artofproblemsolving.com/a/b/9/ab955e58091b3065e509453f7d496e7d01e5762a.png)
See also
| 2006 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
