Difference between revisions of "2006 AIME II Problems/Problem 1"
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== Problem == | == Problem == | ||
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In [[convex polygon|convex]] [[hexagon]] <math>ABCDEF</math>, all six sides are congruent, <math>\angle A</math> and <math>\angle D</math> are [[right angle]]s, and <math>\angle B, \angle C, \angle E,</math> and <math>\angle F</math> are [[congruent]]. The area of the hexagonal region is <math>2116(\sqrt{2}+1).</math> Find <math>AB</math>. | In [[convex polygon|convex]] [[hexagon]] <math>ABCDEF</math>, all six sides are congruent, <math>\angle A</math> and <math>\angle D</math> are [[right angle]]s, and <math>\angle B, \angle C, \angle E,</math> and <math>\angle F</math> are [[congruent]]. The area of the hexagonal region is <math>2116(\sqrt{2}+1).</math> Find <math>AB</math>. | ||
− | == Solution == | + | == Solution 1 == |
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Let the side length be called <math>x</math>, so <math>x=AB=BC=CD=DE=EF=AF</math>. | Let the side length be called <math>x</math>, so <math>x=AB=BC=CD=DE=EF=AF</math>. | ||
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Then we have to solve the equation | Then we have to solve the equation | ||
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− | < | + | <cmath>2116(\sqrt{2}+1)=x^2\sqrt{2}+x^2</cmath> |
+ | <cmath>2116(\sqrt{2}+1)=x^2(\sqrt{2}+1)</cmath> | ||
+ | <cmath>2116=x^2</cmath> | ||
+ | <cmath>x=46</cmath> | ||
Therefore, <math>AB</math> is <math>\boxed{046}</math>. | Therefore, <math>AB</math> is <math>\boxed{046}</math>. | ||
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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,E); |
label("{\tiny $D$}",D,N); | label("{\tiny $D$}",D,N); | ||
label("{\tiny $E$}",E,N); | label("{\tiny $E$}",E,N); | ||
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</asy> | </asy> | ||
+ | == See Also == | ||
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{{AIME box|year=2006|n=II|before=First Question|num-a=2}} | {{AIME box|year=2006|n=II|before=First Question|num-a=2}} | ||
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 18:40, 18 April 2025
Contents
[hide]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 1
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
so
, and
.
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.