Difference between revisions of "2018 AIME II Problems/Problem 9"
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− | We represent | + | 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> | 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> |
Revision as of 23:34, 23 February 2019
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
.
Note that
passes through corresponding vertices of the two heptagons. Also, by centroid properties, our ratio between the sidelengths is
, and their area ratio is hence
. 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 . 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
2018 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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