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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{046}</math>.
+
Therefore, <math>AB</math> is <math>\boxed{46}</math>.
 +
~removal of extraneous zeros by K124659
  
−
== 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>\displaystyle
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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>
−
{{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
 +
<cmath>\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</math>, and <math>x=\boxed{46}</math>.
+
\end{align*}</cmath>so <math>x^2=2116</math>, and <math>x=\boxed{46}</math>.
  
−
[asy]
+
<asy>
−
pair A,B,C,D,I,F;
+
pair A,B,C,D,E,F;
 
A=(0,0);
 
A=(0,0);
 
B=(7,0);
 
B=(7,0);
 +
C=(13,6);
 +
E=(6,13);
 +
D=(13,13);
 
F=(0,7);
 
F=(0,7);
−
I=(6,13);
 
−
D=(13,13);
 
−
C=(13,6);
 
 
dot(A);
 
dot(A);
 
dot(B);
 
dot(B);
 
dot(C);
 
dot(C);
 
dot(D);
 
dot(D);
−
dot(I);
+
dot(E);
 
dot(F);
 
dot(F);
−
draw(A--B--C--D--I--F--cycle,linewidth(0.7));
+
draw(A--B--C--D--E--F--cycle,linewidth(0.7));
−
label("{\tiny <math>A</math>}",A,S);
+
label("{\tiny $A$}",A,S);
−
label("{\tiny <math>B</math>}",B,S);
+
label("{\tiny $B$}",B,S);
−
label("{\tiny <math>C</math>}",C,E);
+
label("{\tiny $C$}",C,dir(0));
−
label("{\tiny <math>D</math>}",D,N);
+
label("{\tiny $D$}",D,N);
−
label("{\tiny <math>E</math>}",I,N);
+
label("{\tiny $E$}",E,N);
−
label("{\tiny <math>F</math>}",F,W);
+
label("{\tiny $F$}",F,W);
−
[/asy]
+
</asy>
 +
 
 +
 
 +
~minor asymptote edit by Yiyj1
 +
~removal of extraneous zeros by K124659
  
 
== See also ==
 
== See also ==

Latest revision as of 11:07, 31 August 2024

Problem

In convex hexagon $ABCDEF$, all six sides are congruent, $\angle A$ and $\angle D$ are right angles, and $\angle B, \angle C, \angle E,$ and $\angle F$ are congruent. The area of the hexagonal region is $2116(\sqrt{2}+1).$ Find $AB$.

Solution

Let the side length be called $x$, so $x=AB=BC=CD=DE=EF=AF$.

2006 II AIME-1.png

The diagonal $BF=\sqrt{AB^2+AF^2}=\sqrt{x^2+x^2}=x\sqrt{2}$. Then the areas of the triangles AFB and CDE in total are $\frac{x^2}{2}\cdot 2$, and the area of the rectangle BCEF equals $x\cdot x\sqrt{2}=x^2\sqrt{2}$

Then we have to solve the equation

$2116(\sqrt{2}+1)=x^2\sqrt{2}+x^2$.

$2116(\sqrt{2}+1)=x^2(\sqrt{2}+1)$

$2116=x^2$

$x=46$

Therefore, $AB$ is $\boxed{46}$. ~removal of extraneous zeros by K124659

Solution 2

Because $\angle B$, $\angle C$, $\angle E$, and $\angle F$ are congruent, the degree-measure of each of them is ${{720-2\cdot90}\over4}= 135$. Lines $BF$ and $CE$ divide the hexagonal region into two right triangles and a rectangle. Let $AB=x$. Then $BF=x\sqrt2$. Thus \begin{align*} 2116(\sqrt2+1)&=[ABCDEF]\\ &=2\cdot {1\over2}x^2+x\cdot x\sqrt2=x^2(1+\sqrt2), \end{align*}so $x^2=2116$, and $x=\boxed{46}$.

[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,dir(0)); label("{\tiny $D$}",D,N); label("{\tiny $E$}",E,N); label("{\tiny $F$}",F,W); [/asy]


~minor asymptote edit by Yiyj1 ~removal of extraneous zeros by K124659

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. AMC logo.png

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