2003 AIME II Problems/Problem 6
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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, ![[ABC]=\frac{1}{2}\cdot12\cdot14=84](http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/e/5/d/e5d76fb52ece36d20f42e0c8e066eab58d11f840.gif)
.
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, ![\left[\Delta ABC \bigcup \Delta A'B'C'\right] = \frac{12}{9}[\Delta ABC]= \frac{4}{3}\cdot84=\boxed{112}](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/3/d/7/3d7fe0187a2e7c7eb9c6365848ed335f8480c2d9.gif)
.
See also
| 2003 AIME II (Problems • Resources) | ||
| Preceded by Problem 5 | Followed by Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
