2006 AIME I Problems/Problem 8
Contents
[hide]Problem
Hexagon is divided into five rhombuses,
and
as shown. Rhombuses
and
are congruent, and each has area
Let
be the area of rhombus
. Given that
is a positive integer, find the number of possible values for
.
Solution 1
Let denote the common side length of the rhombi.
Let
denote one of the smaller interior angles of rhombus
. Then
. We also see that
. Thus
can be any positive integer in the interval
.
and
, so
can be any integer between 1 and 89, inclusive. Thus the number of positive values for
is
.
Solution 2
Call the side of each rhombus w. w is the width of the rhombus. Call the height h, where . The height of rhombus T would be 2h, and the width would be
. Substitute the first equation to get
. Then the area of the rhombus would be
. Combine like terms to get
See also
2006 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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