2018 AIME II Problems/Problem 9
Contents
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.
![[asy] unitsize(6); pair P = (0, 0), Q = (0, 23), R = (27, 23), SS = (27, 0); pair A = (0, 6), B = (8, 0), C = (19, 0), D = (27, 6), EE = (27, 17), F = (19, 23), G = (8, 23), J = (0, 23/2), H = (0, 17); draw(P--Q--R--SS--cycle); draw(J--B); draw(J--C); draw(J--D); draw(J--EE); draw(J--F); draw(J--G); draw(A--B); draw(H--G); real dark = 0.6; filldraw(A--B--P--cycle, gray(dark)); filldraw(H--G--Q--cycle, gray(dark)); filldraw(F--EE--R--cycle, gray(dark)); filldraw(D--C--SS--cycle, gray(dark)); dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); dot(G); dot(H); dot(J); dot(H); defaultpen(fontsize(10pt)); real r = 1.3; label("$A$", A, W*r); label("$B$", B, S*r); label("$C$", C, S*r); label("$D$", D, E*r); label("$E$", EE, E*r); label("$F$", F, N*r); label("$G$", G, N*r); label("$H$", H, W*r); label("$J$", J, W*r); [/asy]](http://latex.artofproblemsolving.com/9/5/8/958670038ab1dd52c3b3fb6095f47bd59ee398d1.png)
Solution 1 (Massive Shoelace)
We represent octagon in the coordinate plane with the lower 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 ![\[= \frac{1}{2}(17+23)\cdot \frac{19}{2}=190.\]](http://latex.artofproblemsolving.com/d/7/3/d730757874637b564313534c9abb92a3de755130.png)
The area of each triangle is ![\[= \frac{1}{2}\cdot 17\cdot 4=34.\]](http://latex.artofproblemsolving.com/9/4/5/94510d3f76b032b5952796362785bee843358f35.png)
Hence, the area of the large heptagon is ![\[2\cdot 190+34=414.\]](http://latex.artofproblemsolving.com/e/c/5/ec5d64f9150029103fe23661f1c8bf08f60af263.png)
Then, from our homothety, the area of the required heptagon is ![\[\frac{4}{9}\cdot 414=\boxed{184}.\]](http://latex.artofproblemsolving.com/0/3/8/0387e0afb514f5f56eacd1387450537385eb6291.png)
~novus677
Video Solution (Mathematical Dexterity)
https://www.youtube.com/watch?v=HUwJqixBLUI
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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