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{ | + | Therefore, <math>AB</math> is <math>\boxed{46}</math>. |
| + | ~removal of extraneous zeros by K124659 | ||
| + | |||
| + | == Solution 2 == | ||
| + | |||
| + | 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> | ||
| + | {{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{46}</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,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 == | == See also == | ||
| Line 26: | Line 64: | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 11:07, 31 August 2024
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 
.
~removal of extraneous zeros by K124659
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*}](http://latex.artofproblemsolving.com/7/b/4/7b4eac99a9d419ae9b76ff89eae925b572923827.png)
so 
, and 
.
![[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]](http://latex.artofproblemsolving.com/9/1/a/91aceb75842732accdf7ebdb228cdc455ed91656.png)
~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. 
