2018 AIME II Problems/Problem 9
Problem
Octagon with side lengths and is formed by removing 6-8-10 triangles from the corners of a rectangle with side on a short side of the rectangle, as shown. Let be the midpoint of , and partition the octagon into 7 triangles by drawing segments , , , , , and . Find the area of the convex polygon whose vertices are the centroids of these 7 triangles.
Solution 1 (Massive Shoelace)
We represent octagon in the coordinate plane with the upper left corner of the rectangle being the origin. Then it follows that . Recall that the centroid is way up each median in the triangle. Thus, we can find the centroids easily by finding the midpoint of the side opposite of point . Furthermore, we can take advantage of the reflective symmetry across the line parallel to going through by dealing with less coordinates and ommiting the in the shoelace formula.
By doing some basic algebra, we find that the coordinates of the centroids of are and , respectively. We'll have to throw in the projection of the centroid of to the line of reflection to apply shoelace, and that point is
Finally, applying Shoelace, we get:
Solution by ktong
Solution 2 (Homothety)
Draw the heptagon whose vertices are the midpoints of octagon except . We have a homothety since:
1. passes through corresponding vertices of the two heptagons.
2.By centroid properties, our ratio between the sidelengths is , and their area ratio is hence .
Compute the area of the large heptagon by dividing into two congruent trapezoids and a triangle. The area of each trapezoid is . The area of each triangle is .
Hence, the area of the large heptagon is . Then, from our homothety, the area of the required heptagon is ~novus677
See Also
2018 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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