2003 AIME II Problems/Problem 6
Contents
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(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
Solution 3 (Rigorous)
can be calculated as 84 using Heron's formula or other methods. Since a rotation is equivalent to reflection through a point, we have a homothety with scale factor from to through the centroid . Let be the midpoint of which maps to and note that implying that Similarly, we have Also let and be the intersections of with and respectively. The homothety implies that we must have so there is in fact another homothety centered at taking to . Since the scale factor of this homothety is 3 and thus We can apply similar reasoning to the other small triangles in not contained within , so our final answer 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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