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 == | + | == 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 |
+ | <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> | |
− | pair A,B,C,D, | + | 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); | ||
− | |||
− | |||
− | |||
dot(A); | dot(A); | ||
dot(B); | dot(B); | ||
dot(C); | dot(C); | ||
dot(D); | dot(D); | ||
− | dot( | + | dot(E); |
dot(F); | dot(F); | ||
− | draw(A--B--C--D-- | + | draw(A--B--C--D--E--F--cycle,linewidth(0.7)); |
− | label("{\tiny | + | label("{\tiny $A$}",A,S); |
− | label("{\tiny | + | label("{\tiny $B$}",B,S); |
− | label("{\tiny | + | label("{\tiny $C$}",C,dir(0)); |
− | label("{\tiny | + | label("{\tiny $D$}",D,N); |
− | label("{\tiny | + | label("{\tiny $E$}",E,N); |
− | label("{\tiny | + | label("{\tiny $F$}",F,W); |
− | + | </asy> | |
+ | |||
+ | |||
+ | ~minor asymptote edit by Yiyj1 | ||
+ | ~removal of extraneous zeros by K124659 | ||
== See also == | == See also == |
Latest revision as of 12:07, 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
.
![$x=46$](http://latex.artofproblemsolving.com/1/7/7/177d9e27be3fbdeb68593a8cef48eed1e0bde215.png)
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
so
, and
.
~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.