Difference between revisions of "2006 AIME II Problems/Problem 1"
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Therefore, <math>AB</math> is <math>\boxed{046}</math>. | Therefore, <math>AB</math> is <math>\boxed{046}</math>. | ||
| − | == Solution == | + | == Solution 2 == |
Because <math>\angle | Because <math>\angle | ||
| − | B</math>, <math>\angle C</math>, <math>\angle E</math>, and <math>\angle F</math> are congruent, the degree-measure of each of them is <math> | + | B</math>, <math>\angle C</math>, <math>\angle E</math>, and <math>\angle F</math> are congruent, the degree-measure of each of them is <math> |
| − | {{720-2\cdot90}\over4}= 135</math>. Lines <math>BF</math> and <math>CE</math> divide the hexagonal region into two right triangles and a rectangle. Let <math>AB=x</math>. Then <math>BF=x\sqrt2</math>. Thus \begin{align*} | + | {{720-2\cdot90}\over4}= 135</math>. Lines <math>BF</math> and <math>CE</math> divide the hexagonal region into two right triangles and a rectangle. Let <math>AB=x</math>. Then <math>BF=x\sqrt2</math>. Thus |
| − | 2116(\sqrt2+1)&=[ABCDEF]\\ | + | \begin{align*} |
| − | &=2\cdot {1\over2}x^2+x\cdot x\sqrt2=x^2(1+\sqrt2), | + | 2116(\sqrt2+1)&=[ABCDEF]\\&=2\cdot {1\over2}x^2+x\cdot x\sqrt2=x^2(1+\sqrt2), |
\end{align*}so <math>x^2=2116</math>, and <math>x=\boxed{46}</math>. | \end{align*}so <math>x^2=2116</math>, and <math>x=\boxed{46}</math>. | ||
Revision as of 01:11, 3 May 2016
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*}so 
, and 
.
[asy]
pair A,B,C,D,I,F;
A=(0,0);
B=(7,0);
F=(0,7);
I=(6,13);
D=(13,13);
C=(13,6);
dot(A);
dot(B);
dot(C);
dot(D);
dot(I);
dot(F);
draw(A--B--C--D--I--F--cycle,linewidth(0.7));
label("{\tiny 
}",A,S);
label("{\tiny 
}",B,S);
label("{\tiny 
}",C,E);
label("{\tiny 
}",D,N);
label("{\tiny 
}",I,N);
label("{\tiny 
}",F,W);
[/asy]
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. 
