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 | ||
− | + | <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 < | + | \end{align*}</cmath>so <math>x^2=2116</math>, and <math>x=\boxed{46}</math>. |
[asy] | [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 <math>A</math>}",A,S); |
− | label("{\tiny < | + | label("{\tiny <math>B</math>}",B,S); |
− | label("{\tiny < | + | label("{\tiny <math>C</math>}",C,E); |
− | label("{\tiny < | + | label("{\tiny <math>D</math>}",D,N); |
− | label("{\tiny < | + | label("{\tiny <math>E</math>}",E,N); |
− | label("{\tiny </math> | + | label("{\tiny <math>F</math>}",F,W); |
[/asy] | [/asy] | ||
Revision as of 15:08, 30 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 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,S); label("{\tiny }",B,S); label("{\tiny }",C,E); label("{\tiny }",D,N); label("{\tiny }",E,N); label("{\tiny }",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.