2003 AIME II Problems/Problem 6
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
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(Doesn’t require good diagram)
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 .
Solution by Someonenumber011
|2003 AIME II (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|