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2006 AIME II Problems/Problem 1: Difference between revisions

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== Problem ==
== Problem ==


The lengths of the sides of a [[triangle]] with positive area are <math>\log_{10} 12</math>, <math>\log_{10} 75</math>, and <math>\log_{10} n</math>, where <math>n</math> is a positive integer. Find the number of possible values for <math>n</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 ==
By the [[Triangle Inequality]]:
<div style="text-align:center;">
<math>\log_{10} 12 + \log_{10} n > \log_{10} 75 </math>


<math>\log_{10} 12n > \log_{10} 75 </math>
Let the side length be called <math>x</math>, so <math>x=AB=BC=CD=DE=EF=AF</math>.


<math> 12n > 75 </math>
[[Image:2006_II_AIME-1.png]]


<math> n > \frac{75}{12} = \frac{25}{4} = 6.25 </math>
The diagonal <math>BF=\sqrt{AB^2+AF^2}=\sqrt{x^2+x^2}=x\sqrt{2}</math>. Then the areas of the triangles AFB and CDE in total are <math>\frac{x^2}{2}\cdot 2</math>,
</div>
and the area of the rectangle BCEF equals <math>x\cdot x\sqrt{2}=x^2\sqrt{2}</math>
Also:
<div style="text-align:center;">
<math>\log_{10} 12 + \log_{10} 75 > \log_{10} n </math>


<math>\log_{10} 12\cdot75 > \log_{10} n </math>
Then we have to solve the equation


<math> n < 900 </math>  
<cmath>2116(\sqrt{2}+1)=x^2\sqrt{2}+x^2</cmath>
</div>
<cmath>2116(\sqrt{2}+1)=x^2(\sqrt{2}+1)</cmath>
Combining these two inequalities:
<cmath>2116=x^2</cmath>
<cmath>x=46</cmath>


<cmath> 6.25 < n < 900 </cmath>
Therefore, <math>AB</math> is <math>\boxed{046}</math>.


The number of possible integer values for <math>n</math> is the number of integers over the interval <math>(6.25 , 900)</math>, which is <math>893</math>.
== Solution 2 ==


== See also ==
Because <math>\angle
{{AIME box|year=2006|n=I|num-b=1|num-a=3}}
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
<cmath>\begin{align*}
2116(\sqrt2+1)&=[ABCDEF]\\
&=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{046}</math>.


<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,S);
label("{\tiny $D$}",D,N);
label("{\tiny $E$}",E,N);
label("{\tiny $F$}",F,W);
</asy>
== See Also ==
{{AIME box|year=2006|n=II|before=First Question|num-a=2}}
[[Category:Intermediate Geometry Problems]]
[[Category:Intermediate Geometry Problems]]
[[Category:Intermediate Algebra Problems]]
{{MAA Notice}}

Latest revision as of 07:46, 16 October 2025

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 1

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

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 $\boxed{046}$.

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 $x=\boxed{046}$.

[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,S); label("{\tiny $D$}",D,N); label("{\tiny $E$}",E,N); label("{\tiny $F$}",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

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