Difference between revisions of "2006 AIME II Problems/Problem 1"
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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 | {{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*} | + | <math>\begin{align*} |
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*}so <math>x^2=2116< | + | \end{align*}so </math>x^2=2116<math>, and </math>x=\boxed{46}<math>. |
[asy] | [asy] | ||
| Line 47: | Line 47: | ||
dot(F); | dot(F); | ||
draw(A--B--C--D--I--F--cycle,linewidth(0.7)); | draw(A--B--C--D--I--F--cycle,linewidth(0.7)); | ||
| − | label("{\tiny <math>A< | + | label("{\tiny </math>A<math>}",A,S); |
| − | label("{\tiny <math>B< | + | label("{\tiny </math>B<math>}",B,S); |
| − | label("{\tiny <math>C< | + | label("{\tiny </math>C<math>}",C,E); |
| − | label("{\tiny <math>D< | + | label("{\tiny </math>D<math>}",D,N); |
| − | label("{\tiny <math>E< | + | label("{\tiny </math>E<math>}",I,N); |
| − | label("{\tiny <math>F | + | label("{\tiny </math>F$}",F,W); |
[/asy] | [/asy] | ||
Revision as of 16:56, 18 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
$\begin{align*}
2116(\sqrt2+1)&=[ABCDEF]\\
&=2\cdot {1\over2}x^2+x\cdot x\sqrt2=x^2(1+\sqrt2),
\end{align*}so$ (Error compiling LaTeX. Unknown error_msg)x^2=2116
x=\boxed{46}$.
[asy] pair A,B,C,D,I,F; A=(0,0); B=(7,0); F=(0,7); I=(6,13); D=(13,13); C=(13,6); dot(A); dot(B); dot(C); dot(D); dot(I); dot(F); draw(A--B--C--D--I--F--cycle,linewidth(0.7)); label("{\tiny$ (Error compiling LaTeX. Unknown error_msg)A$}",A,S); label("{\tiny$ (Error compiling LaTeX. Unknown error_msg)B$}",B,S); label("{\tiny$ (Error compiling LaTeX. Unknown error_msg)C$}",C,E); label("{\tiny$ (Error compiling LaTeX. Unknown error_msg)D$}",D,N); label("{\tiny$ (Error compiling LaTeX. Unknown error_msg)E$}",I,N); label("{\tiny$ (Error compiling LaTeX. Unknown error_msg)F$}",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. 
