Difference between revisions of "2018 AIME II Problems/Problem 9"
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Octagon <math>ABCDEFGH</math> with side lengths <math>AB = CD = EF = GH = 10</math> and <math>BC = DE = FG = HA = 11</math> is formed by removing 6-8-10 triangles from the corners of a <math>23</math> <math>\times</math> <math>27</math> rectangle with side <math>\overline{AH}</math> on a short side of the rectangle, as shown. Let <math>J</math> be the midpoint of <math>\overline{AH}</math>, and partition the octagon into 7 triangles by drawing segments <math>\overline{JB}</math>, <math>\overline{JC}</math>, <math>\overline{JD}</math>, <math>\overline{JE}</math>, <math>\overline{JF}</math>, and <math>\overline{JG}</math>. Find the area of the convex polygon whose vertices are the centroids of these 7 triangles. | Octagon <math>ABCDEFGH</math> with side lengths <math>AB = CD = EF = GH = 10</math> and <math>BC = DE = FG = HA = 11</math> is formed by removing 6-8-10 triangles from the corners of a <math>23</math> <math>\times</math> <math>27</math> rectangle with side <math>\overline{AH}</math> on a short side of the rectangle, as shown. Let <math>J</math> be the midpoint of <math>\overline{AH}</math>, and partition the octagon into 7 triangles by drawing segments <math>\overline{JB}</math>, <math>\overline{JC}</math>, <math>\overline{JD}</math>, <math>\overline{JE}</math>, <math>\overline{JF}</math>, and <math>\overline{JG}</math>. Find the area of the convex polygon whose vertices are the centroids of these 7 triangles. | ||
| + | ==Solution 1 (Massive Shoelace)== | ||
| + | We represent Octagon <math>ABCDEFGH</math> in the coordinate plane with the upper left corner of the rectangle being the origin. Then it follows that <math>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})</math>. Recall that the centroid is <math>\frac{1}{3}</math> way up each median in the triangle. Thus, we can find the centroids easily by finding the midpoint of the side opposite of point <math>J</math>. Furthermore, we can take advantage of the reflective symmetry across the line parallel to <math>BC</math> going through <math>J</math> by dealing with less coordinates and ommiting the <math>\frac{1}{2}</math> in the shoelace formula. | ||
| + | |||
| + | By doing some basic algebra, we find that the coordinates of the centroids of <math>\bigtriangleup JAB, \bigtriangleup JBC, \bigtriangleup JCD, \bigtriangleup JDE</math> are <math>\left(\frac{8}{3}, -\frac{35}{6}\right), \left(9, -\frac{23}{6}\right), \left(\frac{46}{3}, -\frac{35}{6}\right),</math> and <math>\left(18, -\frac{23}{2}\right)</math>, respectively. We'll have to throw in the projection of the centroid of <math>\bigtriangleup JAB</math> to the line of reflection to apply shoelace, and that point is <math>\left( \frac{8}{3}, -\frac{23}{2}\right)</math> | ||
| + | |||
| + | Finally, applying Shoelace, we get: | ||
| + | <math>\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|</math> | ||
| + | <math>=\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|</math> | ||
| + | <math>=\left|-\frac{232}{9}-\frac{1373}{6}-207+\frac{529}{9}+\frac{184}{3}+105+\frac{105}{2}\right|</math> | ||
| + | <math>=\left|\frac{297}{9}-\frac{690}{6}-102\right|=\left| 33-115-102\right|=\left|-184\right|=\boxed{184}</math> | ||
{{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 11:05, 24 March 2018
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:
| 2018 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
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