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Difference between revisions of "2018 AIME II Problems/Problem 9"

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We now compute the area of our large heptagon. We can divide this into two congruent trapezoids and a triangle. The area of each trapezoid is <math>\frac{1}{2}(17+23)\cdot \frac{19}{2}=190</math>. The area of each triangle is <math>\frac{1}{2}\cdot 17\cdot 2=34</math>.
 
We now compute the area of our large heptagon. We can divide this into two congruent trapezoids and a triangle. The area of each trapezoid is <math>\frac{1}{2}(17+23)\cdot \frac{19}{2}=190</math>. The area of each triangle is <math>\frac{1}{2}\cdot 17\cdot 2=34</math>.
  
−
Hence, the area of the large heptagon is <math>2\cdot 190+34=414</math>. Then from our homothety, the area of the required heptagon is <math>frac{4}{9}\cdot 414=\boxed{184}</math>
+
Hence, the area of the large heptagon is <math>2\cdot 190+34=414</math>. Then from our homothety, the area of the required heptagon is <math>\frac{4}{9}\cdot 414=\boxed{184}</math>
 +
~novus677
 
{{AIME box|year=2018|n=II|num-b=8|num-a=10}}
 
{{AIME box|year=2018|n=II|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 05:37, 30 March 2018

Problem

Octagon $ABCDEFGH$ with side lengths $AB = CD = EF = GH = 10$ and $BC = DE = FG = HA = 11$ is formed by removing 6-8-10 triangles from the corners of a $23$ $\times$ $27$ rectangle with side $\overline{AH}$ on a short side of the rectangle, as shown. Let $J$ be the midpoint of $\overline{AH}$, and partition the octagon into 7 triangles by drawing segments $\overline{JB}$, $\overline{JC}$, $\overline{JD}$, $\overline{JE}$, $\overline{JF}$, and $\overline{JG}$. Find the area of the convex polygon whose vertices are the centroids of these 7 triangles.

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

Solution 2 (Homothety)

Draw the heptagon whose vertices are the midpoints of octagon $ABCDEFGH$ except $J$. Note that $J$ passes through corresponding vertices of the two heptagons. Also, by centroid properties, our ratio between the sidelengths is $\frac{2}{3}$, and their area ratio is hence $\frac{4}{9}$. So, we have a homothety.

We now compute the area of our large heptagon. We can divide this 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 $\frac{1}{2}\cdot 17\cdot 2=34$.

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

2018 AIME II (Problems • Answer Key • Resources)
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