2006 AIME II Problems/Problem 1
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
\begin{align*}2116(\sqrt2+1)&=[ABCDEF]\\&=2\cdot {1\over2}x^2+x\cdot x\sqrt2=x^2(1+\sqrt2),
\end{align*}so , and .
[asy] pair A,B,C,D,I,F; A=(0,0); B=(7,0); F=(0,7); I=(6,13); D=(13,13); C=(13,6); dot(A); dot(B); dot(C); dot(D); dot(I); dot(F); draw(A--B--C--D--I--F--cycle,linewidth(0.7)); label("{\tiny }",A,S); label("{\tiny }",B,S); label("{\tiny }",C,E); label("{\tiny }",D,N); label("{\tiny }",I,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 |
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