Difference between revisions of "2006 AIME II Problems/Problem 1"
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Let the side length be called <math>x</math>, so <math>x=AB=BC=CD=DE=EF=AF</math>. | Let the side length be called <math>x</math>, so <math>x=AB=BC=CD=DE=EF=AF</math>. | ||
− | [[Image: | + | [[Image:2006_II_AIME-1.png]] |
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>, | 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>, | ||
Line 20: | Line 20: | ||
<math>x=46</math></div> | <math>x=46</math></div> | ||
− | Therefore, <math>AB</math> is <math> | + | 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 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.