Difference between revisions of "2006 AIME II Problems/Problem 1"
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2116(\sqrt2+1)&=[ABCDEF]\\ | 2116(\sqrt2+1)&=[ABCDEF]\\ | ||
&=2\cdot {1\over2}x^2+x\cdot x\sqrt2=x^2(1+\sqrt2), | &=2\cdot {1\over2}x^2+x\cdot x\sqrt2=x^2(1+\sqrt2), | ||
− | \end{align*}</cmath>so <math>x^2=2116</math>, and <math>x=\boxed{ | + | \end{align*}</cmath>so <math>x^2=2116</math>, and <math>x=\boxed{046}</math>. |
<asy> | <asy> | ||
Line 49: | Line 49: | ||
label("{\tiny $A$}",A,S); | label("{\tiny $A$}",A,S); | ||
label("{\tiny $B$}",B,S); | label("{\tiny $B$}",B,S); | ||
− | label("{\tiny $C$}",C, | + | label("{\tiny $C$}",C,dir(0)); |
label("{\tiny $D$}",D,N); | label("{\tiny $D$}",D,N); | ||
label("{\tiny $E$}",E,N); | label("{\tiny $E$}",E,N); | ||
label("{\tiny $F$}",F,W); | label("{\tiny $F$}",F,W); | ||
</asy> | </asy> | ||
+ | |||
+ | |||
+ | ~minor asymptote edit by Yiyj1 | ||
== See also == | == See also == |
Latest revision as of 03:19, 12 January 2024
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
.
![$x=46$](http://latex.artofproblemsolving.com/1/7/7/177d9e27be3fbdeb68593a8cef48eed1e0bde215.png)
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
so
, and
.
~minor asymptote edit by Yiyj1
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.