Difference between revisions of "2006 AIME II Problems/Problem 1"
(→Solution 2) |
m (→Solution 2) |
||
| Line 49: | Line 49: | ||
label("{\tiny $A$}",A,S); | label("{\tiny $A$}",A,S); | ||
label("{\tiny $B$}",B,S); | label("{\tiny $B$}",B,S); | ||
| − | label("{\tiny $C$}",C, | + | label("{\tiny $C$}",C,dir(0)); |
label("{\tiny $D$}",D,N); | label("{\tiny $D$}",D,N); | ||
label("{\tiny $E$}",E,N); | label("{\tiny $E$}",E,N); | ||
label("{\tiny $F$}",F,W); | label("{\tiny $F$}",F,W); | ||
</asy> | </asy> | ||
| + | |||
| + | |||
| + | ~minor asymptote edit by Yiyj1 | ||
== See also == | == See also == | ||
Revision as of 02:19, 12 January 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
.
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
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.
