2003 AIME II Problems/Problem 6
Contents
[hide]Problem
In triangle
and point
is the intersection of the medians. Points
and
are the images of
and
respectively, after a
rotation about
What is the area of the union of the two regions enclosed by the triangles
and
Solution
Since a triangle is a
triangle and a
triangle "glued" together on the
side,
.
There are six points of intersection between and
. Connect each of these points to
.
There are smaller congruent triangles which make up the desired area. Also,
is made up of
of such triangles.
Therefore,
.
Solution 2
First, find the area of either like the first solution or by using Heron’s Formula. Then, draw the medians from
to each of
and
. Since the medians of a triangle divide the triangle into 6 triangles with equal area, we can find that each of the 6 outer triangles have equal area. (Proof: Since I’m too lazy to draw out a diagram, I’ll just have you borrow the one above. Draw medians
and
, and let’s call the points that
intersects
“
” and the point
intersects
“
”. From the previous property and the fact that both
and
are congruent,
has the same area as
. Because of that, both “half” triangles created also have the same area. The same logic can be applied to all other triangles).
Also, since the centroid of a triangle divides each median with the ratio
, along with the previous fact, each outer triangle has
the area of
and
. Thus, the area of the region required is
times the area of
which is
.
See also
2003 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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