AIME 2018 II · 第 9 题

Solution 1 (Massive Shoelace)

We represent octagon $ABCDEFGH$ in the coordinate plane with the upper left corner of the rectangle being the origin. Then it follows that $A=(0,-6), B=(8, 0), C=(19,0), D=(27, -6), E=(27, -17), F=(19, -23), G=(8, -23), H=(0, -17), J=(0, -\frac{23}{2})$. Recall that the centroid is $\frac{1}{3}$ way up each median in the triangle. Thus, we can find the centroids easily by finding the midpoint of the side opposite of point $J$. Furthermore, we can take advantage of the reflective symmetry across the line parallel to $BC$ going through $J$ by dealing with less coordinates and ommiting the $\frac{1}{2}$ in the shoelace formula.

By doing some basic algebra, we find that the coordinates of the centroids of $\bigtriangleup JAB, \bigtriangleup JBC, \bigtriangleup JCD, \bigtriangleup JDE$ are $\left(\frac{8}{3}, -\frac{35}{6}\right), \left(9, -\frac{23}{6}\right), \left(\frac{46}{3}, -\frac{35}{6}\right),$ and $\left(18, -\frac{23}{2}\right)$, respectively. We'll have to throw in the projection of the centroid of $\bigtriangleup JAB$ to the line of reflection to apply shoelace, and that point is $\left( \frac{8}{3}, -\frac{23}{2}\right)$

Finally, applying Shoelace, we get: $\left|\left(\frac{8}{3}\cdot (-\frac{23}{6})+9\cdot (-\frac{35}{6})+\frac{46}{3}\cdot (-\frac{23}{2})+18\cdot (\frac{-23}{2})+\frac{8}{3}\cdot (-\frac{35}{6})\right) - \left((-\frac{35}{6}\cdot 9) +\\(-\frac{23}{6}\cdot \frac{46}{3})+ (-\frac{35}{6}\cdot 18)+(\frac{-23}{2}\cdot \frac{8}{3})+(-\frac{23}{2}\cdot \frac{8}{3})\right)\right|$ $=\left|\left(-\frac{92}{9}-\frac{105}{2}-\frac{529}{3}-207-\frac{140}{9}\right)-\left(-\frac{105}{2}-\frac{529}{9}-105-\frac{92}{3}-\frac{92}{3}\right)\right|$ $=\left|-\frac{232}{9}-\frac{1373}{6}-207+\frac{529}{9}+\frac{184}{3}+105+\frac{105}{2}\right|$ $=\left|\frac{297}{9}-\frac{690}{6}-102\right|=\left| 33-115-102\right|=\left|-184\right|=\boxed{184}$

Solution by ktong

note: for a slightly simpler calculation, notice that the heptagon can be divided into two trapezoids of equal area and a small triangle.

Solution 2 (Homothety)

Draw the heptagon whose vertices are the midpoints of octagon $ABCDEFGH$ except $J$. We have a homothety since:

1. $J$ passes through corresponding vertices of the two heptagons.

2. By centroid properties, our ratio between the sidelengths is $\frac{2}{3},$ and their area ratio is hence $\frac{4}{9}.$

Compute the area of the large heptagon by dividing into two congruent trapezoids and a triangle. The area of each trapezoid is

$$= \frac{1}{2}(17+23)\cdot \frac{19}{2}=190.$$ The area of each triangle is

$$= \frac{1}{2}\cdot 17\cdot 4=34.$$ Hence, the area of the large heptagon is

$$2\cdot 190+34=414.$$ Then, from our homothety, the area of the required heptagon is

$$\frac{4}{9}\cdot 414=\boxed{184}.$$ ~novus677

Supplement

Note that we use $17.$ A proof of this is as follows:

• First, note that the shaded triangles are 6-8-10 triangles. Because the homothety is defined on the midpoints of the sides of the octagon, the smaller triangle created by that midpoint and the surrounding rectangle has sides 3-4-5. Thus, the sidelength of the octagon (17) is just 23-2*3=17.

~mathboy282

Solution 3 (Less bashy finish than shoelace)

Instead of bashing shoelace, we can find a clever way to calculate the area of the heptagon without using homothety. By connecting the centroids of $\bigtriangleup JAB, \bigtriangleup JCD$, $\bigtriangleup JHG, \bigtriangleup JFE$, and $\bigtriangleup JFE, \bigtriangleup JCD$. This will give us three triangles and a rectangle. The area of the rectangle is $\frac{34}{3}*\frac{38}{3}$, the top and bottom triangles each give $\frac{38}{3}*2*\frac{1}{2}$, and the triangle on the right yields $\frac{34}{3}*\frac{8}{3}*\frac{1}{2}$ by using $\frac{bh}{2}$. Summing them up, we get $\frac{1656}{9}$, which gives us $\boxed{184}$. ~boppitybobii

Video Solution (Mathematical Dexterity)

https://www.youtube.com/watch?v=HUwJqixBLUI