2011 AMC 10B Problems/Problem 20
- The following problem is from both the 2011 AMC 12B #16 and 2011 AMC 10B #20, so both problems redirect to this page.
Contents
Problem
Rhombus has side length
and
°. Region
consists of all points inside the rhombus that are closer to vertex
than any of the other three vertices. What is the area of
?
Solution
Suppose that is a point in the rhombus
and let
be the perpendicular bisector of
. Then
if and only if
is on the same side of
as
. The line
divides the plane into two half-planes; let
be the half-plane containing
. Let us define similarly
and
. Then
is equal to
. The region turns out to be an irregular pentagon. We can make it easier to find the area of this region by dividing it into four triangles:
Since
and
are equilateral,
contains
,
contains
and
, and
contains
. Then
with
and
so
. Multiply this by 4 and it turns out that the pentagon has area
.
Solution 2
We follow the steps shown above until we draw pentagon . We know that rhombus
can be divided into equilateral triangles
and
. Using the
special right triangle rules, we find the height of the equilateral triangles (and the height of the rhombus) to be
. Therefore, the area of
is
. We now have to take off the areas
,
, and
to get the desired shape.
is just half of
and
and
are each
, for a total area of
.
See Also
2011 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2011 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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