Difference between revisions of "2001 AMC 12 Problems/Problem 22"
(New page: == Problem == In rectangle <math>ABCD</math>, points <math>F</math> and <math>G</math> lie on <math>AB</math> so that <math>AF=FG=GB</math> and <math>E</math> is the midpoint of <math>\ov...) |
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− | Let <math>[X]</math> denote the area of the polygon <math> | + | Let <math>[X]</math> denote the area of the polygon <math>EHJ</math>. |
=== Solution 1 === | === Solution 1 === |
Revision as of 22:26, 22 February 2010
Contents
[hide]Problem
In rectangle , points
and
lie on
so that
and
is the midpoint of
. Also,
intersects
at
and
at
. The area of the rectangle
is
. Find the area of triangle
.
Solution
Let denote the area of the polygon
.
Solution 1
Note that the triangles and
are similar, as they have the same angles. Hence
.
Also, triangles and
are similar, hence
.
We can now compute as
. We have:
.
is
of
, as these two triangles have the same base
, and
is
of
, therefore also the height from
onto
is
of the height from
. Hence
.
is
of
, as the base
is
of the base
, and the height from
is
of the height from
. Hence
.
is
of
for similar reasons, hence
.
Therefore .
Solution 2
As in the previous solution, we note the similar triangles and prove that is in
and
in
of
.
We can then compute that .
As is the midpoint of
, the height from
onto
is
of the height from
onto
. Therefore we have
.
See Also
2001 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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 |