Difference between revisions of "2000 AMC 12 Problems/Problem 19"
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== See also == | == See also == |
Revision as of 17:35, 7 August 2017
Contents
Problem
In triangle ,
,
,
. Let
denote the midpoint of
and let
denote the intersection of
with the bisector of angle
. Which of the following is closest to the area of the triangle
?
Solution 1
The answer is exactly , choice
.
We can find the area of triangle
by using the simple formula
. Dropping an altitude from
, we see that it has length
( we can split the large triangle into a
and a
triangle). Then we can apply the Angle Bisector Theorem on triangle
to solve for
. Solving
, we get that
.
is the midpoint of
so
. Thus we get the base of triangle
, to be
units long. Applying the formula
, we get
.
Solution 2
The area of is
where
is the height of triangle
. Using Angle Bisector Theorem, we find
, which we solve to get
.
is the midpoint of
so
. Thus we get the base of triangle
, to be
units long. We can now use Heron's Formula on
.
Therefore, the answer is
.
See also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.